1

EXPERIMENTAL IMPINGEMENT HEAT TRANSFER IN A SMOOTH

CHANNEL WITH

CROSS OVER JETS AND BLEED HOLES

A Thesis Presented

By

Amornrat Nongsaeng

To

The Department of Mechanical and Industrial Engineering

In partial fulfillment of the requirements

For the degree of

Master of Science

In

Mechanical Engineering

In the field of

Thermofluids Engineering

Northeastern University

Boston, Massachusetts

August 2009

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NORTHEASTERN UNIVERSITY

Graduate School of Engineering

Thesis Title: Experimental Impingement Heat Transfer in a Smooth Channel With

Cross Over Jets and Bleed Holes

Author: Amornrat Nongsaeng

Department: Mechanical and Industrial Engineering

Approved for Thesis Requirement of the Master of Science Degree

__________________________________________________ __________________

Thesis Adviser Date

__________________________________________________ __________________

Thesis Reader Date

__________________________________________________ __________________

Thesis Reader Date

__________________________________________________ __________________

Thesis Reader Date

__________________________________________________ __________________

Department Chair Date

Graduate School Notified of Acceptance:

__________________________________________________ __________________

Director of the Graduate School Date

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ABSTRACT

Trailing edge cooling cavities in modern gas turbine airfoils play an important

role in maintaining the trailing edge temperature at levels consistent with airfoil design

life. These cavities often have a trapezoidal cross-sectional shape in order to be

compatible with the external contour of the blade at the trailing edge. The cooling flow

for these geometries generally enters the trailing edge cavity from the adjacent cavity

through a row of race-track shaped slots with mounted holes on the partition wall

between the two cavities. Cross-over jets, impinge on a smooth trailing-edge wall and

exit through the second row of race-track shaped slots. Steady state liquid crystal

technique is used in this experimental investigation to measure the impingement heat

transfer coefficients on the smooth surface of the trapezoidal channel. The experimental

setups were arranged for zero- and five-degree tilt angle for the cross-over jets with zero,

two and four blocked consecutive slots along the trailing-edge. Tests were performed for

two inlet and exit slot arrangements - staggered and inline. These setups were

investigated over a range of Reynolds number. The results showed that the Nusselt

numbers increase monotonically with increasing Reynolds numbers. Pressure ratio

(Psupply/Ptrailing-edge) increased with increasing Reynolds numbers. The four-blocked

consecutive holes along the trailing-edge induced higher pressure ratios than the cases

where all exit holes were opened. Heat transfer coefficients decreased significantly near

the blocked slots while the cross flow created by the blockage of the exit slots increased

the heat transfer coefficients near the open slots.

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TABLE OF CONTENTS

ABSTRACT ........................................................................................................... 3

TABLE OF CONTENTS ..................................................................................... 4

LIST OF FIGURES .............................................................................................. 7

LIST OF TABLES .............................................................................................. 12

NOMENCLATURE ............................................................................................ 13

ACKNOWLEDGEMENTS ............................................................................... 15

1. INTRODUCTION....................................................................................... 16

1.1 Need for Turbine Cooling ......................................................................... 17

1.2 Techniques for Turbine Blade Cooling ..................................................... 19

1.3 Purpose of the Present Investigation ........................................................ 26

2. LITERATURE REVIEW .......................................................................... 28

3. EXPERIMENTAL SETUP AND PROCEDURE .................................... 33

3.1 Experimental Setup ................................................................................... 33

3.2 Experimental Procedure ........................................................................... 54

3.2.1 Digital Pictures.................................................................................. 62

3.2.2 Code Steps ........................................................................................ 62

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4. DATA REDUCTION .................................................................................. 64

5. RESULTS AND DISCUSSIONS ............................................................... 73

5.1 Nusselt Number ......................................................................................... 77

5.1.1 Cases One through Fourteen Analyses ............................................. 77

5.1.2 Plots of Cases 1 through 14 .............................................................. 78

5.1.3 Heat Transfer Results for Areas 1 through 5 .................................... 85

5.2 Pressure Ratios ......................................................................................... 97

6. CONCLUSION ......................................................................................... 100

REFERENCES .................................................................................................. 101

APPENDIX A .................................................................................................... 103

Raw data: Test1-Zero Degree-Inline-Baseline (All Holes Open) ................... 103

APPENDIX B .................................................................................................... 109

Raw data (Cold Test): Test1-Zero Degree-Inline-Baseline ............................ 109

APPENDIX C .................................................................................................... 110

FORTRAN Program Check.f .......................................................................... 110

APPENDIX D .................................................................................................... 112

FORTRAN Program Reduce.f ......................................................................... 112

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APPENDIX E .................................................................................................... 134

FORTRAN Program Integarea.f ..................................................................... 134

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LIST OF FIGURES

Figure 1 Increased turbine inlet temperature dramatically improves cycle power output

(courtesy of Pratt & Whitney; Sautner et al., 1992; AGARD CP 527) .................... 17

Figure 2 Protective Film Layer Made by Coolant Injection ............................................. 20

Figure 3 Typical cooled aircraft gas turbine blade (Gladden and Simoneau, 1988;

collected in Sokolowski, 1988) [1] ........................................................................... 21

Figure 4 General convection cooling technique ............................................................... 22

Figure 5 Several blade designs for convection cooling .................................................... 22

Figure 6 General impingement cooling technique ............................................................ 24

Figure 7 One possible impingement cooling design ......................................................... 24

Figure 8 The trailing-edge of an airfoil is cooled by pin fins (Metzger et al., 1984) ....... 25

Figure 9 Schematic of critical venturi ............................................................................... 36

Figure 10 Setup for the case of all open trailing-edge slots .............................................. 40

Figure 11 Setup for the case of two blocked trailing-edge slots ....................................... 41

Figure 12 Setup for the case of four blocked trailing-edge slots ...................................... 42

Figure 13 Details of the test section .................................................................................. 43

Figure 14 Setup for the inline arrangement with all open trailing-edge slots ................... 44

Figure 15 Setup for the inline arrangement with two blocked trailing-edge slots ............ 45

Figure 16 Setup for the inline arrangement with four blocked trailing-edge slots ........... 46

Figure 17 Setup for the staggered arrangement with all open trailing-edge slots ............ 47

Figure 18 Setup for the staggered arrangement with two blocked trailing-edge slots ...... 48

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Figure 19 Setup for the staggered arrangement with four blocked trailing-edge slots ..... 49

Figure 20 Details of the rig cross-section for zero-degree tilt angle ................................. 50

Figure 21 Details of the rig cross-section for five-degree tilt angle ................................. 51

Figure 22 Details of the trailing-edge slots ....................................................................... 52

Figure 23 Details of the cross-over holes ......................................................................... 52

Figure 24 Stack of elements in the heaters and liquid crystal film ................................... 53

Figure 25 Home-made power distribution panel .............................................................. 55

Figure 26 Micro-manometer ............................................................................................. 56

Figure 27 Nusselt Number versus Reynolds Number for the Baseline case of Zero-degree

tilt, Inline ................................................................................................................... 78

Figure 28 Nusselt Number versus Reynolds Number for the Baseline case of Zero-degree

tilt, Staggered ............................................................................................................ 78

Figure 29 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree

tilt, Inline ................................................................................................................... 79

Figure 30 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree

tilt, Staggered ............................................................................................................ 79

Figure 31 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree

tilt, Opposite- wall, Inline ......................................................................................... 80

Figure 32 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree

tilt, Opposite- wall, Staggered .................................................................................. 80

Figure 33 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of

Zero-degree tilt, Inline .............................................................................................. 81

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Figure 34 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of

Zero-degree tilt, Staggered ........................................................................................ 81

Figure 35 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of

Five-degree tilt, Inline ............................................................................................... 82

Figure 36 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of

Five-degree tilt, Staggered ........................................................................................ 82

Figure 37 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of

Zero-degree tilt, Inline .............................................................................................. 83

Figure 38 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of

Zero-degree tilt, Staggered ........................................................................................ 83

Figure 39 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of

Five-degree tilt, Inline ............................................................................................... 84

Figure 40 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of

Five-degree tilt, Staggered ........................................................................................ 84

Figure 41 Nusselt Number versus Reynolds Number for all cases on Area1 ................... 85

Figure 42 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on

Area1 ......................................................................................................................... 86

Figure 43 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on

Area1 ......................................................................................................................... 86

Figure 44 Nusselt Number versus Reynolds Number for all cases on Area2 ................... 87

Figure 45 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on

Area2 ......................................................................................................................... 87

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Figure 46 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on

Area2 ......................................................................................................................... 88

Figure 47 Nusselt Number versus Reynolds Number for all cases on Area3 ................... 88

Figure 48 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on

Area3 ......................................................................................................................... 89

Figure 49 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on

Area3 ......................................................................................................................... 89

Figure 50 Nusselt Number versus Reynolds Number for all cases on Area4 ................... 90

Figure 51 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on

Area4 ......................................................................................................................... 90

Figure 52 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on

Area4 ......................................................................................................................... 91

Figure 53 Nusselt Number versus Reynolds Number for all cases on Area5 ................... 91

Figure 54 Nusselt Number versus Reynolds Number for all Baseline cases on Area5 .... 92

Figure 55 Nusselt Number versus Reynolds Number for the Two and Four Blocked cases

on Area5 .................................................................................................................... 92

Figure 56 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on

Area5 ......................................................................................................................... 93

Figure 57 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on

Area5 ......................................................................................................................... 93

Figure 58 (Psup/Ptrailing-edge ) versus Reynolds number for all cases ................................... 97

Figure 59 (Psup/Ptrailing-edge ) versus Reynolds number for the baseline cases .................... 98

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Figure 60 (Psup/Ptrailing-edge ) versus Reynolds number for the blocked-slot cases ............. 98

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LIST OF TABLES

Table 1: Reynolds Number, Nusselt Number, and Heat Transfer Coefficient Results for

All Baseline Cases .................................................................................................... 74

Table 2: Reynolds Number, Nusselt Number, and Heat Transfer Coefficient Results for

Zero-Degree tilt angle and Blocked Holes Cases ..................................................... 75

Table 3: Reynolds Number, Nusselt Number, and Heat Transfer Coefficient Results for

Five-Degree tilt angle and Blocked Holes Cases ...................................................... 76

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NOMENCLATURE

D h Cross-over hole hydraulic diameter

A cross Channel cross sectional area

A bleed Total bleed area

A throat Venturi throat area

m Air mass flow rate from the critical venturi

bleedm

Mass flow rate through bleed valve

D e Outer hydraulic diameter of the channel

h0 Outer natural convection heat transfer coefficient

k amb Air thermal conductivity at ambient temperature

k Thermal conductivity

C p Air specific heat at constant pressure

Air dynamic viscosity

gc Proportionality constant in Newton’s second law

g Acceleration of gravity

Air density

Re jet Reynolds number based on cross-over hole hydraulic diameter

Pr Prandtl number

V Voltage

A Amperage

Q Total heat added

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R Thermal resistance

q” Heat flux

lossq"

Heat loss

h w Pressure drop across the orifice

P ven Venturi inlet pressure

P amb Ambient pressure

P f Air pressure at the orifice inlet

T abs Air temperature entering the orifice in degrees Rankine

T ven Venturi inlet temperature

T m Air mixed mean temperature

T w Air mixed wall temperature

T f Air film temperature

T in Inlet temperature

T heater Heater temperature

T liquid crystal Liquid crystals temperature

T surface Surface temperature

T amb Ambient temperature

Nu jet Nusselt number based on cross-over hole hydraulic diameter

h Area-weighted Average heat transfer coefficient

U m Air mean velocity

β Cross-over hole tilt angle

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ACKNOWLEDGEMENTS

First, I would like to thank Northeastern University, specifically the Mechanical

and Industrial Engineering Department for the educational opportunities provided by the

faculty and staff throughout my graduate experience. Most importantly, I would like to

express my gratitude to my advisor Professor Mohammad E. Taslim who helped and

advised me through this study. Without his help and encouragement the completion of this

thesis would not have been possible. Thank you for being such a wonderful advisor

throughout the years who advised me both in research and life in general as an

international student.

I would like to thank my friends Michael Fong and Mehdi Abedi for showing me

how to run the experiment and helping me to set it up. Our many discussions resulted in

exchange of knowledge and skills. My thanks also go to Deepti Dutt and Ghassan J.

Nicolas for their friendship, encouragement and constant support. I would also like to

thank all my friends, both here and back home, for supporting and helping me, while

working on my thesis.

Last, my deepest gratitude goes to my family who provided the item of greatest

worth, the opportunity to leave my country and pursue graduate study. They have not

only been a source of inspiration, support and love but also have stood by me through the

many steps and decisions of my educational career.

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1. INTRODUCTION

Today, gas turbine is widely used. During the period from granting of the first patent

in 1971 to the present [2], the gas turbine has been developed into various reliabilities. It

is used exclusively to operate all new commercial airplanes. Most business aircrafts are

operated by one form or another of a gas turbine. Gas turbine is an engine designed to

convert the energy of a fuel into some form of practical power, such as mechanical

power, the high-speed thrust of jet or industrial applications.

Thermal efficiency and power output of gas turbines increase with increasing turbine

inlet temperatures. This is dramatically illustrated in Figure 1, which plots specific core

power production as a function of turbine inlet temperature. The engines tend to track

fairly close to the ideal performance line, which represents a cycle power output with

100% efficient turbine with no leakage or cooling flows. Clearly, increasing in let

temperature is one of the key technologies in raising gas turbine engine performance. The

inlet temperature in advance gas turbines may be several hundred degrees higher than the

melting point of the blade material [1]. Therefore, the efficient cooling of the gas turbine

airfoils is necessary to keep the metal temperatures below their critical limits.

The goal of the experiment was to investigate the heat transfer coefficient by crating

cross flow and see its effect on the jets heat transfer.

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Figure 1 Increased turbine inlet temperature dramatically improves cycle power output (courtesy of

Pratt & Whitney; Sautner et al., 1992; AGARD CP 527)

1.1 Need for Turbine Cooling

Increasing the thermal efficiency and power (thrust) output of advanced gas

turbine engines which operate at high temperatures (2500-2600°F) is causing a

premature failure of the turbine blades’ life. Thus, gas turbine engine designers try to find

the way to increase the power output from a given engine and prevent the turbine blades

from premature failure.

One way to achieve these improvements is to operate at as high turbine inlet

temperature as possible. The turbine inlet temperature is limited by current materials.

Improvements in materials have allowed the maximum turbine inlet temperature to

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increase slowly with time, but engine designers have had to seek ways of increasing the

turbine inlet temperature at a rate faster than materials will allow. One very common

solution has been the cooling of gas turbine airfoils, which has allowed the turbine

designer to increase the turbine inlet temperature while maintaining a constant blade

(material) temperature [2].

As the turbine inlet temperature increases, the heat transferred to the turbine

blades also increases. The level and variation in the temperature within the blade material

which cause thermal stress must be limited to achieve reasonable durability goals.

The operating temperatures are far above the permissible metal temperatures.

Therefore, there is a need to cool the blades for safe operation. The blades are cooled by

extracted air from the compressor of the engine. Since this extraction incurs a penalty to

the thermal efficiency, it is necessary to understand and optimize the cooling technique,

operating conditions and turbine blade configuration.

The cooling methods implemented in the turbine industry can be classified into

two types: internal and external cooling. Internal cooling is achieved by passing the

coolant through several enhanced serpentine passages inside the blade and extracting the

heat from the outside of the blades. In external cooling, air is bled from the compressor

stage, ducted through internal chambers of the turbine blades and then ejected out

through discrete holes or slots on the blade outer walls to provide a thin, cooler,

insulating film along the external surface of the turbine blade to protect it from hot

combustion gases, due to which the method is called “film cooling”.

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The engine cooling system must be designed to ensure that the maximum blade

surface temperatures and temperature gradients during operations are compatible with the

maximum blade thermal stress for the life of the design. Too little coolant flow results in

higher blade temperatures and reduced component life. Similarly, too much coolant flow

results in reduced engine performance. The engine cooling system must be designed to

minimize the use of compressor bleed air for cooling purposes to achieve maximum

benefits of the high inlet gas temperature.

Highly sophisticated cooling techniques in advanced gas turbine engines include

film cooling, impingement cooling and convection cooling. The optimum combination of

these cooling techniques to meet the highly complex design requirements is the key to

designing air-cooled turbine blades and vanes [1].

1.2 Techniques for Turbine Blade Cooling

Recent development in turbine-cooling technology plays a vital role in increasing

thermal efficiency and power output of advanced gas turbines. In order to improve

convection heat transfer rates in gas turbine engines, serpentine passages are used for

internal cooling of turbine airfoils. Various techniques including film cooling, convection

cooling, jet impingement cooling and pin-fin cooling have been used.

Film cooling, occurs when the coolant flows from inside the airfoil to the outside

environment through manufactured holes in the turbine blade. When, outside, the cooled

air slides along the surface of the airfoil and forms a boundary layer. The boundary layer

acts as an insulator, shielding the airfoil from corrosion and oxidation. To better

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understand film cooling, consider a simple case where the mainstream and air coolant are

mixed up as shown in Figure 2. If there was no coolant, no film, then the heat transfer

will be a simple convection mode and the rate of heat transfer per unit area is

q′′ = h (Tm – Tw), where h is local heat transfer coefficient and (Tm – Tw), is the local

temperature difference between the surface of the airfoil and the mainstream. Heat flux

(q′′) represents the heat exchanged between the hot air and turbine airfoil. It should be

obvious that keeping this value to a minimum is desirable. For this purpose, the film

cooling introduced creates a protection zone between the hot air and surface of the airfoil.

Figure 2 Protective Film Layer Made by Coolant Injection

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Figure 3 Typical cooled aircraft gas turbine blade (Gladden and Simoneau, 1988; collected in

Sokolowski, 1988) [1]

Convection cooling is the simplest way and was the first turbine cooling method

used. With this type of cooling, the coolant air from the base of the turbine blade flows

outward to the end through the internal passages of turbine blade. Figure 4 illustrates the

general concept of convection cooling and Figure 5 shows several possible blade

configurations using convection cooling.

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The effectiveness of convection cooling is limited by size of the internal passages

within the blade and the restriction on the quantity of cooling air available. Air as a

coolant requires large internal surface area and high velocity. One must decide between

the most effective geometry from a heat transfer standpoint and the most economical

from a manufacturing standpoint. Besides, one of the major deficiencies of this method is

failure to cool the thin trailing-edge of the blade effectively [2].

Figure 4 General convection cooling technique

Figure 5 Several blade designs for convection cooling

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Impingement cooling, is a form of convection cooling, the main difference being

that instead of the air flowing radially through one or more sections of the blade, the

cooling air is brought radially through a center core of the blade, then turned normal to

the radial direction, and passed through a series of holes so that it impinges on the inside

of the blade, usually just opposite from the stagnation point of the blade. This is shown

schematically in Figure 6, with one possible design in Figure 7 [2]. Furthermore, among

all heat-transfer enhancement techniques, jet impingement has the most significant

potential to increase the local heat-transfer coefficient. However, the construction of this

flow arrangement weakens the structural strength, and therefore impingement cooling is

used in locations where thermal loads are excessively high. Jet impingement heat transfer

is most suitable for the leading edge of a rotor airfoil (bucket), where the thermal load is

highest and a thicker cross section of this portion of the airfoil can suitably accommodate

impingement cooling [1]. Part of the spent air moves through rows of distributed film

holes on the necessary location for film-cooling surface protection. The other part of the

spent air moves toward the trailing-edge region and exits through tiny holes for trailing-

edge cooling region. Impingement cooling is very effective because the cooling air can be

delivered to impinge on the hot region. However, a spent-air cross-flow effect can reduce

the impingement cooling effect. It is important to determine the impingement heat-

transfer coefficient distributions inside the turbine vane under typical jet configurations

and distributions and engine-cooling flow conditions. Today jet impingement cooling is

used in the trailing and leading-edge region of the turbine blade.

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Figure 6 General impingement cooling technique

Figure 7 One possible impingement cooling design

Pin-fin cooling, pin fins are mostly round projections protruding from the heart-

transfer surface to the coolant flow path. As the name implies, this heat-transfer

enhancement technique uses pin-shaped fins, and these pins are oriented perpendicular to

the flow direction to maximize forced convected fin cooling. Pin-fin cooling analysis is

an interesting combination of internal and external flow analyses. A flow bounded by top

and bottom flat surfaces is a typical internal flow situation, whereas cross-flow over a

tube bundle is a typical external flow situation. In a pin-fin cooled channel, similar to

tube bundles, pins protrude in the flow and the flow is contained in a flow passage. Thus,

both internal and external characteristics are presented in a pin-finned channel. The wake

shed by each pin increases the free stream turbulence, and the boundary layer

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development over the pin-mounted surface gets disturbed. The wakes from upstream pins

also affect both the flow and heat-transfer performance of downstream pins. In addition

to flow disturbances, pins conduct thermal energy away from the heat-transfer surface,

and long pins can increase the effective heat-transfer area [1]. Like cylinders in a cross-

flow, pins shed wake at downstream flow. Besides this wake shedding, a horseshoe

vortex originates just upstream of the base of the pin and wraps the pin around, causing

more flow disturbances. Boundary layer also separates if the pin is placed as a three-

dimensional protrusion. These partial pins or three dimensional protrusions do not extend

to the top surfaces, and heat-transfer enhancement occurs on the pin-mounted surface.

Interactions of all these flow disturbances (wakes, horseshoe vortex, boundary layer

separation) increase heat transfer from the pin-mounted surface. Pins are mostly used in

the narrow trailing-edge of an airfoil where the impingement cannot be accommodated

due to manufacturing constraint. Figure 8 shows typical pin locations in a turbine airfoil.

Figure 8 The trailing-edge of an airfoil is cooled by pin fins (Metzger et al., 1984)

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1.3 Purpose of the Present Investigation

Gas turbines play a significant role generating power to meet increasing demand

in industry. A gas turbine system can be divided into three main components:

compressor, combustion chamber and turbine. The compressor takes in outside-air and

raises it to high pressure in the compression process. The high pressured air is then

passed to the combustion chamber where gas is burned to create a high pressure high

velocity gas. These hot gases with a temperature higher than the melting temperature of

the metallic airfoil can cause severe damage on the turbine blades if the blades are not

properly protected. Consequently, it is imperative to employ effective cooling techniques

in order to keep the blade temperatures at a safe level. In order to prevent premature

failure, it is important to have fairly accurate predictions of the local heat transfer

coefficients and local airfoil metal temperatures. As was mentioned before, impingement

cooling is one effective method to use in the trailing-edge cooling channel of an airfoil

and that is focus of our experimental investigation.

To measure the heat transfer coefficients and pressure drop in a test section that

simulates a typical trailing-edge cooling channel, our scaled-up test section was designed.

The test section had a trapezoidal cross-section with eleven cross-over holes on the wider

base and twelve trailing-edge slots along the narrower base. A fully developed turbulent

flow of cooling air travelled through the cavity adjacent to the trailing-edge cavity. This

air went through the cross over holes and created eleven jets that impinge into the

trailing-edge cavity and then ejected through the trailing-edge slots along the narrower

base. Airfoils with similar cooling arrangements are common in today’s gas turbines.

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Several flow arrangements including the number of the trailing-edge slots, and the angle

of cross-over holes (zero- and five-degree tilt angle), each case with and without the tip

bleed were tested.

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2. LITERATURE REVIEW

Lau et al. (1989) examined the effects of varying the length and the configuration of

the trailing-edge ejection holes on turbulent heat transfer and friction in a pin fin channel.

Experimental results were obtained for two trailing-edge ejection hole lengths, four

ejection hole configurations, and Reynolds numbers ranging from 10,000 to 60,000. The

pins were staggered in the test channel with spacing of 2.5 times the pin diameter in both

straight (radial) flow and the lateral (trailing-edge) flow directions. The results showed

that the overall heat transfer increases when the length of the trailing edge ejection holes

is increased and when the trailing edge ejection holes are configured so that much of the

cooling air is forced to flow farther downstream in the radial flow direction before exiting

the pin-fin channel through ejection holes. Increase in the overall heat transfer is

accompanied generally by an increase in the overall pressure drop, except that the overall

heat transfer was lower and the overall pressure drop was higher when there was no

radial flow ejection. Results also showed that about 15 to 20% of the flow exits through

the tip bleed hole in the cases with both radial and trailing-edge flow ejection [3].

Byerley et al. (1992) presented experimental results on the effects of ejection flow

rate and ejection hole inclination angle on the local heat transfer distribution near the

entrance to a single ejection hole in one wall of a two dimensional smooth duct. They

also included numerically determined flow fields to help describe the mechanisms

responsible for the measured heat transfer distributions. The heat transfer distribution

29

near an ejection hole was dependent on the ejection flow to main flow velocity ratio and

the inclination angle of the hole. The heat transfer was high near the ejection hole,

especially in a region up to several times the hole diameter downstream of the hole. The

heat transfer enhancement upstream of the hole was due to the local acceleration of the

flow entering the hole. Downstream of the hole, the heat transfer enhancement was the

result of the redevelopment of the boundary layers, the downwash from a vortex pair, and

the local acceleration of the reverse flow immediately downstream of the hole, when the

ejection velocity ratio was high [4].

Taslim et al. (1995) investigated the effects of tapered turbulators on heat transfer

coefficient and friction factor in trailing edge passages with and without bleed holes

using the steady state liquid crystals technique. Two low-aspect-ratio trapezoidal test

section geometries, simulating the trailing edge cooling cavities of turbine blades,

roughened with seven different turbulator geometries, were tested for heat transfer and

pressure variations. The test sections were tested both with and without bleed holes. From

this study, it was concluded that:

1) The effects of the geometric parameters (e/Dh, S/e, ARt, etc.) on heat transfer

behavior is consistent with previous results obtained from circular and

rectangular test sections.

2) A large spanwise variation in heat transfer coefficient exists in such type of test

sections due to the major portion of the flow passing through the wider base and

interacting with the high blockage end of the turbulators. Adding bleed holes to

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the small base induced a spanwise velocity component to the main flow and

consequently caused a more uniform distribution of spanwise heat transfer

coefficient.

3) For the “no-bleed” cases, a variation of -40 to +80 percent with respect to the

area-weighted average was recorded. Whereas the variation in local heat transfer

coefficient given in the “no-bleed” cases was reduced to -10 to +41 percent for

the cases with bleed holes.

4) Nusselt numbers measured at two locations along the test sections with bleed

holes correlate well with the local Reynolds number [5].

Ekkad et al. (1998) presented the heat transfer coefficient distributions on the

smooth and rib-roughened walls with ejection holes of two-pass square channels using a

transient liquid crystal technique. Results showed that a higher pressure drop produces

lower bleed mass flow ratio and that the Nusselt numbers in the smooth channel are

enhanced by the presence of bleed holes in the first pass. However, the 180° turn caused a

reduction in the effects of the bleed holes on the second pass due to decreased bleed. For

all ribbed channels, they concluded that bleed holes enhance heat transfer around the hole

edges in the first pass while the effects is reduced in the second pass due to the

combination of the 180° turn and ribs. In addition, the 60° parallel ribs, 60°V, and 60°

inverted ribs produced similar high regional averaged heat transfer enhancement in the

first pass but the 60° inverted V ribs showed the highest heat transfer enhancement in the

second pass. The final conclusion was that, as a result of the increase of the heat transfer

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near the ejection holes, a total loss of 20 to 25% of the main flow through the ejection

holes which is used for external blade surface film cooling did not lower the regional

average heat transfer along the channel [6].

Taslim et al. (1998) applied a Liquid crystal method in an experimental investigation

to measure the local and average heat transfer coefficients on the walls of six test sections

simulating the trailing edge cooling cavity of a modern turbine blade. All test sections

had trapezoidal cross sectional areas with two rows of racetrack-shaped slots on two

opposite bases. Crossover jets, issued from the slots on one base, impinged on the test

section rib-roughened walls and exited from the slots on the opposite wall. The first test

section had all smooth walls and served as a baseline. The remaining five test sections

were rib-roughened on either one wall or two opposite walls simulating the pressure and

suction sides of the blade trailing-edge cooling cavity. In the first four test, the jets issued

into the test section along the test section plane of symmetry. Therefore, the two opposite

walls, simulating pressure and suction sides of the blade, saw the same jet effects. This

symmetric pattern was altered in test sections 5 and 6 in which the jets were tilted

towards one or the other wall at an angle of 6 °. The ribs in the roughened test sections,

covering only 62% of the wall span, were mounted to the surface with an angle of attack

to the jet axis, α, of 30°. The objective of this study was to investigate the effects that

crossover jets have on the heat transfer coefficient and pressure recovery in a cooling

cavity of a modern gas turbine blade. Major conclusions of this study were that

combining the crossover jets with rib-roughened surfaces can be an effective method of

32

cooling the trailing edge cavities and by proper arrangement of the jets and ribs, heat

transfer coefficients on the two opposite walls can be tailored [7].

Taslim et al. (2008) performed an experimental study on the rib-roughened

surface of the trapezoidal channel roughened with tapered turbulators to investigate the

impingement heat transfer coefficients. Several turbulator geometries for different angles

of attack were tested. Two different flow arrangements were tested: 1) all race track

shaped exit holes were opened, and 2) seven consecutive exit holes were blocked. These

geometries and flow arrangements were tested over a range of Reynolds numbers and the

final results (Nusselt numbers) were compared for all cases. Also pressure ratios were

measured for all tests and compared to each other. Major conclusions were: 1) Nusselt

numbers increase monotonically with increasing Reynolds numbers, 2) When the exit

slots were partially blocked, areas one through three had lower heat transfer coefficient

compared with the case of all open exit holes, 3) Areas 4 and 5, the 7 blocked exit holes

cases induced higher Nusselt numbers than the cases where all exit holes were opened, 4)

Both pressure ratios, Pplenum/Ptrailing edge and Ptrailing edge/Pambient, increase with increasing

Reynolds numbers and 5) The 7 blocked exit holes cases induced higher pressure ratios

than the cases where all exit holes were opened [8].

33

3. EXPERIMENTAL SETUP AND PROCEDURE

3.1 Experimental Setup

Detailed schematics of the experimental setup are shown in Figures 10 through

23. Figures 10 through 13 show the overall arrangement of the test sections. Figure 10

shows the set up with all open the trailing-edge (TE) slots, Figure 11 shows the set up

with 2 blocked TE slots and Figure 12 shows the set up with 4 blocked TE slots. Figures

14 through 19 show all flow arrangements. Figures 20 and 21 show the cross sectional

area of the test section with a trapezoidal shape. Figures 22 and 23 show the details of the

cross-over and trailing-edge slot geometries. These slots were drilled on the partition wall

and along the trailing-edge wall.

Air entered the plenum from two inlet ports on two opposite walls of the plenum

which was made from 1-inch-thick clear acrylic plastic with a cubical shape of 20-inches-

length by 20-inches-width by 20-inches-depth. A honeycomb flow straightener was

placed in a groove in the middle of the plenum. Two thermocouples measured the air

inlet temperature to the test section. A pressure tap was also provided to measure the

plenum pressure. A bell-mouth opening machined on the plenum wall introduced the air

into test section.

The supply channel which had a trapezoidal cross-section was made from 0.5-

inches-thick clear acrylic plastic. The front, back and bottom walls were rectangular

while the side wall was in trapezoidal shape. A thermocouple measuring the jet

34

temperature was located in the middle of the bottom wall of the supply channel. Three

pressure taps were provided to measure the pressure in the supply channel. The first

pressure tap was located 17 inches away from the plenum to measure the supply channel

pressure before the air entered the cross-over holes on the removable partition wall,

shown as Psup1 in Figures10 through 13. The next tap was 13.25 inches away from the first

pressure tap to measure the supply channel pressure at the location of the sixth cross-over

slot, shown as Psup2 in Figures 10 through 12. The last pressure tap was located on the

side wall at the end of the supply channel to measure the supply channel pressure at the

end of the channel, shown as Psup3 in Figures 10 through 12.

The 36-inches-length removable partition wall with 11 holes was placed in

between the top of the supply channel and the bottom of the trailing-edge cavity, the first

end which was located 11.5 inches away from the plenum. The first hole of the

removable partition wall was located 8.5 inches from the first end and each hole was 2

inches apart from each other. This partition wall was also made from clear acrylic plastic

in a trapezoidal shape.

The top wall of the trailing edge cavity is known as trailing-edge removable

partition wall. The 38-inches-length with 12 holes of the trailing-edge removable wall

was also made from the clear acrylic plastic with the trapezoid shape. The first hole was

placed 7.5 inches away from the first end of the trailing edge removable partition wall

base on center point and each hole was 2 inches apart from each other.

35

The back wall of the trailing-edge channel is one of the most important parts,

which in this wall contain the stack of elements in the heaters and liquid crystal film is

shown in Figure 24. This wall contains three heaters each measuring 11.1 inches by 2.6

inches in rectangular shape. The heaters were connected to the power supply on the

home-made power distribution panel which controlled the voltages across each heater.

The voltages were adjusted such that all heaters had almost the same heat flux. In this

case, since heaters were identical, we had equal voltages across the heaters1 through 3.

For the liquid crystal, liquid crystals are referred to as thermochromic since they reflect

different colors selectively when subjected to temperature changes. At any particular

temperature, liquid crystals reflect a single wavelength of light. The colors can be

calibrated to particular temperatures since the transition of colors is sharp and precise,

[1]. Where the calibration of this experimental investigation was considering an

appearance of specific green shade which corresponding to a temperature of 93.2 °F.

The three walls of the trailing-edge channel were made of 0.5-inches-thick clear

acrylic plastic. The front wall had a rectangular shape while the two side walls had a

trapezoidal shape. Two thermocouples measured the temperature at the two ends of the

channel (Tend1 and Tend2) as shown in Figures 10 through 12. Two pressure taps were also

mounted to measure the pressure at both ends of the trailing-edge channel (Pend1 and

Pend2) as shown in Figures 10 through 12. A bleed line, made of a 0.5-inch-diameter PVC

pipe, was connected to the end of the trailing-edge side wall. A bleed valve controlled the

amount of bleed. The mass flow rates were measured using two different devices. A

critical venturi with a throat diameter of 0.32 inches, manufactured by Fox Valve

36

Development Corp. measured the total mass flow rate entering the test section. An orifice

plate with a 0.6-inch-hole diameter measured the bleed flow. The critical venturi always

used for measuring the total mass flow rate while the orifice plate was only used when

the bleed valve was open. The mass flow rate at critical venturi was calculated by the

following equation and the Figure 9 is shown the schematic of critical venturi.

Tven Pven

Dthroat=0.32

”

Flow Flow

Figure 9 Schematic of critical venturi

37

460

)(5215.0

ven

throatambven

T

APPm

Where

m : Mass flow rate through the critical venturi (lbm/s)

throatA : Area of critical venturi throat ( 2

4D

)

venP : Pressure at venturi inlet (psig)

ambP : Ambient pressure from the site http://www.weather.com in

Boston area

venT : Air temperature at venturi inlet (˚F)

The correlation for the mass flow rate through the orifice plate bleedm

, provided by the

manufacturer is:

absf

bfw

orificemhableed

TZ

GPhDYFFFKm

)(862509.0 2

1

Where

bleedm

: Mass flow rate through bleed valve

fP : Air pressure at the orifice inlet (psi)

38

wh : Pressure drop across the orifice (psi)

absT : Air temperature entering the orifice in degrees Rankine

fZ = 0.99958

bG = 1

aF = 1.00013

hF = 1.00000

mF = 1.00000

K = 0.80466

1Y = 0.99527

orificeD = 0.6

The camera, Panasonic Lumix model DMC-FZ8 was used to take a photograph of

the reference color display on the liquid crystal film during the experiment. A set of

photographs were taken by increasing incrementally the voltage across the heaters.

Approximately 20 photographs were taken in each air mass flow rate (Reynolds number).

The first picture was taken when the first green patch appeared on the liquid crystal film

and last picture was the last green patch appeared on the liquid crystal film. To remove

all reflections from the surrounding lights in the lab on the photographs, the rectangular

black boxes were created to cover the camera and the test section area. The box one 14

inches long 14 inches wide and 2.5 inches deep. The other one 10 inches long 10 inches

39

wide and 2.5 inches deep, big enough to cover all 5 areas of interested. A smaller box

was used when only one area was covered. The camera viewing area is shown in Figures

13 through 19.

40

Figure 10 Setup for the case of all open trailing-edge slots

41

Figure 11 Setup for the case of two blocked trailing-edge slots

42

Figure 12 Setup for the case of four blocked trailing-edge slots

43

Figure 13 Details of the test section

44

Figure 14 Setup for the inline arrangement with all open trailing-edge slots

45

Figure 15 Setup for the inline arrangement with two blocked trailing-edge slots

46

Figure 16 Setup for the inline arrangement with four blocked trailing-edge slots

47

Figure 17 Setup for the staggered arrangement with all open trailing-edge slots

48

Figure 18 Setup for the staggered arrangement with two blocked trailing-edge slots

49

Figure 19 Setup for the staggered arrangement with four blocked trailing-edge slots

50

Figure 20 Details of the rig cross-section for zero-degree tilt angle

51

Figure 21 Details of the rig cross-section for five-degree tilt angle

52

Figure 22 Details of the trailing-edge slots

Figure 23 Details of the cross-over holes

53

Figure 24 Stack of elements in the heaters and liquid crystal film

54

3.2 Experimental Procedure

Before running the experiment the liquid crystal film needed to be calibrated.

Liquid crystal calibration is the process whereby specific temperatures are assigned to

specific colors. The procedure establishes a one-to-one correspondence between the color

and temperature. This is achieved by creating an isothermal surface of known

temperature, treated with liquid crystal, changing the temperature and assigning a color

band to that temperature [12]. In this experiment, the method used was the isothermal

bath where the calibration surface was coated with liquid crystal and mounted to an

isothermal plate located in a well stirred water bath. The reference temperature

corresponding to a specific shade of green was selected to be 93.2°F.

The compressor which provided the air to the system was turned on and a venturi

pressure was set for a pre-determined Reynolds number. The voltages across different

heaters were adjusted such that to see the first patch of green color on the liquid crystal

film.

In this experiment, since all three heaters had identical resistances, we had

identical voltages across the heaters. This was done using the home-made power

distribution panel. The home-made power distribution panel is shown in Figure 25.

55

Figure 25 Home-made power distribution panel

The control panel contains eight small rheostats. Each rheostat controlled the

voltage across a heater since we had three heaters, we used three rheostats. These heaters,

as we mentioned in the previous section had the same area and the same resistance.

Therefore, to have a uniform heat flux on the heated wall, we set the small rheostats to

the same level. To move the liquid crystal reference color from one location to another,

we had to gradually increase the power to each heater. This was done with the bigger

rheostat which controlled the voltage going to smaller rheostat.

An eight-way switch connects the volt- and amp-meters to one heater at a time.

Therefore, we were able to record the voltage and current for each heater. Eight 2-amp

fuses prevented any malfunction of the rheostat.

Depending on the level of pressure, two manometers were used- one was a micro-

manometer and the second one was an oil manometer. The micro-manometer model 1430

Microtector® Portable Electronic Point Gage with an accuracy of to ±.00015 inches of

56

water column, powered by an AA battery, used a fluid type of A-126 fluoresce in green

color concentrate which had a specific gravity of 1.0. The maximum pressure measured

by this micro-manometer is 2 inches of water column. The micro-manometer needed to

be primed every time it was used. First, the instrument had to be leveled (Figure 26-A).

Next, in Figure 26-B, while both ends of the tube were opened to ambient air the needle

had to be adjusted by turning it up or down until the tip itself made contact with the liquid

surface. This position was set to zero. Any other pressure was measured with respect to

this zero reference. The micro-manometer was used for measuring small pressures (< 2”

of H2O) in most cases especially for ΔP jet and ΔP orifice. Whenever the pressure exceeded

the maximum scale of the micro-manometer the oil manometer was utilized.

Figure 26 Micro-manometer

A

B

57

The oil manometer used in this experiment, had an orange pigment and a specific

gravity of 0.827. This type of a manometer has an accuracy of 0.05 inches of liquid. It is

important that the oil level in the tube when it is not connected to any pressure source be

zero to avoid any further adjustment.

For temperatures, Agilent 34970A Data Acquisition Switch Unit was used to

record the temperature. The temperature was read in degree Fahrenheit from 7 different

on channel. The first reading was at venturi, the next two readings in the plenum as the

inlet temperature, the fourth and fifth readings were at both ends of the trailing edge

cavity and the sixth reading was the ambient air. The final reading was at the middle of

the feed channel where air entered the sixth cross-over have to form a jet in the trailing

edge cavity. This is called jet temperature. Figures10 through 12 show the positions of

the thermocouples.

The experimental investigation was divided into two parts; one was heat transfer

test and the other was the cold test. The heat transfer test was run for six different venturi

inlet pressures of 13, 26, 40, 54, 69 and 80 psig corresponding to six jet Reynolds

numbers. Before turning on the compressor, the water from the condensation in the

system was drained out in order to measure only the air flows in the test section. Then the

compressor was turned on to supply the coolant air. The air, after exited the compressor,

passed through a refrigerated air dryer system located above the compressor where most

of the water vapor in the air was condensed and removed. From there, the air was

transported to a storage tank. After that, the air went through an air filter in the laboratory

where the experimental investigation was performed. The venturi gauge pressure was set

58

to a pre-determined level and the electrical power panel, volt-meter, amp-meter,

acquisition switch unit, micro-manometer, and the camera were turned on. After all

devices were ready and the system reached equilibrium state then the data acquisition

process began. The air was supplied to the plenum from two inlet ports on two opposite

side walls and passed through the flow straightener to the supply channel. Then the air

travelled across the cross-over holes to the trailing-edge cavity and then passed through

the trailing-edge slots to the ambient. While the air was passing through the trailing edge

cavity photographs were taken at the five areas of interest as the voltage across the

heaters was increased by approximately 1 volt at a time.

Several flow and geometry arrangements were studied; the angle of cross over

holes (zero- and five- degree tilt angle), each case with and without tip bleed were tested

and each was done for staggered and inline arrangements. When the cross-over slots and

trailing-edge slots were inline the geometry looked like what is shown in Figures 14

through 16 and when the cross-over slots were staggered with respect to the trailing-edge

slots the geometry looked like what is shown in Figures 17 through 19. All cases have

been investigated are shown in the tree-diagram below.

59

Zero degree tilt angle

- Case-1 Zero degree tilt angle with no blocked holes-Inline

- Case-2 Zero degree tilt angle with no blocked holes-Staggered

- Case-3 Zero degree tilt angle with two blocked holes-Inline

- Case-4 Zero degree tilt angle with two blocked holes-Staggered

Five Degrees

No Blocked Hole (Opposite Wall)

Two Blocked Holes

Four Blocked Holes

In line

Staggered

In line

Staggered

In line

Staggered

No Blocked Hole

Staggered

In line

Zero

Degree

No Blocked Hole

Two Blocked Holes

Four Blocked Holes

In line

Staggered

In line

Staggered

In line

Staggered

60

- Case-5 Zero degree tilt angle with four blocked holes-Inline

- Case-6 Zero degree tilt angle with four blocked holes-Staggered

Five degree tilt angle

- Case-7 Five degree tilt angle with no blocked holes-Inline

- Case-8 Five degree tilt angle with no blocked holes-Staggered

- Case-9 Five degree tilt angle with no blocked holes-Inline-Opposite wall

- Case-10 Five degree tilt angle with no blocked holes-Staggered-Opposite wall

- Case-11 Five degree tilt angle with two blocked holes-Inline

- Case-12 Five degree tilt angle with two blocked holes-Staggered

- Case-13 Five degree tilt angle with four blocked holes-Inline

- Case-14 Five degree tilt angle with four blocked holes-Staggered

Before a geometry was tested it was necessary to do a leakage test to ensure that

the test section did not leak and the setup was sealed with silicon at every separation line

where the air could leak. Once this procedure had done and the silicon dried out then the

experiment was run.

The most important part of the investigation was the blocked-hole cases because

the main objective of this experiment was to see the effects of the cross flow on heat

transfer coefficient. Once two or four holes were blocked, two or four jets could not

61

directly exit the trailing-edge channel. As a result, they turned around and created cross

flow. The heat transfer coefficient and the area corresponding to it can be calculated from

the surface area of the green pixels on each photograph in conjunction with the jet

temperature, liquid crystal reference temperature, the coolant flow rate and the heat flux

measurement.

The cold test was run for 11 different venturi inlet pressures at 5, 10, 15, 20, 25,

30, 40, 50, 60, 70 and 80 psig while the heaters were off. Therefore, the voltages and

amperages were not measured in the cold test. The purpose of running the cold test was

to find the pressure ratio between the supply and trailing-edge channels.

62

3.2.1 Digital Pictures

The photographs were taken with a digital camera in flash off mode. Special non-

reflecting light bulbs were used in order to photograph the temperature distribution along

the liquid crystal film through the transparent Plexiglas with no glare and no reflection.

Once the experiment was done the photographs were collected approximately 20

photographs per case. Then, the images were transferred to a computer and each was

labeled and saved in jpg format. Next step, was using Sigma Scan Pro5 a commercial

software to digitize each picture. In the digitization process, the total number of pixels

converting the green reference color were counted and saved in a file.

3.2.2 Code Steps

After all raw data were collected, several data sheets in electronic format needed

to be prepared; the raw data log sheets for both heat transfer and cold tests were put into

excel format. The surface areas of the green pixels also need to be prepared as mention

previously. The input file in appropriate ASCII format was created to be used in

FORTRAN program to calculate Reynolds numbers, the heat transfer coefficient and the

Nusselt numbers. The raw data can be seen in Appendices A and B. Using the FORTRAN

program check.f, (Appendix C), each raw data set file was checked for typographic

errors. When errors were indicated the file was fixed and the program was re-run until no

errors occurred. When the file passed the check, the energy balance for each photograph

set was iteratively solved, using the FORTRAN program reduce.f (Appendix D). Finally,

the photograph-area data file and the output files from reduce.f were used to calculate the

area-weighted averages of the Nusselt number with respect to Reynolds number, using

63

the program integarea.f (Appendix E).The following chapter describes the data reduction

in more detail, relating the actual equations used in the FORTRAN programs.

64

4. DATA REDUCTION

The data reduction was done by running three different FORTRAN programs; check.f,

reduce.f and integarea.f. The first step was to check that the raw data file did not have

any transposition errors. These errors may have occurred when the hand-recorded data

was input to a computer text file. The inspection of the data files was accomplished by

running the FORTRAN program check.f. Basically, this program read in each column of

the text file and checked the data for major deviations. After reading each row, the

voltage across each heater was divided by its respective current to calculate the heater’s

resistance. If at any point the calculated resistance varied by more than one percent, the

data point was flagged. Then, that row of data was inspected alongside the hand-recorded

data for inaccuracies. Other checks in this program ensured that the temperature and

pressure inputs were not beyond the expected range of variation.

After checking for errors, the reduce.f program was run to do all the necessary

calculations for data reduction. This program calculated the total heat flux corresponding

to each heater as well as the local heat transfer coefficients, total heat losses to the

ambient, radiative fluxes, Nusselt number using an iterative process in conjunction with

thermal resistance equations.

The reduce.f program performed all geometric calculations first. Those geometric

calculations included: The perimeter of the channel, the height of the channel, the cross-

sectional area of the channel, the hydraulic diameter (Dh) determined by multiplying the

cross-sectional area of the bleed hole by four and dividing the result by the cross-

section’s perimeter.

65

P

AD cross

h

4

Where:

Dh Hydraulic Diameter

Across Cross-Sectional Area

P Perimeter of the Cross-Sectional Area

Then the air mass flow rate (lbm/s) through the critical venturi was calculated,

following the correlation provided by the manufacturer.

460

)(5215.0

ven

throatambven

T

APPm

The inlet temperature was taken as an average of the two temperatures recorded

and the ambient pressure was converted from inches of mercury to pounds per square

inch.

)(5.0 21 ininin TTT

Air properties at jet temperature were used to calculate the jet Reynolds number

using the equation below.

jet

jetP

m

4Re

66

Then the total heat added to the coolant by the heaters was determined. Since,

only heater 1, 2 and 3 were designated, the following formula was developed to solve for

power (BTU/hr):

Q=3.4121 (V1A1+V2A2+V3A3)

Therefore, the heat flux (BTU/ (ft2·hr)) to the sidewall at the target zone was:

HeaterArea

VAq sidewall

224121.3"

Next, the losses due to conduction through the walls of the test section were

accounted for using thermal equivalent circuits. The natural convection heat transfer

coefficient on the outer surface of the channel was calculated from the following

correlation [10]:

e

ambo

D

kh

36.0

Where De is the hydraulic diameter of the outside of the channel and the thermal

conductivity of air is found at ambient temperature. Ultimately, the resistance to heat

transfer due to natural convection is equal to the inverse of the convection coefficient.

However, the resistance due to conduction is dependent on the inverse of the material’s

thermal conductivity multiplied by the material’s thickness. Therefore, it was important

to break down the heater assembly into its numerous layers of adhesive, kapton and the

heating element itself. The liquid crystal foil was broken into Mylar protective, liquid

crystals and absorptive black paint background and Plexiglas outer wall and polyurethane

insulation, depending on which wall was analyzed [11]. When looking at the back wall

from the inside of the test section at the center of the heating element to ambient air, it

67

contained the stacks of the following; 0.25 mil inconel heating element, 0.5 mil adhesive,

0.5 mil kapton, 2 mil adhesive, 3 inches polyurethane. And from the center of the heating

element to the air inside the test section have contained 0.25 mil inconel heating element,

1.0 mil adhesive, 2.0 mil kapton, 1.0 mil adhesive, 0.5 mil inconel spreader, 1.0 mil

adhesive, 2.0 mil kapton, 1.5 mil adhesive, 3.0 mil absorptive black background, 2.0 mil

liquid crystal and 5.0 mil mylar. The materials used had the following thermal

conductivities: kkapton = 0.0942 BTU/hr.ft·°F, kpolyurethane= 0.033 BTU/hr.ft·°F,

kplexiglas=0.11 BTU/hr·ft·°F, kmylar=0.085 BTU/hr·ft·°F, kadhesive=0.1272 BTU/hr·ft ·°F,

kinconel=9.0152 BTU/hr·ft·°F, kblack background=0.165 BTU/hr·ft ·°F, kliquid crystal=0.165

BTU/hr·ft ·°F

When looking at the front wall (camera side), the circuit comprises of the air

inside the test section, and a following layer of 0.46 inches Plexiglas to ambient air.

Using an iterative method in conjunction with the data collected and the

resistances calculated, an energy balance was solved and the internal convection

coefficient of the sidewall was determined. Following are the algebraic equations used to

solve the energy balance. The temperature at the center of the heating element was found

in terms of heat flux as given below.

outsidewallinsidewall

outsidewall

amb

insidewall

crystalliquid

sidewall

heater

RR

R

T

R

Tq

T11

"

Knowing the temperature of the heating element center would allow the heat flux

out of the heater, from the heater to the ambient air, as well as the heat flux toward the

68

liquid crystal layer to be determined. The following equations were used to calculate the

heat flux out and the heat flux in.

outsidewall

ambheateroutsidewall

T

TTq

"

insidewall

crystalliquidheater

insidewallT

TTq

"

The percentage of heat loss was found as follows.

sidewall

outsidewall

lossq

qH

"

"100%

The surface temperature of the side wall could be solved by further manipulation

of the thermal equivalent circuit.

mylsidewallcrystalliquidsurface RqTTinsidewall "

Film temperature can also be calculated as given below.

2

msurface

f

TTT

sidewalll

More calculations of unheated and heated channel wall areas were needed as well

as flow temperature in order to solve for the convection coefficient of the sidewall at the

target zone. The total area of unheated surfaces on the top wall, bottom wall, and front

sidewall were calculated as was the total heated sidewall surface area in order to calculate

the radiational losses from the heated wall to the unheated walls. The inlet temperature

average of air was used to supply the air properties which included density, specific heat,

thermal conductivity, viscosity, and Prandtl number. Then, the air mixed mean

69

temperature at the target zone was calculated using an energy balance from the channel

inlet to the point in question.

inp

inm

Cm

QTT

.

3600

Using Newton’s law of cooling, the convective heat transfer coefficient at the

sidewall could then be calculated.

msurface

losssidewallsidewall

sidewallTT

qqh

sidewall

""

Here an initial guess set the convective heat transfer coefficient of the opposite

wall equal to the target wall’s convective heat transfer coefficient. The temperatures of

the top and bottom walls were each set equal to the air mixed mean temperature, which

was also an initial guess. The actual wall temperatures took into consideration the

conductive heat losses and the radiative heat losses as well. This involved calculating the

view factors in and out of the four surfaces, setting the emissivities, finding the

temperatures at each surface, and finding the heat flux into or out of each surface.

Therefore, another thermal equivalent circuit which accounted for radiation had to be

established and solved. The resistances to radiative heat transfer to the ambient air at the

top wall, front sidewall, and bottom wall was equal to the sum of resistances of the

Plexiglas, and natural convection. The resistance to convective heat transfer at each of

those walls was equal to the inverse of the respective wall’s convection coefficient. Thus,

the temperature at each of those walls could be found in terms of the heat flux at each

wall.

70

outin

radiation

outin

bottombottom

bottom

bottom

amb

bottom

m

bottom

RR

qR

T

R

T

T11

"

Once the temperature of the top, front, and bottom surfaces was known, the heat

flux out toward the ambient at each of those walls could be found.

out

out

bottom

ambbottom

bottomR

TTq

"

Then the total heat loss to the ambient air via radiation and conduction could be

quantified.

sidewallsidewallfrontfrontbottombottomtoptopwaste AqAqAqAqQoutoutoutout

Therefore, the net heat added to the air from the heaters from inlet to the target

zone is the difference between the heat supplied by the heaters and the heat wasted.

wasteadded QQQ

Hence an energy balance between the inlet and the point in question was applied

again to find the mixed mean temperature of air. The losses in heat flux at the sidewall

are simply the summation of the losses due to radiation and conduction out to the ambient

air. Then the heat transfer coefficient at the side wall from the Newton’s law of cooling

was obtained as follows.

71

msurface

sidewallsidewallsidewall

sidewallTT

qqqh

sidewall

radiationout

net

...

Air properties at jet temperature were used to calculate the jet Reynolds number

using the equation below.

jet

jet

jetperimeter

m

4

Re

Where jetm.

is the mass flow rate through one cross-over hole and perimeter is

that of one cross-over hole.

The experimental uncertainty in each data point calculation was determined,

following the Kline and McClintock’s [9] method.

The convection coefficient found for the target wall could then be multiplied by

the hydraulic diameter of the cross-over hole and divided by the air thermal conductivity

at jet temperature to determine the Nusselt number.

k

DhNu

hwallett

jet

net

arg

The last step in data reduction was to run the FORTRAN program integarea.f

which found the area weighted average of the Nusselt number and the uncertainty

associated with the average Nusselt number at each flow rate. For example, the average

Nusselt number was calculated from:

72

ni

i

i

i

ni

i

i

A

ANu

Nu

1

1

Where A is the pixel area digitized and n is the number of pictures.

All baseline experiments, where all trailing edge slots were open, showed

repeating color patterns on the area between two adjacent jets. That is the reason why

only digitization of one representative area was needed. In the cases where two or four

consecutive trailing edge slots were closed, distribution of green reference color didn’t

show a repeating pattern along the test section. Different color distributions were

recorded because little air was reaching on the areas under the blocked holes. It took

some time for the green color to widespread in the region where the trailing edge were

opened because the cooling air at a higher velocity was flowing on the target area under

the open trailing-edge slots.

73

5. RESULTS AND DISCUSSIONS

The comparison of fourteen different flow arrangements were made by comparing: a)

zero- and five- degree tilt angles of the cross-over holes, b) inline and staggered

arrangements, and c) 2 or 4 consecutive trailing-edge blocked holes to create the cross

flow. These comparisons are based on the magnitude and behavior of the Nusselt

number, (heat transfer coefficient) and pressure ratio.

The results of all fourteen cases are shown in the Tables 1 through 3. The experimental

uncertainty associated with the Nusselt number was estimated to be 6 % according to

Kline and McClintock’s method [9].

The following Tables contain all reduced results, categorized by Reynolds number,

Nusselt number and heat transfer coefficient.

74

Table 1: Reynolds Number, Nusselt Number, and Heat Transfer Coefficient Results for

All Baseline Cases

Zero-degree tilt angle of cross-over holes

Inline Staggered

Re Nu h Re Nu h

10262 33.659 11.534 10219 32.568 11.146

15086 41.66 14.244 15033 40.931 13.982

20256 52.404 17.885 20219 51.323 17.494

25354 59.515 20.313 25298 59.787 20.387

30725 71.222 24.348 30699 69.609 23.747

35434 77.421 26.443 35389 80.204 27.35

Five-degree tilt angle of cross-over holes

Inline Staggered

Re Nu h Re Nu h

10109 40.531 13.874 10284 36.914 12.647

15049 48.166 16.352 15130 47.696 16.295

20296 59.942 20.289 20290 56.413 19.255

25419 72.221 24.456 25387 69.261 23.681

30800 84.199 28.603 30776 78.191 26.797

35382 94.064 32.011 35390 89.859 30.857

Five-degree tilt angle of cross-over holes (measurements on the opposite wall)

Inline Staggered

Re Nu h Re Nu h

10211 33.646 11.533 10300 35.448 12.178

15053 45.577 15.56 15125 44.696 15.303

20240 59.445 20.348 20253 63.454 21.731

25357 66.23 22.381 25325 78.972 27.08

30729 80.413 27.418 30750 84.066 28.852

35351 84.003 28.933 35398 99.6 34.23

* Note: h (BTU/hr·ft2·°F)

75

Table 2: Reynolds Number, Nusselt Number, and Heat Transfer Coefficient Results

for Zero-Degree tilt angle and Blocked Holes Cases

Zero-degree, Inline, 2 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10196 22.9 7.841 10196 34.97 11.98 10196 39.21 13.43 10196 39.96 13.69 10196 40.12 13.75

15007 28.64 9.784 15007 45.14 15.42 15007 51.28 17.51 15007 51.5 17.59 15007 53.88 18.4

20179 36.61 12.49 20179 51.57 17.59 20179 61.5 20.97 20179 58.01 19.79 20179 60.82 20.74

25298 44.31 15.09 25298 61.62 20.98 25298 74.56 25.39 25298 71.49 24.34 25298 72.55 24.7

30680 51.25 17.48 30680 73.47 25.05 30680 86.77 29.58 30680 85.92 29.29 30680 84.45 28.79

35355 57.38 19.57 35355 78.64 26.81 35355 94.39 32.18 35355 92.98 31.7 35355 96.9 33.04

Zero-degree, Staggered, 2 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10169 22.44 7.673 10169 33.36 11.41 10169 34.68 11.86 10169 36.52 12.49 10169 38.72 13.24

14999 29.63 10.11 14999 44.02 15.02 14999 48.88 16.67 14999 49.79 16.98 14999 51.3 17.5

20188 37.57 12.8 20188 52.44 17.87 20188 59.05 20.12 20188 60.72 20.68 20188 62.91 21.43

25304 45.26 15.42 25304 64.41 21.94 25304 71.18 24.25 25304 71.52 24.36 25304 75.23 25.63

30697 53.21 18.16 30697 77.43 26.4 30697 83.06 28.32 30697 83.83 28.59 30697 84.3 28.75

35425 63.27 21.57 35425 90.17 30.74 35425 98.31 33.52 35425 98.27 33.51 35425 100.3 34.2

Zero-degree, Inline, 4 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10313 21.95 7.511 10313 31.74 10.86 10313 35.41 12.12 10313 38.91 13.32 10313 41.44 14.19

15170 27.6 9.414 15170 37.05 12.64 15170 43.7 14.9 15170 46.23 15.77 15170 52.07 17.76

20350 33.66 11.47 20350 45.81 15.61 20350 53.75 18.31 20350 58.13 19.8 20350 64.98 22.13

25438 39.08 13.33 25438 54.77 18.67 25438 61.01 20.8 25438 65.41 22.3 25438 79.61 27.13

30868 49.28 16.81 30868 65.95 22.5 30868 72.06 24.58 30868 76.97 26.26 30868 88.98 30.35

35544 57.31 19.57 35544 71.44 24.39 35544 79.23 27.05 35544 91.61 31.28 35544 101.7 34.73

Zero-degree, Staggered, 4 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10260 25.11 8.613 10260 34.44 11.81 10260 36.99 12.69 10260 40.45 13.87 10260 43.09 14.78

15097 27.8 9.504 15097 38.43 13.14 15097 43.93 15.02 15097 47.77 16.33 15097 52.97 18.11

20298 35.81 12.2 20298 49.52 16.87 20298 55.85 19.03 20298 60.28 20.54 20298 70.02 23.86

25432 43.88 14.95 25432 59.23 20.18 25432 64.83 22.08 25432 72.42 24.67 25432 79.17 26.96

30914 46.85 15.93 30914 61.57 20.95 30914 67.33 22.92 30914 74.77 25.46 30914 81.44 27.73

35559 54.67 18.63 35559 75.74 25.83 35559 85.48 29.15 35559 92.4 31.51 35559 103.3 35.22

*Note: h (BTU/hr·ft2·°F)

76

Table 3: Reynolds Number, Nusselt Number, and Heat Transfer Coefficient Results for

Five-Degree tilt angle and Blocked Holes Cases

Five-degree, Inline, 2 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10255 20.93 7.149 10255 33.71 11.52 10255 37.89 12.95 10255 39.56 13.52 10255 38.15 13.04

15089 27.96 9.543 15089 43.11 14.71 15089 45.11 15.38 15089 46.47 15.85 15089 47.18 16.09

20277 38.15 12.99 20277 56.35 19.19 20277 57.86 19.7 20277 60.95 20.75 20277 57.54 19.59

25400 46.76 15.93 25400 63.94 21.78 25400 65.96 22.47 25400 66.59 22.69 25400 67.12 22.87

30802 51.95 17.72 30802 75.58 25.77 30802 78.03 26.6 30802 76.89 26.21 30802 81.67 27.84

35307 63.65 21.78 35307 84.15 28.8 35307 87.19 29.84 35307 88.9 30.42 35307 91.82 31.42

Five-degree, Staggered, 2 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10258 20.35 6.945 10258 33.31 11.37 10258 35.17 12.01 10258 38.34 13.09 10258 36.15 12.34

15096 27.57 9.388 15096 43.81 14.91 15096 45.04 15.33 15096 46.43 15.8 15096 46.07 15.68

20292 33.17 11.27 20292 55.47 18.85 20292 56.22 19.1 20292 58.08 19.74 20292 60.49 20.55

25401 48.04 16.33 25401 69.71 23.69 25401 70.82 24.07 25401 72.57 24.66 25401 73.39 24.94

30860 58.7 19.95 30860 90.24 30.66 30860 85.17 28.93 30860 88.6 30.1 30860 91.52 31.09

35538 67.32 22.89 35538 97.06 33 35538 95.06 32.32 35538 94.88 32.26 35538 102 34.68

Five-degree, Inline, 4 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10207 18.85 6.453 10207 28.26 9.678 10207 31.93 10.94 10207 34.03 11.66 10207 39.49 13.53

15056 21.72 7.416 15056 37.04 12.64 15056 39.95 13.63 15056 40.44 13.8 15056 48.58 16.57

20255 30.2 10.29 20255 47.56 16.19 20255 50.01 17.03 20255 52.28 17.8 20255 60.69 20.66

25336 38.37 13.08 25336 62 21.12 25336 63.52 21.64 25336 65.96 22.48 25336 76.33 26

30786 47 16.03 30786 69.95 23.85 30786 71.69 24.44 30786 72.81 24.82 30786 83.81 28.56

35363 57.79 19.74 35363 80.19 27.39 35363 83.76 28.61 35363 88.99 30.39 35363 94.72 32.35

Five -degree, Staggered, 4 Blocked

Area 1 Area 2 Area 3 Area 4 Area 5

Re Nu h Re Nu h Re Nu h Re Nu h Re Nu h

10228 18.35 6.264 10228 31.09 10.62 10228 30.64 10.47 10228 35.26 12.05 10228 41.71 14.25

15089 24.42 8.322 15089 39.59 13.48 15089 43.1 14.68 15089 47 16.01 15089 50.92 17.34

20280 32.06 10.9 20280 50.63 17.21 20280 53.72 18.26 20280 62 21.07 20280 66.34 22.55

25389 39.96 13.6 25389 63.51 21.6 25389 64.99 22.1 25389 71.1 24.18 25389 76.9 26.15

30822 44.1 15.01 30822 64.71 22.01 30822 68.64 23.35 30822 72.85 24.78 30822 74.91 25.48

35516 59.49 20.26 35516 84.38 28.75 35516 85.84 29.24 35516 95.13 32.41 35516 96.06 32.73

*Note: h (BTU/hr·ft2·°F)

77

5.1 Nusselt Number

As mentioned earlier, the target zone in all tested configurations was divided into

five areas as shown in Figures 13 through 19 where Nusselt Numbers were obtained for

each of those areas. For all the cases, those Nusselt Numbers were plotted versus the jet

Reynolds Numbers.

5.1.1 Cases One through Fourteen Analyses

The results show that the Nusselt number increases as the Reynolds number

increases as shown in Figures 27 through 40. This behavior is expected because when the

mass flow rate increases, the convective effects of the flow also increase so Nusselt

number increases. Besides, the heat transfer results show that, for any given area, the

slope of the lines does not change significantly in the plots as the flow arrangement

changes. All flow arrangements, both in baseline and blocked holes cases, show the

maximum Nusselt number on the area 5 as shown in Figures 27 through 40. The

maximum Nusselt number occurred on area 5 which can be explained by the effects of

the cross flow that is created when two or four consecutive cross over holes are blocked.

The results of the baseline cases show that the setup with 5-degree tilt angle of cross-over

jets produced higher Nusselt numbers than the 0-degree-tilt cases. Therefore, it can be

concluded that the 5-degree tilt angle jets can be more effective than 0-degree jets.

78

5.1.2 Plots of Cases 1 through 14

Figure 27 Nusselt Number versus Reynolds Number for the Baseline case of Zero-degree tilt, Inline

Figure 28 Nusselt Number versus Reynolds Number for the Baseline case of Zero-degree tilt,

Staggered

79

Figure 29 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree tilt, Inline

Figure 30 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree tilt,

Staggered

80

Figure 31 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree tilt,

Opposite- wall, Inline

Figure 32 Nusselt Number versus Reynolds Number for the Baseline case of Five-degree tilt,

Opposite- wall, Staggered

81

Figure 33 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of Zero-degree

tilt, Inline

Figure 34 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of Zero-degree

tilt, Staggered

82

Figure 35 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of Five-degree

tilt, Inline

Figure 36 Nusselt Number versus Reynolds Number for the Two Blocked Holes case of Five-degree

tilt, Staggered

83

Figure 37 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of Zero-degree

tilt, Inline

Figure 38 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of Zero-degree

tilt, Staggered

84

Figure 39 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of Five-degree

tilt, Inline

Figure 40 Nusselt Number versus Reynolds Number for the Four Blocked Holes case of Five-degree

tilt, Staggered

85

5.1.3 Heat Transfer Results for Areas 1 through 5

As mentioned previously, the target zone in all the tested geometries was divided

into five areas. In this section, for each area, the Nusselt numbers obtained for different

flow cases are compared.

Figure 41 Nusselt Number versus Reynolds Number for all cases on Area1

86

Figure 42 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on Area1

Figure 43 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on Area1

87

Figure 44 Nusselt Number versus Reynolds Number for all cases on Area2

Figure 45 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on Area2

88

Figure 46 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on Area2

Figure 47 Nusselt Number versus Reynolds Number for all cases on Area3

89

Figure 48 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on Area3

Figure 49 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on Area3

90

Figure 50 Nusselt Number versus Reynolds Number for all cases on Area4

Figure 51 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on Area4

91

Figure 52 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on Area4

Figure 53 Nusselt Number versus Reynolds Number for all cases on Area5

92

Figure 54 Nusselt Number versus Reynolds Number for all Baseline cases on Area5

Figure 55 Nusselt Number versus Reynolds Number for the Two and Four Blocked cases on Area5

93

Figure 56 Nusselt Number versus Reynolds Number for the Two Blocked Holes cases on Area5

Figure 57 Nusselt Number versus Reynolds Number for the Four Blocked Holes cases on Area5

94

Figures 41 through 43 show a monotonic increase and a similar shape in the

Nusselt number. The curves are almost linear. It also shows that on area 1, the 2-blocked-

hole cases, produce higher Nusselt numbers than the 4-blocked-hole cases and the

difference in Nusselt numbers on the first area increases as the Reynolds number

increases. As for the difference between the staggered and inline arrangements, the

Nusselt number only varies slightly in the first area, which means that the staggered and

inline arrangement did not play a significant role in enhancing the heat transfer

coefficients on this area. Therefore, the major reason for the difference in the two cases is

the flow that is exiting through 9 instead of 7 trailing-edge slots. That is why the heat

transfer coefficients were lower in the 4-blocked-hole-case which explains why the green

color spread so fast through it while in the 2-blocked-hole cases it did but at a slower rate.

For the differences between 0-degree tilt angle and 5-degree-tilt angle, on the average, for

2-blocked-hole cases, the highest results are found in 5-degree tilt angle and staggered

arrangement case, while the lowest Nusselt numbers were given in 0-degree tilt angle and

inline arrangement case. For the cases with 4 blocked holes, 5-degree tilt angle with

staggered arrangement case provided the highest Nusselt numbers while the lowest

values corresponding to this case at the lowest flow.

Figures 44 through 46 show that on area 2, the 2-blocked-hole cases gave higher

Nusselt number than the 4-blocked-hole cases. And the highest Nusselt numbers for 2-

blocked-hole cases were found in 5-degree tilt angle with staggered arrangement case,

while the lowest Nusselt numbers were provided by the 5-degree tilt angle with staggered

arrangement case. For the cases with 4 blocked holes, 5-degree tilt angle with staggered

arrangement case generated the highest Nusselt numbers while the 5-degree tilt angle

95

with inline arrangement case showed the lowest Nusselt numbers. Furthermore, Nusselt

numbers increase with increasing Reynolds number.

Figures 47 through 49 show Nusselt numbers versus Reynolds number on area 3.

The 2-blocked-hole cases gave higher Nusselt numbers than the 4-blocked-hole cases.

For the 2-blocked-hole cases, the highest results were found in 0-degree tilt angle with

staggered arrangement case, while the lowest Nusselt numbers were also given by 0-

degree tilt angle with staggered case. As for the cases with 4 blocked holes, the 5-degree

tilt angle with staggered arrangement provided the highest and lowest Nusselt numbers,

at the highest and lowest arrangement jet Reynolds number, respectively.

Figures 50 through 52 show the Nusselt numbers versus Reynolds number on area

4. On the average, the 2-blocked-hole cases gave higher Nusselt numbers than the 4-

blocked-hole cases. But the difference in value between the two cases is very small. For

2-blocked-hole cases, the highest results of the Nusselt numbers were given by 0-degree

tilt angle with staggered arrangement case, also the lowest Nusselt numbers were found

in 0-degree tilt angle with staggered arrangement case. As for the case with 4 blocked

holes, the 5-degree tilt angle with staggered arrangement case provided the highest

Nusselt numbers while the 5-degree tilt angle with inline arrangement showed the lowest

Nusselt numbers.

Finally in Figures 53 through 57, Nusselt number is plotted versus Reynolds

number on area 5. In this part of the test section, the cross flow effects are very strong.

That is why we see Nusselt numbers peaking in this area while compared to any other

previous areas. Besides, many of the 4-blocked-hole cases have induced higher Nusselt

numbers than some of the 2-blocked-hole cases. For 2-blocked-hole cases, the highest

96

results were found in 5-degree tilt angle with staggered arrangement case, while the

lowest Nusselt numbers were given by 5-degree tilt angle with staggered arrangement

case. As for the case with 4 blocked holes, 0-degree tilt angle with staggered arrangement

case provided the highest Nusselt numbers while the 5-degree tilt angle with inline

arrangement showed the lowest Nusselt numbers. Also for the baseline cases, the 0-

degree tilt angle cases gave lower Nusselt numbers than the 5-degree tilt angle cases

including the 5-degree impinged on opposite wall cases because the air flow passing

through a tilted slot directly the heated wall and enhanced the heat transfer coefficient.

But in the blocked cases, the 0-degree tilt angle gave higher Nusselt numbers than the 5-

degree tilt angle cases effects of cross flow were stronger in the case of zero tilt angle.

97

5.2 Pressure Ratios

In this section, we are interested in plotting the pressure ratios Psup/Ptrailing-edge

versus Reynolds number. Psup is the pressure measured in the middle of supply channel

and Ptrailing-edge is the average of two pressures: Pend1 and Pend2 where Pend1 and Pend2 were

measured at both ends of the trailing-edge channel as shown in Figures 10 through 12.

The pressure ratios versus Reynolds number are plotted for all 14 cases in Figures 58. On

two separate graphs, the baseline cases and the 2 and 4-blocked trailing edge slots are

also plotted as shown in Figures 59 and 60.

Figure 58 (Psup/Ptrailing-edge ) versus Reynolds number for all cases

98

Figure 59 (Psup/Ptrailing-edge ) versus Reynolds number for the baseline cases

Figure 60 (Psup/Ptrailing-edge ) versus Reynolds number for the blocked-slot cases

99

When studying these graphs, the following remarks could be noted:

(Psup/Ptrailing-edge) increases with increasing Reynolds number. Also the difference

in pressure ratios between different cases increases with increasing Reynolds number.

Furthermore, staggered or inline arrangement did not play an important role for the

pressure ratios. As for the differences between 0 and 5-degree tilt angle, on the average,

the baseline cases at 5-degree tilt angle gave higher pressure ratios than the 0-degree

cases. For blocked trailing-edge slot cases, the 4-blocked gave higher pressure ratios than

the 2-blocked cases on the average. This could be justified by the fact that air exiting

through a smaller number of exit slots i.e. reduced exit area will have a higher pressure in

the supply channel. Besides, the highest results of pressure ratios are found in 5-degree

tilt angle with inline arrangement, while the lowest pressure ratios were given by 5-

degree tilt angle with staggered arrangement. For blocked cases, the highest results are in

4-blocked-hole, 0-degree tilt angle with staggered arrangement, and the lowest results are

given by 2-blocked-hole, 0-degree tilt angle with inline arrangement.

100

6. CONCLUSION

An experimental investigation was conducted to determine the effect of cross-

flow on cross-over jets impinging into a trailing-edge cooling channel. The cross flow

was created by blocking two or four consecutive trailing-edge slots. Each case was

investigated for two tilt angles of 0 and 5 degrees and flow arrangements of staggered

and inline. The results showed that for all cases the Nusselt numbers increase

monotonically with increasing Reynolds numbers. When compared for the same amount

of coolant mass flow rate in each case, the results showed that the highest Nusselt number

occurred on area 5 due to the cross flow effects when four consecutive trailing-edge slots

were blocked and jets were at 5 degree tilt angles. Heat transfer coefficients decreased

significantly near the blocked slots while the cross flow created by the blockage of the

exit slots increased the heat transfer coefficients near the open slots. Besides, staggered or

inline arrangement did not play an important role in the heat transfer coefficients.

For the pressure ratio results, Psupply/Ptrailing-edge increased with increasing Reynolds

numbers. The four-blocked consecutive holes along the trailing-edge induced higher

pressure ratios than the cases where all exit holes were opened. Furthermore, staggered or

inline arrangement did not play an important role in the pressure ratios.

101

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[2] William W. Bathie, “Fundamental of Gas Turbines,” 1984, pp.1-4, 252-255.

[3] Lau, S.C., Han, J.C., and Batten, T., Heat Transfer, “Pressure Drop and Mass Flow

Rate in Pin Fin Channels with Long and Short Trailing Edge Ejection Holes,” Journal of

Turbomachinery, Vol. 111, No.2, 1989, pp. 117-123.

[4] Byerley, A.R., Jones, T.V., and Ireland, P.T., “Internal Cooling Passage Heat Transfer

near the Entrance to a Film Cooling Hole: Experimental and Computational Results,”

ASME Paper No. 92-GT-241, 1992.

[5] Taslim, M.E., Li, T., and Spring, S.D., “Experimental Study of the Effects of Bleed

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[7] Taslim, M.E., Li, T. and Spring, S.D., “Measurements of Heat Transfer Coefficients

in Rib-Roughened Trailing-Edge Cavities with Crossover Jets,” ASME Paper No. 98-GT-

435, 1998.

[8] Taslim, M.E., Nicolas, G.J., “An Experimental and Numerical Investigation of Jet

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102

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103

APPENDIX A

Raw data: Test1-Zero Degree-Inline-Baseline (All Holes Open)

Pic # Pven Tven Tin1 Tin2 Tend1 Tend2 Tamb Tjet V1 A1 V2 A2 V3 A3 Pplen Psup1 Psup2 Psup3 Pend1 Pend2 Flag1 ΔP

orifice Flag2

Pamb

“Hg

1 13 71.6 73.5 73.6 98.4 95.6 77.4 73.7 18.00 0.4166 18.37 0.4236 18.03 0.4188 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

2 13 72.0 73.6 73.7 98.6 95.3 77.4 77.8 18.62 0.4309 18.95 0.4374 18.67 0.4332 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

3 13 71.8 73.8 73.9 99.1 96.0 77.4 74.0 19.10 0.4417 19.50 0.4492 19.18 0.4448 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

4 13 72.1 74.0 74.1 99.9 97.1 77.5 74.2 19.72 0.4560 20.14 0.4642 19.77 0.4593 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

5 13 71.8 74.1 74.2 100.3 97.7 77.4 74.3 20.80 0.4810 21.16 0.4885 20.85 0.4843 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

6 13 72.2 74.1 74.2 101.2 98.7 77.4 74.3 21.70 0.5018 22.14 0.5096 21.77 0.5052 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

7 13 72.1 74.2 74.2 102.6 100.6 77.4 74.4 22.74 0.5254 23.22 0.5344 22.81 0.5300 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

8 13 72.1 74.2 74.3 104.2 102.8 77.4 74.4 23.66 0.5467 24.16 0.5560 23.80 0.5523 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

9 13 71.9 74.2 74.3 106.3 104.8 77.4 74.5 24.46 0.5647 25.01 0.5753 24.55 0.5690 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

10 13 71.9 74.3 74.4 107.1 105.9 77.5 74.5 25.30 0.5840 25.85 0.5947 25.40 0.5889 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

11 13 72.2 74.3 74.5 108.6 107.2 77.5 74.6 26.18 0.6043 26.72 0.6151 26.31 0.6107 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

12 13 72.1 74.4 74.5 110.6 109.4 77.5 74.6 26.88 0.6199 27.45 0.6314 27.01 0.6258 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

13 13 72.2 74.4 74.5 112.2 110.8 77.4 74.6 27.52 0.6345 28.10 0.6449 27.64 0.6405 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

14 13 72.3 74.4 74.5 113.9 113.1 77.3 74.7 28.11 0.6482 28.75 0.6609 28.26 0.6550 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

15 13 71.9 74.4 74.5 115.2 113.9 77.3 74.7 28.86 0.6652 29.46 0.6771 28.96 0.6713 0.89 0.80 0.85 0.85 0.45 0.35 2 0 1 30.32

1 26 69.8 74.6 74.7 96.9 95.4 77.5 74.2 21.09 0.4877 21.52 0.4956 21.07 0.4890 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

2 26 71.4 72.7 72.8 96.5 95.1 77.5 73.3 21.86 0.5054 22.30 0.5142 21.78 0.5074 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

3 26 72.6 72.5 72.6 97.1 95.9 77.5 73.1 22.46 0.5190 22.99 0.5295 22.37 0.5194 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

104

Pic # Pven Tven Tin1 Tin2 Tend1 Tend2 Tamb Tjet V1 A1 V2 A2 V3 A3 Pplen Psup1 Psup2 Psup3 Pend1 Pend2 Flag1 ΔP

orifice Flag2

Pamb

“Hg

4 26 71.1 72.7 72.8 97.7 96.8 77.4 73.2 23.03 0.5327 23.54 0.5417 22.91 0.5323 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

5 26 72.6 72.9 73.0 98.8 97.9 77.4 73.4 23.58 0.5449 24.05 0.5540 23.62 0.5482 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

6 26 73.2 73.0 73.1 99.2 98.5 77.4 73.5 24.08 0.5564 24.55 0.5650 24.05 0.5576 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

7 26 73.5 73.2 73.4 99.9 99.2 77.4 73.8 24.72 0.5710 25.23 0.5806 24.73 0.5738 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

8 26 68.2 72.4 72.5 101.0 100.9 77.4 73.1 25.52 0.5894 26.01 0.5985 25.56 0.5923 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

9 26 71.3 72.2 72.3 101.6 101.7 77.5 72.7 26.18 0.6046 26.71 0.6149 26.28 0.6094 2.05 1.90 1.95 1.95 1.15 0.95 2 0 1 30.33

10 26 72.3 71.9 72.1 102.6 102.6 77.4 72.7 26.73 0.6173 27.25 0.6274 26.76 0.6208 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

11 26 72.5 72.1 72.2 104.3 104.6 77.3 72.9 27.47 0.6336 28.04 0.6448 27.50 0.6377 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

12 26 72.9 72.6 72.7 105.5 105.8 77.4 73.0 28.08 0.6482 28.72 0.6602 28.14 0.6531 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

13 26 71.7 72.8 72.8 106.1 106.9 77.5 73.1 28.82 0.6650 29.42 0.6773 28.97 0.6716 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

14 26 72.0 72.4 72.5 107.1 107.7 77.4 73.0 29.47 0.6801 30.09 0.6920 29.63 0.6874 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

15 26 72.3 72.6 72.7 108.3 109.6 77.4 73.2 30.03 0.6923 30.61 0.7038 30.20 0.6998 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

16 26 69.7 72.9 73.1 110.1 111.3 77.4 73.4 30.63 0.7064 31.28 0.7191 30.81 0.7150 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

17 26 71.3 72.3 72.4 111.3 112.8 77.4 73.0 31.48 0.7257 32.09 0.7384 31.70 0.7345 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

18 26 70.9 72.2 72.4 113.0 114.6 77.4 73.0 32.07 0.7397 32.76 0.7534 32.33 0.7488 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

19 26 72.1 72.0 72.1 114.8 116.4 77.4 72.7 32.74 0.7545 33.42 0.7690 33.01 0.7651 2.05 1.90 1.95 1.99 1.15 0.95 2 0 1 30.32

1 40 74.6 71.8 71.9 89.6 90.8 77.4 72.4 24.93 0.5763 25.46 0.5858 24.93 0.5791 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

2 40 72.5 71.5 71.6 91.6 92.7 77.2 72.1 25.60 0.5916 26.12 0.6010 25.68 0.5953 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

3 40 73.6 71.9 71.9 93.0 94.4 77.4 72.3 26.20 0.6056 26.72 0.6151 26.29 0.6102 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

4 40 74.6 71.8 71.9 93.9 94.9 77.4 72.3 26.77 0.6179 27.31 0.6281 26.82 0.6224 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

5 40 73.9 71.9 72.0 94.8 96.4 77.4 72.3 27.55 0.6361 28.05 0.6458 27.65 0.6418 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

6 40 73.1 71.6 71.7 96.0 97.1 77.4 72.2 28.23 0.6520 28.80 0.6632 28.35 0.6580 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

105

Pic # Pven Tven Tin1 Tin2 Tend1 Tend2 Tamb Tjet V1 A1 V2 A2 V3 A3 Pplen Psup1 Psup2 Psup3 Pend1 Pend2 Flag1 ΔP

orifice Flag2

Pamb

“Hg

7 40 74.4 71.6 71.6 97.1 99.2 77.5 72.1 28.89 0.6667 29.50 0.6787 29.03 0.6733 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

8 40 73.8 71.4 71.5 98.3 100.1 77.4 72.0 29.59 0.6812 30.11 0.6928 29.67 0.6884 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

9 40 72.6 71.6 71.7 99.8 102.0 77.4 72.1 30.10 0.6954 30.72 0.7072 30.29 0.7024 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

10 40 72.8 71.3 71.4 100.8 103.1 77.4 71.9 30.58 0.7060 31.26 0.7194 30.77 0.7131 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

11 40 73.3 71.2 71.2 101.5 103.8 77.4 71.7 31.31 0.7222 31.96 0.7348 31.46 0.7294 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

12 40 72.1 71.5 71.5 103.0 105.3 77.4 71.9 32.08 0.7403 32.70 0.7525 32.30 0.7490 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

13 40 73.3 71.4 71.5 104.2 106.6 77.4 71.9 32.76 0.7556 33.44 0.7691 32.99 0.7649 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

14 40 72.2 71.4 71.5 105.6 108.3 77.4 71.9 33.43 0.7699 34.10 0.7836 33.69 0.7807 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

15 40 73.4 71.3 71.5 107.1 109.4 77.4 71.9 34.05 0.7847 34.78 0.7996 34.31 0.7945 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

16 40 72.5 71.7 71.8 108.5 110.8 77.3 72.1 34.75 0.8011 35.47 0.8184 35.04 0.8113 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

17 40 73.7 71.5 71.5 110.0 112.7 77.4 71.9 35.31 0.8145 36.10 0.8289 35.63 0.8245 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

18 40 72.6 71.5 71.5 113.6 116.3 77.4 71.7 36.53 0.8415 37.31 0.8570 36.93 0.8546 3.80 3.55 3.65 3.65 2.2 1.75 2 0 1 30.33

1 54 76.5 72.5 72.5 92.3 92.2 77.4 72.9 26.30 0.6074 26.83 0.6177 26.49 0.6151 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

2 54 75.9 71.9 71.9 93.0 93.0 77.4 72.5 26.64 0.6171 27.20 0.6253 26.85 0.6242 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

3 54 76.6 71.9 72.0 93.4 93.5 77.2 72.3 27.31 0.6308 27.81 0.6395 27.51 0.6386 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

4 54 75.8 71.8 71.9 93.9 94.2 77.3 72.3 27.87 0.6439 28.45 0.6547 28.06 0.6517 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

5 54 76.4 71.7 71.8 94.2 94.8 77.3 72.3 28.52 0.6582 29.06 0.6684 28.70 0.6662 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

6 54 76.6 71.6 71.7 95.1 95.7 77.4 72.2 29.13 0.6728 29.70 0.6837 29.38 0.6815 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

7 54 76.3 71.5 71.6 95.5 96.3 77.4 72.0 29.83 0.6878 30.42 0.7004 30.13 0.6984 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

8 54 76.0 71.6 71.6 96.4 97.5 77.4 72.0 30.46 0.7031 31.08 0.7149 30.75 0.7131 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

9 54 76.1 71.6 71.6 97.2 98.7 77.3 72.0 31.24 0.7210 31.82 0.7310 31.46 0.7301 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

10 54 76.1 71.4 71.5 99.2 100.2 77.2 72.1 31.95 0.7377 32.59 0.7499 32.22 0.7471 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

106

Pic # Pven Tven Tin1 Tin2 Tend1 Tend2 Tamb Tjet V1 A1 V2 A2 V3 A3 Pplen Psup1 Psup2 Psup3 Pend1 Pend2 Flag1 ΔP

orifice Flag2

Pamb

“Hg

11 54 76.1 71.5 71.5 99.5 100.9 77.3 72.1 32.60 0.7522 33.23 0.7638 32.89 0.7620 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

12 54 75.7 71.5 71.5 100.6 102.0 77.3 72.0 33.26 0.7670 33.92 0.7800 33.53 0.7774 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

13 54 75.6 71.6 71.7 102.0 103.9 77.4 72.0 34.21 0.7888 34.73 0.7981 34.47 0.7989 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

14 54 76.2 71.5 71.6 103.3 104.8 77.3 72.0 34.77 0.8021 35.26 0.8125 35.10 0.8132 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

15 54 76.3 71.5 71.5 104.3 105.6 77.3 71.9 35.21 0.8110 35.79 0.8225 35.47 0.8215 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

16 54 76.0 71.4 71.4 104.9 106.7 77.3 71.9 35.70 0.8230 36.34 0.8344 36.11 0.8362 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

17 54 76.4 71.5 71.6 106.4 108.6 77.5 72.0 36.36 0.8377 36.94 0.8492 36.67 0.8493 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

18 54 76.3 71.4 71.5 107.6 109.6 77.3 71.8 37.10 0.8542 37.73 0.8665 37.42 0.8669 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

19 54 75.7 71.4 71.5 109.0 111.1 77.3 71.9 37.63 0.8684 38.34 0.8799 38.06 0.8815 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.34

20 54 76.0 71.5 71.6 110.4 112.3 77.4 71.9 38.31 0.8822 39.05 0.8956 38.76 0.8972 6.00 5.60 5.70 5.80 3.55 3.05 2 0 1 30.33

1 69 80.5 72.8 72.8 91.9 90.6 77.2 73.1 28.08 0.6486 28.67 0.6598 28.36 0.6580 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

2 69 80.3 73.1 73.1 92.5 91.3 77.2 73.3 28.69 0.6628 29.26 0.6736 28.92 0.6717 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

3 69 80.2 73.1 73.1 93.0 92.3 77.2 73.3 29.30 0.6767 29.88 0.6874 29.58 0.6868 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

4 69 80.3 72.9 72.9 93.3 92.1 77.2 73.2 29.98 0.6922 30.59 0.7040 30.29 0.7027 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

5 69 79.9 72.7 72.8 93.7 92.7 77.3 73.1 30.62 0.7070 31.19 0.7176 30.93 0.7172 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

6 69 80.1 72.9 73.0 94.6 93.5 77.3 73.2 31.30 0.7227 31.88 0.7339 31.62 0.7337 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

7 69 79.9 73.1 73.1 95.1 94.4 77.3 73.4 32.13 0.7418 32.77 0.7535 32.45 0.7537 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

8 69 80.2 73.0 73.1 96.1 96.0 77.2 73.4 32.59 0.7523 33.20 0.7636 32.97 0.7632 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

9 69 80.2 72.8 72.8 96.7 96.9 77.2 73.2 33.24 0.7668 33.92 0.7802 33.63 0.7803 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

10 69 80.2 72.8 72.9 97.4 97.8 77.2 73.2 34.09 0.7866 34.84 0.8017 34.52 0.8004 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

11 69 80.1 72.4 72.4 96.1 97.3 77.3 72.8 34.94 0.8063 35.57 0.8190 35.26 0.8173 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

12 69 80.1 72.3 72.3 98.2 98.9 77.3 72.8 35.64 0.8223 36.31 0.8344 36.08 0.8353 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

107

Pic # Pven Tven Tin1 Tin2 Tend1 Tend2 Tamb Tjet V1 A1 V2 A2 V3 A3 Pplen Psup1 Psup2 Psup3 Pend1 Pend2 Flag1 ΔP

orifice Flag2

Pamb

“Hg

13 69 80.6 72.3 72.3 99.5 100.3 77.3 72.7 36.33 0.8373 37.07 0.8513 36.79 0.8529 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

14 69 79.9 72.4 72.5 100.9 102.1 77.4 72.8 37.08 0.8549 37.78 0.8672 37.42 0.8676 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

15 69 80.4 72.5 72.5 102.0 102.8 77.4 72.9 37.72 0.8698 38.44 0.8832 38.10 0.8819 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

16 69 80.2 72.1 72.2 103.0 103.7 77.4 72.6 38.36 0.8839 39.05 0.8968 38.79 0.8976 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

17 69 80.1 72.0 72.2 104.8 105.4 77.3 72.6 38.97 0.8968 39.65 0.9103 39.47 0.9132 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

18 69 79.8 72.0 72.0 105.3 106.2 77.3 72.5 39.49 0.9089 40.24 0.9241 39.94 0.9242 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

19 69 79.4 72.0 72.0 106.5 106.8 77.3 72.5 40.10 0.9241 40.87 0.9390 40.54 0.9384 8.85 8.25 8.45 8.55 5.25 4.50 2 0 1 30.37

1 82 82.1 72.3 72.4 87.5 87.4 77.1 72.7 30.02 0.6938 30.63 0.7049 30.48 0.7071 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.39

2 82 82.0 72.1 72.2 88.6 88.2 77.1 72.5 30.55 0.7065 31.10 0.7154 30.95 0.7178 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.39

3 82 82.3 72.0 72.0 89.2 88.9 77.2 72.3 31.12 0.7189 31.67 0.7276 31.48 0.7305 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.39

4 82 81.8 72.2 72.2 90.1 89.2 77.3 72.4 31.52 0.7277 32.08 0.7381 31.84 0.7382 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.39

5 82 81.9 72.0 72.1 90.8 89.8 77.2 72.4 31.97 0.7380 32.56 0.7487 32.30 0.7505 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.39

6 82 82.6 72.1 72.1 91.5 90.4 77.2 72.5 32.48 0.7498 33.02 0.7587 32.83 0.7616 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.39

7 82 82.3 72.1 72.1 92.1 91.2 77.2 72.4 33.07 0.7648 33.62 0.7738 33.51 0.7770 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.39

8 82 82.6 72.1 72.1 93.0 91.5 77.2 72.5 33.45 0.7736 34.08 0.7846 33.96 0.7871 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

9 82 82.5 72.2 72.2 93.6 92.3 77.2 72.6 34.10 0.7875 34.67 0.7963 34.49 0.7997 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

10 82 82.7 72.1 72.1 94.6 93.0 77.2 72.4 34.57 0.7970 35.06 0.8058 34.92 0.8082 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

11 82 82.6 72.4 72.4 95.5 93.9 77.2 72.6 35.22 0.8122 35.76 0.8223 35.57 0.8262 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

12 82 82.1 72.2 72.3 96.0 94.4 77.2 72.6 35.66 0.8220 36.33 0.8344 36.00 0.8357 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

13 82 81.8 71.9 72.0 96.9 95.1 77.2 72.4 36.20 0.8352 36.85 0.8464 36.57 0.8462 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

14 82 82.2 72.1 72.1 97.9 95.8 77.1 72.4 36.75 0.8468 37.41 0.8602 37.17 0.8636 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

15 82 83.0 72.4 72.5 98.8 96.8 77.3 72.7 37.35 0.8615 38.05 0.8736 37.80 0.8759 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

108

Pic # Pven Tven Tin1 Tin2 Tend1 Tend2 Tamb Tjet V1 A1 V2 A2 V3 A3 Pplen Psup1 Psup2 Psup3 Pend1 Pend2 Flag1 ΔP

orifice Flag2

Pamb

“Hg

16 82 83.0 72.4 72.5 99.2 97.1 77.3 72.8 37.74 0.8708 38.41 0.8826 38.26 0.8861 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

17 82 82.8 72.4 72.4 100.3 98.2 77.3 72.6 38.24 0.8815 39.04 0.8963 38.74 0.8973 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

18 82 82.6 72.2 72.3 101.2 99.0 77.2 72.6 38.97 0.8987 39.71 0.9126 39.53 0.9153 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

19 82 83.0 72.3 72.3 101.5 99.7 77.2 72.6 39.56 0.9113 40.28 0.9254 40.05 0.9278 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

20 82 82.9 72.2 72.2 102.3 100.3 77.2 72.6 40.18 0.9253 40.95 0.9413 40.75 0.9430 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

21 82 82.9 72.1 72.0 103.5 101.3 77.3 72.3 40.82 0.9399 41.65 0.9566 41.38 0.9581 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

22 82 82.9 72.0 72.1 104.8 102.5 77.2 72.4 41.56 0.9574 42.39 0.9725 42.19 0.9751 11.45 10.70 11.00 11.10 6.85 5.85 2 0 1 30.37

109

APPENDIX B

Raw data (Cold Test): Test1-Zero Degree-Inline-Baseline

Pven

(psig)

Tven

(°F)

Tin1

(°F)

Tin2

(°F)

Tamb

(°F)

Tjet

(°F) Pplen Psup1 Psup2 Psup3 Pend1 Pend2

Man.

Liquid

Pamb

("Hg)

ΔPjet

(H2O)

ΔP

orifice

5 72.5 75.0 75.1 77.4 75.1 0.39 0.30 0.35 0.35 0.10 0.05 oil 30.23 0.019 0

10 72.3 74.8 74.9 77.4 74.9 0.70 0.60 0.65 0.65 0.30 0.25 oil 30.23 0.028 0

15 73.2 74.4 74.5 77.4 74.5 1.05 0.95 0.99 1.00 0.55 0.40 oil 30.23 0.032 0

20 72.9 73.9 74.0 77.4 74.0 1.45 1.30 1.35 1.39 0.75 0.65 oil 30.22 0.039 0

25 73.9 73.1 73.1 77.3 73.3 1.95 1.80 1.85 1.85 1.10 0.90 oil 30.22 0.046 0

30 75.2 72.8 72.9 77.3 73.2 2.49 2.30 2.35 2.40 1.40 1.19 oil 30.22 0.067 0

40 75.2 72.9 73.0 77.4 73.2 3.75 3.55 3.65 3.70 2.20 1.89 oil 30.22 0.090 0

50 77.3 73.0 73.2 77.3 73.6 5.35 4.99 5.10 5.19 3.15 2.65 oil 30.21 0.123 0

60 77.8 73.4 73.5 77.3 72.8 7.00 6.09 6.75 6.80 4.15 3.05 oil 30.21 0.158 0

70 81.3 71.5 71.8 77.3 72.1 9.05 8.45 8.65 8.75 5.39 4.60 oil 30.21 0.202 0

80 82.1 70.1 70.2 77.3 70.7 11.25 10.50 10.75 10.89 6.75 5.75 oil 30.21 0.255 0

110

APPENDIX C

FORTRAN Program Check.f

character*60 filename

character*80 title

write(6,*)'enter the name of the data file that u',

&' want to check'

read(5,10)filename

10 format(a60)

open(unit=1,file=filename,status='old')

open(unit=2,file='output.dat',status='old')

READ(1,*)NPoints,Tliquid,angle,TEslots

do i=1,12

read(1,221)title

enddo

221 FORMAT(A80,//)

do i=1,NPoints

read(1,*)photo,Pven,Tven,Tin1,Tin2,Tend1,Tend2,Tamb,Tjet,V1,A1,

&V2,A2,V3,A3,Pplen,Psup1,Psup2,Psup3,Pend1,Pend2,flag1,Dporif,

&flag2,Pamb

C2345678901234567890123456789012345678901234567890123456789012345678901

2

if(Pven.lt.10.or.Pven.gt.90)write(6,*)'** CHECK Pven IN

TEST',photo,Pven

if(DPorif.lt.0.or.DPorif1.gt.17)write(6,*)

&'** CHECK DPorif1 IN TEST',photo,Pven

if(Pamb.lt.28.or.Pamb.gt.31)

&write(6,*)'** CHECK Pamb IN TEST #, Pven',photo,Pven

if(Tamb.lt.45.or.Tamb.gt.90)

&write(6,*)'** CHECK Tamb IN TEST #, Pven',photo,Pven

if(Tven.lt.45.or.Tven.gt.90)

&write(6,*)'** CHECK Tven IN TEST #, Pven',photo,Pven

if(Tin1.lt.45.or.Tin1.gt.90)

&write(6,*)'** CHECK Tin1 IN TEST #, Pven',photo,Pven

if(Tin1.lt.45.or.Tin2.gt.90)

&write(6,*)'** CHECK Tin2 IN TEST #, Pven',photo,Pven

if(Tend1.lt.45.or.Tend1.gt.110)

&write(6,*)'** CHECK Tend1 IN TEST #, Pven',photo,Pven

if(Tend2.lt.45.or.Tend2.gt.110)

&write(6,*)'** CHECK Tend2 IN TEST #, Pven',photo,Pven

if(Tjet.lt.45.or.Tjet.gt.90)

&write(6,*)'** CHECK Tjet IN TEST #, Pven ',photo,Pven

if(old1.eq.0)goto 31

111

err1=abs((v1/a1)-old1)/old1

err2=abs((v2/a2)-old2)/old2

err3=abs((v3/a3)-old3)/old3

if(err1.gt..0125)write(6,*)'error in heater 1 entry, test #',i

if(err2.gt..0125)write(6,*)'error in heater 2 entry, test #',i

if(err3.gt..0125)write(6,*)'error in heater 3 entry, test #',i

31 write(6,35)photo,Pven,v1/a1,v2/a2,v3/a3

write(2,35)photo,Pven,v1/a1,v2/a2,v3/a3

if(flag.eq.1)goto 32

old1=v1/a1

old2=v2/a2

old3=v3/a3

flag=1.

32 continue

35 format(1x,f4.0,1x,f5.0,3(1x,f10.6))

enddo

write(6,*)' '

write(6,*)' '

write(6,*)' Resistances are in file : output.dat'

stop

end

112

APPENDIX D

FORTRAN Program Reduce.f

C NOTE: Tm = Tjet is used in h calculation

C

C TRAILING-EDGE SIMULATING CHANNEL WITH CROSS-OVER SLOTS

C AND TRALING-EDGE SLOTS

C IMPINGEMENT ON A SMOOTH WALL

C JUNE 2008

C CONTROL PANEL ARANGEMENT

C CHANNEL 1 Back Heater # 1

C CHANNEL 2 Back Heater # 2

C CHANNEL 3 Back Heater # 3

IMPLICIT REAL*8(A-H,O-Z)

CHARACTER*80 TITLE

REAL*8 Mv,Nuj,K

COMMON TEbmax,TEbmin,TEside,TElength,Hlength,Rgas,Mv,

&Tin,Tamb,Pdown,Tliquid,TEAR

C C R I T I C A L V E N T U R I C O R R E L A T I O N

F1(A,P,T)=0.5215*A*P/SQRT(T) ! Correlation for small critial

! venturi-meters

C O R I F I C E P L A T E C O R R E L A T I O N

F2(Pf,hw,Tabs)=0.862509*K*Fa*Fh*Fm*Y1*(Dorif**2)*

&SQRT(hw*Pf*Gb/(Zf*Tabs))

Zb=0.99961

Zf=0.99958

Fa=1.00013

Fh=1.00000

Fm=1.00000

K=0.80466

Y1=0.99527

Dorif=0.6

Gb=1.

OPEN(Unit=1,FILE='input.dat',STATUS='old')

OPEN(Unit=7,FILE='detailed.out',STATUS='old')

113

OPEN(Unit=5,FILE='uncertain.out',STATUS='old')

OPEN(Unit=9,FILE='error.out',STATUS='old')

OPEN(Unit=2,FILE='on-heater1-Nu-for-plot.out',STATUS='old')

OPEN(Unit=3,FILE='on-heater2-Nu-for-plot.out',STATUS='old')

OPEN(Unit=8,FILE='on-heater1-Nu.out',STATUS='old')

OPEN(Unit=10,FILE='on-heater2-Nu.out',STATUS='old')

OPEN(Unit=4,FILE='bleed.out',STATUS='old')

Hgtopsi= 0.49083935 ! converts inches of Hg to psi

H2otopsi=Hgtopsi/13.6 ! converts inches of H2o to psi

Oiltopsi=0.826*Hgtopsi/13.6 ! converts inches of Oil to psi

FAC1=3.4121 ! converts Watts to BTU/hr

PFAC=248.8*1.4504E-04*144 ! converts inches of H2O to psf

Rgas=53.34 ! gas constant for air

FAC1=3.4121

C V E N T U R I G E O M E T R Y

C PI=3.141592..........

PI=4.*ATAN(1.E00)

Dthroat=0.32 ! 165 EGAN

Athroat=(PI/4.)*(Dthroat*Dthroat)

TElength=36.

PartThick=0.9

SupHeight=4.5

SupBmin=1.1

SupBmax=1.96

SupSide=SQRT(SupHeight**2+((SupBmax-SupBmin)/2.)**2)

SupLength=49.

C H E A T E R S

Hlength=11.1 ! Spreader length

Hlength=Hlength/12. ! ft

Hwidth=2.6 ! Spreader width

Hwidth=Hwidth/12. ! ft

C T R A L I N G - E D G E C H A N N E L

TElength=36.

TEheight=2.5

TEheight=TEheight/12. ! ft

TEbmax=0.78

TEbmax=TEbmax/12. ! ft

114

TEbmin=0.45

TEbmin=TEbmin/12. ! ft

TEside=SQRT(TEheight**2+((TEbmax-TEbmin)/2.)**2)

TEperim=2.*TEside+TEbmax+TEbmin

TEcross=TEheight*(TEbmax+TEbmin)/2.

TEDh=4.*TEcross/TEperim

TEAR=(TEbmax+TEbmin)/(2.*TEside)

C C R O S S - O V E R S L O T S (I N L E T )

NCO=11

SLOTL=0.64

SLOTL=SLOTL/12.

SlotW=0.41

SlotW=SlotW/12.

SLotP=PI*SlotW+ 2.*(SlotL-SlotW)

SLotA=(PI/4.)*SlotW*SlotW + SlotW*(SlotL-SlotW)

SLotDH=4.*SlotA/SlotP

SlotR=0.205

SlotSpacing=2.

SoDh=SlotSpacing/(12.*SlotDh)

Zjet=1.25

ZoDh=Zjet/(12.*SlotDh)

C E X I T S L O T G E O M E T R Y

NE=12

ESlotL=0.68

ESlotL=ESlotL/12.

ESlotW=0.25

ESlotW=ESlotW/12.

ESlotP=PI*ESlotW+2.*(ESlotL-ESlotW)

ESlotA=(PI/4.)*ESlotW*ESlotW+ESlotW*(ESlotL-ESlotW)

ESlotDH=4.*ESlotA/ESlotP

AreaR=(NCO*SlotA)/(NE*ESlotA)

ESlotR=0.125

ESlotSpacing=2.

write(7,999)NCO,144.*SlotA,12.*SlotP,12.*SlotDH,SlotR,SoDh,

&ZoDh,PartThick,SupHeight,SupBmin,SupBmax,SupSide,

&SupLength,12*TEheight,12*TEbmin,12*TEbmax,12*TEside,TElength,

&144.*TEcross,12.*TEDh,TEAR,NE,144.*ESlotA,12.*ESlotP,12.*ESlotDH,

&ESlotR,ESlotSpacing,AreaR,12.*Hlength,12.*Hwidth

999 format(/,

&5x,'# of Cross-Over Jets : ',i2,/,

&5x,'Inlet Slot Cross-Sectional Area',f12.6,' sq.in.',/,

115

&5x,'Inlet Slot Perimeter',f12.6,' inches',/,

&5x,'Inlet Slot Hydraulic Diameter',f12.6,' inches',/,

&5x,'Inlet Slot Corner Radius',f9.3,' inches',/,

&5x,'Jet Spacing, S/Dh',f6.3,/,

&5x,'Jet Spacing, Z/Dh',f6.3,/,

&5x,'Partition Wall Thickness',f6.1,' inches',/,

&5x,'Supply Channel Height',f6.1,' inches',/,

&5x,'Supply Channel Smaller Base',f6.2,' inches',/,

&5x,'Supply Channel Larger Base',f6.2,' inches',/,

&5x,'Supply Channel Side',f12.6,' inches',/,

&5x,'Supply Channel Length',f6.1,' inches',/,

&5x,'TE Channel Height',f6.1,' inches',/,

&5x,'TE Channel Smaller Base',f6.3,' inches',/,

&5x,'TE Channel Larger Base',f6.3,' inches',/,

&5x,'TE Channel Side',f12.6,' inches',/,

&5x,'TE Channel Length',f6.1,' inches',/,

&5x,'TE Channel Cross Sectional Area',f12.6,' square inches',/,

&5x,'TE Channel Hydraulic Diameter',f12.6,' inches',/,

&5x,'TE Channel Hydraulic Aspect Ratio',f12.6,' inches',/,

&5x,'# of Exit Slots : ',i2,/,

&5x,'Exit Slot Cross-Sectional Area',f12.6,' sq.in.',/,

&5x,'Exit Slot Perimeter',f12.6,' inches',/,

&5x,'Exit Slot Hydraulic Diameter',f12.6,' inches',/,

&5x,'Exit Slot Corner Radius',f6.3,' inches',/,

&5x,'Exit Slot Spacing',f6.1,' inches',/,

&5x,'Inlet-to-Exit Flow Area Ratio',f12.6,/,

&5x,'Heater Spreader Length',f6.1,' inches',/,

&5x,'Heater Spreader Width',f6.1,' inches',/)

C HEAT TRANSFER AREA

Area=Hwidth*Hlength ! sq.ft

C READ IN # OF PHOTOS, RPM, HEATER #, DISTANCE OF CAMERA FROM THE

C BEGINNING OF THAT HEATER, # OF TURBS. PER HEATER, TURB. WIDTH,

ASPECT

C RATIO, DISTANCE FROM THE TEST SECTION INLET TO THE CAMERA

C CENTERLINE AND LIQUID CRYSTAL REFERENCE TEMPERATURE

READ(1,*)NP,Tliquid,angle,TEslots

WRITE(8,402)NP

WRITE(10,402)NP

402 FORMAT(I5)

if(TEslots.eq.0.)WRITE(7,111)Angle

if(TEslots.eq.0.)WRITE(8,111)Angle

if(TEslots.eq.0.)WRITE(10,111)Angle

if(TEslots.eq.1.)WRITE(7,112)Angle

if(TEslots.eq.1.)WRITE(8,112)Angle

if(TEslots.eq.1.)WRITE(10,112)Angle

111 FORMAT('Cross-Over Slot Angle is :', F6.1,' Degrees',///,

&'TRALING-EDGE SLOTS ARE IN-LINE',//)

116

112 FORMAT('Cross-Over Slot Angle is :', F6.1,' Degrees',///,

&'TRALING-EDGE SLOTS ARE STAGGERED',//)

IEND=12

DO 333 I=1,IEND

READ(1,10)TITLE

WRITE(7,10)TITLE

WRITE(10,10)TITLE

333 WRITE(8,10)TITLE

10 FORMAT(A80,//)

WRITE(4,*)' Rejet mtot mbleed %(mbleed/mtor)'

WRITE(8,450)

WRITE(10,450)

450 FORMAT(' PHOTO Rej Nuj hj Uucer ',//)

Reold=11200

DO 1 I=1,NP

read(1,*)photo,Pven,Tven,Tin1,Tin2,Tend1,Tend2,Tamb,Tjet,V1,A1,

&V2,A2,V3,A3,Pplen,Psup1,Psup2,Psup3,Pend1,Pend2,flag,Dporif,

&flag2,Pamb

WRITE(7,*)' '

WRITE(7,*)' '

WRITE(7,*)' '

WRITE(7,100)PHOTO

WRITE(7,*)' '

WRITE(7,*)'Collected Data: PHOTO,Pven,Pplen,Psup1,Psup2,Psup3'

WRITE(7,*)'Pend1,Pend2,DPorif,Pamb'

WRITE(7,*)'V1,A1,V2,A2,V3,A3,Tven,Tin1,Tin2,Tjet,Tend1,Tend2,Tamb'

WRITE(7,*)' '

WRITE(7,200)PHOTO,Pven,Pplen,Psup1,Psup2,Psup3,Pend1,Pend2,DPorif,

&Pamb,V1,A1,V2,A2,V3,A3,Tven,Tin1,Tin2,Tend1,Tend2,Tamb

200 FORMAT(4X,F4.1,' ',F7.3,' ',F7.3,' ',F7.3,' ',F7.3,' ',F7.3/,

&4X,F7.3,' ',F7.3,' ',F7.3,' ',F7.2/,

&' ',F5.2,' ',F6.4,' ',F5.2,1x,F6.4,' ',F5.2,' ',F6.4,' ',

&F4.1,' ',F4.1,' ',F4.1,' ',F4.1,' ',F4.1,' ',F4.1,' ',F4.1)

C Oil manometer was 0.25" below the zero line

Pamb =Pamb*Hgtopsi

Pdown=0.5*(Pdown1+Pdown2)

if(flag.eq.1)Pplen=2*Pplen*H2otopsi+Pamb ! psi

if(flag.eq.1)Psup1=2*Psup1*H2otopsi+Pamb ! psi

if(flag.eq.1)Psup2=2*Psup2*H2otopsi+Pamb ! psi

if(flag.eq.1)Psup3=2*Psup3*H2otopsi+Pamb ! psi

if(flag.eq.1)Pend1=2*Pend1*H2otopsi+Pamb ! psi

if(flag.eq.1)Pend2=2*Pend2*H2otopsi+Pamb ! psi

117

if(flag.eq.2)Pplen=(Pplen+0.25)*Oiltopsi+Pamb ! psi

if(flag.eq.2)Psup1=(Psup1+0.25)*Oiltopsi+Pamb ! psi

if(flag.eq.2)Psup2=(Psup2+0.25)*Oiltopsi+Pamb ! psi

if(flag.eq.2)Psup3=(Psup3+0.25)*Oiltopsi+Pamb ! psi

if(flag.eq.2)Pend1=(Pend1+0.25)*Oiltopsi+Pamb ! psi

if(flag.eq.2)Pend2=(Pend2+0.25)*Oiltopsi+Pamb ! psi

Tin=0.5*(Tin1+Tin2)

Tm=Tjet ! NOTE; Tm=Tjet is used for h calculation, not

calculated Tm which is a few degrees higher.

C AIR MASS FLOW RATE FROM THE CRITICAL VENTURI

Mv=F1(Athroat,Pven+Pamb,Tven+460.)

if(flag2.eq.1)Dporif=(2*Dporif)*H2otopsi

if(flag2.eq.2)Dporif=(Dporif+0.25)*Oiltopsi

Pf=Dporif+Pamb

Tabs=Tend2+460.

tipmass=F2(Pf,Dporif,Tabs)

ratiof=tipmass/Mv

C TOTAL HEAT ADDED TO THE AIR BY THE HEATERS FROM THE

C INLET TO THE POINT IN QUESTION

C HEATER 1

write(7,*)' '

write(7,*)' **** ON HEATER 1 ****'

write(7,*)' '

Q=V1*A1+V2*A2+V3*A3

Q=Q*FAC1

C HEAT FLUX, BTU/(sqft.Sec)

Fluxb=V1*A1*FAC1/(Area)

CALL COEFFICIENT(Q,Fluxb,Tm,Tback,hturb,Floss,PHOTO,fluxnet)

C FILM TEMPERATURE

TF=(Tback+Tm)/2.

C DENSITY AT JET TEMPERATURE

RHO=(Pdown*144.)/(Rgas*(Tjet+460.))

C OTHER PROPERTIES AT JET TEMPERATURE

TjR=Tjet+460.

118

CALL AIRPROP(TjR,GAMA,CON,VIS,PR,CP)

VIS=VIS/3600.

C WRITE(6,*)' Ratio=',CON/SlotDH

C JET REYNOLDS NUMBER

Rej=4.*(Mv/NCO)/(SlotP*VIS)

C NUSSELT NUMBER

Nuj=Hturb*SlotDH/Con

C UNCERTAINTY ANALYSIS

CALL UNCERTAIN(V1,A1,V2,A2,V3,A3,

&Area,Tback,Tin,Pven+Pamb,Floss,Uncer)

WRITE(7,300)Tin,Tjet,Tm,3600*Vis,con,Mv,Rej

WRITE(7,401)Hturb,Nuj,Uncer

if((abs(Reold-Rej)/Reold).gt.0.1)WRITE(2,*)' '

WRITE(8,403)PHOTO,Rej,Nuj,Hturb,Uncer

WRITE(2,405)Rej,Nuj,Hturb,Uncer

C HEATER 1

write(7,*)' '

write(7,*)' **** ON HEATER 2 ****'

write(7,*)' '

Q=V1*A1+V2*A2+V3*A3

Q=Q*FAC1

C HEAT FLUX, BTU/(sqft.Sec)

Fluxb=V2*A2*FAC1/(Area)

CALL COEFFICIENT(Q,Fluxb,Tm,Tback,hturb,Floss,PHOTO,fluxnet)

C FILM TEMPERATURE

TF=(Tback+Tm)/2.

C DENSITY AT JET TEMPERATURE

RHO=(Pdown*144.)/(Rgas*(Tjet+460.))

C OTHER PROPERTIES AT JET TEMPERATURE

TjR=Tjet+460.

119

CALL AIRPROP(TjR,GAMA,CON,VIS,PR,CP)

VIS=VIS/3600.

C WRITE(6,*)' Ratio=',CON/SlotDH

C JET REYNOLDS NUMBER

Rej=4.*(Mv/NCO)/(SlotP*VIS)

C Channel REYNOLDS NUMBER if flow was axial

Rechannel=4.*Mv/(TEPerim*VIS)

axialnu=0.023*(Rechannel**0.8)*(PR**0.4)

axialh=axialnu*CON/TEDh

C NUSSELT NUMBER

Nuj=Hturb*SlotDH/Con

C UNCERTAINTY ANALYSIS

CALL UNCERTAIN(V1,A1,V2,A2,V3,A3,

&Area,Tback,Tin,Pven+Pamb,Floss,Uncer)

WRITE(7,300)Tin,Tjet,Tm,3600*Vis,con,Mv,Rej

WRITE(7,401)Hturb,Nuj,Uncer

if((abs(Reold-Rej)/Reold).gt.0.1)WRITE(3,*)' '

WRITE(7,400)tipmass,ratiof

WRITE(4,404)Rej,Mv,tipmass,100*ratiof

WRITE(7,399)Pplen,Psup1,Psup2,Psup3,Pend1,Pend2,

&Rechannel,axialnu,axialh

WRITE(10,403)PHOTO,Rej,Nuj,Hturb,Uncer

WRITE(3,405)Rej,Nuj,Hturb,Uncer

Reold=Rej

doverk=SlotDH/con

1 CONTINUE

403 FORMAT(1X,F3.0,1X,E11.5,1X,E11.5,1X,E11.5,1X,E9.3)

405 FORMAT(1X,E11.5,1X,E11.5,1X,E11.5,1X,E9.3)

100 FORMAT(/,30X,'PHOTO# ',F3.0)

300 FORMAT(/,1X,'Tin=',F6.2,1X,'Tjet=',F6.2,1X,'Tm=',F6.2,1X,

&'vis=',f8.4,' lbm/hr.ft',1x,'con=',f8.4,' BTU/hr.ft.R',

&1X,'Mv=',E12.5,1x,'Rej=',F8.1)

401 FORMAT(1X,' h_jet=',F8.3,2X,' Nu_jet=',F9.4,2X,

&'% Uncer (in h) =',F7.2)

399 FORMAT(1X,'Plenum Pressure=',F9.4,' psi',5X,

&'Supply Channel Pressure 1 =',F9.4,' psi',/,

&'Supply Channel Pressure 2 =',F9.4,' psi',2x,

&'Supply Channel Pressure 3=',F9.4,' psi',/,

&'TE Channel End Pressure 1 =',F9.4,' psi',2x,

120

&'TE Channel End Pressure 2=',F9.4,' psi',/,

&'Axial Flow Reynolds number=',F10.1,/,

&'Axial Flow Smooth Nusselt number=',F12.3,/,

&'Axial Flow Smooth Heat Transfer Coefficient=',F12.3)

400 FORMAT(1X,'Tip Flow rate=',F9.4,' pps',5X,

&'Tip Flow/Total Flow =',F9.4)

404 FORMAT(1X,F9.1,5x,f9.6,5x,f9.6,5x,f5.2)

STOP

END

SUBROUTINE UNCERTAIN(V1,I1,V2,I2,V3,I3,Area,Tback,Tin,

&P1,Floss,Uncer)

IMPLICIT REAL*8(A-H,O-Z)

REAL*8 I1,I2,I3

a=.220174

b=231.182609

dv1=.01

dv2=.01

dv3=.01

di1=.001

di2=.001

di3=.001

da2=1./(32.*32.)

dts=0.5

dti=0.5

dp1=0.5

Ts=Tback

Ti=Tin

FAC=491.355778

Floss=Floss/FAC

DFloss=0.1*Floss

a2=144*Area

H=(V2*I2/A2-Floss)/(Ts-Ti-(V1*I1+V2*I2+V3*I3)/(A+B*P1))

WRITE(5,*)' heat transfer coeff., h, =',H*FAC,'

BUT/hr.sqft.F'

H2=H*H

C

C i2 v2

C ----- - Floss

C a2

C -----------------------------------

C i3 v3 + i2 v2 + i1 v1

C - ----------------------- + ts - ti

C b p1 + a

C

DHDI1=v1*(i2*v2/a2-Floss)/((b*p1+a)*(-(i3*v3+i2*v2+i1*v1)/

&(b*p1+a)+ts-ti)**2)

ZI1=(DI1*DHDI1)**2

121

DHDV1=i1*(i2*v2/a2-Floss)/((b*p1+a)*(-(i3*v3+i2*v2+i1*v1)/

&(b*p1+a)+ts-ti)**2)

ZV1=(DV1*DHDV1)**2

DHDI3=v3*(i2*v2/a2-Floss)/((b*p1+a)*(-(i3*v3+i2*v2+i1*v1)/

&(b*p1+a)+ts-ti)**2)

ZI3=(DI3*DHDI3)**2

DHDV3=i3*(i2*v2/a2-Floss)/((b*p1+a)*(-(i3*v3+i2*v2+i1*v1)/

&(b*p1+a)+ts-ti)**2)

ZV3=(DV3*DHDV3)**2

DHDI2=v2/(a2*(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+ts-ti))+v2*

&(i2*v2/a2-Floss)/((b*p1+a)*(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+

&ts-ti)**2)

ZI2=(DI2*DHDI2)**2

DHDV2=i2/(a2*(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+ts-ti))+i2*

&(i2*v2/a2-Floss)/((b*p1+a)*(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+

&ts-ti)**2)

ZV2=(DV2*DHDV2)**2

DHDA2=-i2*v2/(a2**2*(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+ts-ti))

ZA2=(DA2*DHDA2)**2

DHDTS=-(i2*v2/a2-Floss)/(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+ts-ti)**2

ZTS=(DTS*DHDTS)**2

DHDTI=(i2*v2/a2-Floss)/(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+ts-ti)**2

ZTI=(DTI*DHDTI)**2

DHDP1=-b*(i2*v2/a2-Floss)*(i3*v3+i2*v2+i1*v1)/((b*p1+a)**2*

&(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+ts-ti)**2)

ZP1=(DP1*DHDP1)**2

DHDFloss=-Floss/(-(i3*v3+i2*v2+i1*v1)/(b*p1+a)+ts-ti)

ZFloss=(DFloss*DFloss)**2

Uncer=100*SQRT((ZI1+ZI2+ZI3+ZV1+ZV2+ZV3+ZA2+ZTS+ZTI+

&ZP1+ZFloss)/(H2))

WRITE(5,*)' '

WRITE(5,*)' UNCERTAINTY IN HEAT TRANSFER',

&' COEFFICIENT'

WRITE(5,*)' '

WRITE(5,*)' TOTAL UNCERTAINTY % ',Uncer

WRITE(5,*)' % Uncertainty assoc. with I1',100.*sqrt(ZI1)/H

WRITE(5,*)' % Uncertainty assoc. with V1',100.*sqrt(ZV1)/H

WRITE(5,*)' % Uncertainty assoc. with I2',100.*sqrt(ZI2)/H

WRITE(5,*)' % Uncertainty assoc. with V2',100.*sqrt(ZV2)/H

122

WRITE(5,*)' % Uncertainty assoc. with I3',100.*sqrt(ZI3)/H

WRITE(5,*)' % Uncertainty assoc. with V3',100.*sqrt(ZV3)/H

WRITE(5,*)' % Uncertainty assoc. with A2',100.*sqrt(ZA2)/H

WRITE(5,*)' % Uncertainty assoc. with Ts',100.*sqrt(ZTS)/H

WRITE(5,*)' % Uncertainty assoc. with Tin',100.*sqrt(ZTI)/H

WRITE(5,*)' % Uncertainty assoc. with Pven',100.*sqrt(ZP1)/H

WRITE(5,*)' % Uncertainty assoc. with Floss',100.*sqrt(ZFLOSS)/H

RETURN

END

SUBROUTINE EQSOLVE(A,B,NA,NDIM,NB)

IMPLICIT REAL*8(A-H,O-Z)

DIMENSION A(NDIM,NDIM),B(NDIM,NB)

DO 291 J1=1,NA

C FIND REMAINING ROW CONTAINING LARGEST ABSOLUTE

C VALUE IN PIVOTAL COLUMN.

101 TEMP=0.

DO 121 J2=J1,NA

IF(ABS(A(J2,J1))-TEMP) 121,111,111

111 TEMP=ABS(A(J2,J1))

IBIG=J2

121 CONTINUE

IF(IBIG-J1)5001,201,131

C REARRANGING ROWS TO PLACE LARGEST ABSOLUTE

C VALUE IN PIVOT POSITION.

131 DO 141 J2=J1,NA

TEMP=A(J1,J2)

A(J1,J2)=A(IBIG,J2)

141 A(IBIG,J2)=TEMP

DO 161 J2=1,NB

TEMP=B(J1,J2)

B(J1,J2)=B(IBIG,J2)

161 B(IBIG,J2)=TEMP

C COMPUTE COEFFICIENTS IN PIVOTAL ROW.

201 TEMP=A(J1,J1)

DO 221 J2=J1,NA

221 A(J1,J2)=A(J1,J2)/TEMP

DO 231 J2=1,NB

231 B(J1,J2)=B(J1,J2)/TEMP

IF(J1-NA)236,301,5001

C COMPUTE NEW COEFFICIENTS IN REMAINING ROWS.

236 N1=J1+1

DO 281 J2=N1,NA

TEMP=A(J2,J1)

DO 241 J3=N1,NA

241 A(J2,J3)=A(J2,J3)-TEMP*A(J1,J3)

DO 251 J3=1,NB

251 B(J2,J3)=B(J2,J3)-TEMP*B(J1,J3)

281 CONTINUE

291 CONTINUE

C OBTAINING SOLUTIONS BY BACK SUBSTITUTION.

301 IF(NA-1)5001,5001,311

123

311 DO 391 J1=1,NB

N1=NA

321 DO 341 J2=N1,NA

341 B(N1-1,J1)=B(N1-1,J1)-B(J2,J1)*A(N1-1,J2)

N1=N1-1

IF(N1-1)5001,391,321

391 CONTINUE

5001 CONTINUE

RETURN

END

SUBROUTINE RAD(AR,Width,Length,Tback,Ttop,Tfront,Tbot,

&Frback,Frtop,Frfront,Frbot)

IMPLICIT REAL*8(A-H,O-Z)

REAL*8 Length

DIMENSION A(4,4),B(4,1),E(4),T(4),Q(4)

W=Width

H=AR*Width

H=H/Length

W=W/Length

T(1)=Tback + 460.

T(2)=Ttop + 460.

T(3)=Tfront+ 460.

T(4)=Tbot + 460.

C Emissivities

E(1)=.85 ! Liquid Crystal Foil

E(2)=.85 ! Plexiglas

E(3)=.85 ! Plexiglas

E(4)=.85 ! Plexiglas

N=4

PI=4.*ATAN(1.E00)

SIGMA=0.1712E-08

C WRITE(7,150)

150 FORMAT(//,20X,'SHAPE FACTORS',//)

C Shape Factors

F11=0.

W2=W*W

H2=H*H

Z1=1./(PI*W)

Z2=W*ATAN(1./W)

Z3=H*ATAN(1./H)

Z=SQRT(H2+W2)

Z4=-Z*ATAN(1./Z)

Z=(1.+W2)*(1.+H2)

ZZ=1.+W2+H2

ZZZ=Z/ZZ

Z=W2*ZZ/((1.+W2)*(W2+H2))

Z=Z**W2

124

ZZZ=ZZZ*Z

Z=H2*ZZ/((1.+H2)*(W2+H2))

Z=Z**H2

ZZZ=ZZZ*Z

Z5=.25*LOG(ZZZ)

F12=Z1*(Z2+Z3+Z4+Z5)

F14=F12

F13=1.-F11-F12-F14

F31=F13

F32=F12

F33=0.

F34=F14

DUM=W

W=H

H=DUM

W2=W*W

H2=H*H

Z1=1./(PI*W)

Z2=W*ATAN(1./W)

Z3=H*ATAN(1./H)

Z=SQRT(H2+W2)

Z4=-Z*ATAN(1./Z)

Z=(1.+W2)*(1.+H2)

ZZ=1.+W2+H2

ZZZ=Z/ZZ

Z=W2*ZZ/((1.+W2)*(W2+H2))

Z=Z**W2

ZZZ=ZZZ*Z

Z=H2*ZZ/((1.+H2)*(W2+H2))

Z=Z**H2

ZZZ=ZZZ*Z

Z5=.25*LOG(ZZZ)

F21=Z1*(Z2+Z3+Z4+Z5)

F22=0.

F23=F21

F24=1.-F21-F22-F23

F41=F21

F42=F24

F43=F23

F44=0.

C

C WRITE(7,110)F11,F12,F13,F14

C WRITE(7,120)F21,F22,F23,F24

C WRITE(7,130)F31,F32,F33,F34

C WRITE(7,140)F41,F42,F43,F44

C

110 FORMAT(5X,'F11=',F6.4,5X,'F12=',F6.4,5X,'F13=',F6.4,

&5X,'F14=',F6.4,/)

120 FORMAT(5X,'F21=',F6.4,5X,'F22=',F6.4,5X,'F23=',F6.4,

&5X,'F24=',F6.4,/)

125

130 FORMAT(5X,'F31=',F6.4,5X,'F32=',F6.4,5X,'F33=',F6.4,

&5X,'F34=',F6.4,/)

140 FORMAT(5X,'F41=',F6.4,5X,'F42=',F6.4,5X,'F43=',F6.4,

&5X,'F44=',F6.4,//)

C WRITE(7,160)

160 FORMAT(/,20X,'EMISSIVITIES',//)

C WRITE(7,100)(I,E(I),I=1,N)

C WRITE(7,170)

170 FORMAT(/,20X,'TEMPERATURES IN R',//)

C WRITE(7,100)(I,T(I),I=1,N)

A(1,1)=F11-1./(1.-E(1))

A(1,2)=F12

A(1,3)=F13

A(1,4)=F14

A(2,1)=F21

A(2,2)=F22-1./(1.-E(2))

A(2,3)=F23

A(2,4)=F24

A(3,1)=F31

A(3,2)=F32

A(3,3)=F33-1./(1.-E(3))

A(3,4)=F34

A(4,1)=F41

A(4,2)=F42

A(4,3)=F43

A(4,4)=F44-1./(1.-E(4))

C WRITE(7,180)

180 FORMAT(//,20X,'COEFFICIENT MATRIX',/)

C WRITE(7,200)((A(I,J),J=1,N),I=1,N)

DO I=1,N

B(I,1)=-E(I)*SIGMA*(T(I)**2.)*(T(I)**2.)/(1.-E(I))

ENDDO

C WRITE(7,250)

C WRITE(7,100)(I,B(I,1),I=1,N)

200 FORMAT(1X,4E15.6)

250 FORMAT(/,20X,'RIGHT HAND SIDE ',/)

C WRITE(7,55)

55 FORMAT(//,20X,'GAUSSIAN ELIMINATION METHOD',/)

CALL EQSOLVE(A,B,N,N,1)

C WRITE(7,50)

C WRITE(7,100)(I,B(I,1),I=1,N)

DO I=1,N

Q(I)=E(I)*(SIGMA*(T(I)**2.)*(T(I)**2.)-B(I,1))/(1.-E(I))

ENDDO

Frback= q(1)

Frtop= q(2)

Frfront=q(3)

Frbot= q(4)

126

C WRITE(7,350)

C WRITE(7,100)(I,Q(I),I=1,N)

100 FORMAT(4(I3,E15.6))

50 FORMAT(/,20X,'RADIOCITIES',/)

350 FORMAT(/,20X,'HEAT FLUXES IN BTU/hr.sqft',/)

RETURN

END

SUBROUTINE

COEFFICIENT(Q,Fluxb,Tm,Tback,hback,Floss,PHOTO,fluxnet)

IMPLICIT REAL*8(A-H,O-Z)

REAL*8 kinc,kadh,kkap,kmyl,kpoly,ksty,kblack,kliq,kplex,

&Length,Mv

COMMON Bmax,Bmin,Width,Length,Hlength,Rgas,Mv,

&Tin,Tamb,Pdown,Tliquid,AR

C B A C K W A L L (LIQUID CRYSTAL WALL)

C FROM THE CENTER OF HEATING ELEMENT TO THE AMBIENT AIR

C 0.25 mil INCONEL HEATING ELEMENT ----- 0.5 mil ADHESIVE ----- 0.5

mil

C KAPTON ---- 2 mil ADHESIVE ----- 1.625 inches POLYURETHANE ----

C 2.0 inches STYROFOAM ---- AMBIENT

C tinc1/kinc -- tadh1/kadh -- tkap1/kkap -- tadh4/kadh --

tpoly/kpoly

C -- tsty/ksty -- 1/ho

C FROM THE CENTER OF HEATING ELEMENT TO THE AIR INSIDE THE TEST

SECTION

C 0.25 mil INCONEL HEATING ELEMENT ----- 1.0 mil ADHESIVE ----- 2.0

mil

C KAPTON ---- 1.0 mil ADHESIVE ----- 0.5 mil INCONEL SPREADER ----

1.0 mil

C ADHESIVE ---- 2.0 mil KAPTON ---- 1.5 mil ADHESIVE ---- 3.0 mil

C ABSORPTIVE BLACK BACKGROUND ---- 2.0 mil LIQUID CRYSTAL ---- 5.0

mil

C MYLAR ---- AIR INSIDE THE TEST SECTION

C tinc1/kinc -- tadh2/kadh -- tkap2/kkap -- tadh2/kadh --

tinc2/kinc --

C tadh2/kadh -- tkap2/kkap -- tadh3/kadh -- tblack/kblack --

tliq/kliq --

C tmyl/kmyl -- 1/hi

C F R O N T W A L L (CAMERA SIDE)

127

C TOTAL RESISTANCE

C AIR INSIDE THE TEST SECTION ---- 0.45 inches PLEXIGLAS ----

AMBIENT

C 1/hi -- tplex/kplex -- 1/ho

C Heat transfer coefficient on the outer surface

De=7./12. ! ft, test section side with insulation

TambR=Tamb+460.

CALL AIRPROP(TambR,Gama,CON,VIS,PR,CP)

ho=0.36*con/De ! Ozisik, Page 443

tinc1 = 0.25e-03/12. ! MINCO's fact sheet

tinc2 = 0.50e-03/12.

tadh1 = 0.5e-03/12. ! MINCO's fact sheet

tadh2 = 1.0e-03/12.

tadh3 = 1.5e-03/12.

tadh4 = 2.0e-03/12.

tkap1 = 0.5e-03/12. ! MINCO's fact sheet

tkap2 = 2.0e-03/12.

tpoly = 3./12.

tplex = 0.5/12.

tsty = 2./12.

tblack = 3.0e-03/12. ! absorptive black background (from DAVIS)

tliq = 2.0e-03/12. ! liquid crystal thickness (from DAVIS)

tmyl = 5.0e-03/12. ! MYLAR thickness (from DAVIS)

kkap = 0.0942 ! BTU/hr.ft.F MINCO (0.163 W/m.K) agrees

with

! Sam Spring's 0.095

BTU/hr.ft.F

ksty = 0.02 ! BTU/hr.ft.F Sam Spring

kpoly = 0.0333 ! BTU/hr.ft.F from GOLDENWEST INC. R=2.5-

2.75

! hr.(sqft)/BTU.in

kplex = 0.11 ! BTU/hr.ft.F AIN Plastics k=1.3

BTU/hr.F.sqft/in,

! same given by 1-800-523-

7500

128

kmyl = 0.085 ! BTU/hr.ft.F Abauf's serpentine report,

page 19

kadh = 0.1272 ! BTU/hr.ft.F MINCO (0.220 W/m.K)

kinc = 9.0152 ! BTU/hr.ft.F MINCO (inconel 600 K=15.6

W/m.K)

kblack = 0.165 ! BTU/hr.ft.F Glycerin

kliq = 0.165 ! BTU/hr.ft.F Glycerin

Rinc1 = tinc1/kinc

Rinc2 = tinc2/kinc

Radh1 = tadh1/kadh

Radh2 = tadh2/kadh

Radh3 = tadh3/kadh

Radh4 = tadh4/kadh

Rkap1 = tkap1/kkap

Rkap2 = tkap2/kkap

Rpoly = tpoly/kpoly

Rsty = tsty/ksty

Rplex= tplex/kplex

Rblack = tblack/kblack

Rliq = tliq/kliq

Rmyl = tmyl/kmyl

C write(6,*)Rmyl

Rconv = 1./ho

Rback = 0.5*Rinc1 + Radh1 + Rkap1 + Radh4 + Rpoly + Rconv

Rfront =0.5*Rinc1 + Radh2 + Rkap2 + Radh2 + Rinc2 + Radh2 +

&Rkap2 + Radh3 + Rblack + Rliq

Fluxtop=0.

Fluxf=0.

Fluxbot=0.

Theater = (fluxb+Tamb/Rback+Tliquid/Rfront)/

&(1./Rback+1./Rfront)

C loss from the back side

flback = (Theater-Tamb)/Rback

ffront = (Theater-Tliquid)/Rfront

C Surface temperature, Ts

129

C write(6,*)'tmyl, kmyl,Rmyl,ffront,

DT',12*tmyl,kmyl,Rmyl,ffront,ffront*Rmyl

C write(6,*)'DT',ffront*Rmyl

Tback= Tliquid -ffront*Rmyl

Tf=0.5*(Tm+Tback)

perloss=100.*(flback/fluxb)

WRITE(7,*)' '

WRITE(7,*)' LIQUID CRYTAL SIDE'

WRITE(7,*)' '

WRITE(7,101)Theater,fluxb,flback,ffront,

& perloss,Tliquid,Tback,Tamb,ho

100 FORMAT(/,5X,'HEATER TEMPERATURE = ',F8.3,' F',/,

&5X,'TOTAL HEAT FLUX = ',F8.3,' BTU/hr.sqft',/,

&5X,'HEAT FLUX TO THE BACK = ',F8.3,' BTU/hr.sqft',/,

&5X,'HEAT FLUX TO THE FRONT = ',F8.3,' BTU/hr.sqft',/,

&5X,'% OF HEAT LOST FROM THE BACK SIDE = ',F8.3,/,

&5X,'LIQUID CRYSTAL TEMPERATURE = ',F8.3,' F',/,

&5X,'SURFACE TEMPERATURE = ',F8.3,' F',/,

&5X,'AMBIENT TEMPERATURE = ',F8.3,' F',/,

&5X,'Uinf where camera is located= ',F8.3,' ft/s',/,

&5X,'Re based on the test section outer dimension= ',E13.6,/,

&5X,'Outer heat transfer coefficient= ',F8.3,

&' BTU/hr.sqft.F')

101 FORMAT(/,5X,'HEATER TEMPERATURE = ',F8.3,' F',/,

&5X,'TOTAL HEAT FLUX = ',F8.3,' BTU/hr.sqft',/,

&5X,'HEAT FLUX TO THE BACK = ',F8.3,' BTU/hr.sqft',/,

&5X,'HEAT FLUX TO THE FRONT = ',F8.3,' BTU/hr.sqft',/,

&5X,'% OF HEAT LOST FROM THE BACK SIDE = ',F8.3,/,

&5X,'LIQUID CRYSTAL TEMPERATURE = ',F8.3,' F',/,

&5X,'SURFACE TEMPERATURE = ',F8.3,' F',/,

&5X,'AMBIENT TEMPERATURE = ',F8.3,' F',/,

&5X,'Outer heat transfer coefficient= ',F8.3,

&' BTU/hr.sqft.F')

Abot= 3.*Bmax*Hlength

Atop= 3.*Bmin*Hlength

Aback =3.*Width*Hlength

Afront=Aback

! AIR INLET PROPERTIES

TinR=Tin+460.

CALL AIRPROP(TinR,gamin,CONin,VISin,PRin,CPin)

! FLUX LOSSES FROM TOP, BOTTOM AND FRONT WALLS

! T O P W A L L

C Rtop=Rplex+Rconv

Ttop=Tm

130

C Fltop=(Ttop-Tamb)/Rtop ! Losses from top

! B O T T O M W A L L

C Rbot=Rplex+Rsty+Rconv

Tbot=Tin

C Flbot=(Tbot-Tamb)/Rbot

! F R O N T W A L L

R1=Rplex+Rconv !from surface to ambient

! INITIAL GUESSES

hback=(Fluxb-Flback)/(Tback-Tm)

hfront=hback

Tfront=Tm

DO I = 1,20

R3=1./hfront !convective resistance

! RADIATIONAL LOSSES

call rad(AR,Width,Length,Tback,Ttop,Tfront,Tbot,

&Frback,Frtop,Frfront,Frbot)

TfrontN=((1./R3)*Tm+(1./R1)*Tamb-Frfront)/((1./R1)+(1./R3))

IF(ABS(TfrontN-Tfront).LE.0.001) GO TO 999

Tfront=TfrontN

Flfront=(Tfront-Tamb)/R1

Flbot=0. ! For this geometry

Fltop=0. ! For this geometry

! TOTAL HEAT LOSS TO THE AMBIENT

Qwaste=Fltop*Atop+Flbot*Abot+Flfront*Afront+Flback*Aback

! NET HEAT ADDED TO THE AIR FROM THE INLET TO THE POINT IN

QUESTION

Qadd = Q-Qwaste

! AIR MIXED MEAN ENTHALPY AT THE POINT WHERE THE HEAT

TRANSFER

! COEFFICIENT IS BEING MEASURED

131

TmCAL=Tin+(Qadd)/(3600.*Mv*CPin) ! Energy balance

! HEAT TRANSFER COEFFICIENT FROM THE NEWTON LAW OF COOLING

Floss=Flback+Frback

hback=(Fluxb-Flback-Frback)/(Tback-Tm)

hfront=hback

ENDDO

WRITE(7,*)' did not convergance after 20 iterations'

GO TO 998

999 WRITE(7,*)' convergance after',I,' iterations'

998 write(7,110)Tback,Ttop,Tfront,Tbot

110 FORMAT(5x,'Back, Top, Front and Botom Wall Temperatures: ',

&/,10x,4F10.2,' F')

C For CFD use

fluxnet=(Fluxb-Flback-Frback)

C

write(7,115)TmCAL,Tf

115 FORMAT(5X,'Cal. Air Mixed Mean and Film Temperatures',2F9.3,'

F')

write(7,170)Q

170 format(5x,'Total Elect. Power=',F8.3,' BTU/hr')

write(7,116)Qwaste

116 FORMAT(5X,'Total Heat Loss to Ambient=',F8.3,' BTU/hr')

write(7,190)Fluxb,Fluxtop,Fluxf,Fluxbot

190 FORMAT(5X,'Heat Fluxes Generated by Back, Top, Front and'

&,' Bottom Heaters:',/,5x,4F11.3,' BTU/sqft.hr')

write(7,180)Flback,Fltop,Flfront,Flbot

180 FORMAT(5X,'Flux Losses from Back, Top, Front and'

&,' Bottom Surfaces:',/,10x,4F13.3,' BTU/sqft.hr')

write(7,150)Frback,Frtop,Frfront,Frbot

150 FORMAT(5X,'Radiative Fluxes from Back, Top, Front and'

&,' Bottom Surfaces:',/,10x,4F10.3,' BTU/sqft.hr')

RETURN

END

subroutine AIRPROP(t,gamx,kx,mux,prx,cpx)

IMPLICIT REAL*8(A-H,O-Z)

c physical properties of dry air at one atmosphere

c ref: ge heat transfer handbook

c

c temperature range: 160 to 3960 deg. rankine

c -300 to 3500 deg. fahreinheit

c

c t - temperature, R

c gamx - ratios of specific heats

c kx - thermal conductivity, BTU/hr.ft.R

c mux - viscosity, lbm/hr.ft

c prx - prandtl no.

132

c cpx - specific heat, BTU/lbm.R

c

c

dimension tab(34),gam(34),pr(34),cp(34)

real*8 k(34),mu(34),kx,mux

data nent/34/

data tab/ 160., 260.,

& 360., 460., 560., 660., 760., 860., 960., 1060.,

& 1160., 1260., 1360., 1460., 1560., 1660., 1760., 1860.,

& 1960., 2060., 2160., 2260., 2360., 2460., 2560., 2660.,

& 2760., 2860., 2960., 3160., 3360., 3560., 3760., 3960./

data gam/ 1.417, 1.411,

& 1.406, 1.403, 1.401, 1.398, 1.395, 1.390, 1.385, 1.378,

& 1.372, 1.366, 1.360, 1.355, 1.350, 1.345, 1.340, 1.336,

& 1.332, 1.328, 1.325, 1.321, 1.318, 1.315, 1.312, 1.309,

& 1.306, 1.303, 1.299, 1.293, 1.287, 1.281, 1.275, 1.269/

data k/ 0.0063,0.0086,

& 0.0108,0.0130,0.0154,0.0176,0.0198,0.0220,0.0243,0.0265,

& 0.0282,0.0301,0.0320,0.0338,0.0355,0.0370,0.0386,0.0405,

& 0.0422,0.0439,0.0455,0.0473,0.0490,0.0507,0.0525,0.0542,

& 0.0560,0.0578,0.0595,0.0632,0.0666,0.0702,0.0740,0.0780/

data mu/ 0.0130,0.0240,

& 0.0326,0.0394,0.0461,0.0519,0.0576,0.0627,0.0679,0.0721,

& 0.0766,0.0807,0.0847,0.0882,0.0920,0.0950,0.0980,0.1015,

& 0.1045,0.1075,0.1101,0.1110,0.1170,0.1200,0.1230,0.1265,

& 0.1300,0.1330,0.1360,0.1420,0.1480,0.1535,0.1595,0.1655/

data pr/ 0.7710,0.7590,

& 0.7390,0.7180,0.7030,0.6940,0.6860,0.6820,0.6790,0.6788,

& 0.6793,0.6811,0.6865,0.6880,0.6882,0.6885,0.6887,0.6890,

& 0.6891,0.6893,0.6895,0.6897,0.6899,0.6900,0.6902,0.6905,

& 0.6907,0.6909,0.6910,0.6913,0.6917,0.6921,0.6925,0.6929/

data cp/ 0.247, 0.242,

& 0.241, 0.240, 0.241, 0.242, 0.244, 0.246, 0.248, 0.251,

& 0.254, 0.257, 0.260, 0.264, 0.267, 0.270, 0.272, 0.275,

& 0.277, 0.279, 0.282, 0.284, 0.286, 0.288, 0.291, 0.293,

& 0.296, 0.298, 0.300, 0.305, 0.311, 0.318, 0.326, 0.338/

c

c

if(t.lt.tab(1)) print 510,t,tab(1)

510 format(" in airprop --- temp=",f8.1," is less than min temp",

&" of ",f8.1)

if(t.gt.tab(nent)) print 520, t,tab(nent)

520 format(" in airprop --- temp=",f8.1," is greater than max",

&" temp of ",f8.1)

if(t-tab(1))120,120,100

100 if(tab(nent)-t)130,130,110

110 m=2

go to 140

120 j=1

go to 180

130 j=nent

go to 180

140 if(t-tab(m))160,170,150

133

150 m=m+1

go to 140

c

c -- Linear Interpolation ---

c

160 slp=(t-tab(m-1))/(tab(m)-tab(m-1))

mux= mu(m-1)+(mu(m)-mu(m-1))*slp

prx= pr(m-1)+(pr(m)-pr(m-1))*slp

cpx=cp(m-1)+(cp(m)-cp(m-1))*slp

kx=k(m-1)+(k(m)-k(m-1))*slp

gamx=gam(m-1)+(gam(m)-gam(m-1))*slp

go to 190

170 j=m

go to 180

180 mux=mu(j)

prx=pr(j)

cpx=cp(j)

kx=k(j)

gamx=gam(j)

190 return

end

c

134

APPENDIX E

FORTRAN Program Integarea.f

REAL Nu,Nuave

CHARACTER*80 TITLE

open(unit=1,file='ph.in',status='old')

open(unit=2,file='area.dat',status='old')

open(unit=7,file='pl.dat',status='old')

open(unit=8,file='Nuave-plot.dat',status='old')

open(unit=9,file='have-plot.dat',status='old')

READ(1,*)NP

DO I=1,15

READ(1,100)TITLE

ENDDO

100 FORMAT(A80,//)

N=0

2 N=N+1

Reo=Re

IF(N.GT.NP) GO TO 3

READ(2,*)photo,A

READ(1,*)photo,Re,Nu,h,Uncer

IF(N.EQ.1) THEN

J=1

SUMRe=0.

SUMNu=0.

SUMh=0.

SUMunc=0.

SUMA=0.

GO TO 4

ENDIF

DIFF=ABS((Re-Reo)/Re)

IF(DIFF.GE..05)THEN

3 JTOT=J

J=1

Reave=SUMRe/JTOT

Nuave=SUMNu/SUMA

have=SUMh/SUMA

Uncave=SUMunc/JTOT

WRITE(6,300)Reave,Nuave,have,Uncave,SUMA,JTOT

WRITE(7,300)Reave,Nuave,have,Uncave,SUMA,JTOT

WRITE(8,301)Reave,Nuave,have/Nuave

WRITE(9,301)Reave,have,have/Nuave

300 FORMAT(1X,5E12.5,I4)

301 FORMAT(1X,3E12.5)

IF(N.GT.NP) GO TO 99

SUMRe=0.

135

SUMNu=0.

SUMh=0.

SUMunc=0.

SUMA=0.

ELSE

J=J+1

ENDIF

4 SUMRe=SUMRe+Re

SUMNu=SUMNu+Nu*A

SUMh =SUMh +h*A

SUMunc=SUMunc+Uncer

SUMA=SUMA+A

GO TO 2

99 WRITE(6,200)

WRITE(7,200)

200 FORMAT(5X,'Re',10X,'Nu',13X,'h',7X,'Uncer',

&9X,'Atot',5x,'# OF POINTS',/)

REWIND 1

READ(1,*)NP

DO I=1,14

READ(1,100)TITLE

WRITE(7,100)TITLE

ENDDO

STOP

END

136