Experiment: Plate Heat Exchanger

Experimental Setup

Aim:- To determine Heat Transfer Coefficient by natural convection.

Given Data: 1. Height of calibrated cylinder = 25 cm

2. Area of measuring cylinder = 44 cm2

3. Heat transfer area for heat exchanger = 0.0338 m2

4. Number of parallel channels for test fluid flow = 3

5. Number of parallel channels for cooling water flow = 4

6. Density of test fluid, ρ = 0.835 Kg/m3

7. Specific heat of test fluid, cp = 2.62 J/gK

8. Volume Flow rate of water = 320 LPH.

Calculation Table :

S. No. Time

Mass flow rate (Kg/sec.) Water Temp. (C) Oil Temp. (°C)

Inlet Outlet Inlet Outlet

1 16.24 0.0565579 23.1 24.9 72.6 64.5

2 13.41 0.0684937 24.2 26.5 85 78.2

3 12.82 0.0716459 23.3 25.4 82.5 73.9

4 12.1 0.0759091 23.2 25.2 77.6 70.3

5 11.34 0.0809965 23.4 25.7 85.5 78

6 9.9 0.0927778 24.3 26.8 89.2 82.3

7 9.58 0.0958768 23.4 25.6 84 77.7

8 9.31 0.0986574 24.4 27.1 90.3 84

9 8.41 0.1092152 23.2 25.5 79.5 73.8

10 7.87 0.116709 23.3 25.5 78.9 72.8

Sample Calculations: 1.

2.

3.

(

)

4.

)

For reading 1,

S. No. Mass flow

rate (Kg/sec.) T_m

Q (rate of heat transfer)

W

U (over all heat transfer coefficient)

W/m2-k

(°C) (°C)

1 0.056558 44.5 1200.3 798.4 47.7 41.4

2 0.068494 56.2 1220.3 642.2 58.5 54

3 0.071646 53.8 1614.3 888.0 57.1 50.6

4 0.075909 49.7 1451.8 864.2 52.4 47.1

5 0.080996 57.2 1591.6 823.8 59.8 54.6

6 0.092778 60.2 1677.2 824.7 62.4 58

7 0.095877 56.3 1582.5 831.3 58.4 54.3

8 0.098657 61.4 1628.4 784.9 63.2 59.6

9 0.109215 52.3 1631.0 923.0 54 50.6

10 0.116709 51.4 1865.2 1073.1 53.4 49.5

Plot log(U) Vs log(m)

Slope of best fit line = 0.3339.

According to text, a= 0.3339.

Plot 1/U Vs 1/(dm/dt)a

Intercept on Y axis = 5 * 10-5

so, 1/ho = 5 * 10-5

ho = 2.0 * 104 W/m2K.

y = 0.3339x + 2.0272 R² = 0.3389

2.7

2.75

2.8

2.85

2.9

2.95

3

2.25 2.3 2.35 2.4 2.45 2.5 2.55 2.6 2.65

log

(U)

log(m)

y = 0.0077x + 5E-05 R² = 0.3033

0.0010

0.0011

0.0012

0.0013

0.0014

0.0015

0.0016

0.130 0.135 0.140 0.145 0.150 0.155 0.160 0.165 0.170 0.175

1/

U

1/(dm/dt)a

S. No. U (over all heat transfer

coefficient) (W/m2K) hi (W/m2K)

1 798.4 767.8

2 642.2 622.2

3 888.0 850.3

4 864.2 828.4

5 823.8 791.2

6 824.7 792.0

7 831.3 798.1

8 784.9 755.3

9 923.0 882.3

10 1073.1 1018.5

Error analysis:

S. No. U (over all heat

transfer coefficient) Error in

calculating U % error in

U

1 798.4 40.6 5.1

2 642.2 42.3 6.6

3 888.0 43.7 4.9

4 864.2 50.4 5.8

5 823.8 48.4 5.9

6 824.7 55.3 6.7

7 831.3 59.6 7.2

8 784.9 61.6 7.9

9 923.0 77.2 8.4

10 1073.1 80.5 7.5

Sample calculations for error:

(

)

Results & Conclusions 1. Coefficient of heat transfer for water is found as 20000 W/m2K in this counter flow heat exchanger. It is

higher than typical range observed for water (500 – 10000).

2. A decreasing trend in temperature difference of oil with increasing mass flow rate was observed. This trend

can be explained by considering two conflicting factors

The heat transfer coefficient increases due to increase in fluid velocity.

The heat exchange per unit mass of fluid is reduced as fluid has less time of contact with the

exchange surface.

Among above two factors, second one dominates and hence we observe decreasing trend.

Sources of error 1. The pump was delivering a pulsating flow which may have affected heat transfer coefficient drastically

during flow hence inducting an error. It also created additional error in measuring flow rate during

experiment.

2. There was a slight leakage in the measuring cylinder.

Heat Transfer by Natural Convection

Aim:- To determine Heat Transfer Coefficient by natural convection.

Data:- 1. Diameter of the test cylinder, (d) = 0.038 m

2. Length of the test cylinder (L) = 0.5 m

3. Prandle No. (at 300K) = .707

4. Coefficient of Thermal Conductivity: 26.3e-3 W/mK

Observation and calculation Table :-

Quantity Formulae Case 1 Case 2 Case 3

Voltage across Heater (Volts) - 99 85 74

Current through Heater (Ampere) - 0.70 0.61 0.53

Temp 1 (oC) - 111 109 89

Temp 2 (oC) - 108 106 87

Temp 3 (oC) - 111 108 88

Temp 4 (oC) - 114 111 90

Temp 5 (oC) - 117 114 92

Temp 6 (oC) - 115 112 91

Temp 7 (oC) - 113 110 90

T_avg (oC) (∑T)/7 112.71 110.00 89.57

T_ambient (oC) - 28 28 29

Heat Transfer Coeff., h (W/m2K) V*I/(A*ΔT) 13.70 10.59 10.85

Error Calculation

Absolute Error in Value of h (W/m2K)

(

) 0.67 0.57 0.72

% Error in Value of h 100* Δh/h 4.91% 5.37% 6.65%

Sample Calculation for Reading 1: 1. Surface area for Heat Transfer:

A = πD*L = 3.14*0.038*0.50 = 0.05969 m2

2. Heat Dissipated:

Q = V*I = 99*0.7 = 69.3 W

3. Temperature Difference between Surface and Ambient Air:

ΔT = Tavg – Tamb = 113-28 = 85 oC

4. Heat Transfer Coefficient:

h = Q/(A* ΔT) = 69.3/(0.05969*85) = 13.7 W/m2K

5. Prandlt Number:

Pr from Tables at (301 K) = 0.707

6. Grashof NO.:

Gr = = 1.35*109

7. Theoretical Heat Transfer Coefficient:

htheo = K*NuL /L= 0.56*(Gr*Pr)0.25 *K/L = 5.19 W/m2K

Theoretical Values Calculation

Quantity Case 1 Case 2 Case 3

Grash off No., Gr 1.37E+09 1.32E+09 9.74E+08

Prandle No., Pr From table at Amb Temp 0.707 0.707 0.707

Nussele No., Nu 0.56*(Gr*Pr)0.25 98.72 97.92 90.71

Theoretical Heat Transfer Coeff.,h (W/m2K)

h= Nu*K/L 5.19 5.15 4.77

Result from the Experiment: 1. The Heat transfer coefficient obtained experimentally is 11.72∓0.65 W/m2K with maximum possible

error of 6.7%

2. The theoretically obtained Heat Transfer Coefficient is 5.04 W/m2K which is of same order as compared

to the experimentally obtained value.

3. In the above calculations we have neglected heat loss due to radiation. If radiation was also considered then the obtained h value would have been lesser than the above values.

Other Identified potential sources of error:- 1. As the thermocouples are attached to outer surface of the cylinder, they will hinder the formations of

uniform boundary layer around the cylinder thus affects the Natural convection.

2. The convection B.L. thickness should be checked to make sure that it does not intersects the walls of the

surrounding box.

3. The ambient temperature is measured at a arbitrary position wrt. the cylinder, If this position is within

BL the results will be inaccurate. To remove this error it should place outside BL.

4. Attaining steady state takes considerable time and there can be some error in concluding whether the

system has actually reached its steady state or not.

5. The thermocouples are attached to the cylindrical conductor. Hence there is some thermal contact

resistance between the cylinder and the thermocouple which renders the thermocouple reading

inaccurate.

Date: 22-Jan-2013 Team Members Nakul Nuwal (10003014)

Prateek Nyati (10001012) Shashank Agarwal (10003016)

Dhwanil Shukla (10003011) Prashant Bhatewara (10001028)