reading materials: chapter 7

Chapter 10 CHEM ENG 1007 1

Reading Materials: Chapter 7

LECTURE 22


Shaft Work vs Flow Work


Illustration 1 (from L. 21)Power generation system Steady state, Flow rate of working fluid (eg. Water) is m

QB

m h1

h4

2

3

QLT•

LG•

WG

Qc•

Wp•

•

•

h

h

•

QWT•

Turbine

Generator

Condenser

Pump

Boiler


Boiler


Illustration 22-11000 kg/h water at 75oC and 1 bar is fed to a pump, where the pressure of the water is increased to 20 bar. The high-pressure is fed to the boiler to produce steam at 250oC and 20 bar. The steam fed to a turbine. The exhaust steam exists in the turbine is saturated steam at 1 bar. The exhaust steam is cooled further, leaves the condenser as a subcooled liquid at 75oC, and the cycle continues.

Calculate (a) the work produced by the turbine, (b) the heat input in the boiler, (c) the work required by the pump, (d) the heat removed by the condenser, (e) the efficiency of the heat engine.


Power Generation System


Power Generation System

From Steam Table

1

4

3

2


Solutiona) Work produced by the turbine

5T 4 3

5

T

ˆ ˆW m(H H ) 1000 2674.9 2903.2 2.3x10 kJ/hr

2.3x10 kJ 1 hrWhr

633600s

.4 kW

b) Heat input to the boiler

6B 3 2

6

B

ˆ ˆQ m(H H ) 1000 2903.2 315.8 2.6x10 kJ/hr

2.6x10 kJ 1 hrQhr 3600s

718.7 kW


Solutionc) Work input to the PUMP

3P 2 1

3

P

ˆ ˆW m(H H ) 1000 315.8 314.3 1.5x10 kJ/hr

1.5x10 kJ 1 hrWhr

03600s

.4 kW

d) Heat removed from the condenser

6c 1 4

6

C

ˆ ˆQ m(H H ) 1000 314.3 2674.9 2.4x10 kJ/hr

2.4x10 kJ 1 hrQhr 3600s

656 kW


Solution

e) The efficiency of the heat engine

net

B

63. 0.4W x100%718

48.8% (very

.7Q low)


EB associates with Phase Changes

Chemical processes often include the change of a material between the three phases: solid, liquid, and vapour.

For example, water can be in the form of solid (ice), liquid, or vapour (steam).

Such processes involve the addition or removal of heat.


EB associates with Phase Changes

The three combinations of phases between which a change can occur are as follows:

1. Liquid to vapour (vaporization) or vapour to liquid (condensation)

2. Solid to liquid (melting) or liquid to solid (freezing)

3. Solid to vapour (sublimation) or vapour to solid (solid condensation)


Phase Diagram


, 2H O l 25 C

, 2H O v 250 C

, 2H O l 100 C

, 2H O v 100 C

pLC T

pgC T

ˆfgH???Q

Conversion of water to steam in a boiler

ˆ pL fg pgQ C T H C T


Conversion of water to steam in a boilerA. Initially the water at ambient

temperature is subcooled.

B. As heat is added its temperature rises steadily until it reaches the saturation temperature corresponding with the pressure in the cylinder. The volume of the water hardly changes during this process. At this point the water is saturated.


Conversion of water to steam in a boilerC. As more heat is added, steam is generated and the volume

increases dramatically since the steam occupies a greater space than the water from which it was generated.

D. The temperature however remains the same until all the water has been converted into steam. At this point the steam is saturated.

E. As additional heat is added, the temperature of the steam increases but at a faster rate than when the water only was being heated. The volume of the steam also increases. Steam at temperatures above the saturation temperature is superheated.



If the temperature T is plotted against the heat added Q the three regions namely subcooled water, saturated mixture and superheated steam are clearly indicated.


Three stages are evident:

(i) “Sensible heat”: T of water raised to boiling point.

pLˆq H C T

(ii) “Latent heat”: constant T and P

fg

fg g f

f

g

ˆ ˆq H H

ˆ ˆ ˆwhere H = enthalpy of vaporization = H - H

ˆ H = enthalpy of saturated liquidˆ H = enthalpy of saturated vapor


fgˆQ m H



(iii) “Superheat”: Heat added to dry saturated steam raises its T, at constant P.

pg

pg

ˆq H C T

where C is the specific heat of water vapor


How to determine Phase Enthalpy Change?

Example: Water

Solid Liquid (Melting/Solidification)

m,P 1bar s,P 1barˆ ˆH H 333.5 kJ/kg

Liquid Gas (Vaporisation/Condensation)

v,P 1bar c,P 1barˆ ˆH H 2256.7 kJ/kg


Another Example of steam table

How do we calculate Enthalpy (H)?

Ĥ=

= 2257.6 kJ/kg


Example 10.3

A steady-state boiler produces steam from a waste-water stream. The water enters the boiler as saturated water at 5.7 atm and 430 K (the boiling point of water at 5.7 atm), and the steam exists the boiler as saturated steam at that same temperature and pressure. The properties of importance are:

Mass flow rate : 8150 kg/hr

Heat of vaporization at 5.7 atm and 430 K: 2091 kJ/kg

How much heat must be added to the process?


Example 10.3

6fg

ˆQ m H 8150 2091 17x10 kJ/hg

boilerSaturated water (liquid) 5.7 atm, 430K

Saturated steam (vapour) 5.7 atm, 430KQ

m 8150kg/hr m 8150kg/hr

How much heat must be added to the process?

fgH 2091 kJ/kg

Notices that heat is being added to the process, so Q is a positive valueQ


Conversion of liquid water into “superheated” steam

100 kg/h of water at 25oC & 1 atm is vaporised into steam at 120oC and 2 atm. Calculate the heat input required.Given that:

2 2p,H O(l) p,H O(v)

ofg b

ofg b

kJ kJC 4.18 C 2.02kg.K kg.K

ˆAt 1 atm, H = 2256.7 kJ/kg & T = 100 C

ˆAt 2 atm, H = 2243.5 kJ/kg & T = 105 C

Illustration 22-2


1 pL B,1atm 1 fg pg) 2 B,1atm

1

5

ˆQ (120 C,1atm) m C T T H C T T

Q (120 C,1atm) 100 4.18 100 25 2.02 120 100

2.6x10 kJ/

2256.7

hr 72.5 kW

Solution Path 1: ignore the effect of pressure (assume

outlet pressure =1 atm)

H (L, 25oC) H [V, Tb( 100oC)]H [L, Tb( 100oC)]

H (V, 120oC)


Solution Path 2: ignore the effect of pressure (assume

2 atm at inlet)

H (L, 25oC) H [V, Tb( 105oC)]H [L, Tb( 105oC)]

H (V, 120oC)

2 pL B,2 atm 1 fg,2 atm pg 2 B,2 atm

5

ˆQ (120 C,2 atm) m C T T H C T T

100 4.18 105 25 2.02 120 105

2.6x10 kJ

2243.5

72./hr 4 kW


Path 3 (by using steam table)

oin outh (25 C,1 atm,liquid) h (120 C,2 atm,vapour)

From Steam Table

oin

oout

5

H (25 C,1 atm,liquid) 83.9 kJ/kg

H (120 C,2 atm,vapour) 2706.3 kJ/kg

H 2706.3 83.9 2622.4 kJ/kg

ˆQ m H 100 2622.4 2.62x10 k 72J/hr .8 kW

Solution


Illustration 22-3A small evaporator fitted with three identical electrical heating elements connected in parallel, is used to vaporize alcohol on a continuous basis. Liquid alcohol is fed in at a temperature of 24oC and at a rate of 0.018 kg s. Alcohol vapour is removed at the same rate, at a temperature of 82oC. Because heat is lost to the surrounds only 80% supplied is available to heat the alcohol, determine the required power input (kW)

Data for alcohol:

Mean specific heat 3 kJ kgK

Enthalpy of vaporization at 82oC: 650 kJ kg


Solution

p fgˆQ m C T H

Q 0.018 3 82 24 650

Q 14.83 kW14.83

P0

18..80

54 kW

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