EGR 334 Thermodynamics Chapter 4: Section 10-12 Lecture 18: Integrated Systems and System Analysis Quiz Today?

EGR 334 ThermodynamicsChapter 4: Section 10-12

Lecture 18: Integrated Systems and System Analysis Quiz Today?

Today’s main concepts:• Be able to explain what an integrated system is• Be able to describe the components of some common

integrated systems • Apply mass balance, energy balance, and continuity to

streams of flow through integrated systems.

Reading Assignment:

Homework Assignment:

• No New Reading Assignment for Wed.• Read Chap 5 for Friday.

Problems from Chap 4: 95, 98, 102

3

Terms for Chap 5 Class Discussion :

• Spontaneous Heat Transfer•Clausius Statement•Kelvin-Planck Statement• Entropy Statement• Irreversible vs. Reversible• Internally Reversible Process•Carnot Corollaries•Carnot Efficiency•Max. heat pump efficiency•Max. refrigeration cycle COP•Carnot Cycle•Clausius Inequality

You may want to create a summary sheet to help you discuss each of the concepts.

►Engineers creatively combine components to achieve some overall objective, subject to constraints such as minimum total cost. This engineering activity is called system integration.

System Integration

►The simple vapor power plant of Fig 4.16 provides an illustration.

5Sec 4.11: System Integration

System Integration : Combine components to make a useful cycle Some common systems:

- Power Plant- Refrigerator- Heat Pump

Components:PipesNozzles/DiffusersTurbinesCompressors/PumpsHeat ExchangerThrottling

Integrated Thermodynamics Systems:


Power Plant Cycle

1. Boil WaterBurn somethingNuclear ReactionGeothermal

2. Use the steam as it rises to turn a turbine.


Power Plant Cycle

3.Condense Steam(to recycle water)

using a heat exchanger

4. Pump water back to boiler.


Power Plant Cycle

1. Boiler

2. Turbine

3. Condenser

4. Pump

9

WCV

.

Sec 4.11: System Integration

Power Plant Cycle

1. Boiler

2. Turbine

3. Condenser4. Pump

QCV

QCV

WCV


1.CondenserWe know that condensing something will remove heat from the fluid.

3. Evaporator

Refrigeration cycle

To have something to condense, we must have evaporated something.

(Both are heat exchangers)


Refrigeration cycle

1. Condenser

2. Throttling valve/ Expander

3. Evaporator

4. Compressor


Refrigeration Cycle 1. Condenser

2. Throttling valve/ Expander

3. Evaporator4. Compressor

WCV

QCV

QCV

13

Example: (4.103) A simple gas turbine power cycle operating at steady state with air as the working substance is shown in the figure. The cycle components include an air compressor mounted on the same shaft as the turbine. The air is heated in the high-pressure heat exchanger before entering the turbine. The air exiting the turbine is cooled in the low-pressure heat exchanger before returning to the compressor. KE and PE effects are negligible. The compressor and turbine are adiabatic. Using the ideal gas model for air, determine the


(a) Power required for the compressor, in hp,

(b) Power output of the turbine, in hp,

(c) Thermal efficiency of the cycle

14

state 1 2 3 4

AV (ft3/min)

30,000

p (atm) 1 p2>p1 p3=p2 1

T (°R) 520 650 2000 980

h (BTU/lb) 124.27

155.51

504.71

236.02

Example: (4.103)


(a) Power required for the compressor, in hp,(b) Power output of the turbine, in hp,(c) Thermal efficiency of the cycle

state 1 2 3 4

AV (ft3/min)

30,000

p (atm) 1 p2>p1 p3=p2 p4=p1

T (°R) 520 650 2000 980

h (BTU/lb)

state Q W

Compressor

0 ?

Heat Ex 1 0

Turbine 0 ?

Heat Ex 2 0

From Table A-22E : Ideal Gas Properties of Air Assumptions•Steady State•KE= PE = 0•Turbine and compressor are Adiabatic (QCV= 0)

•No work in Heat Ex. (WCV = 0)

•Air is modeled as an ideal gas

15

Example: (4.103)



state Q W

Compressor

0 ?

Heat Ex 1 0

Turbine 0 ?

Heat Ex 2 0

state 1 2 3 4

AV (ft3/min)

30,000

p (atm) 1 p2>p1 p3=p2 1

T (°R) 520 650 2000 980

h (BTU/lb) 124.27

155.51

504.71

236.02

11

VAm

v

3 2 2

2

(30,000 / min)(14.7 / )(28.97 / ) 144 60 min

(1545 / )(520 ) 1 1f m mol

f mol

ft lb in lb lb inm

ft lb lb R R ft hr

1

1

RTv

P

11

1

AV p Mm

RT

and so

137,394lb

hr

Using Continuity Eq. Ideal Gas Eq.

16

Example: (4.103)



state Q W

Compressor

0 ?

Heat Ex 1 0

Turbine 0 ?

Heat Ex 2 0

state 1 2 3 4

AV (ft3/min)

30,000

p (atm) 1 p2>p1 p3=p2 1

T (°R) 520 650 2000 980

h (BTU/lb) 124.27

155.51

504.71

236.02

2CV V

2CV CV

dEQ W m h gz

dt

Energy Balance (Compressor):

1 2CVW m h h

12545 /137 394 124.27 155.51 1687m

m

lb hpBTUCV hr lb Btu hrW hp

state Q W

Compressor

0 -1687

Heat Ex 1 0

Turbine 0 ?

Heat Ex 2 0

17

Example: (4.103)



state Q W

Compressor

0 -1687

Heat Ex 1 0

Turbine 0 14,505

Heat Ex 2 0

state 1 2 3 4

AV (ft3/min)

30,000

p (atm) 1 p2>p1 p3=p2 1

T (°R) 520 650 2000 980

h (BTU/lb) 124.27

155.51

504.71

236.02

Energy Balance (Same as Compressor): 43 hhmWCV

12545 /(137394 ) 504.71 236.02 14,505m

m

lb hpBTUCV hr lb Btu hrW hp

18

Example: (4.103)



state Q W

Compressor

0 -1687

Heat Ex 1 0

Turbine 0 14,505

Heat Ex 2 0

state 1 2 3 4

AV (ft3/min)

30,000

p (atm) 1 p2>p1 p3=p2 1

T (°R) 520 650 2000 980

h (BTU/lb) 124.27

155.51

504.71

236.02Find Q23 Energy Balance around heat exchanger:

gz

vhmWQ

dt

dECVCV 2

2

CV 3 2CVQ m h h

2545 /137394 504.71 155.51 18,851m

m

lb hpBTUCV hr lb Btu hrQ hp

state Q W

Compressor

0 -1687

Heat Ex 1 18,851 0

Turbine 0 14,505

Heat Ex 2 0

68.0851,18

1687505,14

HeatEx

CompressorTurbne

Q

WW

19

Example: (4.78) As sketched in the figure, a condenser using river water to condense steam with a mass flow rate of 2x105 kg/h from saturated vapor to saturated liquid at a pressure of 0.1 bar is proposed for an industrial plant. Measurements indicate that several hundred meters upstream of the plant, the river has a volumetric flow rate of 2x105 m3/h and a temperature of 15°C. For operation at steady state and ignoring changes in KE and PE, determine the river-water temperature rise, in °C, downstream of the plant traceable to use of such a condenser, and comment.

Sec 4.9: Heat Exchangers (Revisited)

20

Example: (4.78)


state Pl,i Pl,e R,i R,e

Av (m3/h) 2x105

2x105

m (kg/h) 2x105 2x105

x 1 0 0 0

P (bar) 0.1 0.1

T (°C) 15 ?

h (kJ/kg)From Table A-2 : Properties of Saturated Water


Av (m3/h) 2x105

2x105

m (kg/h) 2x105 2x105

x 1 0 0 0

P (bar) 0.1 0.1

T (°C) 45.81 45.81 15 ?

h (kJ/kg)


AV (m3/h) 2x105

2x105

m (kg/h) 2x105 2x105

x 1 0 0 0

p (bar) 0.1 0.1

T (°C) 45.81 45.81 15 ?

h (kJ/kg) 2584.7

191.83

62.99

3

3

58

3R

V 2 102 10 /

1.0009 10

mhR

R mkg

Am kg hr

v

Look up intensive propertiesfor water from Tables based on known property values

21

Example: (4.78)


From Table A-2 : Properties of Saturated Water (p817 & 819)


Av (m3/h) 2x105

2x105

m (kg/h) 2x105 2x105 2x108 2x108

x 1 0 0 0

P (bar) 0.1 0.1

T (°C) 45.81 45.81 15 ?

h (kJ/kg) 2584.7

191.83

62.99

gz

vhmWQ

dt

dECVCV 2

2

CV

, , , ,0 Pl Pl i Pl e R R i R em h h m h h

Using the energy balance simplified for a heat exchanger

, , , ,Pl

R e R i Pl i Pl eR

mh h h h

m

22

Example: (4.78)


From Table A-2 : Properties of Saturated Water (p817 & 819)


Av (m3/h) 2x105

2x105

m (kg/h) 2x105 2x105

x 1 0 0 0

P (bar) 0.1 0.1

T (°C) 45.81 45.81 15 ?

h (kJ/kg) 2584.7

191.83

62.99

ePliPlR

PliReR hh

m

mhh ,,,,

5

, 8

2 10 /62.99 2584.7 191.83 65.38

2 10 /R e

kg hrkJ kJ kJh

kg kg kgkg hr

Using this value, the exit temperature may be found from Table A-2. TR,e = 15.6°C giving a temperature rise of 0.6 °C

Plugging known values:

23

end of Lecture 18 Slides

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EGR 334 Thermodynamics Chapter 4: Section 10-12 Lecture 18: Integrated Systems and System Analysis Quiz Today?