EGR 334 ThermodynamicsChapter 4: Section 6-8

Lecture 16: Control Volume Applications:Day 1 Quiz Today?

Today’s main concepts:• Be able to set up mass and energy balance models for

Turbines Pumps Compressors Boilers Heat Exchangers Nozzles Diffusers Throttle

Reading Assignment:

Homework Assignment:

• Read Chapter 4, Sections 10-12

Problems from Chap 4: 36, 43, 52, 66

3

em

CVdm

dtim

Mass Rate Balance:

Energy Rate Balance:

e em e

CVdE

dti ime

Q W

Review: For a Control Volume:

2 2cv V V

2 2i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

cvin exit

i i

dmm m

dt

4

em

CVdm

dtim

Mass Rate Balance: 1 path, steady state

Energy Rate Balance: 1 path, steady state

e em e

CVdE

dti ime

Q W

Modeling applications with Control Volumes:

2 210 ( ) (V V ) ( )

2cv cv i e i e i eQ W m h h g z z

0 i em m

Many important applications involve one inlet, one exit control volumes at steady state. Today a number of these useful models will be developed using the one inlet, one outlet, steady state forms of the mass balance and energy balance given below.

i em m m

5

Control Volume Applications:

Nozzles Diffuser Turbine

PumpCompressor

Boiler

Heat Exchanger

Throttling Valve

6

If:

a) Outer surface of CV is well insulated…

Common Modeling assumptions:

0Q

Application models generally make use of simplifying assumptions to reduce the complexity of the Energy Balance. By removing terms that do not apply to a particular application or whose impact on the application is generally only minor the models take on simplified, useful, and easy to use forms.

Assumption:

b) Small change of elevation… 0i ez z

c) No mechanical mechanisms present… 0mechanicalW

d) CV maintains same shape and volume… 0pdV

W

e) No electrical effects act on CV… 0electricW

f) Inlet and Outlet have same physical size…2 2V V 0i e

g) Outer surface of CV is small…

h) Small ΔT between CV and environment…i) flow passes through CV in short time…

0Q 0Q

0Q

j) inlet size much larger than outlet size…

V >>V V 0e i i

Nozzles and Diffusers

►Nozzle: a flow passage of varying cross-sectional area in which the velocity of a gas or liquid increases in the direction of flow.

►Diffuser: a flow passage of varying cross-sectional area in which the velocity of a gas or liquid decreases in the direction of flow.

8Sec 4.6: Nozzles and Diffusers

Nozzles and Diffusers are used to change the speed of the mass flow through the control volume.

For continuous flow, changing the size of the cross section alters the speed of the flow.

if

incompressible

What common assumptions may be used to simplify the energy balance?

i em m

Vi Ve Vi Ve

V Vi i e e

i e

A A

v v V Vi i e eA A

if

continuous

e

eeei

iiiCVCV gz

vhmgz

vhmWQ

dt

dE

22

22CV

9Sec 4.6: Nozzles and Diffusers

Typical Energy Balance simplifications,

2 2CV V V

2 2i e

CV CV i i i e e e

dEQ W m h gz m h gz

dt

Horizontal Section (or very short vertical)

No pump/turbines

Even though there is no insulation, the V is high so there may be little heat transfer.

Steady State

Therefore,

2 2V V0

2 2i e

i i e em h m h

2 210 ( ) (V V )

2i e i eh h

10

Example: (4.34) Air with a mass flow rate of 5 lb/s enters a horizontal nozzle operating at steady state at 800°R, 50 psi and a velocity of 10 ft/s. At the exit, the temperature is 570°R and the velocity is 1510 ft/s. Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing.

50 psi800°R10 ft/s

570°R1510 ft/s

Ain = ?, Q = ?

Sec 4.6: Nozzles and Diffusers

.m = 5 lbm/s

11

ideal gas

Example: (4.34) Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing.

50 psi800°R10 ft/s

570°R1510 ft/s

Ain = ?, Q = ?

Sec 4.6: Nozzles and Diffusers

Vin exit

Am m m

v

mass balance

V

V Vi i i

ii i i

mv m RTAm A

v p

296.2 ftAi

2

22

(1545 / ) 800(5 / )

(10 / ) 28.97 / 14450 /f molm

im mol f

ft lb lb R Rlb s ftA

ft s lb lb inlb in

continuity

12

Example: (4.34) Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing.

50 psi800°R10ft/s

570°R1510ft/s

Sec 4.6: Nozzles and Diffusers

Energy balance at SS2 2

02 2i e

CV CV i i i e e e

v vQ W m h gz m h gz

2 2

0 0 0 02 2i e

CV i i e e

v vQ m h m h

2 2 2

2 2

1510 10 1 1 1136.26 191.81

2 1 / 32.2 778fCV

m f

lbQ BTU ft slug Btu

m lb s slug ft s lb ft lb

m

CV

lb

BTU

m

Q01.10

CVQ

5 /mm lb s 2 2

2e iCV

e i

v vQh h

m

reduced using assumptions:

from Table A-22E

Turbines

►Turbine: a device in which power is developed as a result of a gas or liquid passing through a set of blades attached to a shaft free to rotate.

14Sec 4.7: Turbines

Use Mass and Energy Balances still hold:

e

eeei

iiiCVCV gz

vhmgz

vhmWQ

dt

dE

22

22CV

A turbine is a device that develops power from a gas or liquid passing through a set of blades which are attached to a shaft free to rotate.

in exitm m m 0 in exitm m

15

Typical Energy Balance simplifications,

e

eeei

iiiCVCV gz

vhmgz

vhmWQ

dt

dE

22

22CV

Horizontal Section (or very short vertical)

Even though there is no insulation, the V is high so there is no heat transfer.

Steady State

Therefore, eeiiCV hmhmW 0

Then, eiCV hhmW

V 0

Sec 4.7: Turbines

WCV

.min

.mexit

16

Example: (4.50) Steam enters the first stage of a turbine at 40 bar and 500 °C with a volumetric flow rate of 90 m3/min. Steam exits the turbine at 20 bar and 400°C. The steam is then reheated at constant pressure to 500 °C before entering the second stage turbine. Steam leaves the second stage as saturated vapor at 0.6 bar. For operation at steady state, and ignoring stray heat transfer and KE and PE effects, determine the

40 bar500°C90 m3/min

Sec 4.7: Turbines

(a) Mass flow rate of steam, in kg/hr(b) Total power produced by both turbines, in kW(c) The rate of heat transfer to the steam flowing through the reheater, in kW.

WCV,1

Reheater

WCV,2

20 bar400°C

20 bar500°C

Sat. vapor0.6 bar

AssumptionsKE= PE=0QTurbine = 0

17

Example: (4.50)

40 bar500°C90 m3/min

Sec 4.7: Turbines

(a) Mass flow rate of steam, in kg/hr(b) Total power produced by both turbines, in kW(c) The rate of heat transfer to the steam

flowing through the reheater, in kW.

WCV,1

Reheater

WCV,2

20 bar400°C

20 bar500°C

Sat’d vapor0.6 bar

AssumptionsKE= PE=0QTurbine = 0

2) Find intensive properties from Table A-4

state In T1 Ex T1 Ex RH Ex T2

sat. vapor

P (bar) 40 20 20 0.6

T (C) 500 400 500 36.16

v (m3/kg)

0.08643 0.1512 0.1757 23.739

h (kJ/kg) 3445.3 3247.6 3467.6 2567.4

state In T1 Ex T1 Ex RH Ex T2

sat. vapor

P (bar) 40 20 20 0.6

T (C) 500 400 500

v (m3/kg)

h (kJ/kg)

state In T1 Ex T1 Ex RH Ex T2

phase

P (bar)

T (C)

v (m3/kg)

h (kJ/kg)

1) Identify state properties given in problem statement.

18

Example: (4.50)

Sec 4.7: Turbines

(a) Mass flow rate of steam, in kg/hr

(b) Total power produced by both turbines, in kW

(c) The rate of heat transfer to the steam flowing through the reheater, in kW.

state In T1 Ex T1 Ex RH Ex T2

Sat’d vap

P (bar) 40 20 20 0.6

T (C) 500 400 500 36.16

v (m3/kg)

0.08643 0.1512 0.1757 23.739

h (kJ/kg) 3445.3 3247.6 3467.6 2567.4

1 1 11

1 1

VinT

A Vm m

v v

To find mass flow rate from volumetric flow rate:

34

3

90 /min 60 min6.248 10 /

0.08643 / 1

mkg hr

m kg hr

31 90 / minV m

19

Example: (4.50)

Sec 4.7: Turbines

(a) Mass flow rate of steam, in kg/hr

(b) Total power produced by both turbines, in kW

(c) The rate of heat transfer to the steam flowing through the reheater, in kW.

state In T1 Ex T1 Ex RH Ex T2

Sat’d vap

P (bar) 40 20 20 0.6

T (C) 500 400 500 36.16

v (m3/kg)

0.08643 0.1512 0.1757 23.739

h (kJ/kg) 3445.3 3247.6 3467.6 2567.4

2211 eTiTeTinTCV hhhhmW

4 1(6.248 10 / ) 3445.3 3247.6 3467.6 2567.4 /

3600

kW hrkg hr kJ kg

kJ s s

17,565CVW kW

To find the power produced in both Turbines:

20

Example: (4.50)

Sec 4.7: Turbines

state In T1 Ex T1 Ex RH Ex T2

Sat’d vap

P (bar) 40 20 20 0.6

T (C) 500 400 500 36.16

v (m3/kg)

0.08643 0.1512 0.1757 23.739

h (kJ/kg) 3445.3 3247.6 3467.6 2567.4

2 2CV V V

2 2i e

CV CV i i i e e e

dEQ W m h gz m h gz

dt

0 CV i i eQ m h h

4 1(6.248 10 / ) 3467.6 3247.6 / 3,819

3600CV

kW hrQ kg hr kJ kg kW

kJ s s

Heat transferred in the reheater…starting with energy balance

CV i e iQ m h h Simplified with assumptions:

(a) Mass flow rate of steam, in kg/hr

(b) Total power produced by both turbines, in kW

(c) The rate of heat transfer to the steam flowing through the reheater, in kW.

Compressors and Pumps

►Compressors and Pumps: devices in which work is done on the substance flowing through them to change the state of the substance, typically to increase the pressure and/or elevation.

►Compressor : substance is gas

►Pump: substance is liquid

22Sec 4.8: Compressors and Pumps

Compressors and Pumps: Device where work is used to increase pressure and/or elevation of the flow substance.

Pump Model:used for liquids

Compressor Model: used for gases

WCV

m

in

.

mexit

.

.WCV

min

mexit

.

.

.

23

Typical Energy Balance simplifications,

e

eeei

iiiCVCV gz

vhmgz

vhmWQ

dt

dE

22

22CV

Horizontal Section (or very short vertical)

Even though there is no insulation, the V is high so there is no heat transfer.

Steady State

Therefore, eeiiCV hmhmW 0

Then, eiCV hhmW

v0

Sec 4.8: Compressors and Pumps

WCV

min

mexit

.

.

.

24

Example: (4.60) Air is compressed at steady state from 1 bar, 300 K, to 6 bar with a mass flow rate of 4 kg/s. Each unit of mass passing from the inlet to the exit undergoes a process described by pV1.27= constant. Heat transfer occurs at a rate of 46.95 kJ/kg of air flowing to the cooling water circulating in a water jacket enclosing the compressor. If ΔKE and ΔPE of the air are negligible, calculate the compressor power in kW. 1 bar

300 K4 kg/s

6 barQCV /m= 46.95 kJ/kg

Assumptions•Steady State•KE= PE=0•pV1.27= constant•Ideal gas

state In Ex

P (bar) 1 6

T (K) 300

h (kJ/kg) 300.19

Sec 4.8: Compressors and Pumps

Table A-22 (Ideal Gas Properties of Air) h is independent of p

WCV

.

state In Ex

P (bar) 1 6

T (K) 300

h (kJ/kg)

state In Ex

P (bar)

T (K)

h (kJ/kg)

25

Example: (4.60)1 bar300 K4 kg/s

6 barQCV = -46.95 kJ/kg

state In Ex

P (bar) 1 6

T (K) 300

h (kJ/kg) 300.19

Sec 4.8: Compressors and Pumps

From ideal gas equation and polytropic eq.

WCV

.

pV mRT

1 1 2 2

1 2

pV p V

T T

onstantnpV c

1 1 2 2n npV p V

Rearranging:2 1 2

1 2 1

V p T

V p T

1/2 1

1/1 2

n

n

V p

V p

Combining

1/1 2 1

1/2 1 2

n

n

p T p

p T p

11/ 1 (1/ )

2 2 2 2 21/ 1 (1/ )

1 1 1 1 1

nn n n

n n

T p p p p

T p p p p

26

state In Ex

P (bar) 1 6

T (K) 300 439.1

h (kJ/kg) 300.19 440.7

Example: (4.60) 1 bar300 K4 kg/s

6 barQCV = -46.95 kJ/kg

state In Ex

P (bar) 1 6

T (K) 300

h (kJ/kg) 300.19

Sec 4.8: Compressors and Pumps

0 CV CV i i eQ W m h h

1

22 1

1

nnP

T TP

4 / 46.95 / 300.19 440.7 / 750CV

kWW kg s kJ kg kJ kg kW

kJ s

Therefore the exit temperature is

WCV

.

The energy balance can then find the pump work

(Since T2 and p2 are nowknown, h2 may be found on table A22)

1.27 11.27

2

6300 439.1

1T K K

CVCV i i e

i

QW m h h

m

27

end of Lecture 16 Slides