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