6-62
6-90 A room is to be heated by an electric resistance heater placed in a duct in the room. The power rating of the electric
heater and the temperature rise of air as it passes through the heater are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3
Kinetic and potential energy changes are negligible. 4 The heating duct is adiabatic, and thus heat transfer through it is
negligible. 5 No air leaks in and out of the room.
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The specific heats of air at room temperature are cp =
1.005 and c
v
= 0.718 kJ/kg·K (Table A-2).
Analysis (a) The total mass of air in the room is
kg 3.142
)K 288)(K/kgmkPa 0.287(
m 012m 654
3
3
1
1
33
RT
m
V
We first take the entire room as our system, which is a closed system
since no mass leaks in or out. The power rating of the electric heater is
determined by applying the conservation of energy relation to this
constant volume closed system:
energies etc. p otential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
EEE
outin
456 m3
We
200 W
150 kJ/min
6-64
6-92 Problem 6-91 is reconsidered. The effect of the moving velocity of the steel plate on the rate of heat transfer
from the oil bath as the velocity varies from 5 to 50 m/min is to be investigated. Tate of heat transfer is to be plotted against
the plate velocity.
Analysis The problem is solved using EES, and the solution is given below.
“Knowns”
6-65
6-93E Water is heated in a parabolic solar collector. The required length of parabolic collector is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat loss from the tube is negligible so that the entire solar energy
incident on the tube is transferred to the water. 3 Kinetic and potential energy changes are negligible
Properties The specific heat of water at room temperature is cp = 1.00 Btu/lbm.F (Table A-3E).
Analysis We take the thin aluminum tube to be the system, which is a control volume. The energy balance for this steady–
Q
6-94 A house is heated by an electric resistance heater placed in a duct. The power rating of the electric heater is to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific
heats at room temperature. 3 Kinetic and potential energy changes are negligible.
Properties The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg·K (Table A–2)
Analysis We take the heating duct as the system, which is a control volume since mass crosses the boundary. There is only
one inlet and one exit, and thus
m m m
1 2
. The energy balance for this steady-flow system can be expressed in the rate
form as
0
energies etc. potential,
kinetic, internal,in change of Rate
(steady) 0
sy stem
mass and work,heat,by
nsferenergy tranet of Rate
outin
EEE
300 W
6-68
6-98 Water is heated by a 7-kW resistance heater as it flows through an insulated tube. The mass flow rate of water is to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Water is an incompressible substance
with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The tube is
adiabatic and thus heat losses are negligible.
6-70
6-100 Problem 6-99 is reconsidered. The effect of the exit velocity on the mass flow rate and the exit volume flow
rate is to be investiagted.
Analysis The problem is solved using EES, and the solution is given below.
R=0.287 [kJ/kg-K]
“Analysis”
Vel2
[m/s]
m
[kg/s]
Vol2
[m3/s]
5
7.5
10
12.5
15
17.5
20
22.5
25
0.02815
0.02813
0.02811
0.02808
0.02804
0.028
0.02795
0.0279
0.02783
0.02852
0.0285
0.02848
0.02845
0.02841
0.02837
0.02832
0.02826
0.0282
5 9 13 17 21 25
0.02815
0.0282
0.0283
0.02835
0.0284
0.02845
0.0285
0.02855
Vel2 [m/s]
Vol2 [m3/s]
5 9 13 17 21 25
0.0278
0.0279
0.02795
0.028
0.02805
0.0281
0.02815
m [kg/s]
6-71
6-101E The ducts of an air-conditioning system pass through an unconditioned area. The inlet velocity and the exit
temperature of air are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific
heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 There are no work interactions involved.
Properties The gas constant of air is 0.3704 psia.ft3/lbm.R (Table A-1E). The constant pressure specific heat of air at room
temperature is cp = 0.240 Btu/lbm.R (Table A-2E)
2
1
r
A
Then the mass flow rate of air becomes
/lbmft 12.6
psia 15
R 510R/lbmftpsia 0.3704
3
3
1
1
1
v
P
RT
450 ft3/min
AIR
2 Btu/s
D = 10 in
PROPRIETARY MATERIAL. © 2017 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
6-102 Steam in a supply line is allowed to enter an initially evacuated tank. The temperature of the steam in the supply line
and the flow work are to be determined.
Analysis Flow work of the steam in the supply line is converted to sensible internal energy in the tank. That is,
tankline uh
where
MPa 4
tank
u
P
Steam
4 MPa
6-73
6-103 Steam flowing in a supply line is allowed to enter into an insulated tank until a specified state is achieved in the tank.
The mass of the steam that has entered and the pressure of the steam in the supply line are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can
be analyzed as a uniform-flow process since the state of fluid entering the tank remains constant. 2 Kinetic and potential
energies are negligible.
Properties The initial and final properties of steam in the tank are (Tables A-5 and A-6)
kJ/kg 8.2582
/kgm 19436.0
vap.)(sat. 1
MPa 1
1
3
1
1
1
u
x
P
v
kJ/kg 2.2773
/kgm 12551.0
C300
MPa 2
2
3
2
1
2
u
T
P
v
Analysis We take the tank as the system, which is a control volume since mass crosses
the boundary. Noting that the microscopic energies of flowing and nonflowing fluids
are represented by enthalpy h and internal energy u, respectively, the mass and energy
balances for this uniform-flow system can be expressed as
Steam
Sat. vapor
2 m3
1 MPa
400C
6-75
6-105 A rigid tank initially contains superheated steam. A valve at the top of the tank is opened, and vapor is allowed to
escape at constant pressure until the temperature rises to 500C. The amount of heat transfer is to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can
be analyzed as a uniform-flow process by using constant average properties for the steam leaving the tank. 2 Kinetic and
potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank
kJ/kg 3468.3,kJ/kg 3116.9
/kgm 0.17568
C500
MPa 2
kJ/kg 3024.2,kJ/kg 2773.2
/kgm 0.12551
C300
MPa 2
22
3
2
2
2
11
3
1
1
1
hu
T
P
hu
T
P
v
v
Energy balance:
kinetic, internal,in Change
system
nsferenergy traNet
outin
EEE
STEAM
2 MPa
Q
allowed to enter the tank until one-half of the tank is filled with liquid water. The final pressure in the tank, the mass of
steam that entered, and the heat transfer are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can
be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies
are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).
Properties The properties of water are (Tables A-4E through A-6E)
Btu/lbm 1210.9
F400
psia 020
Btu/lbm 1099.8,51.269
/lbmft 4663.6,01745.0
mixture sat.
300
Btu/lbm 1099.8
/lbmft 4663.6
vaporsat.
F030
3
2
F030@1
3
F030@1
1
i
i
i
gf
gf
g
g
h
T
P
uu
FT
uu
T
vv
vv
Analysis We take the tank as the system, which is a control volume since
mass crosses the boundary. Noting that the microscopic energies of
flowing and nonflowing fluids are represented by enthalpy h and internal
energy u, respectively, the mass and energy balances for this uniform-
lbm 20.86232.097.85
/lbmft 6.4663
.ft 1.5
/lbmft 0.01745
ft 1.5
3
3
3
3
2
g
g
f
f
gf mmm
v
V
v
V
Then from the mass balance
464.020.86
12 mmmi
85.74 lbm
(c) The heat transfer during this process is determined from the energy balance to be
Btu 80,900,
out
1122in
Btu 90080
Btu/lbm 1099.8lbm 0.464Btu 23,425Btu/lbm 1210.9lbm 85.74
Q
umumhmQ ii
since
Btu 23,4251099.80.232269.5197.85
222 ggff umumumU
Discussion A negative result for heat transfer indicates that the assumed direction is wrong, and should be reversed.
200 psia
400F
Water
3 ft3
300F
Sat. vapor
Steam
Q
cp = 0.240 Btu/lbm·R and c
v
= 0.171 Btu/lbm·R (Table A–2Ea).
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the
mass and energy balances for this uniform-flow system can be expressed as
Mass balance:
12systemoutin mmmmmm i
1122
1122
energies etc. potential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
outin
TcmTcmTcm
umumhm
EEE
ipi
ii
vv
20 psia
60F
2 ft3
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can
be analyzed as a uniform-flow process since the state of fluid at the exit remains constant. 2 Kinetic and potential energies
are negligible. 3 There are no work interactions involved.
21
12
mmm
e
e
Energy balance:
EEE
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
Liquid R-134a
5 kg
24C