EGR 334 Thermodynamics: Homework 16
Problem 4:36
Methane (CH4) gas enters a horizontal well insulated nozzle operating at steady state at 80 deg. C and a velocity of
10 m/s. Assuming ideal gas behavior for the methane, plot the temperature of the gas exiting the nozzle in deg C
versus the exit velocity ranging from 500 to 600 m/s.
----------------------------------------------------------------------------------------------------------------------------- -------------Methane as ideal gas
State 1: T1 = 80 C= 353 K
V1 = 10 m/s
State 2: V2 = 500 to 600 m/s
Energy Balance for a Nozzle at steady state:
0  Q  W  mi (hi 
Vi 2
V2
 gzi )  me (he  e  gze )
2
2
assuming: no work, no heat loss, constant elevation.
1
0  h1  h2  (V12  V2 2 )
2
Using ideal gas model:
1
0  c p (T1  T2 )  (V12  V2 2 )
2
T2  T1 
1
(V12  V2 2 )
2c p
do unit check:
[K ]  [K ] 
[m / s]2
kJ
N
[kJ / kg  K ] 1000 N  m kg  m / s 2
cp can be found from table A-21 where
cp
R
cp
R
cp
R
   T   T 2   T 3  T 4
 3.826  3.979  103 T  24.558  106 T 2  22.733 109 T 3  6.963 1012 T 4
c p  4.59R
 4.59
where
so
cp 
cp
M
cp 
4.59R 4.59(8.314kJ / kmol  K )

 2.379kJ / kg  K
M
(16.04kg / kmol )
A plot of the T2 vs V2 is shown on the next page which was created using Excel:
Plot of T2 vs. V2 for nozzle.
or solved with IT
x
EGR 334 Thermodynamics: Homework 16
Problem 4:43
Air expands through a turbine from 8 bar, 960 K to 1 bar, 450 K. The inlet velocity is small compared to the exit
velocity of 90 m/s. The turbine operates at steady state and develops a power output of 2500 kW. Heat transfer
between the turbine and its surroundings and potential energy effects are negligible. Modeling air as an ideal gas,
calculate the mass flow rate of air in kg/s and the exit area in sq. m.
----------------------------------------------------------------------------------------------------------------------------- ----Turbine using air as ideal gas
State 1: p1 = 8 bar and T1 = 960 K
V1 << V2 = 90 m/s
State 2: p2 = 1 bar, and T2= 450 K
Turbine power =Wdot = 2500 kW
Assume: Q = 0 and ΔPE = 0
constants:
use
R = 0.2870 kJ/kg-K
cp(@700K) = 1.075 kJ/kg-K
Using ideal gas:
State 1:
v1 
RT1 (0.2870kJ / kg  K )(960K )
bar
1000 N  m

 0.344m3 / kg
5
2
p1
(8bar )
10 N / m
kJ
State 2:
v2 
RT2 (0.2870kJ / kg  K )(450 K )
bar
1000 N  m

 1.2915m3 / kg
5
2
p2
(1bar )
10 N / m
kJ
from the energy balance at steady state:
0  Q  W  mi (hi 
Vi 2
V2
 gzi )  me (he  e  gze )
2
2
Assume Q = 0 and ΔPE = 0
V2
V2
W
 h1  1  h2  2
m
2
2
2
2
V
V
W
 h1  1  h2  2

m
2
2
0
W
1
 c p (T1  T2 )  V12  V2 2 
m
2
W
m
1
c p (T1  T2 )  V12  V2 2 
2


2500kW
m

1 2
kJ
N
2
 (1.075kJ / kg  K )(960  450) K   0  (90m / s ) 
2
1000
N

m
kg

m / s2

from the mass balance at steady state:
0  m1  m2
m1  m2  4.59kg / s
Volumetric flow rate is given by
V2  m v1  (4.59kg / s)(1.2915m3 / kg )  5.93m3 / s
and
V2  A2V2

A2 
V2 5.93m3 / s

 0.066 m2
V2
90m / s


 kJ / s  4.59kg / s
 kW


EGR 334 Thermodynamics: Homework 16
Problem 4:52
Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of 290 kJ/kg and exits at a higher
pressure with a specific enthalpy of 1023 kJ/kg. The mass flow rate is 0.1 kg/s. If the compressor power input is 77
kW, determine the rate of heat transfer between the compressor and its surroundings, in kW. Neglect kinetic and
potential energy effects and assume the ideal gas model.
-----------------------------------------------------------------------------------------------------------------------------Air at steady state as an ideal gas: mdot = 0.1 kg/s
W compressor in = 77 kW
p1 = 1 atm
h1 = 290 kJ/kg
p2 > p1
h2 = 1023 kJ/kg
constants: use R = 0.2870 kJ/kg-K
Assume V1≈V2 and ΔPE = 0
for Air as ideal gas, T can be found from table A-22:
for h1 = 290 kJ/kg……T1 = 289.8 K
h2 = 1023kJ/kg……T2 = 979.8 K
from the mass balance at steady state:
0  m1  m2
m1  m2  0.1kg / s
from the energy balance at steady state:
0  Q  W  mi (hi 
Vi 2
V2
 gzi )  me (he  e  gze )
2
2
Assume V1≈V2 and ΔPE = 0
0  Q  W  m1h1  m2 h2
kJ / s
Q  (77kW )
 (0.1kg / s )(1023  290)kJ / kg   3.7kJ / s  3.7kW
kW
EGR 334 Thermodynamics: Homework 16
Problem 4:66
The figure provides steady state operating data for a pump drawing water from a reservoir and delivering it at a
pressure of 3 bar to a storage tank perched above the reservoir. The mass flow rate of the water is 1.5 kg/s. The
water temperature remains nearly constant at 15 deg C, there is no significant change in kinetic energy from inlet to
exit and heat transfer between the pump and its surroundings is small. Determine the power required by the pump in
kW. Let g = 9.81 m/s2.
--------------------------------------------------------------------------------------------------------------------------------State 1: T1 = 15 C
p1 = 1bar
z1 = 0
state 2: T2 = 15 C
p2 = 3 bar
z2 =15 m
Assume ΔKE = 0 and Q = 0
Steady state:
from the mass balance at steady state:
0  m1  m2
m1  m2
from the energy balance at steady state:
0  Q  W  mi (hi 
Vi 2
V2
 gzi )  me (he  e  gze )
2
2
Assume Q = 0 and ΔKE = 0
for liquid h1 = h2 = hf(T=15) if temperature is constant
0  W  mg ( z1  z2 )
W  mg ( z1  z2 )  (1.5kg / s)(9.81m / s 2 )(0  15)m
N
kJ
kW
 0.221kW
2
kg  m / s 1000 N  m kJ / s
The work is negative since is done on the system which the work that needs to be done by the pump on the system:
Wpump  Wpump  0.221kW
Alternate solution: Using equation 3.13:
W  m (h1  h2 )  m g ( zi  ze )
where
h1  h2  h f (T )  v f (T )[ p1  psat (T )]  h f (T )  v f (T )[ p2  psat (T )]
h1  h2  v f ( p1  p2 )
W  m v f ( p1  p2 )  m g ( zi  ze )

105 N / m2
N
kJ
1000W 
W  (1.5kg / s) (0.0010009m3 / kg )(1  3)bar
 (9.81m / s 2 )(0 15)m
  521W
2
1bar
kg  m / s 1000 N  m kJ / s 

= -0.521 kW