CANKAYA UNIVERSITY
FACULTY OF ENGINEERING AND ARCHITECTURE
MECHANICAL ENGINEERING DEPARTMENT
ME 211 THERMODYNAMICS I
FALL 2015
HW # 4 SOLUTION
1) Air at 10 0C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s.
The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very
small compared with the inlet velocity. Determine (a) the mass flow rate of the air and (b) the
temperature of the air leaving the diffuser.
Solution
(b) SSSF conditions
Conservation of mass
 me   mi
e
i
Single inlet single outlet
m2  m1  m
First Law of Thermodynamics
1
1
QCV  WCV   me (h e  Ve2  gz e )   mi (h i  Vi2  gz i )
2
2
e
i
QCV  WCV  0, V2  0, PE  0
1
0   me h e   mi (h i  Vi2 )
2
e
i
Single inlet single outlet
1
m(h1  V12 )  mh 2
2
1
h 2  (h1  V12 )
2
At the inlet
T1  10 0C 

p1  80 kPa 
2) Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of
the air is 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes
in kinetic and potential energies are negligible, determine the necessary power input to the
compressor.
Solution
(b) SSSF conditions
Conservation of mass
 me   mi
e
i
Single inlet single outlet
m2  m1  m
First Law of Thermodynamics
1
1
QCV  WCV   me (h e  Ve2  gz e )   mi (h i  Vi2  gz i )
2
2
e
i
KE  0, PE  0
QCV  WCV   me h e   mi h i
e
i
Single inlet single outlet
QCV  WCV  m(h 2  h1 )
Qout  (  WC )  m(h 2  h1 )
Qout  mq out
WC  mq out  m(h 2  h1 )
p1  100 kPa 
h1  280.13kJ / kg
T1  280 K 
p1  600 kPa 
h 2  400.98 kJ / kg
T1  400 K 
3) The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions
of the steam are as indicated in Figure given below
(a) Compare the magnitudes of h, KE, PE
(b) Determine the work done per unit mass of the steam flowing through the
turbine.
(c) Calculate the mass flow rate of the steam.
Solution
SSSF
At the inlet:
p1  2 MPa  20 bar  Tsat  212.4 0C
T1  Tsat
At the outlet:
Let us compute h, KE, PE
b)
Conservation of mass
 me   mi
e
i
Single inlet single outlet
m2  m1  m
First Law of Thermodynamics
1
1
QCV  WCV   me (h e  Ve2  gz e )   mi (h i  Vi2  gz i )
2
2
e
i
Single inlet single outlet
1
1
QCV  WCV  me (h e  Ve2  gz e )  m i (h i  Vi2  gz i )
2
2
1


QCV  WCV  m (h e  h i )  (Ve2  Vi2 )  g(z e  z i ) 
2


1


Wt  m (h e  h i )  (Ve2  Vi2 )  g(z e  z i ) 
2


Neglecting heat loss,
4) QCV
0
1


Wt  m (h 2  h1 )  (V22  V12 )  g(z 2  z1 ) 
2


or
W
1


 w t  (h 2  h1 )  (V22  V12 )  g(z 2  z1 )  kJ / kg
m
2


4) Refrigerant-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and
is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state
and the temperature drop during this process.
Solution
5) Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters the condenser
with a mass flow rate of 6 kg/min at 1 MPa and 70 0C and leaves at 35 0C. The cooling water
enters at 300 kPa and 15 0C and leaves at 25 0C. Neglecting any pressure drops, determine (a) the
mass flow rate of the cooling water required and (b) the heat transfer rate from the refrigerant to
water.
SSSF conditions
Conservation of mass
 me   mi
e
i
m 2  m 4  m1  m 3
m3  m4  m R
m1  m 2  m w
m R  mass flow rate of R-134a
m w =mass flow rate of water\
First Law of Thermodynamics
1
1
QCV  WCV   me (h e  Ve2  gz e )   mi (h i  Vi2  gz i )
2
2
e
i
KE  0, PE  0,QCV  0, WCV  0
m h  m h
e
e
e
i
i
i
m4 h 4  m 2 h 2  m1h1  m3h 3
m R (h 4  h 3 )  m W (h1  h 2 )
Now we need to determine the enthalpies at all four states. Water exists as a compressed liquid at
both the inlet and the exit since the temperatures at both locations are below the saturation
temperature of water at 300 kPa.
At inlet 1:
T1  15 0C
p1  300 kPa
p1  300 kPa  Tsat  133.6 0C
T1  Tsat compressed liquid
At outlet 2:
T1  25 0C
p1  300 kPa
p2  300 kPa  Tsat  133.6 0C
T2  Tsat compressed liquid
State-3
p3  1MPa  10 bar  Tsat  39.39 0C
T1  Tsat
State-4
p4  1MPa  10 bar  Tsat  39.39 0C
T4  Tsat
Compressed liquid
So the heat lost by hot fluid is equal to heat gained by cold fluid
See CV on next page
QW  QR
QW  m w (h 2  h1 )
6) An insulated 8-m3 rigid tank contains air at 600 kPa and 400 K. A valve connected to the tank
is now opened, and air is allowed to escape until the pressure inside drops to 200 kPa. The air
temperature during the process is maintained constant by an electric resistance heater placed in
the tank. Determine the electrical energy supplied to air during this process.
Solution
USUF process
Conservation of mass:
 m2  m1 cv   mi   me
i
i
Conservation of energy

𝑄𝐶𝑉 +  mi  hi 
i

=  me  he 
e



Vi 2
 gzi 
2


𝑉22
𝑉2
Ve 2
+ 𝑔𝑧2 ) − 𝑚1 (𝑢1 + 1 + 𝑔𝑧1 )] + Wcv
 gze  +[𝑚2 (𝑢2 +
2
2
2
𝐶𝑉

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 exit conditions remain
constant.
2) Kinetic and potential energies are negligible.
3 ) The tank is insulated and thus heat transfer is negligible.
4) Air is an ideal gas with variable specific heats. For this reason, we will use air tables.
We take the contents of the tank as the system, which is a control volume since mass crosses the
boundary. Noting that the microscopic energies of flowing and non- flowing 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
Conservation of mass.
No mass is entering. So
 m2  m1 cv   mi   me
i
me   m1  m2 cv
i
Conservation of energy
KE  0, PE  0, no mass is leaving since tank is insulated QCV  0,
Electrical work is done on the CV
We  me h e +[𝑚2 (𝑢2 ) − 𝑚1 (𝑢1 )]𝐶𝑉