SECTION IV – FIRST LAW APPLIED TO OPEN SYSTEMS
Refrigerant-12 vapour enters a steady-flow compressor as a saturated vapour at 12C. The outlet
conditions are 0.6 MPa and 50C, and the process is assumed to be adiabatic. Calculate the power
in Kilowatts, required if the refrigerant flow rate is 20 kg/min. Determine the diameter of the inlet
tubing to the compressor if the inlet velocity is not to exceed 3 m/s.
Schematic of compressor
system boundary
ie

Inlet conditions TI = 12C, saturated vapour

Exit Conditions
Pe = 0.6 MPa, Te = 50C
Pe < Osat = 1.2193 MPa corresponding to
Te = 50C  Refrigerant-12 is superheated
at the exit.
e

e

 0
Process is adiabetic  Q
ie

i m
e m

Flow through compressor is steady  m
  20 kg / min  0.333 kg / sec
m

Assumptions
- Properties are uniformly distributed at the inlet and exit.
- epot and ekin are negligible compared to h.

Steady-state, steady-flow form of the conservation of energy for the compressor as an open
system is then:
 m
 (he  hi )
-W
ie
hi 
h ig

h ig  191.602
T  12C
12  10
h e  216.141 kJ / kg

193.644  191.602
 h ig  192.419 kJ / kg
15  10
 m
 (h i  h e )  0.333(192.419  216.141)   7.90 KW
W
ie
Power Input
D i2
1
4(0.333)
 i   i Vi A i  Vi 
m
 D i2 
vi
vi
4
3 
Section IV – First Law Applied to Open Systems
Page 19
v i  0.040914 0.035413  0.040914

 v i  0.0387136 m 3 / kg
12  10
15  10
D i2 
4(0.333)(0.0387136)
 Di  0.074 m  74 mm
3 
Steam enters a turbine at 600C and 6 MPa with a velocity of 300 m/s. The mass flow rate entering
the turbine is 400 kg/min. Assume the turbine is well-insulated and the exhaust steam leaves the
turbine at a low velocity. The steam exits the turbine at 200 KPa and 260C. Calculate the power
developed by the turbine and the inlet-duct area.
Schematic of turbine
system boundary

Inlet Conditions: TI = 600C, PI = 6 MPa,
 i  400 kg / min  6.67 kg / s, Vi  300 m / s
m

Exit Conditions: Te = 260C, Pe = 200 KPa
ie
e

e

 0
Turbine is well-insulated  Q
ie

Assumptions
- flow through turbine is steady i.e. no mass build-up in turbine
- flow properties are uniformly distributed at the inlet and exit to turbine
- changes in potential energy of working fluid are negligible compared to change in enthalpy
- neglect kinetic energy at the exit since the velocity will be low

Steady-state, steady-flow form the conservation of energy for the turbine as an open system is
then:


W ie  m(he  hi  ekini )

steam is superheated at the inlet since T > Tsat = 275.64C corresponding to P = 6 MPa  hI
= 3658.4 kJ
kg

e kini 

1 (300) 2
 45 kJ kg
2 1000
steam is also superheated at the outlet since P < Psat = 4.688 Mpa corresponding to P = 200
KPa
Section IV – First Law Applied to Open Systems
Page 20
he  2971.0 3071.8  2971.0
kJ

 he  2991.2
260  250
300  250
kg


 Wie  m(hi  ekini  he )  6.67(3658.4  45  2991.2)

Wie  4750.4 KW  4.75MW


VA
mv
6.67 x0.06525
m i   iVi Ai  i i  Ai  i i 
vi
Vi
300
A i  0.00145m 2  14.5cm 2
4-64 Boles & Cengel
Refrigerant –12 at 1MPa and 80C is cooled to 1MPa and 30C in a condenser by air. The air
enters at 100KPa and 27C with a volume flow rate of 800 m3/min and leaves at 95KPa and 60C.
Determine the mass flow rate of the refrigerant.
Air

1 = 800 3/min
R-12
P3 = 100 kPa
T3 = 27C
system boundary
P1 = 1 MPa
T1 = 80C
P4 = 95 kPa
T4 = 60C
P2 = 1 MPa
T2 = 30C
Section IV – First Law Applied to Open Systems

Page 21
Can air be treated as an ideal gas under the given conditions?
0.1
 0.027
3.76
300
Tr3 
 2.26
133
0.095
Pr4 
 0.025
3.76
333
Tr4 
 2.50
133
Pr3 
Z3 = Z4  1 (See p. 807 of Text)
Assumptions:



 Flow of Refrigerant –12 is steady  m1  m2  m R12







Flow of air is steady  m 3  m4  m air
Heat lost by R-12 is gained by air  Qie air = -Qie R-12
Properties are uniformly distributed at 1 & 2 and at 3 & 4.
Conservation of energy applied to the flow of air:



m air (h4 – h3)


P 
100  800
m air   3 1  3 1 
 15.486 kg / s
RT3
0.287  300  60
h3  hT 300K  300.19 KJ Kg 
see p. 788 of text
Q ie air W
ie air =
h4  hT 333k 
h4  330.34
333  330

340.42  330.34 340  330
h4  333.36 KJ Kg

Q ie air  15.486(333.36  300.19)  513.7 KJ
 Conservation of energy applied to the flow of Refrigerant –12:



Q
ie R 12  Wie R 12  m R 12 ( h 2  h 1 )


Qie R 12  Q ie air  513.7 KJ

 m R 12 
 513.7
(h2  h1 )
(no work done on or by the Refrigerant –12)
Section IV – First Law Applied to Open Systems
Page 22
T1 = 80C  Tsat = 41.64C corresponding to P1 = 1MPa  R-12 is a superheated vapor at 1
h1 = 232.91 KJ/Kg (see p.784 of Text)
T2 = 30C  Tsat = 41.64C corresponding to P2 = 1MPa  R-12 is a subcooled liquid at 2
h 2  h fT 30 C 

m
vf
(PCL  PSat )  64.59  0.0007739(1000  744.90)  64.79 KJ Kg
T  30C
(see pp. 780-781 of text)
 513.7
Kg
Kg
 3.06
 183.3
(64.79  232.91)
S
min
Section IV – First Law Applied to Open Systems
Page 23
Section IV – First Law Applied to Open Systems
Page 24
Section IV – First Law Applied to Open Systems
Page 25
The operating data from the simple, steady-flow, steam-power plant cycle shown in the
accompanying figure are summarized in the table below. Determine the power output of the
turbine, the heat-transfer rate in the steam generator, the power input to the pump, and the thermal
efficiency of the cycle.
Location
Pressure, Mpa
Quality or Temperature
1
0.01
0.0
2
48°C
3
5
450° C
4
0.95
Other data:
Mass flow rate of steam = 20 kg/s
Negligible pressure drops through steam generator and condenser
Section IV – First Law Applied to Open Systems
Location
1
2
3
4
Page 26
Pressure, Mpa
0.01
5
5
0.01
Quality or Temperature
0.0
48C
450
0.95
Assumptions:
 ep << h, ek << h for all units
 Properties are uniformly distributed at the inlet and exit of each unit of the cycle
- Pressure drop through steam generator is negligible  P2  P3 = 5Mpa
- Pressure drop through condenser is negligible  P4  P1 = 0.01 Mpa = 10KPa

m  20 Kg S

Flow through the steam generator:

system
boundary
First law:
 W
  no
work  h 2 )
Q
23
23  m( h 3 


done


T2 = 48C < Tsat = 263.99C
Corresponding to P2 = 5MPa
 water is a compressed liquid at state (2)

h2  hf
T  48 C
 vf
h f  188.45
209.33  188.45
T  48 C

PCL  Psat 
(48  45)
 h f  200.98 KJ Kg
(50  45)
v f  0.001010
(0.001012  0.001010)

(48  45)
m3
 v f  0.001011
(50  45)
Kg
( Psat  9.593)
(48  45)

 Psat  11.247 KPa
(12.349  9.593) (50  45)
h2  200.98  0.001011(5000  11.247)  206.02
KJ
Kg
Section IV – First Law Applied to Open Systems
Page 27
T3 = 450C > Tsat = 263.99C corresponding to P3 = 5Mpa  water is a superheated
vapor at state(3).
h3 = 3316.2 KJ/Kg

Q 23  20(3316.2  206.02)  62203.6 KW

Flow through the turbine:

First Law:



Q 34  W 34  m(h4  h3 )
 turbine is

 adiabatic
Steam turbine


P 10 KPa
 x 4 h fg
P 10 KPa

W 34  m (h3  h4 )
system boundary
h4  hf



 191.83  (0.95)(2392.8)  2465 KJ Kg

W34  20(3316.2  2465)  17024 KW

Flow through the pump:
First law:

system
boundary


h4  hf
P 10 KPa


Q 12  W12  m(h2  h1 )
 pump is 


 adiabatic 
 191.83 kJ / kg

W12  20(191.83  206.02)  283.8KW
 Power input


 W12  m(h 1  h 2 )
Section IV – First Law Applied to Open Systems

Page 28
Flow through the condenser:
system
boundary
First law:



Q 21  W21  m(h 1  h 4 )

 no work done 


 in condenser 


Q12  20(191.83  2465)  45463.4 KW


Thermal Efficiency:
 th 
 th 
W net
Q steam generator
(17024  283.8)
 0.269 (or 26.9%)
(62203.6)

 Q
Check: W
net
net
 Q

 W
 Q
W
34
12
23
4l
(17024 – 283.8) = (52203.6 – 45463.4)
16740.2
16740.2
Check!