MIET2421 Applied Thermodynamics: Rankine Cycle Power Plant Irrigation

This document presents a comprehensive solution to an applied thermodynamics assignment focusing on the analysis of a Rankine cycle power plant designed for an irrigation system, utilizing solar energy to boil the refrigerant R134a. The solution includes schematic sketches, calculations of specific enthalpy at various points in the cycle (turbine entry and exit, pump entry and exit), and ideal process path diagrams (T-s, H-s, and P-v). It further determines the electric power output of the generator, turbine work, boiler feed pump input power, rates of heat delivered and rejected, and the efficiency of conversion. The assignment also covers unit mass flow rate calculations, the impact of non-ideal conditions, and the determination of mass flow rate, electrical power input to the boiler pump, electrical power output rating of the turbo generator, necessary solar collector area, and necessary water flow rate through the condenser. The solution concludes by identifying suitable heat exchangers and pumps for the system and determining the wet bulb temperature given dry bulb temperature and relative humidity. Desklib provides this solved assignment as a valuable resource for students studying thermodynamics.

Applied Thermodynamics 1
APPLIED THERMODYNAMICS
by [NAME]
Course
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Location of Institution
Date

Applied Thermodynamics 2
Applied Thermodynamics
Question 1: Schematic Sketch
Question 2
a) Specific enthalpy of the refrigerant
The power output from the generator is 1.5kW and the solar energy is incident at 1kW per unit
area.
i) The specific enthalpy at the entry of the turbine
At the entry of the turbine, the conditions of the refrigerator are 1600kPa at 60℃.
Checking from the superheated tables, the specific enthalpy at 1600kPa at 60℃ is given by
h1=280.7 kJ /kg

Applied Thermodynamics 3
ii) The specific enthalpy at the exit of the turbine
The refrigerant is at a state of a saturated vapor at 35℃ at the exit of the turbine. Taking the
value of specific enthalpy of refrigerant saturated vapor state at 35℃ from the saturated tables
we find h2 =269.1kJ / kg
iii) The specific enthalpy at the entry of the pump
The liquid is in a saturated liquid state at 35℃. From the saturated tables, the specific enthalpy
of refrigerant saturated liquid state at 35℃ is given by h3 =100.9 kJ /kg
Also from the tables, the saturated pressure of the refrigerant liquid at the entry of the pump is
given by P3=887.5 kPa
The specific volume of the refrigerant liquid across the pump is given by v=0.0009 m3 /kg .
It is also important to note that as long as the pressure across the solar heater is 1600kPa then the
pressure of the liquid at the exit of the pump is also 1600kPa.
Therefore, work added by the pump is calculated as follows;
w=v (P4 −P1)
¿ 0.0009(1600−887.5)
¿ 0.64125 kJ /kg
Thus, enthalpy of the refrigerant at the exit of the pump is given by;
h4=h3 + w
¿ 100.9+0.64125

Applied Thermodynamics 4
¿ 101.54125 kJ /kg
b) Ideal process paths
i) T-s diagram
ii) H-s diagram
iii) P-v diagram

Applied Thermodynamics 5
c) Determine;
i) The electric power output of the generator
The required electric power, Wre is 1.5kW.
Power output of the generator,
W generator= W ℜ
ngenerator
Where ngenerator=efficiency of the electrical generator .
W generator= 1.5
0.92
¿ 1.630 kW
ii) Net electric output power
W turbine =W g enerator
nturbine
Where nturbine is theefficiency of the electric motor drivng the feed pump .

Applied Thermodynamics 6
W turbine =1.630
0.85
¿ 1.918 kW
But remember, W turbine =m ( h1−h2 ) =1.918 kW
m= 1.918
280.7−269.1
m=0.165 kg /s
iii) Input power of the boiler feed pump
W P=m ( h4−h3 )
W P=m ( w )
W P=0.165 × 0.64125
W P=0.1060 kW
iv) Rate of heat delivered
Q¿=m ( h1−h4 )
¿ 0.165 ( 280.7−101.54125 )
¿ 29.561 kW
v) Rate of heat rejection
Qcon=m ( h2−h3 )
¿ 0.165 ( 269.1−100.9 )

Applied Thermodynamics 7
¿ 27.753 kW
vi) Efficiency of conversion
η=W ele
Qele
¿ 1.5
29.561
¿ 0.0507
¿ 5.07 %
Question 3
a) Sketch
i) T-s

Applied Thermodynamics 8
ii) H-s
iii) P-v

Applied Thermodynamics 9
b) Unit mass flow rate
i) Necessary electrical power input
¿ h4 −h3
ηmotor
¿ 249.48−248.94
0.901
¿ 0.6 kW
ii) Electrical power output of the generator:
¿ h1−h2
ηgenerator
¿ 426.86−415.46
0.8857

Applied Thermodynamics 10
¿ 10.09 kW
iii) Net electrical output power
Remember that 1.5 kW power is used to pump water and charge the battery
Therefore, net electrical power output is given by;
10.09−1.5=7.99
¿ 7.99 kW
iv) Rate of heat delivery
Q¿=m ( h1−h4 )
¿ 0.165 ( 426.86−249.48 )
¿ 29.2677 kW
There is a brief contrast in the values calculated here and those calculated in question 2 because
there is no assumption of ideal conditions.
Question 4
a) Mass flow rate
W generator= W ℜ
ngenerator
Where ngenerator=efficiency of the electrical generator .
W generator= 1.7
0.92

Applied Thermodynamics 11
¿ 1.848 kW
W turbine =W generator
nturbine
Where nturbine is theefficiency of the electric motor drivng the feed pump .
W turbine =1.848
0.85
¿ 2.174 kW
But remember, W turbine =m ( h1−h2 ) =2.174 kW
m= 2.174
280.7−269.1
m=0.187 kg/ s
b) Electrical power input to the boiler pump
¿ 1.7 × h4 −h3
ηmotor
¿ 1.7 × 249.48−248.94
0.901
¿ 1.019 kW
c) Electrical power output rating of turbo generator
¿ 1.7 × h1−h2
ηgenerator
¿ 1.7 × 426.86−415.46
0.8857

Applied Thermodynamics 12
¿ 21.881 kW
d) Necessary solar collector area
Q¿=m(h1−h4 )
¿ 0.187(426.86−249.48)
¿ 33.170 kW
Q¿=0.7 ¿
Therefore;
¿
¿ 33.170
0.7
¿ 47.3858 kW
A=¿ ¿
¿ 47.3858
1
¿ 47.3858 m2
e) Necessary water flow rate through the condenser
mass flow rate=water flow rate(h1−h2 )
Where h1=specific enthalpy of refrigerant entering theturbine
While h2 =specific enthalpy of theliquid leaving the turbine