Applied Thermodynamics: Solved Questions and Analysis

Applied Thermodynamics 1
APPLIED THERMODYNAMICS
by [NAME]
Course
Professor’s Name
Institution
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°C.
Checking from the superheated tables, the specific enthalpy at 1600kPa at 60°C 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°C at the exit of the turbine. Taking the
value of specific enthalpy of refrigerant saturated vapor state at 35°C from the saturated tables
we find h2 =269.1 kJ /kg
iii) The specific enthalpy at the entry of the pump
The liquid is in a saturated liquid state at 35°C. From the saturated tables, the specific enthalpy
of refrigerant saturated liquid state at 35°C 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