MAE 204: Fall 2019 Homework Assignment #7 on Thermodynamics Problems

This document presents the solutions to Homework #7 for MAE 204, a mechanical engineering course, from Fall 2019. The assignment covers several key concepts in thermodynamics. Problem 1 involves calculating work produced and heat rejected by a heat engine, determining the coefficient of performance (COP) of an air conditioner, analyzing a combined air conditioner/heat pump system, and identifying the Kelvin statement. Problem 2 focuses on calculating the total power output of a steady-flow turbine given inlet and outlet conditions, mass flow rate, and heat loss. Problem 3 addresses the calculation of the final temperature and mass of air in a tank, and determining the heat transfer involved. The solutions demonstrate the application of thermodynamic principles, including energy conservation and the use of steam tables. The document provides step-by-step solutions to the problems, including calculations and explanations.

Running head: MAE 204 Fall 2019 HW #7
MAE 204 Fall 2019 HW #7
Name of the Student
Name of the University
Author Note

1MAE 204 Fall 2019 HW #7
Problem 1:
a) Given that, the total heat input of a heat engine = 2 KJ
The thermal efficiency of the engine = 40%
Hence, the amount of work produced by the engine = 2*40/100 = 0.8 KJ.
Hence, the amount of the heat rejected by the engine assuming there are no other losses = 2-
0.8 = 1.2 KJ.
b) Given, the air-conditioner removes the heat at the rate of 1 kW.
Hence, refrigeration rate ( ˙QL) = 1 kW
Also, given the power consumed by the air-conditioner or the power input ( W net ,∈¿¿) = 0.7
kW
Hence, by assuming the air-conditioner works steadily the coefficient of performance (
CO PR ¿ is given by,
CO PR = ˙QL
W net ,∈¿ ¿ = 1/0.7 = 1.429

2MAE 204 Fall 2019 HW #7
Outside Air
A/C
House/Room
0.7 kW
= 1 kW
c)
When a heat pump and an air conditioner are working in combination then the CO PHP is
given by,
CO PHP=CO PR +1
Hence, CO PHP = 5 + 1 = 6
d) The KELVIN statement says that
“It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do
work W. Some amount of heat QC must be exhausted to a cold reservoir. This precludes a
perfect heat engine”.
This statement implies that there is no heat engine which can produce entire work by
accepting heat from only one reservoir. There must be at least two reservoirs working in
synchronism.

3MAE 204 Fall 2019 HW #7
Problem 2:
Given, the inlet conditions of the turbine is
P1 = 7 MPa, T1 = 600 °C
And outlet conditions are
x2 = 0.95 and P2 = 25 kPa
mass flow rate ˙m = 20 kg/sec
heat lost = 20 kJ/kg
Now, given that kinetic and potential energy effects are negligible, hence ˙Q = 0, ˙W = 0
Now, by conservation of energy
(dE/dt) = ˙Q - ˙W + ˙m*Cv∗(T 1−T 2)
20 = 20(600-T2) => 600-T2 = 1 => T2 = 599 °C.
Hence, outlet temperature is 599 °C.
Now, from steam table h1 = 3649.79 kJ/kg
Now, h2 = hf + x 2∗hfg = 271.96 + 0.95*2345.5 = 2500.185 KJ/kg
Hence, the total power output = - ˙m∗(h 2−h1) = -20*(2500.185 - 3649.79) = 22992.1 kW =
22.99 MW.
Hence, the total power produced by the steady flow turbine = 22.99 MW.
Problem 3:
Given, the volume of tank v = 0.3 m^3
Initial temperature T1 = 22 °C = 22+ 273 = 295 K

4MAE 204 Fall 2019 HW #7
Initial pressure P1 = 1000 kPa.
Final pressure P2 = 1500 kPa
Given, 22 °C air is added to tank at 2000 kPa and hence temperature only changes due to
pressure increase.
Final temperature from Charles’ law T2 = P 2∗T 1
P 1 = 1500*295/1000 = 442.5 K = 169.5 °C.
Now, initial mass of air m1 = P1*V/(R*T1) = 1000*0.3/(0.287*295) = 3.54 kg.
Final mass of air m2 = P2*V/(R*T2) = 1500*0.3/(0.287*169.5) = 9.25 kg.
Now, specific heat of air at 22 °C and 1000 kPa pressure C p 1 = 1.02248 kJ/kg-K
Specific heat of air at 169.5 °C and 1500 kPa pressure C p 2 = 1.02884 kJ/kg-K.
Hence, the amount of heat transfer is given by,
Q = (m2-m1)*( C p 2−Cp 1 ¿∗(T 2−T 1) = (9.25-3.54)*(1.02884−1.02248 ¿∗(169.5−22) =
5.36 kJ.