MAE 204 Fall 2019 Homework #2: Thermodynamics Problems Solved

This document provides a complete solution to MAE 204 Fall 2019 Homework #2, a mechanical engineering assignment focused on thermodynamics. The solution covers three main problems: The first problem involves calculating the pressure at which a piston-cylinder assembly begins to rise and determining the work done by the gas as the piston rises. The second problem addresses power generation from a reservoir, calculating mass flow rate, power, specific energy, and maximum power production. The third problem explores the concept of internal energy within a system, differentiating between different types of energy and applying thermodynamic principles to isothermal and adiabatic processes, with an example calculation involving the work of a spring. This assignment showcases the application of fundamental thermodynamics principles to practical engineering problems.

Q1
a)
Mass = 50 kg
Patm = 101.325 kPa
Area = 0.01 m2
g = 9.81 m/s2
∑ F=0
P*Ac = mg + Patm*Ac
P = Patm + mg
Ac
= 101.32 * 103 Pa + 50 kg∗9.81 m/ s2
0.01 m2
= 150370 Pa
= 150.37 kPa
b) Work done = Pressure*Area *d
= 150370 * 0.1
= 15037 J
= 15.037 kJ
Q2
a) H = 100m
V = 100 m3/s
ρ=1000 kg/m3
g=9.81 m/ s2
Mass flow rate = volume flow rate * density = 100 * 1000 = 100000 kg/s
Power = 9.8 * q*h
= 9.8 *100 *100

= 98000 kW
= 98000000 J/s
Specific energy = power/ mass flow rate =
98000000 J
s
100000 kg
s
= 980 J/kg
b) Maximum power production in MW
Power = 9.8∗q∗h
1000 = 9.8∗100∗100
1000 = 98 MW
c) Voltage = 13.5 kv
P = Voltage * Current = V*I
Current = I = Power
voltage = P
V
= 98000 kW
13.5 kV
= 7259.26 A
Q3)
a) Internal energy of the system
b) All thermodynamic energy, which includes; kinetic energy, potential energy, internal
energy and flow energy.
c) Work done = force * distance
= F*d
But w=Fθr=τθ
Power = w
dt = τθ
dt =τw
P = 2 π nrpm T
60
Where; P = power
dt = time taken
n = speed (rev/sec)
w=2 πn=angular velocity( rad
s )

nrpm = speed (rev/min, rpm)
d) NO, since
Adiabatic process, is where a system would exchange no heat with it surrounding, this
means that Q = 0 but∆ T ≠0
Example
Compression of a gas cylinder
If a gas cylinder is compressed there would be no enough time that the system would use
to transfer energy to its environment. Therefore the work done by the system would
change the internal energy of the system.
Isothermal process, is where the change in temperature will be zero and internal
energies would also be zero (ideal gases) but exchange of heat is not equal to zero, this
means ∆ T =0 , ∆U =0 but Q ≠ 0
Example
Carnot heat engine
The temperature of the system would only be maintained if the work would either be
done by the system or done to the system. In the process of doing work internal energy
would increase and therefore temperature would be increased, in a situation where the
temperatures are higher than the level required, then the work would be done by the
system on the surrounding and therefore temperature would decrease since there would
be energy release to the surrounding as heat.
e) Work of spring = 2.079 J
Extension = 10 mm – 1 mm = 9 mm = 0.009 m
Work done = ½*k*e2
2.079 = ½*k*0.0092
K = 51333.33 N/m
K = 51.33 kN/m