# Applied thermodynamics Question Papers Final Midterm Exam with Solution

APPLIED THERMODYNAMICS  Exam paper with solution Applied Thermodynamics sample Old Question Papers Download
Questions and Solutions
Question
a.)
With proper justification, identify system, surrounding and boundary at the
instant you are writing your exam-paper of ME 322: Applied Thermodynamics.

Solution
System: Human body, pen and paper etc
[0.5 for identification and 0.5 for justification]
Surrounding: Air, rest of students and class environment etc
[0.5 for identification and 0.5 for justification]
Boundary: Clothes/Skin and Pen cover etc
[0.5 for identification and 0.5 for justification]

If incorrect or partly correct, student’s creativity based on logical justification may be
considered for awarding marks.

Air at 100 psi and 300°F is contained in a rigid iron cylinder. Temperature and
pressure inside the cylinder drops to 150°F and 60 psi respectively as a result of
heat transfer to surrounding. Find out the work done by air during such heat
transfer process.

Solution
Since the tank is rigid, so there is no change in volume. The boundary work is given as:

[1 for a logical reasoning statement]
[1 for correct formula]
[1 for correct resulting value of Wb]

c.)
A bath tub contains liquid water at 60 C and 5 MPa. Find internal energy of
compressed liquid water using:
i. Compressed Liquid Table
ii. Saturated Liquid data
How much error (%) exists between the two results? Also, draw a T-u diagram
showing magnitude of stated properties.
[4 Marks]
Solution
i. From the compressed liquid table (Table A-7)
u = 250.29 kJ/kg
[1 Mark]
ii. From saturated liquid table (Table A-4)
u = 251.16 kJ/kg
[1 Mark]
%age difference
(ub-ua)/ua = (251.16 – 250.29)/251.16 = 0.00346 = 0.346 %

QUESTION 02
a) Based on proper justification(s), state the value of ΔE for a closed cyclic
process (system). Write down energy balance equation for cyclic system.

SOLUTION

For a closed cyclic system the value of ΔE is zero, Because the initial and final states are
identical in a cycle therefore change in energy will be zero.

Energy Balance Equation for a System

According to closed cyclic process ΔE=0 & energy transfer due to mass is zero for a close
system. After neglecting mass, only heat (Q) and work (W) will left. So above equation
becomes ; Ein-Eout = 0 Ein=Eout Qin=Wout Q=W

b)
In the turbine of a gas turbine unit the gases flow through the turbine at
17Kg/s. Power developed by the turbine is given as 14000KW. Consider
internal energies per unit mass of the gases at I/P and O/P to be
1200KJ/Kg and 360 KJ/Kg respectively. Further, the velocities of gases at
inlet and outlet are given as 60m/s and 50m/s respectively.
1) Calculate the rate at which heat is rejected from the turbine.
[5 Marks]
2) Find the area of the inlet pipe when the specific volume of the gases at
the inlet is 0.5m3
/Kg.
[2 Marks]
[7 Marks]

QUESTION
a.)
What is meant by specific heat of a gas? Justify that Cp (specific heat at
constant pressure) is always greater than Cv (specific heat at constant
volume) for an ideal gas in a freely moveable piston cylinder system?
[3 Marks]
SOLUTION
1) The amount of heat required to raise the temperature of a unit mass of gas to one degree
Celsius, is called specific heat of the gas.
[1 Mark]
2) As gas is heated in a freely moveable piston cylinder device, the pressure tends to increase
but the piston moves itself up to make the pressure constant which in return increases the
volume of the gas. As volume increases temperature tends to decrease so extra amount of heat
is required to accommodate this tendency in the drop of temperature which results in the form
of boundary work. This extra amount of heat is never required in piston cylinder device having
fixed volume as no boundary work is involved. Hence it is justified that Cp > Cv.
[2 Marks]
b.
A working fluid is contained in a piston-cylinder system. The piston has an area of
1m2
. For all of the following cases take initial volume and pressure of the fluid as
1m3
and 100kPa respectively. Find the work done if:
a) 10J energy is added to the system following an isochoric process. [1 Marks]
b) The fluid expands to 2m3
following an isobaric process. [1 Marks]
c) The fluid expands to 2m3
following an isothermal process. [2 Marks]
d) Energy is provided to the cylinder so that piston rises up compressing a
spring (k=100kN/m). Final volume of the fluid is 2m3
. [3 Marks]
[7 Marks]
Solution:
a) Work is defined as As fluid volume is fixed during an isochoric (constant
volume) process, no work is done by the fluid.
[1 Mark]
b) In isobaric process pressure remains constant. So

[1 Mark]
c) As PV=P1V1=P2V2 = Constant

= 69.31 kJ
[2 Marks]
d) V2= 2m3

=1m

As force due to spring is directly proportional to the displacement ‘x’, therefore pressure
is also proportional to the displacement. Similarly volume increase is linear with ‘x’. It
can be safely assumed that the pressure increase from 100kPa to 200kPa is linear with
respect to volume.

Therefore work done against compressing the spring plus 100kPa constant atmospheric
pressure is:

QUESTION
a.)
Define Latent Heat of Fusion? What is vacuum cooling? How is low temperature
achieved in vacuum cooling process? Explain the stages involved.
[3 Marks]

SOLUTION
Latent Heat of Fusion
The amount of energy absorbed during melting or the amount of energy released during
freezing is known as Latent Heat of Fusion.
[1 Mark]
Vacuum Cooling
Vacuum Cooling is based on reducing the pressure of the cooling chamber until the desired low
temperature is achieved and evaporating some water from the products to be cooled. The heat
of vaporization during evaporation is absorbed from the products, which lowers the product
temperature.
There are two stages involved in the process. In the first stage, the products at ambient
temperature are loaded into the chamber, whose temperature is kept constant until saturation
pressure is reached. In the second stage, saturation conditions are maintained at lower
pressures and the corresponding lower temperatures until the desired temperature is reached.
[2 Marks]

B.) A rigid tank contains 10 kg of water. Determine the missing properties and
the phase descriptions in the following table.

Phase
Description
Mark
(i) 15 0 858 (1)
(ii) .01 2783 (2)
(iii) 200 .03 (4)

SOLUTION
(i) As quality x=0, we have compressed liquid therefore using the compressed liquid water
table.
At P=15 MPa
T= 200 °C
u= 840.84 kJ/kg
v= .0011435 m3
/kg
[1 Mark]
(ii) First we need to determine the phase, using saturated water-Pressure Table
At P=.01 MPa , we have
hf=191.81 kJ/kg
hg= 2583.9 kJ/kg
As h> hg we have superheated vapor.
So using the superheated water table, at P=.01 MPa and h=2783kJ/kg we got
T= 150°C
u= 2587.9 kJ/kg
v= 19.513 m3
/kg
x=1, for superheated vapors
[2 Marks]

(iii) We first need to find the phase, Therefore using the saturated water-Temperature Table
At 200°C we have
vf= 0.001157 m3
/kg
vg=.12721 m3
/kg As vf<v< vg
We have saturated water, Therefore
x= 0.23
Using the saturated water-Temperature Table
At T=200°C
We have
P= 1554.9kPa
uf= 850.46 kJ/kg, ufg= 1743.7 kJ/kg
hf= 852.26 kJ/kg, hfg= 1939.8 kJ/kg
u= uf + x.(ufg)
u=1251.511 kJ/kg
h= hf + x.(hfg)
h= 1298.414 kJ/kg

Description
(i) 200 15 840.84 0 .0011435 858 Compressed
water
(ii) 150 .01 2587.9 1 19.513 2783 Superheated
water
(iii) 200 1.5549 1251.51 .23 .03 1298.414 Saturated
water-vapor • Title : Applied thermodynamics Question Papers Final Midterm Exam with Solution
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