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

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