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Thermodynamics & Power Plant - 1 - Mechanical Engineering MCQ


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15 Questions MCQ Test - Thermodynamics & Power Plant - 1

Thermodynamics & Power Plant - 1 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Thermodynamics & Power Plant - 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Thermodynamics & Power Plant - 1 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Thermodynamics & Power Plant - 1 below.
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Thermodynamics & Power Plant - 1 - Question 1

Gas from a cylinder of compressed Argon is used to inflate an inelastic balloon, originally folded completely to a volume of 1 m3. If the barometer reads 760 mm of Hg. What is the amount of work done upon the atmosphere by the balloon?

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 1

Thermodynamics & Power Plant - 1 - Question 2

An ideal gas undergoes an isothermal expansion from state R to state S in a turbine as shown in the diagram given below

The area of shaded region is 1000 N-m, what is the amount of turbine work done during the process?

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 2

Turbine work = area of PQSR

= area of RMS + area of MSPQ

= 1000 + (2 - 1) x 105 (0.2 - 0.1)

= 1000 + 10,000

= 11000 N-m

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Thermodynamics & Power Plant - 1 - Question 3

The work done by a closed system in a reversible process is always ___ that done in an irreversible process.

Thermodynamics & Power Plant - 1 - Question 4

A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa the pressure remains constant throughout the process. There is a transfer of 37.6 kJ of heat from the gas during the process. How much does the internal energy of the gas changes?

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 4


= 0.105 (0.15 - 0.30)MJ

w1_2 = -15.75 kJ

Q1-2 = 37.6 kJ 

- 37.6 = Δu - 15.75

u2 - u1 = -21.85 kJ

Thermodynamics & Power Plant - 1 - Question 5

A refrigerator having COP 5 is used for heating purpose in winter so that atmosphere becomes the cold body and room to be heated becomes hot body, how much heat would be available for heating for each kW input to the driving motor. Electromechanical efficiency of motor is 90%.

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 5

W= 0.9 x 1 kW
= 0.9 kW

COP =

Q2 = 5 X 0.9

= 4.5 kW

Q1= Q2 + W = 4.5 + 0.9

= 5.4 kW

*Answer can only contain numeric values
Thermodynamics & Power Plant - 1 - Question 6

Two reversible engines operate between thermal reservoirs at 1200 K, T 2 K and 300 K such that first engine receives heat from 1200 K reservoir and rejects heat to thermal reservoir at T2K, while the second engine at 300 K. The efficiency of both the engines is equal then, the value of temperature T2 is _________ k.


Detailed Solution for Thermodynamics & Power Plant - 1 - Question 6

When the efficiency of both the engines are equal then intermediate temperature

= 600 K

Thermodynamics & Power Plant - 1 - Question 7

Saturated liquid at a higher pressure P1 having h11 = 1000 kJ/kg is throttled to a lower pressure P2 . The enthalpy of saturated liquid and saturated vapour are 800 kJ/kg and 2800 kJ/kg respectively. Find the dryness fraction of vapour after throttling.

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 7

In throttling process enthalpy remains constant.

h1 = h2

1000 = 800 + x(2800 — 800)

x = 0.1.

Thermodynamics & Power Plant - 1 - Question 8

The system is partitioned in two equal parts. An intensive property of each part will

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 8

An intensive property is not depended on mass of the system.

Thermodynamics & Power Plant - 1 - Question 9

The molar heat capacity of a certain substance varies with temperature according to the empirical equation C = (27.2 + 4 × 10–3 T) J/mol-K How much heat is necessary to change the temperature of 2 mole of this substance from 27°C to 427°C.

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 9

T i = 300 k

T f = 700 k

= 23360 J = 23.3 kJ 

Thermodynamics & Power Plant - 1 - Question 10

For a real thermodynamic cycle, which one of the following is correct?

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 10

From the Clausius inequality for real thermodynamic cycle,

Thermodynamics & Power Plant - 1 - Question 11

When a system is in equilibrium, any conceived change in entropy would be

*Answer can only contain numeric values
Thermodynamics & Power Plant - 1 - Question 12

The loss of available energy associated with the transfer of 1000 kJ of heat from a constant temperature system of 600 k to another at 400 k when the atmospheric temperature is 300 k, is ________ kJ.


Detailed Solution for Thermodynamics & Power Plant - 1 - Question 12

Loss of available energy =

= 1000 × 300 =

= 250 kJ

Thermodynamics & Power Plant - 1 - Question 13

One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 13

For every irreversible process entropy of the universe increases.

Thermodynamics & Power Plant - 1 - Question 14

A thermal power plant operates on a regenerative cycle with a single open feed water heater, as shown in figure. For the state points shown, the specific enthalpies are: h1 = 2800 kJ/kg and h2 = 200 kJ/kg. The bleed to the feed water heater is 20% of the boiler steam generation rate. The specific enthalpy at state 3 is 

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 14

Enthalpy at state 3 = h3

= 0.2 h 1 + 0.8 h2

= (0.2 × 2800) + (0.8 × 200)

= 506 + 160 = 720 kJ/kg

Thermodynamics & Power Plant - 1 - Question 15

Two vapour power cycles having efficiencies 0.50 and 0.40 are coupled in series then the efficiency of coupled cycle will be

Detailed Solution for Thermodynamics & Power Plant - 1 - Question 15

We know that for two cycle coupled in series η = 1 – (1 + η1 ) (1 – η2 )
= 1 – [1 – η1 – η2 + η1 η2 )
= η1 + η2 – η1 η2
= 0.5 + 0.4 – 0.5 × 0.4 = 0.70

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