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Test: Thermodynamic Properties & Compressibility - Mechanical Engineering MCQ


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10 Questions MCQ Test - Test: Thermodynamic Properties & Compressibility

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

 If there is no exchange of heat between system and surrounding where system comprises of a compressible fluid but the heat is generated due to friction, the process is an adiabatic process.

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 1

Explanation: For process to be adiabatic, there is no heat exchange and no heat generation within fluid.

Test: Thermodynamic Properties & Compressibility - Question 2

For a compressible fluid, if there is no change in specific volume at constant temperature, what type of process it is?

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 2

Explanation: As, specific volume remains constant, density remains constant. Therefore for given temperature there is no change in volume. hence, the process is isothermal.

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Test: Thermodynamic Properties & Compressibility - Question 3

 If the fluid is incompressible, do thermodynamic properties play an important role in its behaviour at varying temperature and pressure?

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 3

Explanation: If fluid is incompressible there is not much change in observed properties with variation in temperature and pressure. Hence, no perceivable change.

Test: Thermodynamic Properties & Compressibility - Question 4

If for same temperature and pressure change, the value of bulk modulus is compared for isothermal process and adiabatic process, which one would be higher?

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 4

Explanation: For isothermal process
K=p
For adiabatic process
K=kp
where K=Bulk modulus
k=Polytropic constant
p=Pressure.

Test: Thermodynamic Properties & Compressibility - Question 5

The value of gas constant is same for all the gases

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 5

Explanation: The value of gas constant depends on molecular weight. As the molecular weight is different, gas constant will be different.

Test: Thermodynamic Properties & Compressibility - Question 6

Calculate the pressure exerted by 9 kg of air at a temperature of 20℃ if the volume is 0.8m3. Assuming ideal gas laws are applicable. 

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 6

Explanation: Ideal gas Law: PV=nRT
n=M/m
P=(9*8314*293)/28.97=946 kN/m2.

Test: Thermodynamic Properties & Compressibility - Question 7

A gas weighs 16 N/m3 at 30℃ and at an absolute pressure of 0.35 N/mm2. Determine the gas constant. 

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 7

Explanation: R=P/(ρ*T)=3500000*9.81/16*303=708.23.

Test: Thermodynamic Properties & Compressibility - Question 8

A cylinder of 0.8 m3 in volume contains superheated steam at 70℃ and .4 N/m2 absolute pressure. The superheated steam is compressed to .3 m3 . Find pressure and temperature. 

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 8

Explanation: For polytropic process,

P2=(v1/v2)n *P1
=(0.8/0.3)1.3 * 0.4 ……..(for superheated stream n=1.3)
=.74 N/m2
T1=P1v1/nR=422.3℃.

Test: Thermodynamic Properties & Compressibility - Question 9

Determine the compressibility of an incompressible fluid, if the pressure of the fluid is changed from 70 N/m2 to 130 N/m2. The volume of the liquid changes by 0.15 percent. 

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 9

Explanation: Compressibility=1/Bulk Modulus
=1/K
K=(dp*V/dv)
=60/0.15
=400
Compressibility=.0025.

Test: Thermodynamic Properties & Compressibility - Question 10

What is the variation of cp, cv and k in case of gases when the temperature increases?

Detailed Solution for Test: Thermodynamic Properties & Compressibility - Question 10

Explanation: cp is molar heat capacity at constant pressure. As temperature is increased, enthalpy increases, heat capacity increases.
Same is for cv, cp is molar heat capacity at constant volume.
However cp-cv=R and cp/cv = R
Hence, as cp, cv increases R decreases.

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