Quiz: Mechanical Engineering 26th April |_00.1
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Quiz: Mechanical Engineering 26th April

Quiz: Mechanical Engineering
Exam: AE/JE
Topic: Thermodynamics

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 10 Minutes

Q1. Any process is likely to in a given direction only when
(a) ∆S = 0
(b) ∆S < 0
(c) ∆S ≥ 0
(d) ∆S ≤ 0

Q2. The entropy change for any closed system which undergoes an irreversible adiabatic process
(a) may be positive, negative or zero
(b) may be positive or zero but not negative
(c) is always positive
(d) is always negative

Q3. In isothermal dissipation of work
(a) ∆S_system=0 & (∆S)_surrounding=W/T
(b) (∆S)_system=0 & (∆S)_surrounding=0
(c) (∆S)_system=W/T & (∆S)_surrounding=0
(d) (∆S)_system=W/T& (∆S)_surrounding=(W/T)

Q4. Neglecting changes in kinetic energy and potential energy, for unit mass the availability of a non-flow process becomes a = ϕ-ϕ_0, where ϕ is the availability function of the
(a) open system
(b) closed system
(c) isolated system
(d) steady flow process

Q5. Match List- I with List – II & select the correct answer using the codes given below the lists:
List – 1 List – II
A. Mechanical work 1. Low grade energy
B. Heat 2. Concept of temperature
C. ∮dQ/T≤0 3. High grade energy
D. Zeroth law 4. Clausius inequality
Codes
      A B C D
(a) 3 1  2  4
(b) 3 1  4  2
(c) 1 3  4  2
(d) 1 3  2  4

Q6. Consider the following statements. In an irreversible process:
Entropy always increases
The sum of the entropy of all bodies taking part in a process always increases
Once created, entropy cannot be destroyed.
Of these statements
(a) 1 and 2 are correct
(b) 1 and 3 are correct
(c) 2 and 3 are correct
(d) 1, 2 & 3 are correct

Q7. Consider the following statement and choose the correct one
(a) Steady flow energy equation and Bernoulli equation are equivalent
(b) Bernoulli equation is a special limiting case of steady flow energy equation
(c) Steady flow energy equation is not valid for compressible flow
(d) Steady flow energy equation and Bernoulli’s equation are unrelated

Q8. The air with enthalpy of 100 kJ/kg from the compressor as the air passes through it. Neglecting kinetic and potential energies, the power required for an air mass flow of 0.5 kg/s is
(a) 30 kW
(b) 50 kW
(c) 70 kW
(d) 90 kW

Q9. Which one of the following sets of thermodynamics laws/relations is directly involved in determining the final properties during an adiabatic mixing process?
(a) The first and second laws of thermodynamics
(b) The second law of thermodynamics and steady flow relations
(c) Perfect gas relationship and steady flow relations
(d) The first law of thermodynamics and perfect gas relationship

Q10. “Heat cannot be transported from a system at low temperature to another system at high temperature without the aid of external agency”. This statement of second law is attributed to
(a) Clausius
(b) Joule Thomson
(c) Max – Planck
(d) Gay-Lussac

Solutions

S1. Ans.(c)
Sol. the entropy of an isolated system can never decreases. It always increases and remains constant only when the process is reversible. So, any process is reversible. So, any process is likely to be in a given direction only when ∆S ≥ 0.

S2. Ans.(c)
Sol. in an adiabatic process, as there is no heat transfer, there is no entropy transfer and hence the change in entropy is due to entropy generation.
dS=(dQ/T)_irrev+δS_gen
So, for irreversible adiabatic process, dS > 0, i.e. the entropy change of the system is always positive.

S3. Ans.(a)
Sol. In isothermal dissipation of work.
(∆S)_system=0
(∆S)_surrounding=(W/T)
While in adiabatic dissipation of work, this is opposite.

S4. Ans.(b)
Sol. if KE and PE changes are neglected and for unit mass, the availability becomes
a=u=u_0+p_0 (V-V_0 )-T_0 (S-S_0 )
=(u+p_0 V-T_0 S)
-(u_0+p_0 V_0-T_0 S_0 )
=ϕ-ϕ_0
Where ϕ is the available function of the closed system.
For steady flow system, availability per unit mass, is
a=(h-T_0 S+V^2/2+gZ)-(h_0-T_0 S_0+gZ)
=Ψ-Ψ_0
Where Ψ is availability function for a steady flow system.

S5. Ans.(b)
Sol. Mechanical work – High grade energy
Heat – low grade energy
∮▒dQ/T≤0- Clausius inequality
Zeroth law – concept of temperature

S6. Ans.(b)
Sol. in an irreversible process, entropy always increases. Once created, entropy cannot be destroyed. More the irreversibility, more the entropy generation.

S7. Ans.(b)
Sol. Bernoulli equation is a special limiting case of steady flow energy equation. When
U_1=U_2
Steady state
q = 0
W = 0
ρ_1=ρ_2=ρ,
Then from SFEE, Bernoulli equation is obtained

S8. Ans.(c)
Sol. using S.F.E.E.,
mh_1+(mq)=mh_2+W_cv
0.5 × 100 + 0.5 × (-40) = 0.5 × 200 + W_cv
W_CV = -70 kW
Thus, power required = 70 kW

S9. Ans.(d)
Sol. during an adiabatic mixing process, the final properties are determined using the first law of thermodynamic and perfect gas relationship.

S10. Ans.(a)
Sol. Clausius statement of the second law gives:
Heat cannot be transported from a system at low temperature to another system at high temperature without the aid of external agency.

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