Quiz: Mechanical Engineering
Exam: DFCCIL
Topic: Thermal Engineering
Each question carries 2 mark
Negative marking: 1/3 mark
Time: 10 Minutes
Q1. Triple point of water at 4.587 mm of Hg is.
(a) 273 K
(b) 0°C
(c) 273.16 K
(d) 300 K
Q2. A standard vapour is compressed to half its volume without changing its temperature. The result is that:
(a) All the vapour condenses to liquid
(b) some of liquid evaporates and the pressure does not change
(c) the pressure is double its initial value.
(d) Some of vapour condenses and the pressure does not change
Q3. The internal energy of a gas obeying vander waals equation (P+a/v^2 ) (v-b)=RT depends on its
(a) temperature
(b) temperature & pressure
(c) temperature & specific volume
(d) pressure & specific volume
Q4. The numerical value of slope of an iso-enthalpy on a T-p at any point known as
(a) Volume expansivity
(b) isothermal compressibility
(c) Adiabatic compressibility
(d) Joule-Kelvin coefficient
Q5. In an air-conditioning process 5 kJ/min heat is extracted from a room. If the sensible heat factor is 0.8, then latent heat extracted will be
(a) 4 kJ/min
(b) 2 kJ/min
(c) 1 kJ/min
(d) 0.25 kJ/min
Q6. The efficiency of the vapor power Rankine cycle can be increased by
- Increasing the temperature of the working fluid at which heat is added.
- Increasing the pressure of the working fluid at which heat is added.
- Decreasing the temperature of the working fluid at which heat is rejected.
Which of these statements is/ are correct?
(a) 2 and 3
(b) 1 alone
(c) 1 and 2
(d) 1, 2 and 3
Q7. Which of the following is wrong statement about the Curtis turbine?
(a) Curtis turbine consists of a set of nozzles followed by a set of moving blades.
(b) In conventional design only two rows of moving blades are used.
(c) There is no restriction on the number of rows of moving blades.
(d) Curtis stage is also known as velocity compounding of impulse turbine.
Q8. Increase in compression ratio in CI engines
(a) decreases turbulence
(b) increases mechanical efficiency
(c) decreases mechanical efficiency
(d) decreases thermal efficiency
Q9. A refrigerator working on reversed Carnot cycle absorbs 10 kW power and has a COP of 4. What will be the quantity of heat delivered to a building if it works as a heat pump?
(a) 30 kW
(b) 40 kW
(c) 50 kW
(d) 60 kW
Q10. Two reversible refrigerator (a) and (b) are arranged in series, the COP of (a) is 3 and COP of (b) is 4. The COP of composite refrigerator system is
(a) 1.5
(b) 3
(c) 4
(d) 2.25
Solutions
S1. Ans.(c)
Sol. The triple point of water is at 4.587 mm Hg and 273.16 K.
S2. Ans.(d)
Sol. As shown in figure, from state 1 – 2, keeping temperature constant vapour is compressed to half its volume resulting in some of vapour condensed and the pressure remains constant.
S3. Ans.(c)
Sol. The Internal energy of a gas obeying vander waals equation depends on its temperature and specific volume.
S4. Ans.(d)
Sol. Joule – Kelvin coefficient, μ_J=(∂T/∂p)_h i.e., the numerical value of slope of an isenthalpic on a T – p diagram at any point is known as joule – kelvin coefficient.
S5. Ans (c)
Sol. We know that
SHF=0.8=SH/(LH+SH)=SH/5
SH=4 kJ/min
LH=5 kJ/min-4 kJ/min
LH=1 kJ/min
S6. Ans. (d)
Sol. By increasing the mean temperature of heat addition and decreasing the mean temperature of heat rejection, efficiency of vapor power Rankine cycle can be increased.
S7. Ans. (c)
Sol. Curtis stage is also known as velocity compounding of impulse turbine. Curtis turbine consists of a set of nozzles followed by a set of moving blades and in conventional design only two rows of moving blades are used.
S8. Ans (c)
Sol. When we increase compression ratio in CI engines then thermal efficiency of the engine increases and mechanical efficiency of the engine decreases due to increase in weight of the engine parts as well as frictional losses between the mating parts.
S9. Ans (c)
Sol. Given,
W_in=10 kW and 〖(COP)〗_R=4
〖(COP)〗_R=4=Q_1/W_in
4=Q_1/10
Q_1=40 kW
Hence, Q_2=40+10=50kW (Heat supplied to building when we use refrigerator as heat pump)
S10. Ans (a)
Sol.
We know that
COP of a composite system COP=(COP_1×COP_2)/(1+COP_1+COP_2 )
COP=(3×4)/(1+3+4)=1.5
