Quiz: Mechanical Engineering
Exam: NHPC JE
Each question carries 1 mark
Negative marking: 1/4 mark
Time: 6 Minutes
Q1. For a simple closed system of constant composition, the difference between the net heat and work interactions is identifiable as the change in
(c) Flow energy
(d) Internal energy
Q2. A closed system undergoes a process 1-2 for which the values of Q_(1-2) and W_(1-2) are + 20kJ and + 50kJ, respectively. If the system is returned to state 1 and Q_(2-1) is –10kJ, what is the value of the work W_(2-1) ?
(a) +20 kJ
(b) -40 kJ
(c) -80 kJ
(d) +40 kJ
Q3. Consider the following statements:
A real gas obeys perfect gas law at very
1. high temperatures
2. high pressures
3. low pressures
Which of these statements is/are correct?
(a) 1 only
(b) 1 and 3
(c) 2 only
(d) 3 only
Q4. At critical point the enthalpy of vaporization is
(a) dependent on temperature only
Q5. An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at 27°c, then the source temperature is
Q6. The irreversibility is defined as the difference of the maximum useful work and actual work : I=W_maxuseful -W_actual. How can this be alternatively expressed?
(a) I=T_0 (∆S_system+∆S_surrounding)
(b) I=T_0 (∆S_System-∆S_surrounding)
(c) I=T_0 (√(∆S_system )+√(∆S_surrounding ))
(d) I=T_0 (√(∆S_system )-√(∆S_Surroundings ))
Sol. The given statement is for first law of Thermodynamics i.e.
Sol. For process 1 -2
For process 2 – 1
Sol. A real gas starts behaving as ideal gas at low pressure and high temperature.
Sol. At critical point enthalpy of vaporization is zero.
Sol. Sink temperature: T_2=27°C = 27 + 273 = 300K
It is given that engine rejects 40% of absorbed heat from the source
For a Carnot cycle engine
Q_1/T_1 =(0.4 Q_1)/300