Thermodynamics Questions and Answers – Other types of Work Transfer-2

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This set of Thermodynamics Interview Questions and Answers focuses on “Other types of Work Transfer-2”.

1. Find the rate of conduction heat transfer through a 1.5 cm thick board(k = 0.16 W/m K), with a temperature difference of 20°C between the two sides.
a) 113 W/m2
b) 413 W/m2
c) 313 W/m2
d) 213 W/m2
View Answer

Answer: c
Explanation: The rate of conduction heat transfer = k(∆T/∆x)
= (0.16 *20)/0.015 = 213 W/m2.
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2. A window having area of 2m^2 has a surface temperature of 15°C and the air is blowing at 2°C across it with convection heat transfer coefficient of h = 125 W/m2K. Find the total heat transfer loss?
a) 3250 W
b) 2250 W
c) 4250 W
d) 5250 W
View Answer

Answer: a
Explanation: Total heat transfer loss = h A ∆T
= 125*2*(15-2) = 3250 W.

3. A radiant heating lamp has a temperature of 1000 K with ε = 0.8. What should be the surface area to provide 250 W of radiation heat transfer?
a) 0.0035 m2
b) 0.0055 m2
c) 0.0075 m2
d) 0.0095 m2
View Answer

Answer: b
Explanation: Radiation heat transfer = εσA(T4)
A = 250/[0.8 × 5.67 × 10-8× 10004] = 0.0055 m2.
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4. A piston of mass 2 kg is lowered by 0.5 m. Find the work involved in the process.
a) 7.805 J
b) 8.805 J
c) 9.805 J
d) 10.805 J
View Answer

Answer: c
Explanation: F = ma = 2 kg × 9.80665 m/s2 = 19.61 N
W = ∫ F dx = F ∫ dx = F ∆x = 19.61 N × 0.5 m = 9.805 J.

5. An escalator raises a 100 kg bucket of water 10 m in 60 seconds. Determine the amount of work done during the process.
a) 9807 J
b) 9307 J
c) 9507 J
d) 9107 J
View Answer

Answer: a
Explanation: F = mg
W = ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s2 × 10 m
= 9807 J.
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6. A hydraulic cylinder has a piston(cross sectional area 25cm2) and a fluid pressure of 2 MPa. If piston moves by 0.25m, how much work is done?
a) 0.25 kJ
b) 1.25 kJ
c) 2.25 kJ
d) 3.25 kJ
View Answer

Answer: b
Explanation: W = ∫ F dx = ∫ PA dx = PA ∆x
= 2000 kPa × 25 × 10(-4) m2 × 0.25 m = 1.25 kJ.

7. In a thermally insulated kitchen, a refrigerator with 2 kW motor for running the compressor provides 6000 kJ of cooling to the refrigerated space during 30 min operation. If the condenser coil placed behind the refrigerator rejects 8000 kJ of heat to the kitchen during the same period, calculate the change in internal energy of the kitchen.
a) 3600 kJ
b) 2400 kJ
c) 4800 kJ
d) none of the mentioned
View Answer

Answer: a
Explanation: QKitchen = 0 (Insulated!),
W(Electrical) = – P*∆τ = – 2 kW*30*60 sec = – 3600 kJ
(It is negative because work is done on to the system)
Change in internal energy of the kitchen (∆UKitchen) = QKitchen – WElectrical
= 0 – (–3600) = 3600 kJ.
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8. An escalator raises a 100 kg bucket of sand 10 m in a minute. Determine the rate of work done during the process.
a) 143 W
b) 153 W
c) 163 W
d) 173 W
View Answer

Answer: c
Explanation: The work is force with a displacement and force is F = mg, which is constant
W = ∫ F dx = F ∫ dx = F ∆x = 100 kg × 9.80665 m/s2 × 10 m = 9807 J
The rate of work is work per unit time = W/∆t = 9807 J / 60 s
= 163 W.

9. A crane lifts a bucket of cement with a mass of 450 kg vertically up with a constant velocity of 2 m/s. Find the rate of work.
a) 8.83 kW
b) 8.33 kW
c) 8.53 kW
d) 8.63 kW
View Answer

Answer: a
Explanation: Rate of doing work = FV = mg × V = 450 kg × 9.807 ms^(−2) × 2 ms^(−1)
= 8826 J/s = 8.83 kW.
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10. A battery is well insulated while being charged by 12.3 V at a current of 6 A. Take the battery as a control mass and find the instantaneous rate of work.
a) 63.8 W
b) 73.8 W
c) 83.8 W
d) 93.8 W
View Answer

Answer: b
Explanation: Battery thermally insulated ⇒ Q = 0
For constant voltage E and current i, Power = E i = 12.3 × 6 = 73.8 W.

11. A current of 10 amp runs through a resistor with resistance of 15 ohms. Find the rate of work that heats the resistor up.
a) 1200 W
b) 1300 W
c) 1400 W
d) 1500 W
View Answer

Answer: d
Explanation: Power = E i = R i^(2) = 15 × 10 × 10 = 1500 W.

12. A pressure of 650 kPa pushes a piston of radius 0.125 m with V = 5 m/s. What is the transmitted power?
a) 139.5 kW
b) 149.5 kW
c) 159.5 kW
d) 169.5 kW
View Answer

Answer: c
Explanation: A = π/4(D)2 = 0.049087 m2;
Volume flow rate = AV = 0049087 m^2 × 5 m/s = 0.2454 m3/s
Power = F V = P(Volume flow rate) = 650 kPa × 0.2454 m3/s = 159.5 kW.

13. Air at a constant pressure in a piston-cylinder is at 300 K, 300 kPa and V=0.1 m^3. It is heated to 600 K in 30 seconds in a process with constant piston velocity. Find the power delivered to the piston.
a) 1 kW
b) 2 kW
c) 3 kW
d) 4 kW
View Answer

Answer: a
Explanation: Process: P = constant : dW = P dV
V2 = V1× (T2/T1) = 0.1 × (600/300) = 0.2
Rate = P (∆V / ∆t) = 300 × (0.2-0.1)/30 = 1 kW.

14. A torque of 650 Nm rotates a shaft of radius 0.125 m with ω = 50 rad/s. What is the transmitted power?
a) 22.5 kW
b) 32.5 kW
c) 42.5 kW
d) 52.5 kW
View Answer

Answer: b
Explanation: V = ωr = 50 × 0.125 = 6.25 m/s
Power = Tω = 650 × 50 Nm/s = 32 500 W = 32.5 kW.

Sanfoundry Global Education & Learning Series – Thermodynamics.
To practice all areas of Thermodynamics for Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter