MCQOPTIONS
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MCQOPTIONS
This section includes 12 Mcqs, each offering curated multiple-choice questions to sharpen your Refrigeration knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
C.O.P. of the refrigerator is always __________ the C.O.P. of the heat pump when both are working between the same temperature limits. |
| A. | less than |
| B. | greater than |
| C. | equal to |
| D. | inverse of |
| Answer» B. greater than | |
| 2. |
A heat pump which runs (1/3)rd of time removes on an average 2400 kJ/hr of heat. If power consumed is 0.25 kW, what is the value of the C.O.P.? |
| A. | 4 |
| B. | 2 |
| C. | 8 |
| D. | 6 |
| Answer» D. 6 | |
| 3. |
A heat pump is used to maintain a hall at 30 C when the atmospheric temperature is 15 C. The heat loss from the hall is 1200 kJ/min. Calculate the power required to run the heat pump if its C.O.P. is 40% of the Carnot machine working between the same temperature limits. |
| A. | 0.495 |
| B. | 4.04 |
| C. | 0.247 |
| D. | 8.08 |
| Answer» D. 8.08 | |
| 4. |
If the coefficient of performance of the refrigerator is 4.67, then what is the value of the coefficient of performance of the heat pump operating under the same conditions? |
| A. | 3.67 |
| B. | 5.67 |
| C. | 0.214 |
| D. | 9.34 |
| Answer» C. 0.214 | |
| 5. |
A reversed Carnot cycle is operating between temperature limits of (-) 33 C and (+) 27 C. If it acts as a heat engine gives an efficiency of 20%. What is the value of C.O.P. of a heat pump operating under the same conditions? |
| A. | 6.5 |
| B. | 8 |
| C. | 5 |
| D. | 2.5 |
| Answer» D. 2.5 | |
| 6. |
If the reversed Carnot cycle operating as a heat pump between temperature limits of 364 K and 294 K, then what is the value of C.O.P? |
| A. | 4.2 |
| B. | 0.19 |
| C. | 5.2 |
| D. | 0.23 |
| Answer» D. 0.23 | |
| 7. |
For the systems working on reversed Carnot cycle, what is the relation between C.O.P. of Refrigerator i.e. (C.O.P.)R and Heat Pump i.e. (C.O.P)P? |
| A. | (C.O.P.)<sub>R</sub> + (C.O.P)<sub>P</sub> = 1 |
| B. | (C.O.P.)<sub>R</sub> = (C.O.P)<sub>P</sub> |
| C. | (C.O.P.)<sub>R</sub> = (C.O.P)<sub>P</sub> 1 |
| D. | (C.O.P.)<sub>R</sub> + (C.O.P)<sub>P</sub> + 1 = 0 |
| Answer» D. (C.O.P.)<sub>R</sub> + (C.O.P)<sub>P</sub> + 1 = 0 | |
| 8. |
C.O.P. of the heat pump is always _____ |
| A. | one |
| B. | less than One |
| C. | greater than One |
| D. | zero |
| Answer» D. zero | |
| 9. |
How is the Relative coefficient of performance represented? |
| A. | Theoretical C.O.P. / Actual C.O.P. |
| B. | Actual C.O.P. / Theoretical C.O.P. |
| C. | Theoretical C.O.P. x Actual C.O.P. |
| D. | 1 / Theoretical C.O.P. x Actual C.O.P. |
| Answer» C. Theoretical C.O.P. x Actual C.O.P. | |
| 10. |
What is the equation between efficiency of Heat engine and C.O.P. of heat pump? |
| A. | <sub>E</sub> = (C.O.P.)<sub>P</sub> |
| B. | <sub>E</sub> = 1 / (C.O.P.)<sub>P</sub> |
| C. | <sub>E</sub> / (C.O.P.)<sub>P</sub> = 1 |
| D. | <sub>E</sub> x (C.O.P.)<sub>P</sub> = 0 |
| Answer» C. <sub>E</sub> / (C.O.P.)<sub>P</sub> = 1 | |
| 11. |
What is the difference between Heat Pump and Refrigerator? |
| A. | Heat Pump Gives efficiency and refrigerator gives C.O.P. |
| B. | Both are similar |
| C. | Both are almost similar, just the desired effect is different |
| D. | Work is output in refrigerator and work is input in heat pump |
| Answer» D. Work is output in refrigerator and work is input in heat pump | |
| 12. |
For a standard system with temperatures T1 and T2, where T1 < Ta < T2 (Ta Atmospheric Temperature). Q1 is the heat extracted from a body at temperature T1, and Q2 is heat delivered to the body at temperature T2. What is the C.O.P. of the heat pump for given conditions? |
| A. | Q<sub>2</sub> / (Q<sub>2</sub> Q<sub>1</sub>) |
| B. | (Q<sub>2</sub> Q<sub>1</sub>) / Q<sub>1</sub> |
| C. | (Q<sub>2</sub> Q<sub>1</sub>) / Q<sub>2</sub> |
| D. | Q<sub>1</sub> / (Q<sub>2</sub> Q<sub>1</sub>) |
| Answer» B. (Q<sub>2</sub> Q<sub>1</sub>) / Q<sub>1</sub> | |