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Vapour Compression Cycle: Questions and Solutions
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In this lesson, one will learn how to solve the problems on vapour compression cycle. Here Harshit solves some sample questions which will help one to recollect and remember all the topics and equations which were discussed in the previous lesson.

Harshit Aggarwal
Cleared UPSC ESE twice with Rank 63 and 90 in mechanical engg. Got 99 percentile in GATE. Cracked ONGC, BHEL,ISRO, SAIL, GAIL successfully

Unacademy user
Sir, in the solution for the 3rd question. 1st point is shown in mixture state(Before Entering compressor) but should it not be saturated vapour?
I didn’t able to understand solution no 3 please make this video in details
Sir how someone can find whether it is wet compression or dry compression when temperature limits are given
Raja Raha
3 years ago
If no condition is given
please try to upload the video in old version because if we pause this video there is dark in surrounding and pause symbol also occupy some space in the middle .so it difficult specially for numerical problem.
Amit Phoenix
4 years ago
Yes! I was about to say the same..
Harshit Aggarwal
3 years ago
OK :)

  2. ABOUT ME Graduated from NIT Nagpur in 2008 . Cleared Indian Engineering Services (IES-UPSC) Exam . Got the offer letter from most of the Maharatna and Navratna Companies Cleared GATE Exanmm Course fee: CONTRIBUTE . Follow me on Unacademy at:

  3. NUMERICAL 1 In Vapor Compression Cycle, following data is obtained: Enthalpy at the inlet to the compressor is 180 kj/Kg and at the exit of compressor is 210 KJ/kg. Enthalpy at the exit of condenser is 80KJ/Kg. Find the COP. hi=180 kJ/kg h2- 2lo kJ/ Sol: Given, 80 kJ/ COPs hi-h3= 180-80 = 3.33 h2-h 210-180

  4. NUMERICAL 2 In a 5KW cooling capacity refrigeration system operating on V C Cycle. Refrigerant enters evaporator with an enthalpy of 75 kJ/kg and leaves with an enthalpy of 183 kl/kg. Enthalpyof refrige . 1) COP tafter compression is 21 0 K/kg. Calculate 2) Power input to the compressor . 3) Rate of heat transfer at the condensor in K/sec

  5. SOLUTION 2 Sot: Given, hu5ha h, = 183 kJ/k . COPs hi-ha 183-75 = 4 Ans ha-h 210-183 And we also know that, L-25 5= heat transfer heat transer at the condeser condenser condense im 4 - * Pin -- = 1.25 kW Ans |NOTE: Higher taup side Condenser hest tronsfer but lower temp. side. Evaporator het tronsfer

  6. NUMERICAL 3 . A refrigerator based on Vapour Compression Cycle operates between temperature limits of -20 C and 40 C.The refrigerantenters the condensoras saturated vapour and leaves as saturated liquid.The enthalpy and entropy of saturated liquid and saturated vapour at these temp are given in the table. If the refrigerant flow rate is 0.025 Kg/sec. Find Refrigeration capacity in KW and COP of the system? 2018o 0 7366 o-1366 40 80 200.30.67 o802003 0.67

  7. SOLUTION 3 3 h2-h 2. = 20+11 C180-2.) But -2 is isenropic process 20 4 ek 067 = 007 +rCo 365-07)

  8. SOLUTION 3. h.-20+-.9 C180-20) = 164 kJ/ h2_h 200-164