wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Byju's Answer
Standard XII
Chemistry
Enthalpy
A refrigerato...
Question

A refrigerator based on ideal vapour compression cycle operates between the temperature limits of −20∘C and 40∘C. The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in table below.

Thfhgsfsg(∘C)(kJ/kg)(kJ/kg)(kJ/kg K)(kJ/kg K)−20201800.070.736640802000.30.67

The COP of the refrigerator is

A
2.0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2.33
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
5.0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6.0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2.33
COP=Refrigeration effectWork input

=REW=2.1m(h2−h1)

=2.10.025(200−164)=2.33

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
Q. A refrigerator based on ideal vapour compression cycle operates between the temperature limits of −20∘C and 40∘C. The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in table below.

Thfhgsfsg(∘C)(kJ/kg)(kJ/kg)(kJ/kg K)(kJ/kg K)−20201800.070.736640802000.30.67

If refrigerant circulation rate is 0.025 kg/s, the refrigeration effect is equal to
Q. In the vapour compression cycle shown in the figure, the evaporating and condensing temperature are 260 K and 310 K, respectively. the compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). the specific heat of the liquid refrigerant is 4.8 kJ/kg−K may be treated as constant. The enthalpy of evaporation for the refrigerator at 310 K is 1054 kJ/kg.


The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is
Q. A refrigerator operates between 120 kPa and 800 kPa in an ideal vapor compression cycle with R−134a as the refrigerant. The refrigerant enters the compressor as saturated vapor and leaves the condenser as saturated liquid. The mass flow rate of the refrigerant is 0.2kg/s. Properties for R−134a are as follows:

Saturated R- 134aP(kPa)T(∘C)hf(kJ/kg)hs(kJ/kg)sf(kJ/kgK)sg(kJ/kgK)120−22.3222.52370.0930.9580031.3195.5267.30.3540.918Superheated R- 134aP(kPa)T(∘C)h(kJ/kg)s(kJ/kgK)80040276.450.95

The power required for compressor in kW is
Q. A refrigerator operates between 120 kPa and 800 kPa in an ideal vapor compression cycle with R−134a as the refrigerant. The refrigerant enters the compressor as saturated vapor and leaves the condenser as saturated liquid. The mass flow rate of the refrigerant is 0.2kg/s. Properties for R−134a are as follows:

Saturated R- 134aP(kPa)T(∘C)hf(kJ/kg)hs(kJ/kg)sf(kJ/kgK)sg(kJ/kgK)120−22.3222.52370.0930.9580031.3195.5267.30.3540.918Superheated R- 134aP(kPa)T(∘C)h(kJ/kg)s(kJ/kgK)80040276.450.95

The rate at which heat is extracted, in kJ/s from the refrigerated space is
Q. A refrigerator uses R−134a as its refrigerant and operates on an ideal vapour compression refrigeration cycle between 0.14MPa and 0.8MPa. If the mass flow rate of the refrigerant is 0.05kg/s, the rate of heat rejection to the environment is \(kW\).
Given data :
At P=0.14MPa,h=236.04 kJ/kg,s=0.9322 kJ/kgK
At P=0.8 MPa,h=272.05 kJ/kg(superheated vapour)
At P=0.8 MPa,h=93.42 kJ/kg(saturated liquid)
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
CHEMISTRY
Watch in App
explore_more_icon
Explore more
Enthalpy
Standard XII Chemistry
Join BYJU'S Learning Program
book_image
AI Tutor
book_image
Textbooks
book_image
Question Papers
book_image
Install app
CrossIcon