A commercial refrigerator with refrigerant -134a as the working fluid is used to keep the refrigerated space at -30C by rejectin
g its waste heat to cooling water that enters the condenser at 18C at a rate of 0.25 kg/s and leaves at 26C. The refrigerant enters the condenser at 1.2 MPa and 65C and leaves at 42C. The inlet state of the compressor is 60 kPa and -34C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine
a. the quality of the refrigerant at the evaporator inlet,
b.the refrigeration load,
c. the COP of the refrigerator, and
d. the theoretical maximum refrigeration load for the same power input to the compressor.
a. the quality of the refrigerant at the evaporator inlet,
b.the refrigeration load,
c. the COP of the refrigerator, and
d. the theoretical maximum refrigeration load for the same power input to the compressor.
3 answers:
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Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂