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1 answer:
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Answer:
(1) 5.74
(2) 5.09
(3) 3.05×10⁻⁵ kg/s
(4) 0.00573 kW
Explanation:
The parameters given are;
Working temperature, = -15°C = 258.15 K
Temperature of the cooling water, = 30°C = 303.15 K
(1) The Carnot coefficient of performance is given as follows;
(2) For ammonia refrigerant, we have;
s₂ = s₁ = 4.9738 kJ/(kg·K)
0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738
∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89
h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg
(3) The circulation rate is given by the mass flow rate, as follows
The refrigeration capacity = 105 kJ/h
The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg
Therefore;
= 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s
(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg
The rating power = Work done per second = W×
∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.
Answer:
6222.22 sec
Explanation:
Given data the power input to the refrigerator is 450 W
The COP of refrigerator is 1.5
Temperature
mass of watermelon =10 kg
specific heat =4.2 KJ/kg°C
The amount of heat removed from 5 watermelon
We know that
W=0.675 KW
so time required to cool the watermelon is