Answer to Question #208384 in Mechanical Engineering for aman

Question

the capacity of a refrigeration is 200tr when working between -6°C and 25°C also find the power requirement to drive the unit

Answer 1

Power input = $\frac{RC * (T_H-T_L)}{T_L}$

But RC = 200TR x 335 KJ/kg = 703.371 kW

$P_{in}=\frac{703.371*(25--6)}{267}$

$P_{in}=81.664 kW$