1 Determine the amount of heat that must be removed to cool

1) Determine the amount of heat that must be removed to cool 20 kilograms of air at 40 degrees C and 20% to 20 degrees C.

Heat Removed = XXXX kJ of Heat

2) What volume of water must be added to air in 30 mins if the air is initially at 80 F and 20% RH to get it to a final condition of 80 F and 70%? Assume the air flow rate is currently at 100 lbs DA/hr.

Water Volume Added = XXXX lbs H₂O

Solution

1.

The amount of heat that must be removed to cool=ms*difference in temperature                                           

heat removeed= 20*20*1.006 =402.5kj

2.

At 80 F 20 % RH w = 0.0043246 lb /lb at 80 F and 70 % RH w = 0.0154 lb/lb change in humidity ratio = 0.0154 - 0.0043246 = 0.0110754 lb/lb ampount of water added = 0.0110754 * 100 = 1.10754 lb/hr So in 30 mins we need 2.21508 lbm of water