A coolant fluid has a latent heat of evaporation of 234 kj/kg . If the pump rate is 2250g per minute. How much thermal heat is removed each hour?
To find the amount of thermal heat removed each hour, we need to calculate the mass flow rate and then multiply it by the latent heat of evaporation.
Step 1: Convert the pump rate to kg/s.
Given that the pump rate is 2250 g (grams) per minute and there are 60 seconds in a minute, we can convert the pump rate to kg/s by dividing it by 1000 (since there are 1000 grams in a kilogram).
2250 g/min / 1000 = 2.25 kg/s
Step 2: Calculate the thermal heat removed each hour.
The amount of thermal heat removed is given by the product of the mass flow rate and the latent heat of evaporation.
Thermal heat removed = mass flow rate * latent heat of evaporation
Using the given values:
Thermal heat removed = 2.25 kg/s * 234 kJ/kg
To convert kJ to J, we multiply the value by 1000.
Thermal heat removed = 2.25 kg/s * 234 kJ/kg * 1000 J/kJ
Finally, we need to convert seconds to hours, multiplying by 3600 seconds:
Thermal heat removed = 2.25 kg/s * 234 kJ/kg * 1000 J/kJ * 3600 s/1 h
Calculating this expression will give us the amount of thermal heat removed each hour.