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 thermal heat removed per hour, we need to know the mass flow rate of the coolant fluid. Given that the pump rate is 2250 grams per minute, we can convert it to kilograms per hour.
Step 1: Convert pump rate from grams per minute to kilograms per hour.
1 minute = 1/60 hours (since there are 60 minutes in an hour)
2250 grams per minute * (1 kg / 1000 grams) * (60 minutes / 1 hour) = 2.25 kg per hour
Therefore, the mass flow rate is 2.25 kg per hour.
Step 2: Calculate the thermal heat removed per hour.
To calculate the thermal heat, we can use the formula:
Thermal heat = mass flow rate * latent heat of evaporation
Thermal heat = 2.25 kg/hr * 234 kJ/kg
= 526.5 kJ/hour
Therefore, the thermal heat removed each hour is 526.5 kJ.