Question

Water is to be boiled at 100 oC at the atmospheric pressure of 101.42 kPa in a 25 cm diameter stainless steel pan placed on top of a 3 kW electric burner. If 70 % of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water in kg/h. Given the property of water: hfg = 2256.4 kJ/kg at 100 oC and 101.42 kPa.

Answer 1

The rate of evaporation of water in the given scenario is 3.35 kg/h.

Step-by-step explanation:

To determine the rate of evaporation of water in kg/h, we need to calculate the amount of heat transferred to the water and use the enthalpy of vaporization. First, we need to calculate the heat transferred to the water using the power of the electric burner and the efficiency of heat transfer. The power of the electric burner is 3 kW, and 70% of this heat is transferred to the water. So, the heat transferred to the water is (3 kW) x (0.70) = 2.1 kW.

Next, we need to convert the power into heat by using the conversion factor of 1 kW = 3.6MJ/h. Thus, the heat transferred to the water is (2.1 kW) x (3.6 MJ/h) = 7.56 MJ/h.

Finally, we can calculate the rate of evaporation of water by dividing the heat transferred to the water by the enthalpy of vaporization. The enthalpy of vaporization of water at 100°C and 101.42 kPa is 2256.4 kJ/kg. So, the rate of evaporation of water is (7.56 MJ/h) / (2256.4 kJ/kg) = 3.35 kg/h.