Page 422 - Handbook of Energy Engineering Calculations

INDUSTRIAL-USE ELECTRIC HEATER SELECTION AND
SIZING

Choose the heating capacity of an electric heater to heat a pot containing 600
lb (272.2 kg) of lead from the charging temperature of 70°F (21.1°C) to a
temperature of 750°F (398.9°C) if 600 lb (272.2 kg) of the lead is to be
melted and heated per hour. The pot is 30 in (76.2 cm) in diameter and 18 in

(45.7 cm) deep.

Calculation Procedure:

1. Compute the heat needed to reach the melting point
When a solid is melted, first it must be raised from its ambient or room
temperature to the melting temperature. The quantity of heat required is H =

(weight of solid, lb) [specific heat of solid, Btu/(lb · °F)] (t – t ), where H =
m
t
Btu required to raise the temperature of the solid, °F; t = room, charging, or
1
initial temperature of the solid, °F; t = melting temperature of the solid, °F.
m
For this pot with lead having a melting temperature of 620°F (326.7°C)

and an average specific heat of 0.031 Btu/(lb · °F) [0.13 kJ/(kg · °C)], H =
(600)(0.031)(620 – 70) = 10,240 Btu/h (3.0 kW), or (10,240 Btu/h)/(3412
Btu/kWh) = 2.98 kWh.

2. Compute the heat required to melt the solid

The heat H Btu required to melt a solid is H = (weight of solid melted, lb)
m
m
(heat of fusion of the solid, Btu/lb). Since the heat of fusion of lead is 10
Btu/lb (23.2 kJ/kg), H = (600)(10) = 6000 Btu/h, or 6000/3412 = 1.752
m
kWh.

3. Compute the heat required to reach the working temperature

Use the same relation as in step 1, except that the temperature range is
expressed as t – t , where t = working temperature of the melted solid.
w
m
w
Thus, for this pot, H = (600)(0.031)(750 – 620) = 2420 Btu/h (709.3 W), or
2420/3412 = 0.709 kWh.