Question

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has specific heat capacity 950 J/kg K melting point 680*C, latent heat of fusion 450 kJ/kg, the efficiency of furnace is 85% and cost of energy is 20 cents/kWhr Calculate: (a) Power required (b) The cost of operating the furnace for 30 hours

Answer 1

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Step-by-step explanation:

Part A)

heat required to melt the aluminium is given by

`Q = ms\Delta T + mL`

here we have

`Q = 40(950)(680 - 32) + 40(450 * 10^3)`

`Q = 24624 kJ + 18000 kJ`

`Q = 42624 kJ`

Since this is the amount of aluminium per hour

so power required to melt is given by

`P = (Q)/(t)`

`P = (42624)/(3600) kW`

`P = 11.84 kW`

Since the efficiency is 85% so actual power required will be

`P = (11.84)/(0.85) = 13.93 kW`

Part B)

Total energy consumed by the furnace for 30 hours

`Energy = power * time`

`Energy = 13.93 kW* 30 h`

`Energy = 417.9 kWh`

now the total cost of energy consumption is given as

`R = P * 20 (Cents)/(kWh)`

`R = 417.9 kWh*  20 (cents)/(kWh)`

`R = 8357.6 Cents`