Question

A furnace of 250 MW rating is used to melt and raise the temperature of aluminium from 25\(^{\circ}\)C to 900\(^{\circ}\)C. Aluminium has a solid-state specific heat, latent heat, and liquid-state specific heat of 0.9 kJ/kg-K, 390 kJ/kg, and 1.108 kJ/kg-K, respectively, and the furnace has 70% efficiency. The melting point of aluminium is 660\(^{\circ}\)C. The amount of aluminium that can be processed per hour is ________ kg (rounded off to 1 decimal place).

Hint:

Add up the heat needed per kg for solid heating, melting, and liquid heating, then divide the furnace's hourly useful output (rating times efficiency times 3600 s) by that value.

Answer 1

Correct Answer: 513271.7

Step 1: Break the heating process into three stages.
To take 1 kg of aluminium from 25 C to 900 C, it must (a) be heated as a solid from 25 C to its melting point 660 C, (b) melt at 660 C (latent heat), and (c) be heated as a liquid from 660 C to 900 C.

Step 2: Heat needed for solid-state heating.
\[ q_1 = c_{p,solid} \times (660-25) = 0.9 \times 635 = 571.5\ \text{kJ/kg} \]

Step 3: Heat needed for melting.
\[ q_2 = L = 390\ \text{kJ/kg} \]

Step 4: Heat needed for liquid-state heating.
\[ q_3 = c_{p,liquid} \times (900-660) = 1.108 \times 240 = 265.92\ \text{kJ/kg} \]

Step 5: Total heat needed per kg of aluminium.
\[ q = q_1+q_2+q_3 = 571.5+390+265.92 = 1227.42\ \text{kJ/kg} \]

Step 6: Find the useful heat delivered by the furnace per hour.
The furnace rating is the input power, so the useful (output) power is 70% of 250 MW:
\[ P_{useful} = 0.70 \times 250{,}000\ \text{kW} = 175{,}000\ \text{kJ/s} \]
Over 1 hour (3600 s):
\[ Q_{hour} = 175{,}000 \times 3600 = 630{,}000{,}000\ \text{kJ} \]

Step 7: Divide total heat available by heat needed per kg.
\[ m = \frac{Q_{hour}}{q} = \frac{630{,}000{,}000}{1227.42} \approx 513{,}271.7\ \text{kg} \]

Final Answer:
About 513,271.7 kg of aluminium can be processed per hour.
\[ \boxed{m \approx 513271.7\ \text{kg/hr}} \]