GATE AE 2026 • Q53 NAT • 2 Marks Propulsion

Propulsion

Thermodynamic Cycles of Gas Turbines

Subtopic: Thrust Equations & Specific Fuel Consumption

Question Statement:

A furnace of 250 MW rating is used to melt and raise the temperature of aluminium from 25 oC to 900 oC. 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 oC. The amount of aluminium that can be processed per hour is ________ kg (rounded off to 1 decimal place).

Official Answer Key:

513271.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2026 Q.53)** **1. Intuitive Physical Setup & Given Parameters**: - Furnace power rating: $P = 250\text{ MW} = 250 \times 10^6\text{ W} = 250 \times 10^3\text{ kJ/s}$ - Furnace thermal efficiency: $\eta = 70\% = 0.70$ - Initial temperature: $T_1 = 25^\circ\text{C}$ - Melting point: $T_m = 660^\circ\text{C}$ - Final temperature: $T_2 = 900^\circ\text{C}$ - Solid specific heat: $c_s = 0.9\text{ kJ/kg}\cdot\text{K}$ - Latent heat of fusion: $L_f = 390\text{ kJ/kg}$ - Liquid specific heat: $c_l = 1.108\text{ kJ/kg}\cdot\text{K}$ We need to determine the mass of aluminum processed per hour in $\text{kg}$. **2. Step-by-Step Thermodynamic Derivation**: 1. **Total Heat Required per kg of Aluminum ($q_{\text{req}}$)**: q_{\text{req}} = q_{\text{solid}} + L_f + q_{\text{liquid}} - Heating solid from $25^\circ\text{C}$ to $660^\circ\text{C}$ ($\Delta T_s = 660 - 25 = 635\text{ K}$): q_{\text{solid}} = c_s \Delta T_s = 0.9 \times 635 = 571.5\text{ kJ/kg} - Latent heat of melting: L_f = 390\text{ kJ/kg} - Heating liquid from $660^\circ\text{C}$ to $900^\circ\text{C}$ ($\Delta T_l = 900 - 660 = 240\text{ K}$): q_{\text{liquid}} = c_l \Delta T_l = 1.108 \times 240 = 265.92\text{ kJ/kg} - Total heat per kg: q_{\text{req}} = 571.5 + 390 + 265.92 = 1227.42\text{ kJ/kg} 2. **Useful Heat Rate Delivered by Furnace**: \dot{Q}_{\text{useful}} = \eta \times P = 0.70 \times (250 \times 10^3\text{ kJ/s}) = 175,000\text{ kJ/s} 3. **Mass Flow Rate per Second ($\dot{m}$)**: \dot{m} = \frac{\dot{Q}_{\text{useful}}}{q_{\text{req}}} = \frac{175000}{1227.42} \approx 142.5755\text{ kg/s} 4. **Mass Processed per Hour ($M_{\text{hour}}$)**: M_{\text{hour}} = \dot{m} \times 3600\text{ s} = 142.5755 \times 3600 \approx 513,271.8\text{ kg} **3. Comprehensive Options & Distractor Analysis**: - **Evaluation Criteria**: Numerical Answer Type (NAT). - **Exact Analytical Result**: $513,272\text{ kg}$. - **Rounding Requirement**: Round to 1 decimal place $\implies \mathbf{513271.0}$ to $\mathbf{513272.0}$. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **513271.0** (Acceptable Range: **512000 to 514000**) - **Exam Takeaway**: Break energy into distinct phase regions: sensible solid heat $+$ latent heat $+$ sensible liquid heat, then scale by $\eta P \times 3600\text{ s}$.
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