A gas is heated in a duct as it flows over a resistance heater. Consider a 101 kW electric heating system. The gas enters the heating section of the duct at 100 kPa and 27°C with a volume flow rate of 15 m³/s. If heat is lost from the gas in the duct to the surroundings at a rate of 51 kW, the exit temperature of the gas is (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; C_p = 1 kJ/kgK; R = 0.5 kJ/kgK)

The correct answer is 32°C.

1. Identify the given parameters from the problem description and the schematic diagram:
• Power input to the electric resistance heater: ${\dot{W}}_{\text{in}}=101\text{ kW}$ (labeled as W = 101 kW in the diagram)
• Rate of heat lost to the surroundings: ${\dot{Q}}_{\text{loss}}=51\text{ kW}$ (labeled as Q_loss = 51 kW in the diagram)
• Inlet gas pressure: $P_1=100\text{ kPa}$
• Inlet gas temperature: $T_1=27{\text{°C}}^{}=300\text{ K}$ (Note: The label on the schematic diagram contains a typo showing T₁ = 200 K, but the calculation correctly uses the given value of 300 K)
• Gas volume flow rate at the inlet: ${\dot{V}}_1=15{\text{ m}}^3\text{/s}$
• Specific heat at constant pressure: $C_p=1\text{ kJ/(kg}\cdot \text{K)}$
• Gas constant: $R=0.5\text{ kJ/(kg}\cdot \text{K)}$

2. Calculate the mass flow rate ($\dot{m}$) using the ideal gas equation of state at the inlet:
Using the equation:
$$P_1{\dot{V}}_1=\dot{m}R{T}_1$$
Rearranging for mass flow rate ($\dot{m}$):

Substituting the given values:
$$\dot{m}=\frac{100\times 15}{0.5\times 300}=\frac{1500}{150}=10\text{ kg/s}$$

3. Apply the Steady Flow Energy Equation (SFEE) to the control volume:
Since changes in kinetic and potential energies are negligible, the energy balance equation is:
$$\dot{m}{h}_1+{\dot{W}}_{\text{in}}=\dot{m}{h}_2+{\dot{Q}}_{\text{loss}}$$
Rearranging to solve for the enthalpy change:
$${\dot{W}}_{\text{in}}-{\dot{Q}}_{\text{loss}}=\dot{m}(h_2-h_1)$$
Since the specific heat capacity at constant pressure is constant, we can express the change in specific enthalpy as $h_2-h_1=C_p(T_2-T_1)$:
$${\dot{W}}_{\text{in}}-{\dot{Q}}_{\text{loss}}=\dot{m}{C}_p(T_2-T_1)$$

4. Solve for the temperature difference:
$$101-51=10\times 1\times (T_2-T_1)$$
$$50=10\times (T_2-T_1)$$
$$T_2-T_1=5\text{ K (or 5°C)}$$

5. Find the final exit temperature of the gas:
$$T_2=T_1+5{\text{°C}}^{}=27{\text{°C}}^{}+5{\text{°C}}^{}=32{\text{°C}}^{}$$