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 option is 32°C.

Step-by-Step Explanation:

First, we list the given parameters of the system:
- Electric power input, ${\dot{W}}_e=101\text{kW}$
- Inlet pressure, $P_1=100\text{kPa}$
- Inlet temperature, $T_1=27°\text{C}=27+273=300\text{K}$
- Inlet volume flow rate, ${\dot{V}}_1=15{\text{m}}^3\text{/s}$
- Heat loss rate, ${\dot{Q}}_{\text{loss}}=51\text{kW}$
- Specific heat capacity at constant pressure, $C_p=1\text{kJ/(kg}·\text{K)}$
- Gas constant, $R=0.5\text{kJ/(kg}·\text{K)}$

Step 1: Calculate the density and mass flow rate of the gas at the inlet
Using the ideal gas equation:

$P_1={\rho }_{1}R{T}_1$

Rearranging to solve for density (${\rho }_1$):

${\rho }_1=\frac{P_1}{R{T}_1}$

Substituting the given values:

${\rho }_1=\frac{100}{0.5\times 300}=\frac{100}{150}=\frac{2}{3}$

Now, we calculate the mass flow rate ($\dot{m}$):

$\dot{m}={\rho }_1{\dot{V}}_1=\frac{2}{3}\times 15=10\text{kg/s}$

Step 2: Apply the First Law of Thermodynamics for steady-flow system
For a steady-flow system with negligible changes in kinetic and potential energies, the energy balance equation is:

${\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}$

$\dot{m}{h}_1+{\dot{W}}_e=\dot{m}{h}_2+{\dot{Q}}_{\text{loss}}$

Rearranging the terms:

${\dot{W}}_e-{\dot{Q}}_{\text{loss}}=\dot{m}(h_2-h_1)$

Since specific heat ($C_p$) is constant, the change in enthalpy is $h_2-h_1=C_p(T_2-T_1)$:

${\dot{W}}_e-{\dot{Q}}_{\text{loss}}=\dot{m}{C}_p(T_2-T_1)$

Step 3: Calculate the exit temperature of the gas ($T_2$)
Substitute the given values into the equation:

$101-51=10\times 1\times (T_2-27)$

$50=10\times (T_2-27)$

$5=T_2-27$

$T_2=32°\text{C}$

Thus, the exit temperature of the gas is 32°C.