Question

Consider two insulated tubes with air flowing steadily through them. The flow rate and temperature of the air are 0.1 kg/s, 20 \(^{\circ}\text{C}\) in tube 1 and 0.2 kg/s, 25 \(^{\circ}\text{C}\) in tube 2.
If the two streams are allowed to mix adiabatically, the steady-state temperature of the mixed stream is ______ \(^{\circ}\text{C}\) (rounded off to one decimal place).
Assume a constant \(C_p\) for air.

Hint:

Use a steady-flow energy balance with constant Cp, it reduces to a mass-weighted average of the two inlet temperatures.

Answer 1

Correct Answer: 23.3

Step 1: Write the steady-flow energy balance.
Both tubes are insulated and the mixing is adiabatic, so no heat leaves the system and no work is done, the total enthalpy flow in equals the total enthalpy flow out.
With constant \(C_p\), enthalpy flow of a stream is \( \dot{m} C_p T \), so \( \dot{m}_1 C_p T_1 + \dot{m}_2 C_p T_2 = (\dot{m}_1 + \dot{m}_2) C_p T_{mix} \).

Step 2: Cancel \(C_p\) and solve for the mixed temperature.
\(C_p\) is constant and common to every term, so it cancels out, leaving a simple mass-weighted average of temperature: \( T_{mix} = \dfrac{\dot{m}_1 T_1 + \dot{m}_2 T_2}{\dot{m}_1 + \dot{m}_2} \).
Substituting \( \dot{m}_1 = 0.1 \), \(T_1 = 20\), \( \dot{m}_2 = 0.2 \), \(T_2 = 25\) gives \( T_{mix} = \dfrac{0.1 \times 20 + 0.2 \times 25}{0.3} = \dfrac{2 + 5}{0.3} \).

Final Answer:
The mixed stream settles at the mass-weighted average temperature of the two inlet streams.
\[ T_{mix} = \frac{7}{0.3} = 23.33\overline{3} \approx 23.3\ ^{\circ}\text{C} \] \[ \boxed{T_{mix} = 23.3\ ^{\circ}\text{C}} \]