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

When two streams of the same fluid mix adiabatically in a steady flow with no work done, energy conservation means the total enthalpy carried in by both streams must equal the enthalpy carried out by the combined stream. There is no heat loss to account for since both tubes are insulated.

For an ideal gas with constant $C_p$, the enthalpy flow of a stream is just $\dot{m} C_p T$. Writing the balance for the two inlet streams and the single outlet stream gives $\dot{m}_1 C_p T_1 + \dot{m}_2 C_p T_2 = (\dot{m}_1 + \dot{m}_2) C_p T_{mix}$, and since $C_p$ appears in every term it drops out completely.

What remains is a mass flow weighted average of temperature. Plugging in $\dot{m}_1 = 0.1$ kg/s at $20$ C and $\dot{m}_2 = 0.2$ kg/s at $25$ C gives $T_{mix} = (0.1 \times 20 + 0.2 \times 25)/(0.1+0.2) = 7/0.3$.

Working that out gives $T_{mix} = 23.33$ C, which rounds to $23.3$ C, the steady state temperature of the combined air stream.