ChemicalEnergy BalanceseasyFE ExamCoursework

Temperature of Two Mixed Streams

Two streams of the same liquid mix adiabatically at steady state:

Stream 1: $3.0\ \text{kg/s}$ at $80^\circ\text{C}$
Stream 2: $2.0\ \text{kg/s}$ at $20^\circ\text{C}$

Specific heat is constant and equal for both. Find the outlet temperature.

Given

m1	3 kg/s	Stream 1 flow
T1	80 degC	Stream 1 temperature
m2	2 kg/s	Stream 2 flow
T2	20 degC	Stream 2 temperature

Hint 1

Write an energy balance. With equal specific heats, what cancels?

Hint 2

The result is a mass-weighted average of the inlet temperatures.

Hint 3

The hot stream is larger, so the answer must exceed the simple midpoint of 50 °C.

Worked solution — try the problem first

Energy balance, adiabatic and steady, with constant equal $c_p$ :

$\dot m_1 c_p T_1 + \dot m_2 c_p T_2 = (\dot m_1 + \dot m_2)c_p T_{out}$

$c_p$ cancels throughout, leaving a mass-weighted average:

$T_{out} = \frac{\dot m_1 T_1 + \dot m_2 T_2}{\dot m_1 + \dot m_2} = \frac{3.0(80) + 2.0(20)}{5.0}$

$= \frac{240 + 40}{5.0} = \frac{280}{5.0} = 56^\circ\text{C}$

$\boxed{T_{out} = 56^\circ\text{C}}$

Not 50°C. The arithmetic mean would apply only to equal flows. The hot stream is larger, so the mixture sits above the midpoint — and 56 > 50, as expected.

Celsius is safe here because the relation is linear and the units cancel; the same computation in kelvin gives 329.15 K = 56°C.

Concepts:Energy balancesAdiabatic mixingWeighted averages

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