Chapter 5.5, Problem 88P

Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surroundings. The mixing process takes place at constant pressure with no work and negligible changes in kinetic and potential energies. Assume the gas has constant specific heats.

1. (a)   Determine the expression for the final temperature of the mixture in terms of the rate of heat transfer to the mixing chamber and the inlet and exit mass flow rates.
2. (b)   Obtain an expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber.
3. (c)   For the special case of adiabatic mixing, show that the exit volume flow rate is the sum of the two inlet volume flow rates.

To determine

The expression for the final temperature of the mixture in terms of the rate of the heat transfer to the mixing chamber and the inlet and exit mass flow rate.

Answer to Problem 88P

The expression for the final temperature of the mixture in terms of the rate of the heat transfer to the mixing chamber and the inlet and exit mass flow rate is shown below.

$T_3=\frac{{\dot{m}}_1{T}_1}{{\dot{m}}_3}+\frac{{\dot{m}}_2{T}_2}{{\dot{m}}_3}+\frac{{\dot{Q}}_{in}}{{\dot{m}}_3{c}_p}$

Explanation of Solution

Here, the two streams (comparatively hot and cold) of ideal gases are mixed in a rigid mixing chamber and operates at steady state. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1+{\dot{m}}_2={\dot{m}}_3$ (I)

Write the energy rate balance equation for two inlet and one outlet system.

$\left\{\begin{array}{l}\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+{\dot{m}}_1\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]\\ +\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+{\dot{m}}_2\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]\\ -\left[{\dot{Q}}_{\text{3}}+{\dot{W}}_{\text{3}}+{\dot{m}}_3\left(h_3+\frac{V_3{}^2}{2}+g{z}_3\right)\right]\end{array}\right\}=\Delta {\dot{E}}_{\text{system}}$ (II)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 indicates the hot water stream inlet, 2 indicates the cold water stream inlet and 3 indicates the mixed water stream outlet.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Neglect the heat transfer, work transfer, kinetic and potential energies.

The Equation (II) reduced as follows.

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2-{\dot{m}}_3{h}_3=0\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\end{array}$ (III)

It is given that the mixing chamber receives energy by heat transfer from the surrounding. Then the Equation (III) will become as follows.

${\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2+{\dot{Q}}_{in}={\dot{m}}_3{h}_3$ (IV)

The enthalpy is expressed as follows.

$h=c_{p}T$

Here, the specific heat is $c_p$ and the temperature is $T$.

Rewrite the Equation (IV) in terms of specific heat and temperature.

${\dot{m}}_1{c}_p{T}_1+{\dot{m}}_2{c}_p{T}_2+{\dot{Q}}_{in}={\dot{m}}_3{c}_p{T}_3$ (V)

Rearrange the Equation (V) to obtain the exit temperature $\left(T_3\right)$ of final mixture.

$\begin{array}{l}{\dot{m}}_1{c}_p{T}_1+{\dot{m}}_2{c}_p{T}_2+{\dot{Q}}_{in}={\dot{m}}_3{c}_p{T}_3\\ T_3=\frac{{\dot{m}}_1{c}_p{T}_1+{\dot{m}}_2{c}_p{T}_2+{\dot{Q}}_{in}}{{\dot{m}}_3{c}_p}\\ T_3=\frac{{\dot{m}}_1{c}_p{T}_1}{{\dot{m}}_3{c}_p}+\frac{{\dot{m}}_2{c}_p{T}_2}{{\dot{m}}_3{c}_p}+\frac{{\dot{Q}}_{in}}{{\dot{m}}_3{c}_p}\\ T_3=\frac{{\dot{m}}_1{T}_1}{{\dot{m}}_3}+\frac{{\dot{m}}_2{T}_2}{{\dot{m}}_3}+\frac{{\dot{Q}}_{in}}{{\dot{m}}_3{c}_p}\end{array}$

Thus, the expression for the final temperature of the mixture in terms of the rate of the heat transfer to the mixing chamber and the inlet and exit mass flow rate is shown below.

$T_3=\frac{{\dot{m}}_1{T}_1}{{\dot{m}}_3}+\frac{{\dot{m}}_2{T}_2}{{\dot{m}}_3}+\frac{{\dot{Q}}_{in}}{{\dot{m}}_3{c}_p}$

To determine

The expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber.

Answer to Problem 88P

The expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber is shown below.

${\dot{V}}_3={\dot{V}}_1+{\dot{V}}_2+\frac{R{\dot{Q}}_{in}}{P{c}_p}$

Explanation of Solution

Write the formula for exit volume flow rate $\left({\dot{V}}_3\right)$.

${\dot{V}}_3=\frac{{\dot{m}}_{3}R{T}_3}{P_3}$ (VI)

Here, the mass flow rate is $\dot{m}$, the gas constant is $R$, the temperature is $T$ and the pressure is $P$; the suffix 3 indicates the outlet stream.

Refer part (a) answer.

Substitute $\frac{{\dot{m}}_1{T}_1}{{\dot{m}}_3}+\frac{{\dot{m}}_2{T}_2}{{\dot{m}}_3}+\frac{{\dot{Q}}_{in}}{{\dot{m}}_3{c}_p}$ for $T_3$ in Equation (VI).

$\begin{array}{c}{\dot{V}}_3=\frac{{\dot{m}}_{3}R}{P_3}\left(\frac{{\dot{m}}_1{T}_1}{{\dot{m}}_3}+\frac{{\dot{m}}_2{T}_2}{{\dot{m}}_3}+\frac{{\dot{Q}}_{in}}{{\dot{m}}_3{c}_p}\right)\\ =\frac{{\dot{m}}_{1}R{T}_1}{P_3}+\frac{{\dot{m}}_{2}R{T}_2}{P_3}+\frac{R{\dot{Q}}_{in}}{P_3{c}_p}\end{array}$ (VII)

Here, the mixing occurs at constant pressure.

$P_1=P_2=P_3=P$

Rewrite the Equation (VII) as follows.

${\dot{V}}_3=\frac{{\dot{m}}_{1}R{T}_1}{P_1}+\frac{{\dot{m}}_{2}R{T}_2}{P_2}+\frac{R{\dot{Q}}_{in}}{P_3{c}_p}$ (VIII)

From Equation (VIII),

$\frac{{\dot{m}}_{1}R{T}_1}{P_1}$ is the volumetric flow rate  of stream-1, i.e. ${\dot{V}}_1=\frac{{\dot{m}}_{1}R{T}_1}{P_1}$.

$\frac{{\dot{m}}_{2}R{T}_2}{P_2}$ is the volumetric flow rate  of stream-2, i.e. ${\dot{V}}_2=\frac{{\dot{m}}_{2}R{T}_2}{P_2}$.

Hence, Substitute ${\dot{V}}_1$ for $\frac{{\dot{m}}_{1}R{T}_1}{P_1}$, ${\dot{V}}_2$ for $\frac{{\dot{m}}_{2}R{T}_2}{P_2}$ and $P$ for $P_3$ in Equation (VIII).

${\dot{V}}_3={\dot{V}}_1+{\dot{V}}_2+\frac{R{\dot{Q}}_{in}}{P{c}_p}$

Thus, the expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber is shown below.

${\dot{V}}_3={\dot{V}}_1+{\dot{V}}_2+\frac{R{\dot{Q}}_{in}}{P{c}_p}$

To determine

To show that the exit volume flow rate is the sum of the two inlet volume flow rates for the adiabatic process.

Answer to Problem 88P

The exit volume flow rate is the sum of the two inlet volume flow rates.

${\dot{V}}_3={\dot{V}}_1+{\dot{V}}_2$

Explanation of Solution

Refer part (b) answer.

The exit volume flow rate is,

${\dot{V}}_3={\dot{V}}_1+{\dot{V}}_2+\frac{R{\dot{Q}}_{in}}{P{c}_p}$ (IX)

When, the mixing process is said to be an adiabatic process, the rate of heat in and out of the system become negligible.

${\dot{Q}}_{in}={\dot{Q}}_{out}=0$

Substitute $0$ for ${\dot{Q}}_{in}$ in Equation (IX).

$\begin{array}{c}{\dot{V}}_3={\dot{V}}_1+{\dot{V}}_2+\frac{R\left(0\right)}{P{c}_p}\\ ={\dot{V}}_1+{\dot{V}}_2+0\\ ={\dot{V}}_1+{\dot{V}}_2\end{array}$

Thus, the exit volume flow rate is the sum of the two inlet volume flow rates.

${\dot{V}}_3={\dot{V}}_1+{\dot{V}}_2$