Chapter 5.5, Problem 83P

An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at 7°C and 105 kPa at a rate of 0.55 m³/s while warm air enters at 34°C and 105 kPa. The air leaves the room at 24°C. The ratio of the mass flow rates of the hot to cold airstreams is 1.6. Using variable specific heats, determine (a) the mixture temperature at the inlet of the room and (b) the rate of heat gain of the room.

FIGURE P5–83

To determine

The mixture temperature at the inlet of the room.

Answer to Problem 83P

The mixture temperature at the inlet of the room is $23.6342°\text{C}$.

Explanation of Solution

Here, the warm and cold streams of fluids 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 cold air stream inlet, 2 indicates the warm air stream inlet and 3 indicates the mixed air 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)

Substitute ${\dot{m}}_1+{\dot{m}}_2$ for ${\dot{m}}_3$.

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2=\left({\dot{m}}_1+{\dot{m}}_2\right)h_3\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2={\dot{m}}_1{h}_3+{\dot{m}}_2{h}_3\\ {\dot{m}}_1{h}_1-{\dot{m}}_1{h}_3={\dot{m}}_2{h}_3-{\dot{m}}_2{h}_2\\ {\dot{m}}_1\left(h_1-h_3\right)={\dot{m}}_2\left(h_3-h_2\right)\end{array}$ (IV)

Here, the ratio of mass flow rate of hot to cold air streams is $1.6$.

$\begin{array}{l}\frac{{\dot{m}}_2}{{\dot{m}}_1}=1.6\\ {\dot{m}}_2=1.6{\dot{m}}_1\end{array}$

Substitute $1.6{\dot{m}}_1$ for ${\dot{m}}_2$ in Equation (IV).

$\begin{array}{l}{\dot{m}}_1\left(h_1-h_3\right)=1.6{\dot{m}}_1\left(h_3-h_2\right)\\ h_1-h_3=\frac{1.6{\dot{m}}_1\left(h_3-h_2\right)}{{\dot{m}}_1}\\ h_1-h_3=1.6h_3-1.6h_2\\ h_1+1.6h_2=1.6h_3+h_3\end{array}$

$h_3=\frac{h_1+1.6h_2}{1.6}$ (V)

Refer Table A-17, “Ideal-gas properties of air”

Obtain the enthalpy $\left(h_1\right)$ of cold air corresponding to the temperature of $7°\text{C}\left(280\text{K}\right)$.

$h_{@280\text{K}}=h_1=280.13\text{kJ/kg}$

Obtain the enthalpy $\left(h_2\right)$ of warm air corresponding to the temperature of $34°\text{C}\left(307\text{K}\right)$ - using interpolation method.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (VI)

Show the temperature and enthalpy values from the Table A-17 as in below table.

S.No.	x	y
	Temperature $\left(T\right)$,in $\text{K}$	enthalpy $\left(h\right)$, in $\text{kJ/kg}$
1	305	305.22
2	307	?
3	310	310.24

Substitute $305$ for $x_1$, $307$ for $x_2$, $310$ for $x_3$, $305.22$ for $y_1$, and $310.24$ for $y_3$ in Equation (VI).

$\begin{array}{c}{y}_2=\frac{\left(307-305\right)\left(310.24-305.22\right)}{\left(310-305\right)}+305.22\\ =307.23\text{kJ/kg}\end{array}$

Thus, the enthalpy $\left(h_2\right)$ of warm air corresponding to the temperature of $34°\text{C}\left(307\text{K}\right)$ is,

$h_2=307.23\text{kJ/kg}$.

Similarly,

Obtain the enthalpy $\left(h_{room}\right)$ of air leaving the room corresponding to the temperature of $24°\text{C}\left(297\text{K}\right)$ - using interpolation method.

$h_{room}=297.18\text{kJ/kg}$

Conclusion:

Substitute $280.13\text{kJ/kg}$ for $h_1$ and $307.23\text{kJ/kg}$ for $h_2$ in Equation (V).

$\begin{array}{c}{h}_3=\frac{280.13\text{kJ/kg}+1.6\left(307.23\text{kJ/kg}\right)}{2.6}\\ =\frac{771.698}{2.6}\\ =296.8069\text{kJ/kg}\end{array}$

Refer Table A-17, “Ideal-gas properties of air”

Obtain the temperature $\left(T_3\right)$ of mixed air entering the room corresponding to the enthalpy of $296.8069\text{kJ/kg}$ - using interpolation method.

Show the temperature and enthalpy values from the Table A-17 as in below table.

S.No.	x	y
	Enthalpy $\left(h\right)$, in $\text{kJ/kg}$	Temperature $\left(T\right)$,in $\text{K}$
1	295.17	295
2	296.8096	?
3	298.18	298

Substitute $295.17$ for $x_1$, $296.8096$ for $x_2$, $298.18$ for $x_3$, $295$ for $y_1$, and $298$ for $y_3$ in Equation (VI).

$\begin{array}{c}{y}_2=\frac{\left(296.8096-295.17\right)\left(298-295\right)}{\left(298.18-295.17\right)}+295\\ =296.6342\text{K}\end{array}$

The temperature $\left(T_3\right)$ of mixed air entering the room is,

$\begin{array}{c}{T}_3=296.6342\text{K}-273\\ =23.6342°\text{C}\end{array}$

Thus, the mixture temperature at the inlet of the room is $23.6342°\text{C}$.

To determine

The rate of heat gain of the room.

Answer to Problem 83P

The rate of heat gain of the room is $0.6921\text{kW}$.

Explanation of Solution

Write the formula for mass flow rate of cold air $\left({\dot{m}}_1\right)$.

${\dot{m}}_1=\frac{{\dot{V}}_1{P}_1}{R{T}_1}$ (VII)

Here, the volumetric flow rate of air is $\dot{V}$, the pressure is $P$, the gas constant of air is $R$ and the temperature is $T$; the suffix one indicates the inlet condition of cold air.

Express the formula for heat gain by the room.

${\dot{Q}}_{gain}={\dot{m}}_3\left(h_{room}-h_3\right)$ (VIII)

Here, the mass flow rate of mixed air entering the room is ${\dot{m}}_3$, the enthalpy of air leaving the room is $h_{room}$ and the enthalpy of the mixed air entering the room is $h_3$.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant of air is $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$.

Conclusion:

Substitute $0.55{\text{m}}^3/\text{s}$ for ${\dot{V}}_1$, $105\text{kPa}$ for $P_1$, $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$ and $7\text{°C}$ for $T_1$ in Equation (VII).

$\begin{array}{c}{\dot{m}}_1=\frac{\left(0.55{\text{m}}^3/\text{s}\right)\left(105\text{kPa}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(7\text{°C}\right)}\\ =\frac{57.75\text{kPa}\cdot {\text{m}}^3/\text{s}}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(7\text{+273}\right)\text{K}}\\ =\frac{57.75\text{kPa}\cdot {\text{m}}^3/\text{s}}{80.36\text{kPa}\cdot {\text{m}}^3/\text{kg}}\\ =0.7186\text{kg/s}\end{array}$

Substitute $1.6{\dot{m}}_1$ for ${\dot{m}}_2$ and $0.7186\text{kg/s}$ for ${\dot{m}}_1$ in Equation (I).

$\begin{array}{l}0.7186\text{kg/s}+1.6{\dot{m}}_1={\dot{m}}_3\\ 0.7186\text{kg/s}+1.6\left(0.7186\text{kg/s}\right)={\dot{m}}_3\\ 0.7186\text{kg/s}+1.1498={\dot{m}}_3\\ {\dot{m}}_3=1.8684\text{kg/s}\end{array}$

Substitute $1.8684\text{kg/s}$ for ${\dot{m}}_3$, $297.18\text{kJ/kg}$ for $h_{room}$ and $296.8096\text{kJ/kg}$ for $h_3$ in Equation (VIII).

$\begin{array}{c}{\dot{Q}}_{gain}=1.8684\text{kg/s}\left(297.18\text{kJ/kg}-296.8096\text{kJ/kg}\right)\\ =1.8684\text{kg/s}\left(0.3704\text{kJ/kg}\right)\\ =0.6921\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =0.6921\text{kW}\end{array}$

Thus, the rate of heat gain of the room is $0.6921\text{kW}$.