Question 6.E.83: In a steam heating system, air is heated by being passed ove......
In a steam heating system, air is heated by being passed over some tubes through which steam flows steadily. Steam enters the heat exchanger at 30 \mathrm{psia} and 400^{\circ} \mathrm{F} at a rate of 15 \mathrm{lbm} / \mathrm{min} and leaves at 25 psia and 212^{\circ} \mathrm{F}. Air enters at 14.7 \mathrm{psia} and 80^{\circ} \mathrm{F} and leaves at 130^{\circ} \mathrm{F}. Determine the volume flow rate of air at the inlet.
Air is heated in a steam heating system. For specified flow rates, the volume flow rate of air at the inlet is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. \mathrm{3} There are no work interactions. \mathrm{4} Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature.
Properties The gas constant of air is 0.3704 \mathrm{psia}^{\mathrm{ft}} / \mathrm{lbm}.R (Table A-1E). The constant pressure specific heat of air is c_{\mathrm{p}}=0.240 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F} (Table A-2E). The enthalpies of steam at the inlet and the exit states are (Tables A-4E through A-6E)
\begin{aligned} & \left.\begin{array}{l} P_{3}=30 \mathrm{psia} \\ T_{3}=400^{\circ} \mathrm{F} \end{array}\right\} h_{3}=1237.9 \mathrm{Btu} / \mathrm{lbm} \\ & \left.\begin{array}{l} P_{4}=25 \mathrm{psia} \\ T_{4}=212^{\circ} \mathrm{F} \end{array}\right\} h_{4} \cong h_{f @ 212^{\circ} \mathrm{F}}=180.21 \mathrm{Btu} / \mathrm{lbm} \end{aligned}
Analysis We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as
Mass balance ( for each fluid stream):
\begin{aligned} \dot{m}_{\text {in }}-\dot{m}_{\text {out }} & =\Delta \dot{m}_{\text {system }} \stackrel{\swarrow 0 (\text { steady)}}{ }=0 \\ \dot{m}_{\text {in }} & =\dot{m}_{\text {out }} \\ \dot{m}_{1} & =\dot{m}_{2}=\dot{m}_{a} \text { and } \dot{m}_{3}=\dot{m}_{4}=\dot{m}_{s} \end{aligned}
Energy balance (for the entire heat exchanger):
\begin{aligned} \underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c} \text { Rate of net energy transfer } \\ \text { by heat, work, and mass } \end{array}} & =\underbrace{\Delta \dot{E}_{\text {system }} {}^{\nearrow 0 \text { (steady) }}}_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc. energies } \end{array}}=0 \\ \dot{E}_{\text {in }} & =\dot{E}_{\text {out }} \\ \dot{m}_{1} h_{1}+\dot{m}_{3} h_{3} & =\dot{m}_{2} h_{2}+\dot{m}_{4} h_{4} \quad(\text { since } \dot{Q}=\dot{W}=\Delta \mathrm{ke} \cong \Delta \mathrm{pe} \cong 0) \end{aligned}
Combining the two,
\dot{m}_{a}\left(h_{2}-h_{1}\right)=\dot{m}_{s}\left(h_{3}-h_{4}\right)
Solving for \dot{m}_{a} :
\dot{m}_{a}=\frac{h_{3}-h_{4}}{h_{2}-h_{1}} \dot{m}_{s} \cong \frac{h_{3}-h_{4}}{c_{p}\left(T_{2}-T_{1}\right)} \dot{m}_{s}
Substituting,
\dot{m}_{a}=\frac{(1237.9-180.21) \mathrm{Btu} / \mathrm{lbm}}{\left(0.240 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)(130-80)^{\circ} \mathrm{F}}(15 \mathrm{lbm} / \mathrm{min})=1322 \mathrm{lbm} / \mathrm{min}=22.04 \mathrm{lbm} / \mathrm{s}
Also,
{v}_{1}=\frac{R T_{1}}{P_{1}}=\frac{\left(0.3704 \mathrm{psia} \cdot \mathrm{ft}^{3} / \mathrm{lbm} \cdot \mathrm{R}\right)(540 \mathrm{R})}{14.7 \mathrm{psia}}=13.61 \mathrm{ft}^{3} / \mathrm{lbm}
Then the volume flow rate of air at the inlet becomes
\dot{v}_{1}=\dot{m}_{a} {v}_{1}=(22.04 \mathrm{lbm} / \mathrm{s})\left(13.61 \mathrm{ft}^{3} / \mathrm{lbm}\right)=\mathrm{3 0 0} \mathrm{ft}^{3} / \mathrm{s}
