A steady flow machine (Fig. E3.20) takes in air at section 1 and discharges it at sections 2 and 3. The properties at each section are as follows:
| Section | A, ft^2 | Q, ft^3/s | T, °F | p, lbf/in^2 abs | z, ft |
| 1 | 0.4 | 100 | 70 | 20 | 1.0 |
| 2 | 1.0 | 40 | 100 | 30 | 4.0 |
| 3 | 0.25 | 50 | 200 | ? | 1.5 |
Work is provided to the machine at the rate of 150 hp. Find the pressure p_3 in lbf/in^2 absolute and the heat transfer \dot{Q} in Btu/s. Assume that air is a perfect gas with R = 1716 and c_p = 6003 ft-lbf/(slug · °R).
• System sketch: Figure E3.20 shows inlet 1 (negative flux) and outlets 2 and 3 (positive fluxes).
• Assumptions: Steady flow, one-dimensional inlets and outlets, ideal gas, negligible shear work. The flow is not incompressible. Note that Q_1 ≠ Q_2 + Q_3 because the densities are different.
• Approach: Evaluate the velocities and densities and enthalpies and substitute into Eq. (3.67). Use BG units for all properties, including the pressures. With Q_i given, we evaluate V_i = Q_i/A_i:
V_1 = \frac{Q_1}{A_1} = \frac{100 ft^3/s}{0.4 ft^2} = 250 \frac{ft}{s} V_2 = \frac{40 ft^3/s}{1.0 ft^2} = 40 \frac{ft}{s} V_3 = \frac{50 ft^3/s}{0.25 ft^2} = 200 \frac{ft}{s}\dot{Q} – \dot{W}_s – \dot{W}_{\nu} = \frac{\partial}{\partial t} \left[\int_{CV}{\left(\hat{u} + \frac{1}{2} V^2 + gz\right)\rho d^{\circ}\mathcal{V}}\right] +\int_{CS}{\left(\hat{h} + \frac{1}{2}V^2 + gz\right) \rho(V \cdot n) dA} (3.67)
The densities at sections 1 and 2 follow from the ideal gas law:
\rho_1 = \frac{p_1}{RT_1} = \frac{(20 \times 144) lbf/ft^2}{[1716 ft-lbf/(slug^{\circ}R)] [(70 + 460)^{\circ}R]} = 0.00317 \frac{slug}{ft^3}\rho_2 = \frac{(30 \times 144)}{(1716)(100 + 460)} = 0.00450 \frac{slug}{ft^3}
However, p_3 is unknown, so how do we find \rho_3? Use the steady flow continuity relation:
\dot{m}_1 = \dot{m}_2 + \dot{m}_3 or \rho_1 Q_1 = \rho_2 Q_2 + \rho_3 Q_3 (1)
\left(0.00317 \frac{slug}{ft^3}\right) \left(100 \frac{ft^3}{s}\right) = 0.00450(40) + \rho_3(50) solve for \rho_3 = 0.00274 \frac{slug}{ft^3}Knowing \rho_3 enables us to find p_3 from the ideal-gas law:
p_3 = \rho_3 RT_3 = \left(0.00274 \frac{slug}{ft^3}\right) \left(1716 \frac{ft-lbf}{slug^{\circ}R}\right) (200 + 460^{\circ}R) = 3100 \frac{lbf}{ft^2} = 21.5 \frac{lbf}{in^2}• Final solution steps: For an ideal gas, simply approximate enthalpies as h_i = c_p T_i. The shaft work is negative (into the control volume) and viscous work is neglected for this solid-wall machine:
\dot{W}_{\nu} \approx 0 \dot{W}_s = (-150 hp)\left(550 \frac{ft-lbf}{s-hp}\right) = -82,500 \frac{ft-lbf}{s} (work on the system)
For steady flow, the volume integral in Eq. (3.67) vanishes, and the energy equation becomes
\dot{Q} – \dot{W}_s = -\dot{m}_1 (c_p T_1 + \frac{1}{2} V^2_1 + gz_1) + \dot{m}_2 (c_p T_2 + \frac{1}{2}V^2_2 + gz_2) + \dot{m}_3 (c_p T_3 + \frac{1}{2} V^2_3 + gz_3) (2)
From our continuity calculations in Eq. (1) above, the mass flows are
\dot{m}_1 = \rho_1 Q_1 = (0.00317)(100) = 0.317\frac{slug}{s} \dot{m}_2 = \rho_2 Q_2 = 0.180 \frac{slug}{s} \dot{m}_3 = \rho_3 Q_3 = 0.137\frac{slug}{s}It is instructive to separate the flux terms in the energy equation (2) for examination:
Enthalpy flux = c_p(-\dot{m}_1 T_1 + \dot{m}_2 T_2 + \dot{m}_3 T_3) = (6003)[(-0.317)(530) + (0.180)(560) + (0.137)(660)] = -1,009,000 + 605,000 + 543,000 ≈ +139,000 ft-lbf/s
Kinetic energy flux = \frac{1}{2}(-\dot{m}_1 V^2_1 + \dot{m}_2 V^2_2 + \dot{m}_3 V^2_3) = \frac{1}{2} [-0.317(250)^2 + (0.180)(40)^2 + (0.137)(200)^2] = -9900 + 140 + 2740 \approx -7000 ft-lbf/s
Potential energy flux = g(-\dot{m}_1 z_1 + \dot{m}_2 z_2 + \dot{m}_3 z_3) = (32.2)[-0.317(1.0) + 0.180(4.0) + 0.137(1.5)] = -10 + 23 + 7 \approx +20 ft-lbf/s
Equation (2) may now be evaluated for the heat transfer:
\dot{Q} – (-82,500) = 139,000 – 7,000 + 20
or \dot{Q} \approx \left(+49,520 \frac{ft-lbf}{s}\right) \left(\frac{1 Btu}{778.2 ft-lbf}\right) = +64 \frac{Btu}{s}
• Comments: The heat transfer is positive, which means into the control volume. It is typical of gas flows that potential energy flux is negligible, enthalpy flux is dominant, and kinetic energy flux is small unless the velocities are very high (that is, high subsonic or supersonic).