Question 7.16: A small air compressor is running steadily at an inlet flowra...

We must determine the heat transfer rate for a compressor operating under the given conditions. Figure 7.28 shows a fixed CV containing the compressor and the air within it. We will assume that all property distributions at the inlet and outlet ports are uniform, that potential energy changes are negligible, and that the air may be treated as a perfect gas. For this steady flow device, a mass balance gives \dot M = ρ_1A_1\bar V_1 = ρ_2A_2\bar V_2, or equivalently \dot M = ρ_1Q_1 = ρ_2Q_2. Upon applying Eq. 7.36, and noting that for this mixed CV we have a shaft power input and heat transfer but no fluid power or other energy input, we write \dot M[(u_2 + p_2/ρ_2 +\frac{1}{2}\bar V^2_ 2 )−(u_1 + p_1/ρ_1 +\frac{1}{2}\bar V^2_1 )]=\dot W_{shaft}+\dot Q_C . For a perfect gas we can introduce the enthalpy h = u + (p/ρ) and use Eq. 2.21b to write h_2 −h_1 =c_P(T_2 −T_1). Thus the energy balance becomes

We can also write this result on a per-unit- mass basis of air passing through the machine as

\frac{\dot{ W}_{shaft}}{\dot M}=\left[c_P(T_2 −T_1)+\frac{1}{2}\left(\bar V^2_2-\bar V^2_1\right) \right]-\frac{\dot Q_C}{\dot M}                                (B)

Using thermodynamic properties for air from Table 2–4, and the given data, we now calculate

\rho _1=\frac{p_1}{RT_1}=\frac{(14.7\ lb_\mathrm{f} /in.^2)\left(\frac{12\ in.}{1\ \mathrm{ft} }\right) ^2}{[53.3(\mathrm{ft} -lb_\mathrm{f} )/(lb_\mathrm{f} -°R)][(70+460)°R]}=7.49×10^{−2}\ lb_m/\mathrm{ft} ^3

\rho _2=\frac{p_2}{RT_2}=\frac{(114.7\ lb_\mathrm{f} /in.^2)\left(\frac{12\ in.}{1\ \mathrm{ft} }\right) ^2}{[53.3(\mathrm{ft} -lb_\mathrm{f} )/(lb_\mathrm{f} -°R)][(200+460)◦R]}=0.470\ lb_m/\mathrm{ft} ^3

\dot M = ρ_1Q_1 = (7.49×10^{−2}\ lb_m/\mathrm{ft} ^3)\left(\frac{4\ \mathrm{ft} ^3}{min} \right)\left(\frac{1\ min}{60\ s} \right)=5.0×10^{−3}\ lb_m/s

\bar V_1=\frac{\dot M}{\rho _1A_1}=\frac{5.00×10^{−3}\ lb_m/s }{(7.49×10^{−2}\ lb_m/\mathrm{ft} ^3)(2\ in.^2)\left(\frac{1\ \mathrm{ft} }{12\ in.} \right)^2}=4.8\ \mathrm{ft} /s

\bar V_2=\frac{\dot M}{\rho _2A_2}=\frac{5.00×10^{−3}\ lb_m/s }{(0.470\ lb_m/\mathrm{ft} ^3)(0.05\ in.^2)\left(\frac{1\ \mathrm{ft} }{12\ in.} \right)^2}=30.6\ \mathrm{ft} /s

The individual terms in (B) are now calculated as

\frac{\dot {W}_{shaft}}{\dot M}=\frac{(2\ hp)\left[\frac{550(\mathrm{ft} -lb_\mathrm{f} )/s}{1\ hp}\right]\left(\frac{1\ Btu}{778\ \mathrm{ft} -lb_\mathrm{f}} \right) }{5.00×10^{−3}\ lb_m/s}=283\ Btu/lb_m

c_P(T₂ − T₁) = [0.241 Btu/(lb_m-°R)][(660 − 530)(°R)] = 31.3 Btu/lb_m

\frac{1}{2}\bar V^2_1=\frac{1}{2}(4.8\ \mathrm{ft} /s)^2\left(\frac{1\ lb_f-s^2}{32.2\ lb_m-\mathrm{ft} } \right)\left(\frac{1\ Btu}{778\ \mathrm{ft} -lb_\mathrm{f} } \right)=4.6\times 10^{-4}\ Btu/lb_m

\frac{1}{2}\bar V^2_2=\frac{1}{2}(30.6\ \mathrm{ft} /s)^2\left(\frac{1\ lb_\mathrm{f} -s^2}{32.2\ lb_m-\mathrm{ft} } \right)\left(\frac{1\ Btu}{778\ \mathrm{ft}-lb_\mathrm{f} } \right)=1.9\times 10^{-2}\ Btu/lb_m

\frac{1}{2}\left(\bar V^2_2-\bar V^2_1\right) = (1.9 × 10⁻² Btu/lb_m − 4.6 × 10⁻⁴ Btu/lb_m) = 1.85 × 10⁻² Btu/lb_m

It is evident that the kinetic energy terms are very small compared with the enthalpy and shaft work terms. You will find that in most problems involving gas flows, the kinetic energy terms are negligible unless velocities are very high.

Using (A) to calculate the heat transfer rate, we have

\dot Q_C=\dot M\left[c_P(T_2 −T_1)\frac{1}{2}\left(\bar V^2_2-\bar V^2_1\right)\right]-\dot W_{shaft}

where

\dot W_{shaft}=(2\ hp)\left[\frac{550(\mathrm{ft} -lb_\mathrm{f} )/s}{1\ hp}\right]\left(\frac{1\ Btu}{778\ \mathrm{ft} -lb_\mathrm{f} } \right)=1.41\ Btu/s.

Thus we find

\dot Q_C = 5.00 × 10⁻³ lb_m/s[31.3 Btu/lb_m + (1.85 × 10⁻²) Btu/lb_m] − (1.41 Btu/s)

= − 1.25 Btu/s

Since \dot Q_C is negative, heat is leaving the CV. It is worthwhile to consider a slightly different approach to this problem by rearranging the original energy balance to obtain

\dot M \left[(u_2 −u_1)+\left(\frac{p_2}{\rho _2}-\frac{p_1}{\rho _1} \right) +\frac{1}{2} \left(\bar V^2_ 2 −\bar V^2_ 1\right) \right]=\dot W_{shaft}+ \dot Q_C

Introducing the perfect gas relationships, we have

\dot W_{shaft} =\dot M \left[c_V(T_2 −T_1)+R(T_2 −T_1)+\frac{1}{2} \left(\bar V^2_ 2 −\bar V^2_ 1\right) \right] − \dot Q_C                        (C)

The first two terms on the right-hand side represent the internal energy and pressure potential energy increase in the gas. We see from this equation that the ratio of the internal energy increase to the pressure potential energy increase is

\frac{\dot M c_V(T_2 −T_1)}{\dot M R(T_2 −T_1)}=\frac{c_V}{R}=\frac{R/(\gamma -1)}{R}=\frac{1}{\gamma -1}

where we have made use of the perfect gas relation Eq. 2.23c to write c_V = R/(\gamma − 1). Since the specific heat ratio for air is \gamma = 1.4, we find 1/(\gamma − 1) = 2.5. Now recall that of the shaft work input of 1.41 Btu/s, 1.25 Btu/s escapes as heat. Neglecting the negligible kinetic energy increase, of the remaining 0.156 Btu/s, an amount [(2.5/3.5)(0.156 Btu/s) = 0.111 Btu/s] increases the internal energy of the gas, and only the remainder [(1/3.5)(0.156 Btu/s) = 0.045 Btu/s] increases the pressure potential energy of the gas. Unlike the case of a liquid, some of the internal energy of a gas can be extracted by devices to produce useful work.
TABLE 2.4 Perfect Gas Parameters for Several Common Gases