A small air compressor is running steadily at an inlet ﬂowrate of Q1 = 4 ft³/min. The inlet conditions for the air are p1 = 14.7 psia, T1 = 70°F, A1 = 2 in.²; and the outlet conditions are p2 = 100 psig = 114.7 psia, T2 = 200°F, A2 = 0.05 in.². If the power required to run the compressor is 2 hp,

Question

A small air compressor is running steadily at an inlet ﬂowrate of Q₁ = 4 ft³/min. The inlet conditions for the air are p₁ = 14.7 psia, T₁ = 70°F, A₁ = 2 in.²; and the outlet conditions are p₂= 100 psig = 114.7 psia, T₂ = 200°F, A₂ = 0.05 in.². If the power required to run the compressor is 2 hp, what is the heat transfer rate between the compressor and the atmosphere?

Accepted Answer

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 [latex]\dot M = ρ_1A_1\bar V_1 = ρ_2A_2\bar V_2[/latex], or equivalently [latex]\dot M = ρ_1Q_1 = ρ_2Q_2[/latex]. 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 [latex]\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[/latex] . For a perfect gas we can introduce the enthalpy h = u + (p/ρ) and use Eq. 2.21b to write [latex]h_2 −h_1 =c_P(T_2 −T_1)[/latex]. Thus the energy balance becomes

[latex]\dot M\left[\left(u+\frac{p}{ ho }+\frac{1}{2}\bar V^2+gz ight)_{out}−\left(u+\frac{p}{ ho }+\frac{1}{2}\bar V^2+gz ight)_{in} ight] =\dot W_{power}+\dot W_{shaft}+\dot Q_C+\dot S[/latex] (7.36)

[latex]\dot W_{shaft} =\dot M \left[c_P(T_2 −T_1)+ \frac{1}{2} \left(\bar V^2_ 2 −\bar V^2_ 1 ight) ight] − \dot Q_C[/latex] (A)

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

[latex]\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 ight) ight]-\frac{\dot Q_C}{\dot M}[/latex] (B)

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

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

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

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

[latex]\bar V_1=\frac{\dot M}{ ho _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.} ight)^2}=4.8\ \mathrm{ft} /s[/latex]

[latex]\bar V_2=\frac{\dot M}{ ho _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.} ight)^2}=30.6\ \mathrm{ft} /s[/latex]

The individual terms in (B) are now calculated as

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

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

[latex]\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} } ight)\left(\frac{1\ Btu}{778\ \mathrm{ft} -lb_\mathrm{f} } ight)=4.6 imes 10^{-4}\ Btu/lb_m[/latex]

[latex]\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} } ight)\left(\frac{1\ Btu}{778\ \mathrm{ft}-lb_\mathrm{f} } ight)=1.9 imes 10^{-2}\ Btu/lb_m[/latex]

[latex]\frac{1}{2}\left(\bar V^2_2-\bar V^2_1 ight)[/latex] = (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

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

where

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

Thus we ﬁnd

[latex]\dot Q_C[/latex] = 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 [latex]\dot Q_C[/latex] 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

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

Introducing the perfect gas relationships, we have

[latex]\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 ight) ight] − \dot Q_C[/latex] (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

[latex]\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}[/latex]

where we have made use of the perfect gas relation Eq. 2.23c to write c_V= R/([latex]\gamma [/latex] − 1). Since the speciﬁc heat ratio for air is [latex]\gamma [/latex] = 1.4, we ﬁnd 1/([latex]\gamma [/latex] − 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

			In SI Units: J/(kg-K)			In BG Units: (ft-lb_f)/(slug-°R)
Gas	MW	[latex]\gamma [/latex]	R	[latex]c_p[/latex]	[latex]c_v[/latex]	R	[latex]c_p[/latex]	[latex]c_v[/latex]
Air	29	1.40	287	1,004	717	1,716	6,006	4,290
CO₂	44	1.30	189	819	630	1,130	4,897	3,767
He	4	1.66	2077	5,224	3,147	12,420	31,238	18,818
H₂	2	1.41	4124	14,180	10,060	24,660	84,806	60,146
Methane	16	1.31	518	2,190	1,672	3,099	13,096	9,997
N₂	28	1.40	297	1,039	742	1,775	6,213	4,438
O₂	32	1.40	260	909	650	1,130	5,439	3,885

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

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