Air compressor, as shown in Figure 5.15, takes in air at atmospheric pressure and at a temperature of 280 K, and the air exiting the compressor leaves at a pressure and a temperature of 600 kPa and 400 K. The air mass flow rate of the air being compressed is 1 kg/s; the compressor loses heat

Question

Air compressor, as shown in Figure 5.15, takes in air at atmospheric pressure and at a temperature of 280 K, and the air exiting the compressor leaves at a pressure and a temperature of 600 kPa and 400 K. The air mass flow rate of the air being compressed is 1 kg/s; the compressor loses heat through its walls equivalent to an amount of 40% of the total work rate running the compressor. (a) Write the mass, energy, entropy, and exergy balance equations, (b) calculate the compressor work input and heat rejection rate, and (c) find the entropy generation rate and exergy destruction rate.

Accepted Answer

One may start with some assumptions: (i) the heat losses occur in the compression process, (ii) the changes in the kinetic and potential energies of air are negligible, and (iii) the air is treated as an ideal gas.

a) Write the mass, energy, entropy, and exergy balance equations for the air compressor.

[latex]MBE : \dot{m}_{1}=\dot{m}_{2}[/latex]

[latex]EBE : \dot{m}_{1} h_{1}+\dot{W}_{ ext {in }}=\dot{m}_{2} h_{2}+\dot{Q}_{ ext {out }}[/latex]

[latex]EnBE : \dot{m}_{1} S_{1}+\dot{S}_{g e n}=\dot{m}_{2} S_{2}+\dot{Q}_{o u t} / T_{b}[/latex]

[latex]ExBE : \dot{m}_{1} e x_{1}+\dot{W}_{ ext {in }}=\dot{m}_{2} e x_{2}+\dot{E} x_{d}+\dot{E} x_{\dot{Q}_{ ext {out }}}[/latex]

b) CalcuThe power consumed during the compression process needs to be calculated in this section. Using the EBE to calculate the specific power consumed by the compressor is obtained as follows:late the compressor work input and heat rejection rate.

[latex]\dot{W}_{ ext {in }}=\dot{Q}_{ ext {out }}+\dot{m}_{2} h_{2}-\dot{m}_{1} h_{1}[/latex]

Here, the properties of the stream entering and leaving the compressor are obtained from air properties tables (such as Appendix E-1) or using the EES, which contains the database for the properties of most of the fluids through a large range of temperatures and pressures, and the properties are tabulated in Table 5.3:

Table 5.3 The properties of the inlet and exit streams.

State Point	P (kPa)	T ( °C)	h (kJ/kg)	s (kJ/kgK)	ex (kJ/kg)
Reference state	101.3	25	298.6	5.696	0
1	100	7·0	280.5	5.637	0.509
2	600	127	401.4	5.482	166.6

The work input rate needed for the compressor is calculated as:

[latex]\dot{W}_{ ext {in }}=0.4 imes \dot{W}_{ ext {in }}+\dot{m}_{2} h_{2}-\dot{m}_{1} h_{1}[/latex]

[latex]\dot{W}_{ ext {in }}=0.4 imes \dot{W}_{ ext {in }}+1 imes 401.4-1 imes 280.5[/latex]

[latex]\dot{W}_{ ext {in }}=201.5 k W[/latex]

Since 40% of this compressor work is lost as through heat rejection, the heat rejection rate becomes:

[latex]\dot{Q}_{ ext {out }}=0.4 imes 201.5 kW =80.6 kW[/latex]

c) Find the entropy generation rate and exergy destruction rate.
Follows the methodology:

[latex]\dot{m}_{1} S_{1}+\dot{S}_{g e n}=\dot{m}_{2} S_{2}+\frac{\dot{Q}_{o u t}}{T_{b}}[/latex]

The entropy change of the system when rearranged is equal to [latex]\dot{S}_{ ext {gen }}[/latex] as follows:

[latex]\dot{S}_{ ext {gen }}=\dot{m}_{2} S_{2}-\dot{m}_{1} S_{1}+\frac{\dot{Q}_{ ext {out }}}{T_{b}}[/latex]

The question provided enough state point properties in Table 5.1 in order to obtain all the entropy values for the state points. One more thing is that the boundary or surface temperature of the turbine is not specified. Thus, it is fine to take the average temperature of the inlet and exit states, 340 K. Hence:

Table 5.1 Mass, energy, entropy, and exergy balance equations for various types of closed systems.
System	Balance equations
Fixed boundary
	[latex] ext { MBE: } m_{1}=m_{2}= ext { constant }[/latex]
	[latex] ext { EBE: } Q_{ ext {in }}+W_{e}+W_{s h}+m_{1} u_{1}=m_{2} u_{2}+Q_{l}[/latex]
	[latex] ext { EnBE: } Q_{ ext {in }} / T_{S}+m_{1} S_{1}+S_{ ext {gen }}=m_{2} S_{2}+Q_{1} / T_{ b }[/latex]
	[latex]\operatorname{ExBE}: E x_{Q_{i n}}+W_{e}+W_{s h}+m_{1} e x_{1}=m_{2} e x_{2}+E x_{Q_{l}}+E x_{d}[/latex]
	with [latex]E x_{Q_{i n}}=\left(1-\frac{T_{o}}{T_{s}} ight) Q_{i n} ; E x_{Q_{l}}=\left(1-\frac{T_{o}}{T_{b}} ight) Q_{l}[/latex]
	[latex]E x_{d}=T_{o} S_{g e n}[/latex]
	where[latex] W_e[/latex] is electrical work, [latex] W_sh[/latex] is shaft work, [latex] T_s [/latex] is source temperature since heat provided from a source, [latex] T_b[/latex] is immediate boundary temperature since heat is leaving the system and rejected into the immediate boundary. One may also use surface temperature or average temperature of the substance [latex]\left(T_{a v}=\left(T_{1}+T_{2} ight) / 2 ight)[/latex]or ambient (or reference state) temperature [latex]\left(T_{a v}=\left(T_{1}+T_{2} ight) / 2 ight)[/latex] or ambient (or reference state) temperature [latex]\left(T_{o} ight)[/latex] if [latex]T_{b}[/latex] is not known.
	[latex] ext { MBE: } m_{w, 1}=m_{w, 2}=m_{w}[/latex]
	[latex]m_{s, 1}=m_{s, 2}=m_{s}[/latex] and
	[latex]m_{c, 1}=m_{c, 2}=m_{c}[/latex]
	[latex] ext { EBE: } m_{w} u_{w, 1}+m_{s} u_{s, 1}+m_{c} u_{c, 1}+W_{ ext {in }}+Q_{ ext {in }}=m_{w} u_{w, 2}+m_{s} u_{s, 2}+m_{c} u_{c, 2}+Q_{o u t}[/latex]
	[latex] ext{EnBE: } m_{w} s_{w, 1}+m_{s} s_{s, 1}+m_{c} s_{c, 1}+Q_{i n}/T_s+S_{gen}= m_{w} s_{w, 2}+m_{s} s_{s, 2}+m_{c} s_{c, 2}+Q_{o u t}/T_b[/latex]
	ExBE: [latex]m_{w} e x_{w, 1}+m_{s} e x_{s, 1}+m_{c} e x_{c, 1}+W_{i n}+E x_{Q_{i n}}=m_{w} e x_{w, 2}+m_{s} e x_{s, 2}+m_{c} e x_{c, 2}+E x_{Q_{o u t}}+E x_{d}[/latex]
	with [latex]E x_{Q_{ ext {in }}}=\left(1-\frac{T_{o}}{T_{s}} ight) Q_{ ext {in }} ; E x_{Q_{ ext {out }}}=\left(1-\frac{T_{o}}{T_{b}} ight) Q_{ ext {out }}[/latex]
	[latex]E x_{d}=T_{o} S_{g e n}[/latex]
	where s denotes steel, c is copper, w is water. Also,[latex]W_{in}[/latex] is shaft work, [latex]T_s[/latex] is source temperature since heat provided from a source, [latex]T_b[/latex] is immediate boundary temperature since [latex]Q_{out}[/latex] is leaving the system and rejected into the immediate boundary. One may also use surface temperature or average temperature of the substance [latex]\left(T_{a u}=\left(T_{1}+T_{2} ight) / 2 ight)[/latex] or ambient (or reference state)temperature ([latex]T_o[/latex]) if [latex]T_b[/latex] is not known. Note that in this rigid tank there are two metallic (steel and copper) balls where their masses are included.
Moving boundary
	[latex] ext { MBE: } m_{1}=m_{2}= ext { constant }[/latex]
	[latex] ext { EBE: } m_{1} u_{1}+W_{e}+W_{ ext {in }}=m_{2} u_{2}+Q_{1}[/latex]
	or one may write this EBE in the following form by considering the boundary conditions directly (where there is no need to include boundary work in the equation):
	[latex] m_{1} u_{1}+m_{1} P v_{1}+W_{e}=m_{2} u_{2}+m_{2} P v_{2}+Q_{1}[/latex]
	Both of these equations result in:
	[latex] m h_{1}+W_{e}=m h_{2}+Q_{1}[/latex]
	with [latex]W_{i n}=W_{b}=m P\left(v_{1}-v_{2} ight)[/latex] and h = u + Pv
	[latex] ext { EnBE: } m_{1} S_{1}+S_{ ext {gen }}=m_{2} S_{2}+\left(Q_{1} / T_{ b } ight)[/latex]
	[latex] ext { ExBE: } W_{e}+W_{i n}+m_{1} e x_{1}=m_{2} e x_{2}+E x_{Q_{l}}+E x_{d}[/latex]
	with [latex]E x_{Q_{l}}=\left(1-\frac{T_{o}}{T_{b}} ight) Q_{l} ; E x_{d}=T_{o} S_{g e n}[/latex]
	where [latex]W_e[/latex] is electrical work, [latex]W_{in} [/latex] is work input (boundary work which changes the boundary) to the system to compress the fluid or gas, [latex] T_b [/latex] is immediate boundary temperature since heat is leaving the system and rejected into the immediate boundary. One may also use surface temperature or average temperature of the substance [latex]\left(T_{a v}=\left(T_{1}+T_{2} ight) / 2 ight)[/latex] or ambient (or reference state) temperature [latex](T_o)[/latex] if[latex] T_b[/latex] is not know.
	[latex] ext { MBE: } m_{1}=m_{2}= ext { constant }[/latex]
	[latex] ext { EBE: } m_{1} u_{1}+Q_{i n}+W_{ e }=m_{2} u_{2}+W_{b}[/latex]
	or one may write this EBE in the following form by considering the boundary conditions directly (where there is no need to include boundary work in the equation):
	[latex]m_{1} u_{1}+m_{1} P v_{1}+Q_{i n}+W_{e}=m_{2} u_{2}+m_{2} P v_{2}[/latex]
	Both of these equations result in:
	[latex]m h_{1}+Q_{i n}+W_{e}=m h_{2}[/latex]
	with [latex]W_{ ext {out }}=W_{b}=m P\left(v_{2}-v_{1} ight) ext { and } h=u+P v[/latex]
	[latex] ext { EnBE: } m_{1} S_{1}+Q_{ ext {in }} / T_{S}+S_{ ext {gen }}=m_{2} S_{2}[/latex]
	[latex] ext { ExBE: } m_{1} e x_{1}+E x_{Q_{i n}}+W_{e}=m_{2} e x_{2}+W_{b}+E x_{d}[/latex]
	with [latex]E x_{Q_{i n}}=\left(1-\frac{T_{o}}{T_{s}} ight) Q_{i n} ; E x_{d}=T_{o} S_{g e n}[/latex]
	where [latex]W_e [/latex] is electrical work, [latex]W_b[/latex] is boundary movement work (which is work done by the system), Ts is source temperature. Note that no heat transfer leaving the system is shown due to the negligible heat losses.
	[latex] ext { MBE: } m_{1}=m_{2}= ext { constant }[/latex]
	[latex] ext { EBE: } m_{1} u_{1}+Q_{ ext {in }}=m_{2} u_{2}+W_{b}[/latex]
	or one may write this EBE in the following form by considering the boundary conditions directly (where there is no need to include boundary work in the equation):
	[latex]m_{1} u_{1}+m_{1} P v_{1}+Q_{i n}=m_{2} u_{2}+m_{2} P v_{2}[/latex]
	Both of these equations result in:
	[latex]m h_{1}+Q_{i n}=m h_{2}[/latex]
	with [latex]W_{o u t}=W_{b}=m P\left(v_{2}-v_{1} ight)[/latex] and h = u + Pv
	[latex] ext { EnBE: } Q_{i n} / T_{S}+m_{1} S_{1}+S_{g e n}=m_{2} S_{2}[/latex]
	[latex] ext { ExBE: } m_{1} e x_{1}+E x_{Q_{i n}}=m_{2} e x_{2}+W_{b}+E x_{d}[/latex]
	with [latex]E x_{Q_{i n}}=\left(1-\frac{T_{o}}{T_{s}} ight) Q_{i n} ; E x_{d}=T_{o} S_{g e n}[/latex]
	where [latex]W_b[/latex] is boundary movement work (which is work done by the system), [latex]T_s [/latex] is source temperature. Note that no heat transfer leaving the system is shown due to the negligible heat losses.

[latex]\dot{S}_{g e n}=1 \frac{k g}{S}\left(5.482-5.637 \frac{k J}{k g K} ight)+\left(\frac{80.6 k W}{340 K} ight)[/latex]

[latex]\dot{S}_{ ext {gen }}=0.082 k W / K[/latex]

In addition, the exergy destruction rate is calculated from the ExBE as follows:

[latex]\dot{m}_{1} e x_{1}+\dot{W}_{i n}=\dot{m}_{2} e x_{2}+\dot{E} x_{d}+\dot{E} x_{\dot{Q}_{ ext {out }}}[/latex]

[latex]1 \frac{ kg }{ s } imes\left(-0.509 \frac{ kJ }{ kg } ight)+201.5 \frac{ kJ }{ s }=1 \frac{ kg }{ s } imes 166.6 \frac{ kJ }{ kg }+\dot{E} x_{d}+\left(1-\frac{298 K }{340 K } ight) imes\left(80.6 \frac{ kJ }{ s } ight)[/latex]

[latex]\dot{E} x_{d}=24.43 kW[/latex]

where [latex]T_o[/latex] is the ambient temperature and [latex]T_b [/latex]is the boundary temperature of system, which can be approximated as the average temperature of the compressor since the heat escapes the compressor from its walls to the surrounding environment. However, the other approximation is to consider the highest temperature in the compressor. In this example, the average temperature of the compressor is taken as the sink (boundary) temperature.

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