Exergy Analysis of Vapor-Compression Refrigeration Cycle A vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is used to maintain a space at -13° C by rejecting heat to ambient air at 27°C. R-134a enters the compressor at 100 kPa superheated by 6.4°C at a rate of 0.05

Question

Exergy Analysis of Vapor-Compression Refrigeration Cycle
A vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is used to maintain a space at -13° C by rejecting heat to ambient air at 27°C. R-134a enters the compressor at 100 kPa superheated by 6.4°C at a rate of 0.05 kg/s. The isentropic efficiency of the compressor is 8 5 percent.
The refrigerant leaves the condenser at 39.4°C as a saturated liquid. Determine (a) the rate of cooling provided and the COP of the system, (b) the exergy destruction in each basic component, (c) the minimum power input and the second-law efficiency of the cycle, and id) the rate of total exergy destruction.

Accepted Answer

A vapor-compression refrigeration cycle is considered. The cooling rate, the COP, the exergy destructions, the minimum power input, the second law efficiency, and the total exergy destruction are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) The T-s diagram of the cycle is given in Fig. 11-10 . The properties of R -134a are (Tables A-11 through A-13)

[latex]\left.\begin{array}{rl}P_{1} & =100 \mathrm{kPa} \\quad\T_{1} & =T_{ ext {sat@ } 100 \mathrm{kPa}}+\Delta T_{ ext {superheat }} \\quad\& =-26.4+6.4=-20^{\circ} \mathrm{C}\end{array} ight\} \qquad\begin{aligned}&h_{1}=239.50 \mathrm{~kJ} / \mathrm{kg} \\quad\&s_{1}=0.9721 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\end{aligned}[/latex]

[latex]P_{3}=P_{\mathrm{sat} @ 39.4^{\circ} \mathrm{C}}=1000 \mathrm{kPa}[/latex]

[latex]\left.\begin{array}{l}P_{2}=P_{3}=1000 \mathrm{kPa} \\quad\s_{2}=s_{1}=0.9446 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\end{array} ight\} \qquad h_{2 s}=289.11 \mathrm{~kJ} / \mathrm{kg}[/latex]

[latex]\left.\begin{array}{l}P_{3}=1000 \mathrm{kPa} \\quad\x_{3}=0\end{array} ight\} \qquad\begin{aligned}&h_{3}=107.32 \mathrm{~kJ} / \mathrm{kg} \\quad\&s_{3}=0.3919 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\end{aligned}[/latex]

[latex]h_{4}=h_{3}=107.32 \mathrm{~kJ} / \mathrm{kg}[/latex]

[latex]\left.\begin{array}{l}P_{4}=100 \mathrm{kPa} \\quad\h_{4}=107.32 \mathrm{~kJ} / \mathrm{kg}\end{array} ight\}\qquad s_{4}=0.4367 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}[/latex]

From the definition of isentropic efficiency,

[latex]\eta_{C} =\frac{h_{2 s}-h_{1}}{h_{2}-h_{1}} [/latex]

[latex]0.85 =\frac{289.11-239.50}{h_{2}-239.50} \longrightarrow h_{2}=297.86 \mathrm{~kJ} / \mathrm{kg}[/latex]

[latex]\left.\begin{array}{l}P_{2}=1000 \mathrm{kPa} \\quad\h_{2}=297.86 \mathrm{~kJ} / \mathrm{kg}\end{array} ight\} \qquad s_{2}=0.9983 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}[/latex]

The refrigeration load, the rate of heat rejected, and the power input are

[latex]\dot{Q}_{L} =\dot{m}\left(h_{1}-h_{4} ight)=(0.05 \mathrm{~kg} / \mathrm{s})(239.50-107.32) \mathrm{kJ} / \mathrm{kg}=6.609 \mathrm{~kW} [/latex]

[latex]\dot{Q}_{H} =\dot{m}\left(h_{2}-h_{3} ight)=(0.05 \mathrm{~kg} / \mathrm{s})(297.86-107.32) \mathrm{kJ} / \mathrm{kg}=9.527 \mathrm{~kW} [/latex]

[latex]\dot{W}_{ ext {in }} =\dot{m}\left(h_{2}-h_{1} ight)=(0.05 \mathrm{~kg} / \mathrm{s})(297.86-239.50) \mathrm{kJ} / \mathrm{kg}=2.918 \mathrm{~kW}[/latex]

Then the COP of the refrigeration cycle becomes

[latex]\mathrm{COP}_{\mathrm{R}}=\frac{\dot{Q}_{L}}{\dot{W}_{\mathrm{in}}}=\frac{6.609 \mathrm{~kW}}{2.918 \mathrm{~kW}}=2.265[/latex]

(b) Noting that the dead-state temperature is [latex]T_{0}=T_{H}=27+273=300 \mathrm{~K}[/latex], the exergy destruction in each component of the cycle is determined as follows:

Compressor:

[latex]\dot{X}_{ ext {dest,1-2 }} =T_{0} \dot{S}_{ ext {gen1 }-2}=T_{0} \dot{m}\left(s_{2}-s_{1} ight) [/latex]

[latex]=(300 \mathrm{~K})(0.05 \mathrm{~kg} / \mathrm{s})(0.9983-0.9721) \mathrm{k} . / \mathrm{kg} \cdot \mathrm{K} [/latex]

[latex]=0.3940 \mathrm{~kW}[/latex]

Condenser:

[latex]\dot{X}_{ ext {dest, } 2-3} =T_{0} \dot{S}_{ ext {gen}, 2-3}=T_{n}\left[\dot{m}\left(s_{3}-s_{2} ight)+\frac{\dot{Q}_{H}}{T_{H}} ight] [/latex]

[latex]=(300 \mathrm{~K})\left[(0.05 \mathrm{~kg} / \mathrm{s})(0.3919-0.9983) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}+\frac{9.527 \mathrm{~kW}}{300 \mathrm{~K}} ight] [/latex]

[latex]=0.4303 \mathrm{~kW}[/latex]

Expansion valve:

[latex]\dot{X}_{ ext {dest,} 3-4} =T_{0} \dot{S}_{ ext {gen }, 3-4}=T_{0} \dot{m}\left(s_{4}-s_{3} ight) [/latex]

[latex]=(300 \mathrm{~K})(0.05 \mathrm{~kg} / \mathrm{s})(0.4367-0.3919) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K} [/latex]

[latex]=0.6725 \mathrm{~kW}[/latex]

Evaporator:

[latex]\dot{X}_{ ext {dest }, 4-1} =T_{0} \dot{S}_{\mathrm{gen}, 4-1}=T_{0}\left[\dot{m}\left(s_{1}-s_{4} ight)-\frac{\dot{Q}_{L}}{T_{L}} ight] [/latex]

[latex]=(300 \mathrm{~K})\left[(0.05 \mathrm{~kg} / \mathrm{s})(0.9721-0.4367) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}-\frac{6.609 \mathrm{~kW}}{260 \mathrm{~K}} ight] [/latex]

[latex]=0.4043 \mathrm{~kW}[/latex]

(c) Exergy flow associated with heat transferred from the low-temperature medium is

[latex]\dot{X}_{Q_{L}}=\dot{Q}_{L} \frac{T_{0}-T_{L}}{T_{L}}=(6.609 \mathrm{~kW}) \frac{300 \mathrm{~K}-260 \mathrm{~K}}{260 \mathrm{~K}}=1.017 \mathrm{~kW}[/latex]

This is also the minimum or reversible power input for the cycle:

[latex]\dot{W}_{\min , \mathrm{in}}=\dot{X}_{\dot{Q}_{L}}=1.017 \mathrm{~kW}[/latex]

The second-law efficiency of the cycle is

[latex]\eta_{ ext {II }}=\frac{\dot{X}_{Q_{L}}}{\dot{W}_{ ext {in }}}=\frac{1.017 \mathrm{~kW}}{2.918 \mathrm{~kW}}=0.3485=34.9 \%[/latex]

This efficiency may also be determined from [latex]\eta_{II}=\mathrm{COP}_{\mathrm{R}} / \mathrm{COP}_{\mathrm{R,rev}}[/latex] where

[latex]\mathrm{COP}_{\mathrm{R}, \mathrm{rev}}=\frac{T_{L}}{T_{H}-T_{L}}=\frac{(-13+273) \mathrm{K}}{[27-(-13)] \mathrm{K}}=6.500[/latex]

Substituting,

[latex]\eta_{\mathrm{II}}=\frac{\mathrm{COP}_{\mathrm{R}}}{\mathrm{COP}_{\mathrm{R}, \mathrm{reV}}}=\frac{2.265}{6.500}=0.3485=34.9 \%[/latex]

The results are identical, as expected.

(d) The total exergy destruction is the difference between the exergy expended (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium):

[latex]\dot{X}_{ ext {dest,total }}=\dot{W}_{ ext {in }}-\dot{X}_{Q_{L}}=2.918 \mathrm{~kW}-1.017 \mathrm{~kW}=1.901 \mathrm{~kW}[/latex]

The total exergy destruction can also be determined by adding exergy destruction in each component:

[latex]\dot{X}_{ ext {dest,total }} =\dot{X}_{ ext {dest,} 1-2}+\dot{X}_{ ext {dest }, 2-3}+\dot{X}_{ ext {dest }, 3-4}+\dot{X}_{ ext {dest, } 4-1} [/latex]

[latex]=0.3940+0.4303+0.6725+0.4043 [/latex]

[latex]=1.901 \mathrm{~kW}[/latex]

The two results are again identical, as expected.

Discussion The exergy input to the cycle is equal to the actual work input, which is 2.92 kW. The same cooling load could have been accomplished by only 34.9 percent of this power (1.02 kW) if a reversible system were used. The difference between the two is the exergy destructed in the cycle (1.90 kW). The expansion valve appears to be the most irreversible component, which accounts for 35.4 percent of the irreversibilities in the cycle. Replacing the expansion valve by a turbine would decrease the irreversibilites while decreasing the net power input. However, this may or may not be practical in an actual system. It can be shown that increasing the evaporating temperature and decreasing the condensing temperature would also decrease the exergy destruction in these components.

Question 11.3: Exergy Analysis of Vapor-Compression Refrigeration Cycle A v...

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