Analyzing an Ideal Brayton Refrigeration Cycle
Air enters the compressor of an ideal Brayton refrigeration cycle at 1 atm, 480°R, with a volumetric flow rate of 50 ft ^{3} / s. If the compressor pressure ratio is 3 and the turbine inlet temperature is 540°R, determine (a) the net power input, in Btu/min, (b) the refrigeration capacity, in Btu/min, (c) the coefficient of performance.
Known An ideal Brayton refrigeration cycle operates with air. Compressor inlet conditions, the turbine inlet temperature, and the compressor pressure ratio are given.
Find Determine the net power input, in Btu/min, the refrigeration capacity, in Btu/min, and the coefficient of performance.
Schematic and Given Data:
Engineering Model
1. Each component of the cycle is analyzed as a control volume at steady state. The control volumes are indicated by dashed lines on the accompanying sketch.
2. The turbine and compressor processes are isentropic.
3. There are no pressure drops through the heat exchangers.
4. Kinetic and potential energy effects are negligible.
5. The working fluid is air modeled as an ideal gas.
Analysis The analysis begins by determining the specific enthalpy at each numbered state of the cycle. At state 1, the temperature is 480°R. From Table A-22E, h_{1}=114.69 Btu / lb , p_{ r 1}=0.9182. Since the compressor process is isentropic, h_{2 s } can be determined by first evaluating p_{ r } at state 2s. That is,
p_{ r 2}=\frac{p_{2}}{p_{1}} p_{ r 1}=(3)(0.9182)=2.755
Then, interpolating in Table A-22E, we get h_{2 s }=157.1 Btu/lb.
The temperature at state 3 is given as T_{3}=540^{\circ} R. From Table A-22E, h_{3}=129.06 Btu / lb , p_{ r 3}=1.3860. The specific enthalpy at state 4s is found by using the isentropic relation
p_{ r 4}=p_{ r 3} \frac{p_{4}}{p_{3}}=(1.3860)(1 / 3)=0.462
Interpolating in Table A-22E, we obtain h_{4 s }=94.1 Btu/lb.
a. The net power input is
\dot{W}_{\text {cycle }}=\dot{m}\left[\left(h_{2 s }-h_{1}\right)-\left(h_{3}-h_{4 s }\right)\right]
This requires the mass flow rate\dot{m} , which can be determined from the volumetric flow rate and the specific volume at the compressor inlet:
\dot{m}=\frac{( AV )_{1}}{v_{1}}
Since v_{1}=(\bar{R} / M) T_{1} / p_{1}
\dot{m}=\frac{( AV )_{1} p_{1}}{(\bar{R} / M) T_{1}}
=\frac{\left(50 ft ^{3} / s \right)|60 s / min |\left(14.7 lbf / in .^{2}\right)\left|144 in .^{2} / ft ^{2}\right|}{\left(\frac{1545 ft \cdot lbf }{28.97 lb \cdot{ }^{\circ} R }\right)\left(480^{\circ} R \right)}
= 248 lb/min
Finally
\begin{aligned}\dot{W}_{\text {cycle }} &=(248 lb / min )[(157.1-114.69)-(129.06-94.1)] Btu / lb \\&=1848 Btu / min\end{aligned}
b. The refrigeration capacity is
\begin{aligned}\dot{Q}_{\text {in }} &=\dot{m}\left(h_{1}-h_{4 s }\right) \\&=(248 lb / min )(114.69-94.1) Btu / lb \\&=5106 Btu / min\end{aligned}
c. The coefficient of performance is
1 \beta=\frac{\dot{Q}_{\text {in }}}{\dot{W}_{\text {cycle }}}=\frac{5106}{1848}=2.76
1 Irreversibilities within the compressor and turbine serve to decrease the coefficient of performance significantly from that of the corresponding ideal cycle because the compressor work requirement is increased and the turbine work output is decreased. This is illustrated in Example 10.6.
Skills Developed
Ability to…
• sketch the T–s diagram of the ideal Brayton refrigeration cycle.
• fix each of the principal states and retrieve necessary property data.
• calculate net power input, refrigeration capacity, and coefficient of performance.
Quick Quiz
Determine the refrigeration capacity in tons of refrigeration. Ans. 25.53 ton.
