Evaluating Performance of a Brayton Refrigeration Cycle with Irreversibilities
Reconsider Example 10.5, but include in the analysis that the compressor and turbine each have an isentropic efficiency of 80%. Determine for the modified cycle (a) the net power input, in Btu/min, (b) the refrigeration capacity, in Btu/min, (c) the coefficient of performance, and discuss its value.
Known A Brayton refrigeration cycle operates with air. Compressor inlet conditions, the turbine inlet temperature, and the compressor pressure ratio are given. The compressor and turbine each have an isentropic efficiency of 80%.
Find Determine the net power input and the refrigeration capacity, each in Btu/min. Also determine the coefficient of performance and discuss its value.
Schematic and Given Data:
Engineering Model
1. Each component of the cycle is analyzed as a control volume at steady state.
2. The compressor and turbine are adiabatic.
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
a. The power input to the compressor is evaluated using the isentropic compressor efficiency, \eta_{ c }. That is,
\frac{\dot{W}_{ c }}{\dot{m}}=\frac{\left(\dot{W}_{ c } / \dot{m}\right)_{ s }}{\eta_{ c }}
The value of the work per unit mass for the isentropic compression, \left(\dot{W}_{ c } / \dot{m}\right)_{ s }, is determined with data from the solution in Example 10.5 as 42.41 Btu/lb. The actual power required is then
\dot{W}_{ c }=\frac{\dot{m}\left(\dot{W}_{ c } / \dot{m}\right)_{ s }}{\eta_{ c }}=\frac{(248 lb / min )(42.41 Btu / lb )}{(0.8)}
= 13,147 Btu/min
The turbine power output is determined using the turbine isentropic efficiency \eta_{ t } \text {. Thus, } \dot{W}_{ t } / \dot{m}=\eta_{ t }\left(\dot{W}_{ t } / \dot{m}\right)_{\text {s }}. Using data from the solution to Example 10.5 gives \left(\dot{W}_{ t } / \dot{m}\right)_{ s }=34.96 Btu/lb. The actual power output of the turbine is then
\begin{aligned}\dot{W}_{ t } &=\dot{m} \eta_{ t }\left(\dot{W}_{ t } / \dot{m}\right)_{ s }=(248 lb / min )(0.8)(34.96 Btu / lb ) \\&=6936 Btu / min\end{aligned}
The net power input to the cycle is
\dot{W}_{\text {cycle }}=13,147-6936=6211 Btu / min
b. The specific enthalpy at the turbine exit, h_{4}, is required to evaluate the refrigeration capacity. This enthalpy can be determined by solving \dot{W}_{ t }=\dot{m}\left(h_{3}-h_{4}\right) \text { to obtain } h_{4}=h_{3}-\dot{W}_{ t } / \dot{m}. Inserting known values
h_{4}=129.06-\left(\frac{6936}{248}\right)=101.1 Btu / lb
The refrigeration capacity is then
\dot{Q}_{\text {in }}=\dot{m}\left(h_{1}-h_{4}\right)=(248)(114.69-101.1)=3370 Btu / min
c. The coefficient of performance is
\beta=\frac{\dot{Q}_{\text {in }}}{\dot{W}_{\text {cycle }}}=\frac{3370}{6211}=0.543
The value of the coefficient of performance in this case is less than unity. This means that the refrigeration effect is smaller than the net work required to achieve it. Additionally, note that irreversibilities in the compressor and turbine have a significant effect on the performance of gas refrigeration systems. This is brought out by comparing the results of the present example with those of Example 10.5. Irreversibilities result in an increase in the work of compression and a reduction in the work output of the turbine. The refrigeration capacity is also reduced. The overall effect is that the coefficient of performance is decreased significantly.
Skills Developed
Ability to…
• sketch the T–s diagram of the Brayton refrigeration cycle with irreversibilities in the turbine and compressor.
• 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 coefficient of performance for a Carnot refrigeration cycle operating between reservoirs at 480°R and 540°R. Ans. 8.
