Evaluating Thermal Efficiency of a Brayton Cycle with Regeneration
A regenerator is incorporated in the cycle of Example 9.4. (a) Determine the thermal efficiency for a regenerator effectiveness of 80%. (b) Plot the thermal efficiency versus regenerator effectiveness ranging from 0 to 80%.
Known A regenerative gas turbine operates with air as the working fluid. The compressor inlet state, turbine inlet temperature, and compressor pressure ratio are known.
Find For a regenerator effectiveness of 80%, determine the thermal efficiency. Also plot the thermal efficiency versus the regenerator effectiveness ranging from 0 to 80%.
Schematic and Given Data:
Engineering Model
1. Each component is analyzed as a control volume at steady state. The control volumes are shown on the accompanying sketch by dashed lines.
2. The compressor and turbine processes are isentropic.
3. There are no pressure drops for flow through the heat exchangers.
4. The regenerator effectiveness is 80% in part (a).
5. Kinetic and potential energy effects are negligible.
6. The working fluid is air modeled as an ideal gas.
Analysis
a. The specific enthalpy values at the numbered states on the T–s diagram are the same as those in Example 9.4: h_{1}=300.19 kJ / kg, h_{2}=579.9 kJ / kg , h_{3}=1515.4 kJ / kg , h_{4}=808.5 kJ / kg.
To find the specific enthalpy h_{ x }, the regenerator effectiveness is used as follows: By definition
\eta_{ reg }=\frac{h_{ x }-h_{2}}{h_{4}-h_{2}}
Solving for h_{ x }
\begin{aligned}h_{ x } &=\eta_{ reg }\left(h_{4}-h_{2}\right)+h_{2} \\&=(0.8)(808.5-579.9)+579.9=762.8 kJ / kg\end{aligned}
With the specific enthalpy values determined above, the thermal efficiency is
1 \eta=\frac{\left(\dot{W}_{ t } / \dot{m}\right)-\left(\dot{W}_{ c } / \dot{m}\right)}{\left(\dot{Q}_{ in } / \dot{m}\right)}=\frac{\left(h_{3}-h_{4}\right)-\left(h_{2}-h_{1}\right)}{\left(h_{3}-h_{ x }\right)}
=\frac{(1515.4-808.5)-(579.9-300.19)}{(1515.4-762.8)}
2 = 0.568 (56.8%)
b. The IT code for the solution follows, where \eta_{\text {reg }} is denoted as etareg, η is eta, \dot{W}_{\text {comp }} / \dot{m} is Wcomp, and so on.
Using the Explore button, sweep etareg from 0 to 0.8 in steps of 0.01. Then, using the Graph button, obtain the following plot:
3 From this plot, we see that the cycle thermal efficiency increases from 0.456, which agrees closely with the result of Example 9.4 (no regenerator), to 0.567 for a regenerator effectiveness of 80%, which agrees closely with the result of part (a). Regenerator effectiveness
is seen to have a significant effect on cycle thermal efficiency.
1 The values for work per unit of mass flow of the compressor band turbine are unchanged by the addition of the regenerator. Thus, the back work ratio and net work output are not affected by this modification.
2 Comparing the present thermal efficiency value with the one determined in Example 9.4, it should be evident that the thermal efficiency can be increased significantly by means of regeneration.
3 The regenerator allows improved fuel utilization to be achieved by transferring a portion of the exergy in the hot turbine exhaust gas to the cooler air flowing on the other side of the regenerator.
Skills Developed
Ability to…
• sketch the schematic of the regenerative gas turbine and the T–s diagram for the corresponding air-standard cycle.
• evaluate temperatures and pressures at each principal state and retrieve necessary property data.
• calculate the thermal efficiency.
Quick Quiz
What would be the thermal efficiency if the regenerator effectiveness were 100%? Ans. 60.4%.
// Fix the states
T1 = 300//K
p1 = 100//kPa
h1 = h_T(“Air”, T1)
s1 = s_TP(“Air”, T1, p1)
p2 = 1000//kPa
s2 = s_TP(“Air”, T2, p2)
s2 = s1
h2 = h_T(“Air”, T2)
T3 = 1400//K
p3 = p2
h3 = h_T(“Air”, T3)
s3 = s_TP(“Air”, T3, p3)
p4 = p1
s4 = s_TP(“Air”, T4, p4)
s4 = s3
h4 = h_T(“Air”, T4)
etareg = 0.8
hx = etareg*(h4 − h2) + h2
// Thermal efficiency
Wcomp = h2 − h1
Wturb = h3 − h4
Qin = h3 − hx
eta = (Wturb − Wcomp) / Qin