 Talamancoeb

2022-01-06

A gas-turbine power plant operates on the simple Brayton cycle with air as the working fluid and delivers 32 MW of power. The minimum and maximum temperatures in the cycle are 310 and 900 K, and the pressure of air at the compressor exit is 8 times the value at the compressor inlet. Assuming an isentropic efficiency of 80 percent for the compressor and 86 percent for the turbine, determine the mass flow rate of air through the cycle. Account for the variation of specific heats with temperature. Step 1
The enthalpy and the relative pressure at the initial state are taken from A-17 for the given temperature:
${h}_{1}=310.24\frac{kj}{kg}$
${P}_{r1}=1.5546$
The pressure ratio and the relative pressure at state 1 are used to calculate the relative pressure at state 2:
${P}_{r2}=\frac{{P}_{2}}{{P}_{1}}{P}_{r1}$
$=8\cdot 1.5546$
$=12.44$
From this the enthalpy at state 2 for an isentropic process can be obtained using interpolation with data from A-17:
${h}_{2s}=562.58\frac{kj}{kg}$
The relative pressure and the enthalpy at state 3 are determined from A-17 for the given temperature:
${h}_{3}=932.93\frac{kj}{kg}$
${P}_{r3}=75.29$
Inferring from the relative pressure in state 3, one may calculate the relative pressure at state 4.
${P}_{r4}=\frac{{P}_{4}}{{P}_{3}}{P}_{r3}$
$=\frac{1}{8}\cdot 75.29$
$=9.41$
From this value the isentropic enthalpy at state 4 can be obtained using interpolation with data from A-17:
${h}_{4s}=519.3\frac{kj}{kg}$
Step 2
The power input of the compressor and the power output of the turbine can be used to calculate the mass flow rate:
$W=Wtask,outPSK-{W}_{c}$, in
$Q=m{n}_{T}\left({h}_{3}-{h}_{4s}\right)-\frac{m}{{n}_{C}}\left({h}_{2}-{h}_{1}\right)$
$m=\frac{W}{{n}_{T}\left({h}_{3}-{h}_{4s}\right)-\frac{1}{{n}_{C}}\left({h}_{2s}-{h}_{1}\right)}$
$=\frac{32000}{0.86\left(932.93-519.3\right)-\frac{1}{0.8}\left(562.58-310.24\right)}$
$=794\frac{kg}{s}$ sirpsta3u

A 32-MW has-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The mass flow rate of air through the cycle is to be determined.
Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with variable specific heats.
Analysis. Using variable specific heats,
${T}_{1}=310K⇒{h}_{1}=310.24\frac{kJ}{kg}$
$⇒{P}_{r1}=1.5546$
${P}_{r2}=\frac{{P}_{2}}{{P}_{1}}P{r}_{1}=\left(8\right)\left(1.5546\right)=12.44⇒{h}_{2s}=562.26\frac{kJ}{kg}$
${T}_{3}=900K⇒{h}_{3}=932.93\frac{kJ}{kg}$
$⇒{P}_{r3}=75.29$
${P}_{r4}=\frac{{P}_{4}}{{P}_{3}}{P}_{r3}=\left(\frac{1}{8}\right)\left(75.29\right)=9.411⇒{h}_{4s}=519.32\frac{kJ}{kg}$
${w}_{\ne t,out}={w}_{T,out}-{w}_{c,\in }={n}_{T}\left({h}_{3}-{h}_{4s}\right)-\frac{{h}_{2s}-{h}_{1}}{{n}_{C}}$
$=\left(0.86\right)\left(932.93-519.32\right)-\frac{562.26-310.24}{0.80}=40.68\frac{kJ}{kg}$
and $m=\frac{{W}_{\ne t,out}}{{w}_{\ne t,out}}=\frac{32.000\frac{kJ}{s}}{40.68\frac{kJ}{kg}}=786.6k\frac{g}{s}$ karton

Use the gas table for initial state,
$\begin{array}{}{h}_{1}=310.24kJ/kg,{P}_{r1}=1.5546\\ \text{Calculate the relative pressure at state 2}\\ {P}_{r2}=\frac{{P}_{2}}{{P}_{1}}⇒8×1.5546⇒{P}_{r2}=12.44\\ \text{use the table at state 2 and 3,}\\ {h}_{2s}=562.58kJ/kg,{h}_{3}=932.93kJ/kg\\ {P}_{r3}=75.29\\ \text{Calculate the relative pressure at state 4,}\\ {P}_{r4}=\frac{{P}_{4}}{{P}_{3}}×{P}_{r3}⇒\frac{1}{8}×75.29⇒9.41\\ \text{use table at reative pressure of state 4,}\\ {h}_{4s}=519.3kJ/kg\\ \text{Calculate the mass flow rate.}\\ W=\left({W}_{t}{\right)}_{out}-\left({W}_{c}{\right)}_{in}\\ m{n}_{t}\left({h}_{3}-{h}_{4s}\right)-\frac{m}{{n}_{c}}\left({h}_{2}-{h}_{1}\right)\\ m=\frac{w}{{n}_{T}\left({h}_{3}-{h}_{4}\right)-\frac{1}{{n}_{c}}\left({h}_{2s}-{h}_{1}\right)}\\ =\frac{32000}{0.86\left(932.93-519.3\right)-\frac{1}{0.8}\left(562.56-310.24\right)}\\ m=794kg/s\end{array}$

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