Question

8. An ideal air-standard Brayton cycle operates at steady state with compressor inlet conditions of 250 K and 25 kPa. The compressor pressure ratio is 10. The turbine inlet temperature is 1800 K. For the cycle: (a) the heat addition and work done in each process, in kJ/kg, (b) the thermal efficiency (c) the back work ratio (35 points)

Answer 1

GIVEN:

COMPRESSOR INLET TEMPATURE (T₁) = 250 K

INLET PRESSURE (P₁) = 25 Kpa

COMPRESSOR RATIO (r_p) = 10

TURBINE INLET TEMPERATURE (T₃) = 1800 K

SOLUTION:

2 Combustion chamber 3 Turbine Compressor Cooler

1-2: isentropic compression

2-3: constant pressure energy addition

3-4: isentropic expansion

4-1: constant pressure energy rejection

3 N Isentropic T P2 Pa 1 S (a) (b)

FROM PROCESS 1-2: isentropic compression

P2 = rp P1

P2 10 = 25

$\small {P_{2}}=10*{25}$

$\small {P_{2}}=10*{25}$

$\small {P_{2}}=250 Kpa$

also, For Air   $\small \gamma =1.4$

$\small \frac{T_{2}}{T_{1}}=\left ( \frac{P_{2}}{P_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

$\small \frac{T_{2}}{250}=\left ( \frac{250}{25} \right )^{\frac{1.4 -1}{1.4 }}$

$\small T_{2}=250* ( 10)^{\frac{0.4}{1.4 }}$

T2 = 482.67K

FOR PROCESS 3-4: isentropic expansion

$\small r_{p} = \frac{P_{3}}{P_{4}}=\frac{P_{2}}{P_{1}}=10$

$\small \frac{T_{3}}{T_{4}}=\left ( \frac{P_{3}}{P_{4}} \right )^{\frac{\gamma -1}{\gamma }}$

$\small \frac{1800}{T_{4}}=( 10)^{\frac{1.4 -1}{1.4 }}$

$\small \frac{1800}{T_{4}}=( 10)^{\frac{1.4 -1}{1.4 }}$

$\small {T_{4}}=932.50 K$

B)THERMAL EFFICIENCY

$\small \eta _{thermal}=1-\left ( \frac{1}{( r_{p})}\right )^{\frac{\gamma -1}{\gamma }}$

$\small \eta _{thermal}=1-\left ( \frac{1}{(10)}\right )^{\frac{1.4 -1}{1.4 }}$

$\small \eta _{thermal}=1-0.517$

$\small \eta _{thermal}=0.482$

_{$\small \eta _{thermal}=48.20%$}%

now

1-2: isentropic compression(REVERSIBLE ADIBATIC PROCESS)

Here mass of air  is 1 kg , C_p of air = 1.005 kJ/kgk

work done (W_c) = m c_p(T₂ - T₁)

= 1*1.005*(482.67-250)

(W_c) = 233.83 kJ/kg

Q₁₂ = 0 ( since it is reversible and adibatic process)

2-3: constant pressure energy addition

Here mass of air  is 1 kg , C_p of air = 1.005 kJ/kgk

Heat Addition (Q₂₃) = mc_p (T₃ - T₂)

= 1*1.005*(1800 - 482.67)

(Q₂₃) = 1323.91 kJ/kg

3-4: isentropic expansion

Here mass of air  is 1 kg , C_p of air = 1.005 kJ/kgk

work done on turbine (W_T) = m c_p(T₃- T₄)

= 1*1.005*(1800 - 932.50

(W_T) = 871.83 KJ/kg

Q₃₄= 0 ( since it is reversible and adibatic process)

4-1: constant pressure energy rejection

Heat Addition (Q₄₁) = mc_p (T₄- T₁)

= 1*1.005*(932.50 - 250)

(Q₄₁) = 685.91 kJ/kg

C) Back work ratio

$\small \gamma _{wb}=\frac{W_{c}}{W_{T}}$

$\small \gamma _{wb}=\frac{233.83}{871.83}$

$\small \gamma _{wb}=0.268$

  

RESULT

a)

1-2: isentropic compression

(W_c) = 233.83 kJ/kg

Q₁₂ = 0 ( since it is reversible and adibatic process)

2-3: constant pressure energy addition

(Q₂₃) = 1323.91 kJ/kg

3-4: isentropic expansion.

(W_T) = 871.83 KJ/kg

Q₃₄= 0 ( since it is reversible and adibatic process)

4-1: constant pressure energy rejection

(Q₄₁) = 685.91 kJ/kg

b)THERMAL EFFICIENCY

_{$\small \eta _{thermal}=48.20%$}%

C) Back work ratio

$\small \gamma _{wb}=0.268$