Question

Air enters the compressor of a gas-turbine engine at 300 K and 100 kPa, where it is compressed to 700 kPa and 580 K.Heat is transferredto air in the amount of 950 kJ/kgbefore it enters the turbine. For a turbine efficiency of 86 percent, determine (a) the fraction of the turbine work output used to drive the compressor and (b) the thermal efficiency. Assume variable specific heats for air.

Answer 1

Answer:

- the fraction of the turbine work output used to drive the compressor is 50%

- the thermal efficiency is 29.6%

Explanation:

given information:

$T_{1}$ = 300 k

$P_{1}$ = 700 kPa

$T_{2}$ = 580 K

Qin = 950 kj/kg

efficiency, η = 86%  = 0.86

Assume, Ideal gas specific heat capacities of air, $C_{p}$ = 1.005

k = 1.4

(a) the fraction of the turbine work output used to drive the compressor

Qin = $C_{p}$ ($T_{3}$ - $T_{2}$)

$T_{3}$ - $T_{2}$ = Qin / $C_{p}$

$T_{3}$ = Qin / $C_{p}$ + $T_{2}$

thus,

$T_{3}$ = $\frac{950 kJ/kg}{1.005 kj/kgK} + 580 K [tex]T_{3}$ = 1525 K

$\frac{T_{4} }{T_{3} }$ = $(\frac{P_{4} }{P_{3} }) ^{k-1/k}$\

$\frac{P_{4} }{P_{3} }$  = $\frac{P_{1} }{P_{2} }$

so,

$\frac{T_{4} }{T_{3} }$ = $(\frac{P_{1} }{P_{2} }) ^{k-1/k}$\

$T_{4}$ = $T_{3}$ $(\frac{P_{1} }{P_{2} }) ^{k-1/k}$\

                        = 1525 K  $(\frac{100}{700} )^{1.4-1/1.4}$

                        = 875 K

$W_{in}$ = $C_{p}$ ($T_{2}$ - $T_{1}$)

                          = 1.005 kj/kgK  (580 K - 300 K)

                          = 281.4 kJ/kg

$W_{out}$ = η $C_{p}$ ($T_{3}$ - $T_{4}$)

                            = 0.86 (1.005 kj/kgK) (1525 k - 875 K)

                            = 562.2 kJ/kg

therefore,

the fraction of turbine = $\frac{W_{in} }{W_{out} }$

                                     = $\frac{281.4}{562.2}$

                                     = 50%

(b) the thermal efficiency

η = ΔW/$Q_{in}$

  = ($W_{out}$ - $W_{in}$)/$Q_{in}$

  = $\frac{562.2}{281.4}$/950

  = 29.6%