Question

A gas turbine plant operates on the regenerative Brayton cycle with two stages of reheating and two stages of intercooling between the pressure limits of 100 and 1200 kPa. The working fluid is air. The air enters the first and the second stages of the comprrssor at 300 K and 350 K, respectively, and the first and the second stage stages of the turbine at 1400 K and 1300 K, respectively. Assuming both the compressor and the turbine have an isentropic efficiency of 80% and the regenerator has an effectiveness of 75% and using variable specific heats, determine:
(a) The back work ratio and the net work output.

(b) The thermal efficiency.

(c) The second-lawnefficiency of the cycle.

(d) The exergies at the exits of the combustion chamber (stage 6) and the regenerator (stage 10).

Answer 1

Answer:

Explanation:

Given

T₁ = 300 K,

T₃ = 350 K

P₁, P₃ = 100 kPa

P₂, P₄ = 1200 kPa

T₆ = 1400 K

T₈ = 1300 K

$\eta_{Compressor}$ , $\eta_{Turbine}$ = 80 %

Effectiveness of regenerator = 75 %

Step by step solution is given as

Point 1

T₁= 300 K

Point 2

$\eta_{c} = \frac{h_{2s}- h_{1} }{h_{2} -h_{1} } = \frac{c_{p}( T_{2s}- T_{1} }{c_{p}(T_{2}- T_{1})}$ from which T₂ = T₁ +$\frac{( T_{2s}- T_{1} )}{\eta_{c} }$

From the above therefore to find T₂, we look for the value of $T_{2s}$ as follows

$\frac{T_{2s} }{T_{1} } = (\frac{P_{2} }{P_{1} })^{\frac{k-1}{k} }$  from which $T_{2s} = 300 K(\frac{1200 kPa}{100kPa} )^{\frac{0.4}{1.4} } =$ 610.18 K

Therefore T₂ = T₁ +$\frac{( T_{2s}- T_{1} )}{\eta_{c} }$ = 300 K + $\frac{610.18-300}{0.8}$ = 687.726 K

Point 3

T₃ = 350 K, inlet condition of 2nd compressor

Point 4

The value of T₄ is calculated as follows

$\eta_{c} = \frac{h_{4s}- h_{3} }{h_{4} -h_{3} } = \frac{c_{p}( T_{4s}- T_{3} }{c_{p}(T_{4}- T_{3})}$ from which T₄ = T₃ +$\frac{( T_{4s}- T_{3} )}{\eta_{c} }$

also since

$\frac{T_{4s} }{T_{3} } = (\frac{P_{4} }{P_{3} })^{\frac{k-1}{k} }$ we have   $T_{4s} = 350 K(\frac{1200 kPa}{100kPa} )^{\frac{0.4}{1.4} } =$ 711.87 K

T4 = 802.34 K

Point 5

$\epsilon_{regen} =\frac{q_{regen,act} }{q_{regen,max}} = \frac{h_{5}- h_{4} }{h_{9} -h_{4} } = \frac{c_{p}( T_{5}- T_{4} }{c_{p}(T_{9}- T_{4})}$ from which T₅ =T₄ +  $\epsilon_{regen}$×(T₉ - T₄)

Therefore we look for T₉ first before determining T₅

Point 6

T₆ is given as = 1400 K

Point 7

Here we also have

$\eta_{t} = \frac{h_{6}- h_{7} }{h_{6} -h_{7s} } = \frac{c_{p}( T_{6}- T_{7} }{c_{p}(T_{6}- T_{7s})}$ from which T₇ = T₆ -$\eta_{t}{( T_{6}- T_{7s} )}$

also $\frac{T_{7s} }{T_{6} } = (\frac{P_{7} }{P_{6} })^{\frac{k-1}{k} }$ which gives  $T_{7s} = 1400 K(\frac{P_{7} }{P_{6} } )^{\frac{0.4}{1.4} }$ here we assume efficient cycle with pressure ratio = √12 thus we have

$T_{7s} = 1400 K(\frac{1}{12 } } )^{\frac{0.4}{1.4} }$  = 688.32 K

and T₇ = T₆ -$\eta_{t}{( T_{6}- T_{7s} )}$ = 1400 - 0.8(1400-688.32) = 830.656 K

Point 8

T₈ = 1400 K

Point 9

Here we have

$\eta_{t} = \frac{h_{8}- h_{9} }{h_{8} -h_{9s} } = \frac{c_{p}( T_{8}- T_{9} }{c_{p}(T_{8}- T_{9s})}$ from which T₉ = T₈ -$\eta_{t}{( T_{8}- T_{9s} )}$

and $\frac{T_{9s} }{T_{8} } = (\frac{P_{9} }{P_{8} })^{\frac{k-1}{k} }$→  $T_{9s} = 1400 K(\frac{1}{12 } } )^{\frac{0.4}{1.4}}$ = 688.32 K

Therefore T₉ = 830.656 K

Therefore for point 5 we have T₅ =T₄ +  $\epsilon_{regen}$×(T₉ - T₄) = 711.87 K + 0.75×(830.656 K-711.87 K) = 914.21 K

T₅ = 800.9595 K

(a)  Back work ratio is given as T₁/T₉ = 300K/981.66 K = 0.3056

(b) Thermal efficiency under cold air standard is given as

$\eta_{th,regen}$ = $1-(\frac{T_{1} }{T_{6} })(r_{p} )^{\frac{k-1}{k} }$ =$1-\frac{300}{1300}\frac{1200}{100}^{\frac{0.4}{1.4} }$ =0.5606 or 53.06 %

(c) The second law efficiency is given by

$\eta _{turbine} = \frac{W_{real} }{W_{rev} }$ = Wrev = 1 - TL/TH = 1 -300/1400 = 0.786

Therefore we have

$\eta_{II} =$ 0.53/0.786 =0.675

(d) The exergy is given by $\epsilon = \frac{E_{out}-E_{in} }{W }$ for the compressor

and for the turbine we have

$\epsilon =\frac{W_{net} +W_{c} }{E_{in} - E_{out} }$