Respuesta :
Answer:
Explanation:
Given
T₁ = 300 K,
T₃ = 350 K
P₁, P₃ = 100 kPa
P₂, P₄ = 1200 kPa
T₆ = 1400 K
T₈ = 1300 K
[tex]\eta_{Compressor}[/tex] , [tex]\eta_{Turbine}[/tex] = 80 %
Effectiveness of regenerator = 75 %
Step by step solution is given as
Point 1
T₁= 300 K
Point 2
[tex]\eta_{c} = \frac{h_{2s}- h_{1} }{h_{2} -h_{1} } = \frac{c_{p}( T_{2s}- T_{1} }{c_{p}(T_{2}- T_{1})}[/tex] from which T₂ = T₁ +[tex]\frac{( T_{2s}- T_{1} )}{\eta_{c} }[/tex]
From the above therefore to find T₂, we look for the value of [tex]T_{2s}[/tex] as follows
[tex]\frac{T_{2s} }{T_{1} } = (\frac{P_{2} }{P_{1} })^{\frac{k-1}{k} }[/tex] from which [tex]T_{2s} = 300 K(\frac{1200 kPa}{100kPa} )^{\frac{0.4}{1.4} } =[/tex] 610.18 K
Therefore T₂ = T₁ +[tex]\frac{( T_{2s}- T_{1} )}{\eta_{c} }[/tex] = 300 K + [tex]\frac{610.18-300}{0.8}[/tex] = 687.726 K
Point 3
T₃ = 350 K, inlet condition of 2nd compressor
Point 4
The value of T₄ is calculated as follows
[tex]\eta_{c} = \frac{h_{4s}- h_{3} }{h_{4} -h_{3} } = \frac{c_{p}( T_{4s}- T_{3} }{c_{p}(T_{4}- T_{3})}[/tex] from which T₄ = T₃ +[tex]\frac{( T_{4s}- T_{3} )}{\eta_{c} }[/tex]
also since
[tex]\frac{T_{4s} }{T_{3} } = (\frac{P_{4} }{P_{3} })^{\frac{k-1}{k} }[/tex] we have [tex]T_{4s} = 350 K(\frac{1200 kPa}{100kPa} )^{\frac{0.4}{1.4} } =[/tex] 711.87 K
T4 = 802.34 K
Point 5
[tex]\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})}[/tex] from which T₅ =T₄ + [tex]\epsilon_{regen}[/tex]×(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
[tex]\eta_{t} = \frac{h_{6}- h_{7} }{h_{6} -h_{7s} } = \frac{c_{p}( T_{6}- T_{7} }{c_{p}(T_{6}- T_{7s})}[/tex] from which T₇ = T₆ -[tex]\eta_{t}{( T_{6}- T_{7s} )}[/tex]
also [tex]\frac{T_{7s} }{T_{6} } = (\frac{P_{7} }{P_{6} })^{\frac{k-1}{k} }[/tex] which gives [tex]T_{7s} = 1400 K(\frac{P_{7} }{P_{6} } )^{\frac{0.4}{1.4} }[/tex] here we assume efficient cycle with pressure ratio = √12 thus we have
[tex]T_{7s} = 1400 K(\frac{1}{12 } } )^{\frac{0.4}{1.4} }[/tex] = 688.32 K
and T₇ = T₆ -[tex]\eta_{t}{( T_{6}- T_{7s} )}[/tex] = 1400 - 0.8(1400-688.32) = 830.656 K
Point 8
T₈ = 1400 K
Point 9
Here we have
[tex]\eta_{t} = \frac{h_{8}- h_{9} }{h_{8} -h_{9s} } = \frac{c_{p}( T_{8}- T_{9} }{c_{p}(T_{8}- T_{9s})}[/tex] from which T₉ = T₈ -[tex]\eta_{t}{( T_{8}- T_{9s} )}[/tex]
and [tex]\frac{T_{9s} }{T_{8} } = (\frac{P_{9} }{P_{8} })^{\frac{k-1}{k} }[/tex]→ [tex]T_{9s} = 1400 K(\frac{1}{12 } } )^{\frac{0.4}{1.4}}[/tex] = 688.32 K
Therefore T₉ = 830.656 K
Therefore for point 5 we have T₅ =T₄ + [tex]\epsilon_{regen}[/tex]×(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
[tex]\eta_{th,regen}[/tex] = [tex]1-(\frac{T_{1} }{T_{6} })(r_{p} )^{\frac{k-1}{k} }[/tex] =[tex]1-\frac{300}{1300}\frac{1200}{100}^{\frac{0.4}{1.4} }[/tex] =0.5606 or 53.06 %
(c) The second law efficiency is given by
[tex]\eta _{turbine} = \frac{W_{real} }{W_{rev} }[/tex] = Wrev = 1 - TL/TH = 1 -300/1400 = 0.786
Therefore we have
[tex]\eta_{II} =[/tex] 0.53/0.786 =0.675
(d) The exergy is given by [tex]\epsilon = \frac{E_{out}-E_{in} }{W }[/tex] for the compressor
and for the turbine we have
[tex]\epsilon =\frac{W_{net} +W_{c} }{E_{in} - E_{out} }[/tex]
