Design a stepped-impedance low-pass filter having a cutoff frequency of 3 GHz and a fifth-order 0.5 dB equal-ripple response. As
1 answer:
You might be interested in
Answer:
a) 764.45K
b) 210.48 kJ/kg
c) 30.14%
Explanation:
pressure ratio = 10
minimum temperature = 295 k
maximum temperature = 1240 k
isentropic efficiency for compressor = 83%
Isentropic efficiency for turbine = 87%
<u>a) Air temperature at turbine exit </u>
we can achieve this by interpolating for enthalpy
h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air
T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) = 764.45K
<u>b) The net work output </u>
first we determine the actual work input to compressor
Wc = h2 - h1 ( calculated values )
= 626.57 - 295.17 = 331.4 kJ/kg
next determine the actual work done by Turbine
Wt = h3 - h4 ( calculated values )
= 1324.93 - 783.05 = 541.88 kJ/kg
finally determine the network output of the cycle
Wnet = Wt - Wc
= 541.88 - 331.4 = 210.48 kJ/kg
<u>c) determine thermal efficiency </u>
лth = Wnet / qin ------ ( 1 )
where ; qin = h3 - h2
<em>equation 1 becomes </em>
лth = Wnet / ( h3 - h2 )
= 210.48 / ( 1324.93 - 626.57 )
= 0.3014 = 30.14%
Answer:
See attached picture.
Explanation:
See attached picture for explanation.
Answer:
(b) 56%
Explanation:
the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature
here we have given T₁ (Higher temperature)= 600+273=873
lower temperature T₂=110+273=383
Efficiency of power cycle is given by =1-
=1-
=1-0.43871
=.56
=56%