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%
a) Air temperature at turbine exit
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 ) [tex](\frac{783.05-778.18}{800.13-778.18} )[/tex] = 764.45K
b) The net work output
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
c) determine thermal efficiency
лth = Wnet / qin ------ ( 1 )
where ; qin = h3 - h2
equation 1 becomes
лth = Wnet / ( h3 - h2 )
= 210.48 / ( 1324.93 - 626.57 )
= 0.3014 = 30.14%
35 points and brainiest. A, B, C, D
Which of the following identifies what carbide-tipped bits, lines on machine bolts, and lock washers all have in common?
A. All are very expensive items.
B. All are very recent inventions.
C. All are rather outdated inventions.
D. All provide an added level of strength.
Answer:c
Explanation: