Answer:
isentropic efficiency = 0.818
Explanation:
given data
pressure P1 = 95 kPa
temperature = 27°C
pressure P2 = 600 kPa
temperature = 277°C
to find out
isentropic efficiency of the compressor and exit temperature of the air
solution
we know from ideal gas of properties of air is
Pr1 at 27°C = 1.3860
and h1 at 300 K = 300.19 kJ/kg
and h2 at 550 K = 555.74 kJ/kg
and
we know equation for isentropic process that is
.........................1
put here value we get
solve we get Pr2
Pr2 = 8.75
by ideal gas of properties of air will be at Pr2
h2s = 508.66
T2s = 505.5 K
'so
isentropic efficiency will be here as
isentropic efficiency =
isentropic efficiency =
isentropic efficiency = 0.818
Answer:
1+1×1 multiplay then you get the answer
Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084
I’m sorry i don’t know the answers but i would try looking up the name of the worksheet online! :) hope you find it