6. After removing the vacuum hose from the fuel-pressure regulator, you see gasoline has dripped out of the hose. Technician A s
1 answer:
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Answer:
hello your question is incomplete attached below is the complete question
A) optimum compressor ratio = 9.144
B) specific thrust = 2.155 N.s /kg
C) Thrust specific fuel consumption = 1670.4 kg/N.h
Explanation:
Given data :
Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k
γ = 1.4
attached below is the detailed solution
Answer:
Ts =Ta E)- 300(
569.5 K
5
Q12-W12 = -4014.26
Mol
AU2s = Q23= 5601.55
Mol
AUs¡ = Ws¡ = -5601.55
Explanation:
A clear details for the question is also attached.
(b) The P,V and T for state 1,2 and 3
P =1 bar Ti = 300 K and Vi from ideal gas Vi=
10
24.9x10 m
=
P-5 bar
Due to step 12 is isothermal: T1 = T2= 300 K and
VVi24.9 x 10x-4.9 x 10-3 *
The values at 3 calclated by Uing step 3l Adiabatic process
B-P ()
Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=
14
Ps-1x(4.99 x 103
P-1x(29x 10)
9.49 barr
And Ts =Ta E)- 300(
569.5 K
5
(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and
Work done for Isotermal process define as
8.314 x 300 In =4014.26
Wi2= RTi ln
mol
And fromn first law of thermodynamic
AU12= W12 +Q12
Q12-W12 = -4014.26
Mol
F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and
Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-
AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53
Inol
Now from first law of thermodynamic the Q23
AU2s = Q23= 5601.55
Mol
For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0
and
AH
C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18
mol
AU=C, (T¡-T)= x 8.314 (300
-5601.55
569.5)
mol
Now from first law of thermodynamie the Ws1
J
mol
AUs¡ = Ws¡ = -5601.55
