Answer:
hi-he = 0
pi-pe = positive
ui-ue = negative
ti-te = negative
Explanation:
we know that fir the sub cool liquid water is
dQ = Tds = du + pdv ............1
and Tds = dh - v dP .............2
so now for process of throhling is irreversible when v is constant
then heat transfer is = 0 in irreversible process
so ds > 0
so here by equation 1 we can say
ds > 0
dv = 0 as v is constant
so that Tds = du .................3
and du > 0
ue - ui > 0
and
now by the equation 2 throttling process
here enthalpy is constant
so dh = 0
and Tds = -vdP
so ds > 0
so that -vdP > 0
as here v is constant
so -dP =P1- P2
so P1-P2 > 0
so pressure is decrease here
Answer:
Explanation:
Attached is the solution to the question
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
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