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:
percentage change in volume = 0.00285 %
Explanation:
given data
bulk modulus = 3.5 ×
N/m²
bulk stress =
N/m²
solution
we will apply here bulk modulus formula that is
bulk modulus =
...............1
put here value and we get
3.5 ×
=
solve it we get
bulk strain = 2.8571 ×
and
bulk strain =
so that percentage change in volume is = 2.8571 ×
× 100
percentage change in volume = 0.00285 %
Answer:
radius = 9.1 ×
m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius =
.................1
here F is applied load and is length
put here value and we get
radius =
solve it we get
radius = 9.1 ×
m
Answer:
Heat gain of 142 kJ
Explanation:
We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.
Δ
⇒
Δ

Therefore, the gas gained heat by an amount of 142 kJ.