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where R is a gas constant for a particular gas (as given in C&B Tables A-1 and A-2). An Isentropic Process for an Ideal Gas. Given: • constant specific heats ...
Typology: Summaries
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Reading Problems
The defining equation for a ideal gas is
P v
= constant = R
Knowing that v = V /m
T m
= constant = R
where R is a gas constant for a particular gas (as given in C&B Tables A-1 and A-2).
Given:
(
∂u
)
V
dT ≡ c p
(
∂h
)
P
Gibb’s equation can be written as
T ds = du + P dv = cvdT + P dv = 0 (1)
where ds = 0 because we have assumed an isentropic process.
The definition of enthalpy is
h = u + P v
Taking the derivative yields
dh = du + P dv
︸ ︷︷ ︸
≡T ds
+vdP
dh = T ds + vdP ⇒ T ds = 0 = dh − vdP
c p
dT − vdP = 0 (2)
Equating Eqs. (1) and (2) through the dT term gives
dP
c p
c v
dv
v
Integrating (3) from its initial state to a final state
1
v
k
1
2
v
k
2
= constant = P v
k
where
k =
cp
cv
The product of P · v
k remains constant for an ideal gas when:
1
and P 2
for an isentropic process
- determine Pr 1 at T 1 from Table A- - calculate Pr 2 , where
(
2
1
)
s=const
r 2
r 1
- read T 2
from Table A-17 for the calculated value of P r 2
(
v 2
v 1
)
s=const
v r 2
v r 1
For an ideal gas with constant cp and cv
P v = RT
u 2
− u 1
= c v
2
1
h 2
− h 1
= c p
2
1
There are 3 forms of a change in entropy as a function of T & v, T & P , and P & v.
s 2
− s 1
= c v
ln
2
1
v 2
v 1
= c p
ln
2
1
− R ln
2
1
= c p
ln
v 2
v 1
ln
1
R = c p
− c v
Steady State, Steady Flow in a Flow Channel of Arbitrary Cross-section with Work and Heat
Transfer
d
f inal
initial
x+dx
x
where
E = m˙(e + P v)
= m˙(u +
(v
∗ )
2
From the 1st law
rate of energy
storage
rate of
work
rate of
heat transf er
net rate of energy
leaving the system
dE CV
dt
= d
W − d
Q − d
Reversible implies ⇒
S
CV
Equation 4 becomes
d
m ˙
= −T ds + du + d(P v)
︸ ︷︷ ︸
=du + P dv
︸ ︷︷ ︸
=T ds
+d
(
(v
∗
)
2
)
Therefore
d
m ˙
= vdP + d
(
(v
∗
)
2
)
Integrating Eq. (6) between the inlet and the outlet
m ˙
∫ out
in
vdP +
(v
∗ )
2
∣
∣
∣
∣
∣
out
in ︸ ︷︷ ︸
∆KE
∣
∣
∣
∣
∣
out
in ︸ ︷︷ ︸
∆P E
but ∆KE and ∆P E are usually negligible.
If ∆KE + ∆P E = 0
m ˙
∫ out
in
vdP (8)
Equation can be used for a reversible, SS-SF flow in a liquid or a gas.
If we keep in mind
ρ liq
ρ gas
⇒ v liq
<< v gas
i.e. water @ 25
◦ C ρ = 997 kg/m
3 and air @ 25
◦ C ρ = 1. 18 kg/m
3
Therefore
m ˙
liq
m ˙
gas
For example: the work required to operate a pump is much less that that required to operate a
compressor.
This is a special case of Eq. (8) where v = constant = v in
− v out
From Equation (8)
m ˙
= v in
out
in
The work term represents the minimum work required to pump a liquid from P in
to P out
with
negligible ∆KE and ∆P E.
From Eq. (6)
vdP + d
(
(v
∗ )
2
)
Therefore
d
(
ρ
)
(
(v
∗ )
2
)
m ˙
(
k
k − 1
)
(P v) in
(
out
in
) (k−1)/k
= c p
out
in
The right side of Eq. (14) is based on the fact that ∆KE + ∆P E = 0 and dh = du + dP v
and du = 0. Which leads to h =
∫
vdP.
Note: for the same inlet state and pressure ratio
m ˙
rev.,isothermal
m ˙
rev.,adiabatic
