
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Transcribed Image Text:EXAMPLE 5–10
Bernoulli Equation for Compressible Flow
Derive the Bernoulli equation when the compressibility effects are not negli-
gible for an ideal gas undergoing (a) an isothermal process and (b) an isen-
tropic process.
SOLUTION The Bernoulli equation for compressible flow is to be obtained
for an ideal gas for isothermal and isentropic processes.
Assumptions 1 The flow is steady and frictional effects are negligible. 2 The
fluid is an ideal gas, so the relation P = pRT is applicable. 3 The specific
heats are constant so that Plp* = constant during an isentropic process.
Analysis (a) When the compressibility effects are significant and the flow
cannot be assumed to be incompressible, the Bernoulli equation is given by
Eq. 5-40 as
dP 4 V².
v?
+ gz = constant
(along a streamline)
(1)
The compressibility effects can be properly accounted for by expressing p in
terms of pressure, and then performing the integration S dPlp in Eq. 1. But
this requires a relation between P and p for the process. For the isothermal
expansion or compression of an ideal gas, the integral in Eq. 1 can be per-
formed easily by noting that T = constant and substituting p = PIRT. It
gives
dP
= RT In P
PIRT
Substituting into Eq. 1 gives the desired relation,
Isothermal process:
RT In P +
+ gz = constant
(2)
(b) A more practical case of compressible flow is the isentropic flow of ideal
gases through equipment that involves high-speed fluid flow such as nozzles,
diffusers, and the passages between turbine blades. Isentropic (i.e.,
reversible and adiabatic) flow is closely approximated by these devices, and
it is characterized by the relation Plp* = C = constant, where k is the spe-
cific heat ratio of the gas. Solving for p from Plp* = C gives p = C-/k pl./k.
Performing the integration,
P- +1
- 1/k + 1
dP = C
(3)
P - 1/k + 1
Substituting, the Bernoulli equation for steady, isentropic, compressible flow
of an ideal gas becomes
k P
Isentropic flow:
+ gz = constant
(4a)
- 1
+ gz, =
+ gz
(4b)
A common practical situation involves the acceleration of a gas from rest
(stagnation conditions at state 1) with negligible change in elevation. In
that case we have z
= z, and V, = 0. Noting that p = PIRT for ideal
.......
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