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MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 4.05
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
Consider the frictionless, steady flow of a compressible fluid in an infinitesimal stream tube.
(a) Demonstrate by the continuity and momentum theorems that
dρ dA dV
ρ + A + V = 0
dp + ρV dV + ρgdz = 0
(b) Determine the integrated forms of these equations for an incompressible fluid.
(c) Derive the appropriate equations for unsteady frictionless, compressible flow, in a stream tube of cross
sectional area which depends on both space and time.
c
2.25 Advanced Fluid Mechanics 1 Copyright © 2013, MIT
Inviscid Flows A.H. Shapiro and A.A. Sonin 4.05
Solution:
A+dA ρ+dρ
V+dV p+dp
CS CS
3 2
CV
ds dz
A
θ
ρ ρgdV
V
p CS
1
(a) Here we consider an arbitrary control volume, CV , sitting along a streamline of length ds. For steady
flow, we may write the integral mass conservation equation as
�
CS ρu · ndAˆ = 0 (4.05a)
To evaluate this integral we must decompose it into three integrals for the three subcontrol surfaces of this
volume. For CS located at the upstream portion of the CV, the integral is
1
� ρu · ndAˆ = −ρV A (4.05b)
CS
1
For CS2 the result is
� d d d d
d d d d
d d d d
ρu·ndAˆ =(ρ+dρ)(V +dV )(A+dA) = ρV A+ρV dA+ρAdV +ρdV dA+V Adρ+V dρdA+AdρdV +dρdV dA
d d d d
CS
2 (4.05c)
where we have neglected higher order terms. There is no flow across CS3 so
� ρu · ndAˆ = 0 (4.05d)
CS3
Combining Eq. (4.05b), (4.05c) and (4.05d) into Eq. (4.05a) we obtain
c
2.25 Advanced Fluid Mechanics 2 Copyright © 2013, MIT
Inviscid Flows A.H. Shapiro and A.A. Sonin 4.05
−ρV A + ρV A + ρV dA + ρAdV + V Adρ = 0
Dividing this result by ρV A, we have
dρ + dA + dV = 0 (4.05e)
ρ A V
Fs,CS2 =(p+dp)(A+dA)
F ≈1(2p+dp)dA
F = pA s,CS3 2
s,CS1
For steady flow, the integral momentum conservation equation is
�Z
CS ρu u · nˆ dA = F (4.05f)
To calculate the left hand side of Eq. (4.05f), we calculate the momentum flux across CS
1
�
Z 2
CS ρu u · nˆ dA = −ρV A (4.05g)
1
For CS2 the result is
Z� 2 2 2 2
CS ρu u · nˆ dA = (ρ + dρ)(V + dV ) (A + dA) ≈ ρV A + 2ρV AdV + V Adρ + ρV dA (4.05h)
2
when we neglect higher order terms. There is no momentum flux across CS3.
Now we must calculate the sum of the forces acting along the streamline direction. Since the flow is friction/
less, the streamwise forces come only from pressure and gravity, hence
c
2.25 Advanced Fluid Mechanics 3 Copyright © 2013, MIT
Inviscid Flows A.H. Shapiro and A.A. Sonin 4.05
F ·sˆ= F + F
gravity,s pressure,s
The gravitational force is
F = −(ρ)d∀g sin θ
gravity,s
1 1
where the angled brackets indicate the average value. Setting (ρ) = 2 ρ+(ρ+dρ) and ∀ = 2 A+(A+dA) ds
dz
and sin θ = , we obtain
ds
1 1 1 d
d
d (4.05i)
F = − 2ρ + dρ 2A + dA gdz = −ρAgdz − (ρdA + dρA)gdz − dρdAgdz
gravity,s d
4 2 4d
where we neglect all terms higher than first order. The force arising from the pressure acting on the control
volume is
Fpressure,s = pA −(p + dp)(A + dA)+ (p)ACS sin θ
3
where we set (p) = 1 p + (p + dp) and ACS sin θ = dA. Having made these substitutions into the above
equation we have 2 3
1 1 d
F = pA −(p + dp)(A + dA)+ 2p + dp dA = −dpA − d
dpdA (4.05j)
pressure,s d
2 2
where again we neglect the higher order term.
Combining Eq. (4.05g), (4.05h), (4.05i) and (4.05j) into Eq. (4.05f) we obtain
2 2 2 2
−ρV A + ρV A +2ρV AdV + V Adρ + ρV dA = −ρAgdz −dpA
Eliminating terms and rearranging this result, we have
2 dV dρ dA
ρAV dV + ρV A V + ρ + A = −ρAgdz −dpA (4.05k)
Substituting Eq. (4.05e) into this result yields
ρAV dV = −ρAgdz −dpA (4.05l)
Diving by A and rearranging we obtain
dp + ρV dV + ρgdz = 0 (4.05m)
c
2.25 Advanced Fluid Mechanics 4 Copyright © 2013, MIT
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