Example 21:
Consider steady, incompressible, laminar flow of a Newtonian fluid of viscosity, µ, in a
pipe of radius R as shown. Assume flow is far downstream of the entrance so the flow is
essentially axial or fully developed (both Vr = 0 and Vθ = 0). The fluid is driven along the
pipe by an applied pressure drop, ∆p, over a length of pipe,  . The effect of gravity is
negligible for this flow. Solve for the following:
(a) velocity distribution as a function of the maximum velocity,
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(b) mean velocity,
(c) volumetric flow rate,
(d) shear stress distribution, and
(e) wall shear stress.
Known: Pressure driven pipe flow, µ, R, Vr = 0, Vθ = 0, ∆p, 
Assumptions: Steady, incompressible, laminar flow of a Newtonian fluid, constant
properties, fully-developed, negligible gravity effects
r
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Find: (a) u(r), (b) V , (c) Q, (d) τrx, and (e) τw
Solution:
2R
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Answers:
(a)
⎡ ⎛ r ⎞ 2 ⎤
u( r) = umax ⎢1− ⎜ ⎟ ⎥
⎣ ⎝ R ⎠ ⎦
(b)
V =
umax R 2 Δp
=
2
8µ 
(c)
Q=
π R 4 Δp π D4 Δp
=
8 µ  128 µ 
(d)
τ rx =
(d)
τ w = τ rx ( r = R) =
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R Δp ⎛ r ⎞
⎜ ⎟
2  ⎝ R ⎠
R Δp
2 
x
Analysis:
(a) Calculate velocity profile, u(r)
Conservation of mass for incompressible flow:
 1 ∂ ( r Vr ) 1 ∂V ∂u
θ
∇⋅V =
+
+
=0
r ∂r
r ∂θ ∂x
For fully developed flow, Vr = Vθ = 0, thus
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∂u
=0
∂x
(which satisfies the fully developed condition)
Momentum equation for steady state, incompressible, constant property flow
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

DV

ρ
= ρ g − ∇p + µ∇ 2V
Dt
For fully-developed flow in a pipe there is NO fluid acceleration
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




DV ∂V
ap =
=
+ V ⋅∇ V =0
Dt
∂t
(
)

and for negligible gravity effects ( ρ g ≈ 0 ) momentum equation reduces to
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
µ∇ 2V = ∇p (balance between forces due to viscosity and pressure).
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r-component of momentum equation for cylindrical coordinates:
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µ∇ 2Vr =
∂p
∂p
⇒
= 0 and
∂r
∂r
p ≠ f ( r)
θ-component of momentum equation for cylindrical coordinates:
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€1 ∂p
∂
p
⇒
µ∇ 2Vθ =
= 0 and p ≠ f (θ )
r ∂θ
∂θ
thus, p( x ) and can write
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∂p dp Δp
=
=
∂x dx

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x-component of momentum equation for cylindrical coordinates:
⎡1 ∂ ⎛ ∂u ⎞ 1 ∂ 2 u ∂ 2 u ⎤ ∂p
µ ⎢
⎜ r ⎟ + 2 2 + 2 ⎥ =
∂x ⎦ ∂x
⎣ r ∂r ⎝ ∂r ⎠ r ∂θ
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µ d ⎛ du ⎞
Δp
⎜ r ⎟ = −
r dr ⎝ dr ⎠

(where for u(r) we can use ordinary derivative)
Integrate once to get:
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⎛ du ⎞
r
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1 Δp

∫ d⎜⎝ r dr ⎟⎠ = − µ
∫ r dr
du
1 Δp r 2
=−
+ C1
dr
µ  2
Integrate again to obtain:
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1 Δp

∫ du = − 2 µ
u=−
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∫ r dr + ∫
r 2 Δp
+ C1 ln( r) + C2
4µ L
Impose boundary conditions:
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C1
dr
r
du
= 0 by symmetry and u( r = R) = 0 (no-slip at wall):
dr r= 0
du
1 Δp r C1
=−
+
dr
µ  2 r
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1 Δp 0 C1
=−
+
=0
µ  2 0
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du
dr r =0
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R 2 Δp
u( r = R) = −
+ C2 = 0
4 µ €
⇒ C1 = 0
R 2 Δp
⇒ C2 =
4µ 
Substituting back into our solution:
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€2
R 2 Δp ⎡ ⎛ r ⎞ ⎤
u( r) =
⎢1 − ⎜ ⎟ ⎥ which is a parabolic velocity profile
4 µ  ⎣ ⎝ R ⎠ ⎦
Write in terms of the maximum velocity umax at the center of the pipe at r = 0
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2
R 2 Δp ⎡ ⎛ 0 ⎞ ⎤
u( r = 0) =
⎢1 − ⎜ ⎟ ⎥ = umax ⇒
4 µ  ⎣ ⎝ R ⎠ ⎦
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⎡ ⎛ r ⎞ 2 ⎤
u( r) = umax ⎢1− ⎜ ⎟ ⎥
⎣ ⎝ R ⎠ ⎦
umax =
R 2 Δp
4µ 
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(b) Calculate the mean velocity, V
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V =
1
ρA
∫
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1
V =
π R2
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2 umax
V =
R2
V =
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A
ρ u dA
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2π
R
0
0
∫ ∫
∫
R
0
2
u( r) r dr dθ = 2
R
⎛
r 3 ⎞
2 umax
⎜ r − 2 ⎟ dr =
R ⎠
R2
⎝
∫
R
0
umax
⎡ ⎛ r ⎞ 2 ⎤
⎢1 − ⎜ ⎟ ⎥ r dr
⎣ ⎝ R ⎠ ⎦
R
⎛ r 2
⎛ 1 1 ⎞
r 4 ⎞
− ⎟
⎜ −
2 ⎟ = 2 umax ⎜
⎝ 2 4 ⎠
⎝ 2 4 R ⎠ 0
umax R 2 Δp
D2 Δp
=
=
2
8µ 
32 µ 
(c) Calculate volumetric flow rate, Q
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⎛ R 2 Δp ⎞
2
Q = V A = ⎜
⎟ (π R )
8
µ

⎝
⎠
Q=
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π R 4 Δp π D4 Δp
=
8 µ  128 µ 
(d) Calculate shear stress distribution, τrx
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2
⎛ du ⎞
⎛ −2 r ⎞
d ⎡ ⎛ r ⎞ ⎤
τ rx = µ ⎜ ⎟ = µ umax
⎢1 − ⎜ ⎟ ⎥ = µ umax ⎜ 2 ⎟
⎝ dr ⎠
⎝ R ⎠
dr ⎣ ⎝ R ⎠ ⎦
τ rx =
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R Δp ⎛ r ⎞
⎜ ⎟
2  ⎝ R ⎠
(e) Calculate wall shear stress, τw = τrx(r = R)
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τw =
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R Δp D Δp
=
2 
4 