BME 50200 - HW 1 - Anthony Mathai
Problem 1 (shear-thinning / power-law): Vx (y) and Q–pressure
What we’re doing. Fluid between two flat plates, gap h, width w (into the page), length L. Let
y ∈ [−h/2, h/2] with y = 0 in the middle.
Assumptions Steady, incompressible, laminar, fully developed. Velocity is V = (Vx (y), 0, 0),
so Vy = Vz = 0. No slip at the plates: Vx (±h/2) = 0. Gravity in x is tiny, so we ignore it.
Start from the x-momentum equation. In components:
∂τ
∂p
∂Vx
∂Vx
∂Vx
x
ρ ∂V
+
V
+
V
+
V
= − ∂x
+ ∂yyx + (terms in x, z).
x
y
z
∂t
∂x
∂y
∂z
Because it’s steady, fully developed, and only changes with y, all time/x/z derivatives drop out
and Vy , Vz = 0. So we get
∂p dτyx
0=−
+
.
∂x
dy
Write the pressure gradient as
τyx (0) = 0:
∂p
∆p
= −
≡ −G with G > 0. Integrate and use symmetry
∂x
L
τyx (y) = −G y.
Fluid rule (power law). For shear-thinning (0 < n < 1):
n−1 dVx
dy ,
x
τyx = K dV
dy
K > 0.
Solve for Vx (y). On the top half (y ≥ 0), τyx = −Gy < 0, so
1/n
G
K
1/n
dVx
=−
dy
y 1/n .
Integrate from 0 to y; let H = h/2:
Vx (y) = Vx (0) −
G
K
n
y (n+1)/n .
n+1
Use Vx (H) = 0 (no slip) to get the centerline speed:
n
Vmax = Vx (0) =
n+1
G
K
1/n
H (n+1)/n .
Final profile (use |y| for both sides):
Vx (y) =
n
n+1
G
K
1/n H (n+1)/n − |y|(n+1)/n ,
1
H = h2 .
Flow rate Q and pressure gradient.
Z H
Q = 2w
Vx (y) dy =
0
2n w
2n + 1
G
K
1/n (2n+1)/n
h
.
2
With G = ∆p/L,
∆p
(2n + 1)Q n 2 2n+1
=K
.
L
2n w
h
Wall shear + a couple handy ratios.
τw = τyx (H) =
∆p h
,
L 2
Vavg =
Vavg
Q
n+1
=
.
,
wh Vmax
2n + 1
(As n gets smaller, the profile gets flatter.)
Quick Newtonian check. Put n = 1, K = µ:
Q=
wh3 ∆p
,
12µ L
τw =
6µQ
,
wh2
Vavg /Vmax = 2/3.
Newtonian case (clear answer)
Constitutive law. Newtonian means
τyx = µ
dVx
.
dy
From momentum to the profile. With steady, fully developed flow that only varies with y:
0=−
∂p
d2 Vx
+µ
,
∂x
dy 2
∂p
∆p
=−
.
∂x
L
Integrate, use dVx /dy|y=0 = 0 (symmetry) and Vx (±h/2) = 0 (no slip):
Vx (y) =
∆p h2
− y2 ,
2µL 4
Vmax = Vx (0) =
∆p h2
.
8µL
Flow rate and pressure gradient.
Q=
wh3 ∆p
12 µ L
∆p
12 µ Q
=
.
L
w h3
⇐⇒
Wall shear and the go-to design formulas.
∆p h
τw =
L 2
⇒
6µQ
τw =
w h2
Vavg =
Q
,
wh
r
⇒
Average vs. max speed:
Vavg
2
= .
Vmax
3
2
h=
6µQ
.
w τw
n−1
dVx
x
How this lines up with power-law. Power-law says τyx = K dV
dy
dy . Set n = 1, K = µ
and you get Newtonian. Plugging n = 1, K = µ into those power-law results gives:
3
2(1) w
∆p
h
wh3 ∆p
Q=
=
,
2(1) + 1 Lµ
2
12µ L
2
2
(2 · 1 + 1)Q
6µQ
τw = µ
=
,
2 · 1w
h
wh2
which exactly matches the Newtonian stuff above.
Problem 2 (Newtonian): from Navier–Stokes to Vx (y), Q, τw , and h
Setup (same channel). Newtonian fluid with viscosity µ. Steady, incompressible, laminar, fully
developed. V = (Vx (y), 0, 0), no slip at y = ±h/2, ignore gravity along x.
Start from x-momentum and cross out what dies.
2
∂p
∂Vx
∂Vx
∂Vx
∂ Vx
∂ 2 Vx
∂ 2 Vx
x
ρ ∂V
+ ρgx .
+
V
+
V
+
V
=
−
+
µ
+
+
x
y
z
2
2
2
∂t
∂x
∂y
∂z
∂x
∂x
∂y
∂z
Steady ⇒ ∂/∂t = 0; fully developed ⇒ ∂/∂x = 0; no z-variation ⇒ ∂/∂z = 0; Vy = Vz = 0; gx ≈ 0.
So we land on
∂p
d2 Vx
0=−
+µ
.
∂x
dy 2
∂p
∆p
Let
=−
.
∂x
L
Get Vx (y). Integrate twice; use dVx /dy|0 = 0 and Vx (±h/2) = 0:
Vx (y) =
∆p h2
− y2 ,
2µL 4
Vmax =
∆p h2
.
8µL
⇐⇒
∆p
12µQ
=
.
L
wh3
τw =
∆p h
.
L 2
Q and ∆p/L.
Z h/2
Q=w
Vx (y) dy =
−h/2
wh3 ∆p
12µ L
Wall shear and link to Q, h.
τyx (y) = µ
dVx
∆p
=−
y ⇒
dy
L
Eliminate ∆p/L:
6µQ
τw =
wh2
r
⇒
h=
6µQ
.
w τw
Numbers (cgs, quick example). w = 2 cm, µ = 0.0087 g cm−1 s−1 , target τw = 20 dyn cm−2 ,
pump limit Q ≤ 2 cm3 /s.
At the max flow Q = 2:
s
6(0.0087)(2) √
h=
= 0.00261 ≈ 0.0511 cm .
(2)(20)
3