3.185 - Recitation Notes
December 5, 2003
Topics Covered
• Drag Force
• Turbulence
• Forced Convection
Drag force can generally be written as (for both laminar and turbulent flow)
1
Fd = ∫ kf x dA = kf L A where k = ρu 2
2
If flow is laminar, the drag force can be derived from τ, which is given by the NavierStoke’s equation.
Friction Factor for Certain Geometries
Flat Plate
Laminar flow (Re < 105)
Boundary layer thickness: δ =
fx =
0.664
Re x
fL =
1.328
Re L
Friction factor:
5.0 x
νx
= 5.0
u∞
Re x
Drag force: Fd = 0.664W ρµu ∞3 L
Transition flow (105 < Re < 107)
Boundary layer thickness: δ =
0.37 x
Re 0x.2
Turbulent flow (107 < Re)
fL =
Friction factor:
0.455
(log Re L )2.58
0.146
fL =
Re 0L.2
(either one)
When flow is fully developed (i.e. x > entrance length), fx = fL.
Entrance length: Le =
u av H 2
where H: distance between plates
ν 100
Sphere
24
for Stoke’s flow (Re < 0.1)
Re
f = 0.44 for Re >> 1 (see fig. 12.4 in textbook)
Drag force: FD = 6πµRu ∞
Friction factor: f =
Tubes
16
for laminar flow, see eqn. (14-12) – (14-15) in textbook
Re
for friction factor on turbulent flow and various pipe roughness
Friction factor: f =
Circular disc and square plates
Friction factor: f = 1 for Re >> 1 (see fig. 12.4 in textbook)
Forced Convection
Small Pr (Pr < 0.1)
• δ c or δ T >> δ u
• Weakly coupled: heat transfer does not affect fluid flow
• For heat transfer, ux is assumed constant
• Solution similar to boundary layer problem in moving solid
Boundary layer thickness: δ T = 3.6
Boundary layer thickness ratio:
Temperature profile:
T − Tenv
T∞ − Tenv
αx
u∞
1
−
δT
= 0.72 Pr 2
δu
⎛
⎞
⎜
⎟
⎜ y ⎟
= erf ⎜
⎟
⎜ 2 αx ⎟
⎜ u ⎟
∞ ⎠
⎝
q = hx (Ts − T∞ )
Heat/Mass flux: hx =
hL = 2
J = hD (C s − C∞ )
kρC P u ∞
πx
hD , x =
kρC P u ∞
πL
Du ∞
πx
hD , L = 2
Du ∞
πL
High Pr (Pr > 5)
Boundary layer thickness ratio:
1
−
δ
δT
or C = 0.975 Pr 3
δu
δu
Heat transfer coefficient h for forced convection can be found by computing the Nusselt
number.
Nu x k
Nu L k
hL =
x
L
1
1
⎫
1
Re x2 Pr 2 ⎪
Nu x =
π
⎪
for small Pr (Pr < 0.1)
1
1⎬
2
⎪
Nu L =
Re L2 Pr 2 ⎪
π
⎭
1
1
⎫
0.564 Re x2 Pr 2 ⎪
Nu x =
1 + 0.90 Pr ⎪⎪
for medium Pr
1
1 ⎬
1.128 Re L2 Pr 2 ⎪⎪
Nu L =
1 + 0.90 Pr ⎪⎭
1
⎫
Nu x = 0.332 Re x2 Pr 0.343 ⎪
⎬ for large Pr (Pr > 0.6)
1
⎪
0
.
343
Nu L = 0.664 Re L2 Pr
⎭
* also see class handout for others Nu relations
Nusselt Number: hx =