Methodology
Geometry
kf, νf, u∞, L or D
Re, Pr
Nu Correlation
h=
k f Nu
Nu
Nu
h=
L
h
q” = h (Ts - T∞)
h=
q”, T
1
1 L
hdA
(
=
As
o hdx)
s
As
L
q, T
h
k f Nu
L
q = h A(Ts - T∞)
Tf =
Ts + T
2
Nusselt Correlations
Flat plate in Parallel Flow
1. Constant surface temperature (Ts)
1/3
1/2
1/3
(7.30)
1) Laminar: For Pr > 0.6,Nu x = 0.332 Re1/2
x Pr (7.23) Nu x = 0.664 Rex Pr
2) Turbulent: Nux = 0.0296 Rex4/5 Pr1/3
0.6  Pr  60
(7.36)
Assuming Rex ,c = 5  105 , Nu L = (0.037 ReL4/5 − 871) Pr1/3
2. Constant heat flux (q”) Turbulent (0.6 < Pr < 60)
1/3
(7.45) Nux = 0.0308 Rex4/5 Pr1/3 (7.46)
1) Local: Laminar (Pr > 0.6) Nux = 0.453 Re1/2
Pr
x
1/3
(7.49)
2) Average Laminar Nu x = 0.680 Re1/2
x Pr
Cylinders of Circular and Noncircular Cross Section:
Tables 7.2 and 7.3 & (7.52), (7.53), and (7.54)
Zukauskas
Hilpert
Nu D = C ReDm Pr1/3 (7.52)
Churchill and Bernstein: For all ReDPr > 0.2
Nu D = 0.3 +
1/2
D
0.62 Re Pr
1/3
[1 + ( 0.4 / Pr ) ]
2/3 1/4
5/8
  Re


D
1 + 
 
282,000
 
 
Properties at the T∞ except for Prs (use Ts)
4/5
(7.54)
Nu D k f
Sphere in Cross Flow Whitaker Nu D = 2 + 0.4Re1/2 + 0.06Re2/3 Pr 0.4 (  /  )1/4
( D
(7.56) h =
D )
s
Properties at T except  s (at Ts )
D
0.71  Pr  380


4
3.5

Re

7.6

10
D


1.0  (  /  s )  3.2 


HW# 4–1 (Local and average convection coefficient & Nu number) For laminar flow over a flat plate, the
local Nu number Nux is proportional to Rex1/2, where x is the distance from the leading edge (x = 0) of the
plate, and fluid properties do not change. Complete the below relations:
(a) Local Nu at x Nu x  x
(b) Local convection coefficient at x
hx  x
(c) Ratio of the average Nu between the leading edge and location x on the plate to the local Nu at x
(d) Ratio of the average convection coefficient between the leading edge and location x on the
plate to the local coefficient at x
Nu x
=
Nu x
hx
=
hx
HW# 4–2 (12 pt, Flat plate in parallel flow) A flat plate of width 1 m is maintained at a uniform surface
temperature of Ts = 150°C by using independently controlled, heat-generating rectangular modules of thickness a =
10 mm and length b = 50 mm. Each module is insulated from its neighbors, as well as on its back side. Atmospheric
air at 25°C flows over the plate at a velocity of 30 m/s. The thermophysical properties of the module are k = 5.2
W/m-K, cp = 320 J/kg-K, and p = 2300 kg/m3.
(a)Find the required power generation 𝑞ሶ (W/m3), in a module positioned at 700 mm from the leading edge [700,750]
(b) Find the maximum temperature Tmax in the heat generating module.
HW# 4–3 (Flat plate in parallel flow) Steel (AISI 1010) plates of
thickness δ = 6 mm and length L = 1 m on a side are conveyed from a heat
treatment process and are concurrently cooled by atmospheric air of
velocity u∞ = 10 m/s and T∞ = 20°C in parallel flow over the plates. For an
initial plate temperature of Ti = 300°C, what is the rate of heat transfer
from the plate? What is the corresponding rate of change of the plate
temperature? The velocity of the air is much larger than that of the plate.
HW# 4–4 (18 pt, Cylinder and sphere in cross flow)
(a) Suppose that a person is standing outside in one cold morning [Figure (a)]. Wind is blowing at a speed (u∞,c) of 2
m/s, and the air temperature (T∞,c) is -8oC (= 17.6oF = 265 K). We model the person's clothed body as a cylinder in
cross flow. The diameter (Dc) and height (H) are 0.4 m and 1.7 m, respectively. Calculate the heat loss, qc(W) from
the clothed body to the cold air. Neglect the heat transfer from the ends of the cylinder.
(b) In the afternoon, air temperature (T∞,h) goes up to 2oC (= 35.6oF = 275 K), but he feels the same cold as in part (a)
due to a stronger wind. What is the wind speed (u∞,h) in this case?
[Hint: our feeling depends on how much heat we lose. Thus, heat loss should be the same as in part (a)]
(c) Going back to the morning scenario (T∞,c = -8oC
and u∞,c = 2 m/s), this person can curl up
(crouching as compared to standing up) to reduce
heat loss [Figure (b)]. In this case, we can model
this heat transfer as convection by cross ow over a
sphere. The diameter of the sphere (Ds) is 0.74 m,
which results in almost the same volume as part
(a). Calculate the heat transfer rate, qs(W), and by
how much the heat loss is reduced. Discuss where
this reduction originates from.
In all parts, the clothed surface temperature (Ts) is assumed to be uniform at 12oC.
Here, use Eqs. (7.52) and (7.56) for Nu correlations.