PROBLEM 7.
13
KNOWN: Dimensions and surface temperatures of a flat plate. Velocity and temperature of air
and water flow parallel to the plate.
FIND: (a) Average convective heat transfer coefficient, convective heat transfer rate, and drag
force when L = 2 m, w = 2 m. (b) Average convective heat transfer coefficient, convective heat
transfer rate, and drag force when L = 0.1 m, w = 0.1 m.
SCHEMATIC: Ts = 50°C or 80°C
u∞ = 5 m/s u∞ = 5 m/s
T∞ = 20°C T∞ = 20°C
w=2m w = 0.1 m
Air, 1 atm Water
L=2m L = 0.1 m
(a) Air (b) Water
ASSUMPTIONS: (1) Steady-state conditions, (2) Boundary layer assumptions are valid, (3)
Constant properties, (4) Transition Reynolds number is 5 × 105.
PROPERTIES: Using IHT, Air (p = 1 atm, Tf = 35°C = 308 K): Pr = 0.706, k = 26.9 × 10-3
W/m⋅K, ν = 1.669 × 10-5 m2/s, ρ = 1.135 kg/m3. Air (p = 1 atm, Tf = 50°C = 323 K): Pr = 0.704,
k = 28.0 × 10-3 W/m⋅K, ν = 1.82 × 10-5 m2/s, ρ = 1.085 kg/m3. Water (Tf = 308 K): Pr = 4.85,
k = 0.625 W/m⋅K, ν = 7.291 × 10-7 m2/s, ρ = 994 kg/m3. Water (Tf = 323 K): Pr = 3.56,
k = 0.643 W/m⋅K, ν = 5.543 × 10-7 m2/s, ρ = 988 kg/m3.
ANALYSIS:
(a) We begin by calculating the Reynolds numbers for the two different surface temperatures:
u L 5 m/s × 2 m
Re L1 = ∞ = -5 2
= 5.99 × 105
ν1 1.669 × 10 m /s
u∞L 5 m/s × 2 m
Re L2 = = -5 2
= 5.49 × 105
ν2 1.82 × 10 m /s
Therefore, in both cases the flow is turbulent at the end of the plate and the conditions in the
boundary layer are “mixed.”
The average drag coefficient can be calculated from Equation 7.40. For the first case,
Cf,L1 = 0.074 Re-1/5
L1 - 1742 Re L1
-1
= 0.074(5.99 × 105 )-1/5 - 1742(5.99 × 105 )-1 = 2.27 × 10-3
Then
1 2
FD1 = Cf,L1 ρu ∞ As
2
1
= 2.27 × 10-3 × × 1.135 kg/m3 × (5 m/s)2 × 8 m 2 = 0.257 N
2
= 0.257 N <
Continued…
PROBLEM 7.13 (Cont.)
The average Nusselt number is calculated from Equation 7.38, with A = 871 for a transition
Reynolds number of 5 × 105.
Nu L1 = (0.037 Re4/5
L - 871) Pr
1/3
= ⎡0.037(5.99 × 105 ) 4/5 - 871⎤ (0.706)1/3 = 604.
⎣ ⎦
Then
h L1 = Nu L1k/L = 604 × 26.9 × 10-3 W/m ⋅ K/2 m = 8.13 W/m 2 ⋅ K <
and
q1 = h L1As (Ts - T∞ ) = 8.13 W/m 2 ⋅ K × 8 m 2 × (50°C - 20°C) = 1950 W <
Similarly for Ts = 80°C we find
FD2 = 0.227 N, h L2 = 7.16 W/m2·K, q2 = 3440 W <
(b) Repeating the calculations for water
u L 5 m/s × 0.1 m
Re L1 = ∞ = = 6.86 × 105
ν 7.291 × 10-7 m 2 /s
Re L2 = 9.02 × 105
The flow is turbulent at the end of the plate in both cases.
Cf,L1 = 0.074(6.86 × 105 )-1/5 - 1742(6.86 × 105 )-1 = 2.49 × 10-3
FD1 = 2.49 × 10-3 × 1/2 × 994 kg/m3 × (5 m/s) 2 × 0.02 m 2 = 0.620 N <
Nu L = ⎡ 0.037(6.86 × 105 ) 4/5 - 871⎤ (4.85)1/3 = 1450
⎣ ⎦
h L1 = 1450 × 0.625 W/m ⋅ K/0.1 m = 9050 W/m 2 ⋅ K <
q1 = 9050 W/m 2 ⋅ K × 0.02 m 2 × (50°C - 20°C) = 5430 W <
For the higher surface temperature,
FD2 = 0.700 N, h L2 = 12,600 W/m2·K, q2 = 15,100 W <
COMMENTS: (1) For air, kinematic viscosity increases with increasing temperature. This
decreases the Reynolds number which causes the transition to turbulence to move downstream,
thereby decreasing the drag force and average heat transfer coefficient. The heat transfer rate
increases for the higher surface temperature, however, because of the greater temperature
difference between the surface and air. (2) For water, kinematic viscosity decreases with
increasing temperature, causing the opposite trends as for air. The heat transfer rate increases
dramatically for the higher surface temperature because of the increases in both the heat transfer
coefficient and temperature difference. (3) Even though the water flows over a plate that is 400
times smaller, the drag force and heat transfer rate are larger than for air because of the smaller
viscosity and greater density, thermal conductivity, and Prandtl number. The discrepancy is
particularly great for the hear transfer rate. (4) The problem highlights the importance of carefully
accounting for the temperature dependence of thermal properties.
