HW #4 /Tutorial # 4
WRF Chapter 18; WWWR Chapter 19
ID Chapter 6
• Correction
Tutorial # 4
• Question WWWR #
WWWR #19.1,19.4.
19.19
ID # 6.33. 6.37.

 3.94 Pr (Pr  0.954) Gr
To be discussed on Feb. x
Nu  0.508Pr (Pr  0.954) Gr ; h
16, 2016.
• By either volunteer or
class list.
•
•
•
•
1/ 2
1/ 4
1/ 4
x
1/ 2
x
1/ 4
1/ 4
x
x
 Nu x
k
x
Convective Heat Transfer
Fundamental Considerations In
Convective Heat Transfer
• Two main classifications of convective heat
transfer
• These have to do with the driving force causing
fluid to flow
Natural or free convection
Fluid motion results from heat transfer
Fluid heated/ cooled -> density change/ buoyancy effect ->
natural circulation in which affected fluid moves of its own
accord past the solid surface
- fluid it replaces is similarly affected by the energy transfer process is repeated
Forced convection
Fluid circulation is produced by external agency (fan or pump)
Analytical Methods
(a) Dimensional Analysis
(b) Analogy between Energy and Momentum Exchange
Significant Parameters In
Convective Heat Transfer
A. Both have same dimensions L2/t; thus their ratio must be
dimensionless
B. This ratio, that of molecular diffusivity of momentum to the
molecular diffusivity of heat, is designed the Prandtl number
n
mcp
Pr  a = k
Prandtl number
•observed to be a combination of fluid properties;
•thus Pr itself may be thought of as a property.
•Primarily a function of temperature
s
A ratio of conductive thermal resistance to the convective thermal
resistance of the fluid
Nusselt number
Nu  hL
k
Where the thermal conductivity of the fluid as opposed to that of the
solid, which was the case in the evaluation of the Biot modulus.
Dimensional Analysis of
Convective Energy Transfer
Forced Convection
Dimensional
Analysis for
Forced Convection
Natural Convection
Dimensional
Analysis for
Natural Convection
Courtesy Contribution
by ChBE Year
Representative, 2004.
Energy and Momentum Transfer
Analogies
The Colburn analogy expression is
St Pr 2/3 = Cf
(19-37)
2
8)
9)
s
on
er
Example 1
o
Water at 50 F enters a heat-exchanger tube having an inside diameter of 1o in and a length of 10
ft. The water flows at 20 gal/min. For a constant wall temperature of 210 F estimate the
=(90+210)/2 = 150
= (50+130)/2
Film temperature = (water mean bulk temperature + pipe wall temperature)/2
Mean bulk temperature of water = (inlet + outlet)/2
Second iteration is required since if |TL – 130| >
o
3 F?