Convective Heat Transfer Analysis and Problem Solving Assignment

This document provides a detailed solution to a convective heat transfer assignment, addressing several key problems. It includes calculations for air velocity, utilizing formulas involving gravitational acceleration, vertical distance, and temperature differences. The solution also calculates heat transfer using the formula for a cylindrical object, considering parameters like diameter, dynamic viscosity, and temperatures of water and the plate. Furthermore, the assignment analyzes the Nusselt number in a 2-dimensional air flow over a square cylinder, emphasizing its relationship with the Reynolds number and Prandtl number. The document also discusses the conditions necessary for purely forced convective heat transfer, assessing the validity of the Reynolds number and other parameters. References to relevant literature are included to support the analysis.

Running head: CONVECTIVE HEAT TRANSFER
CONVECTIVE HEAT TRANSFER
Name of Student
Institution Affiliation

CONVECTIVE HEAT TRANSFER 2
Question 9.2
Air velocity = 0.65 √ gldt
273+te
Where g is the gravitational acceleration
L is the vertical distance
Dt is ts – te
ts is the uniform surface
te is the air temperature
V= 0.65 √ 9.81∗0.2∗30
273+ 40
V= 0.65 √ 58.86
313
V=0.65*0.43364
V=0.29197 m/S
Question 9.6
Q= [ π D2 h(T p−Tw ) ]
And from Re= ρ D
μwater we can obtain the value of D
Where ρ is the density of water.
D is the diameter
μwater Is dynamic viscosity = 8.9*10 -4
Q is the mean heat transfer
Tw is the temperature of water
Tp is the temperature of the plate

CONVECTIVE HEAT TRANSFER 3
1000 D
8.9∗10−4 = 50000
D=0.0445m
Q= [ π D2 h(Tp−Tw ) ]
Q= [ 3.142∗0.04452∗0.3(35−15) ]
Q= 0.03733
Question 9.12
At a 2 dimensional air flow over a square cylinder, the formula of the Nusselt number at a fixed
Grashof number of 10000 is given as below;
Nu=0.037 Re0.8 Pr 1/3
Where
Re is Reynold number
Pr is the Prandtl number
Nu is the Nusselt number
It hence shows that from the equation, the Nusselt number will increase when the Reynold
number increases. Nusselt number will reduce if the Reynold number reduces as long as Grashof
number is kept constant at 1000. In short Nusselt number is directly proportional to the Reynolds
number and this can be expressed mathematically as below;
Nuα Re
Nu=KRe
Where K is constant value which is Grashof number of 10000.
Question 9.18

CONVECTIVE HEAT TRANSFER 4
To have a purely forced convective heat transfer the following conditions must be met. This are
what are known as the VALIDITY
0.6≤ Pr ≤ 160
Re ¿ 10000
Re is 150
L
D >10,
Where L is the horizontal length of the pipe and D is the diameter of the pipe
1.8
0.024 =75, this is more than 10
This flow cannot be a purely forced convective since the value of Re is very low as compared to
the required validity which is 10000.

CONVECTIVE HEAT TRANSFER 5
References
Bejan, A. (2013). Convection Heat Transfer. Stoke: John Wiley & Sons.
Bradshaw, C. (2013). Physical and Computational Aspects of Convective Heat Transfer.
Liverpool: Springer Science & Business Media.