The heating element assembly solutions 2022

1
Heat Transfer
1. Resistive heater is embedded in a ceramic rod that is 100mm in diameter. The
heating element assembly is exposed to two different conditions:
a. Water with a temperature of 300K flowing over the entire surface with
a velocity of 1m/s.
b. Air ,also at 300K but flown at a velocity of 10m/s over the entire
surface
To achieve this, the element requires 2.7 kW for (a) and 2.5 W for (b) to maintain
a spatially uniform surface temperature of 370K.Determine heat transfer
coefficient for each situation and compare them (Shuh 2014).
Solution
Dimensions of the ceramic rod : Length=100 cm,Diameter=30 mm
Water :
temperature=300 k velocity=1 m
s power applied=2.7 Kw
Uniform temperature=370 K
ˇh= qk
A (T2 −T1 )
Where ˇh−heat transfer coefficient .
qk−Power applied
A−area
T −temperature
A=1.0 x 0.3=0.3 m2
ˇh= 2.7 x 103
0.3(370−300)
¿ 2.7 x 103
0.3(70) =128.57 W /m2 K

2
Air
temperature=300 k velocity=10 m
s power applied=2.5 w
Uniform temperature=370 K
ˇh= qk
A (T2 −T1 )
A=1.0 x 0.3=0.3 m2
ˇh= 2.5
0.3(370−300)
¿ 2.5
0.3(70) =0.125 W / m2 K
Heat transfer coefficient for case 1 at 1m/s is higher compared to heat transfer
coefficient for case 2 at a velocity 10m/increase in velocity reduces the heat
transfer coefficient
2. Consider a pair of long fins with the same uniform circular cross section, but
one is constructed out of material A and other out of material B.The fins are
attached to a large plate of conductive metal such that the temperature of each
base is known to be 125o C, while the surface of the rods are exposed to
ambient air at 23 o C. It is observed that at the distance x A=0.15 m away from
the base for Fin A, the temperature was the same as at x A=0.075 m for Fin
B.If the thermal conductivity of material A is k A =70 W /m . K ,what is the
value of k A for rod B (Janna 2018).
Solution
Material A

3
Temperature=12 5oC Ambient temperature=2 3oC x A=0.15 m
k A = 70 W
m . k
Material B
Temperature=12 5oC Ambient temperature=2 3oC x A=0.075 m
k = QL
A ∆ T
Where k −thermal conductivity
Q−heat transferred
L−distance
A−area
∆ T −change of temperature
∆ T =125−23=103
Assuming that amount of heat transferred∧are are similar for the 2 fins .
70=Q x 0.15
A x 103
70 ( 103 A ) =0.15 Q
Q= 7210 A
0.15
= 48 067 A
For fin B
k = QL
A ∆ T
k = 48 067 A x 0.075
A x 103
k = 48 067 A x 0.075
A x 103
¿ 35 W /m . K