11. Three rods of identical geometry but different.
materials are welded to form the english alphabet
letter Y as shown. If their conductivity are K, 2K and
3K and end temperature are 20°C. 20°C and 100°C
respectively, then their junction temperature T is
20°C
100°C
ЗК
2K
20°C
(2) 60°C
(4) 40°C
(1) 70°C
(3) 50°C
Answers
Answer:
Explanation:
3k (100-T)/L=k (T-20)/L+2k (T-20)/L
All the L and K will cancel out.
3 (100-T)=(T-20)+2 (T-20)
300-3T=T-20+2T-40
300+20+40=T+3T+2T
360=6T
360/6=T
60=T
Therefore,T=60°C
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Therefore the junction temperature is 60°C. ( Option-2 )
Given:
For rod-1:
Thermal Conductivity = k₁ = k
End Temperature = T₁ = 20°C
For rod-2:
Thermal Conductivity = k₂ = 2k
End Temperature = T₂ = 20°C
For rod-3:
Thermal Conductivity = k₃ = 3k
End Temperature = T₃ = 100°C
To Find:
The Junction temperature ( T°C )
Solution:
The given question can be solved very easily as shown below.
For rod-1:
Thermal Conductivity = k₁ = k
End Temperature = T₁ = 20°C
For rod-2:
Thermal Conductivity = k₂ = 2k
End Temperature = T₂ = 20°C
For rod-3:
Thermal Conductivity = k₃ = 3k
End Temperature = T₃ = 100°C
Heat flow Q = kA ΔT
Where A = Cross-sectional Area
ΔT = Temperature Difference
⇒ Heat is flowing from rod-3 to rod-1 and rod-2.
⇒ Q₃ = Q₁ + Q₂
⇒ k₁ A₁ ΔT₁ + k₂ A₂ ΔT₂ = k₃ A₃ ΔT₃
As the Area of the cross-section is constant for all rods,
⇒ k₁ ΔT₁ + k₂ ΔT₂ = k₃ ΔT₃
⇒ k ( T - 20 ) + 2k ( T - 20 ) = 3k ( 100 - T )
⇒ 3k ( T-20 ) = 3k( 100 - T )
⇒ T-20 = 100-T
⇒ 2T = 120
⇒ T = 60
Therefore the junction temperature is 60°C.
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