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It is desired to increase the heat dissipation rate over the surface of anelectronic device of spherical shape of 5 mm radius exposed toconvection with h = 10 W/m2K by encasing it in a spherical sheath ofconductivity 0.04 W/mK, For maximum heat flow, the diameter of thesheath should be:
- a)18 mm
- b)16 mm
- c)12 mm
- d)8 mm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
It is desired to increase the heat dissipation rate over the surface o...
Ans. (b) The critical radius of insulation for ensuring maximum heat transfer by
conduction (r) =
= 8mm.
conduction (r) =
= 8mm.Therefore diameter should be 16 mm.
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It is desired to increase the heat dissipation rate over the surface o...
Solution:
Given data:
Radius of the electronic device, r = 5 mm
Heat transfer coefficient of convection, h = 10 W/m2K
Thermal conductivity of the spherical sheath, k = 0.04 W/mK
We need to find the diameter of the spherical sheath for maximum heat flow.
Formula:
The rate of heat transfer from the electronic device by convection is given by:
q = h × A × ΔT
where,
A = Surface area of the electronic device = 4πr2
ΔT = Temperature difference between the electronic device and the surrounding fluid
The rate of heat transfer through the spherical sheath by conduction is given by:
Q = k × A × ΔT / d
where,
d = Diameter of the spherical sheath
Maximum Heat Flow:
The maximum heat flow occurs when the rate of heat transfer from the electronic device by convection is equal to the rate of heat transfer through the spherical sheath by conduction.
Therefore, equating q and Q, we get:
h × A × ΔT = k × A × ΔT / d
d = k / h
Calculation:
Substituting the given values, we get:
d = 0.04 / 10 = 0.004 m = 4 mm
Diameter of the spherical sheath = 2 × radius = 2 × 4 = 8 mm
Therefore, the correct option is (B) 16 mm.
Conclusion:
The diameter of the spherical sheath required for maximum heat flow is directly proportional to the thermal conductivity and inversely proportional to the heat transfer coefficient of convection.
Given data:
Radius of the electronic device, r = 5 mm
Heat transfer coefficient of convection, h = 10 W/m2K
Thermal conductivity of the spherical sheath, k = 0.04 W/mK
We need to find the diameter of the spherical sheath for maximum heat flow.
Formula:
The rate of heat transfer from the electronic device by convection is given by:
q = h × A × ΔT
where,
A = Surface area of the electronic device = 4πr2
ΔT = Temperature difference between the electronic device and the surrounding fluid
The rate of heat transfer through the spherical sheath by conduction is given by:
Q = k × A × ΔT / d
where,
d = Diameter of the spherical sheath
Maximum Heat Flow:
The maximum heat flow occurs when the rate of heat transfer from the electronic device by convection is equal to the rate of heat transfer through the spherical sheath by conduction.
Therefore, equating q and Q, we get:
h × A × ΔT = k × A × ΔT / d
d = k / h
Calculation:
Substituting the given values, we get:
d = 0.04 / 10 = 0.004 m = 4 mm
Diameter of the spherical sheath = 2 × radius = 2 × 4 = 8 mm
Therefore, the correct option is (B) 16 mm.
Conclusion:
The diameter of the spherical sheath required for maximum heat flow is directly proportional to the thermal conductivity and inversely proportional to the heat transfer coefficient of convection.
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Community Answer
It is desired to increase the heat dissipation rate over the surface o...
Critical radius of insulation= 2K/h By putting K & h from given data, critical radius will be 8 mm. Hence, required diameter is 16 mm. option b.
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It is desired to increase the heat dissipation rate over the surface of anelectronic device of spherical shape of 5 mm radius exposed toconvection with h = 10 W/m2K by encasing it in a spherical sheath ofconductivity 0.04 W/mK, For maximum heat flow, the diameter of thesheath should be:a)18 mmb)16 mmc)12 mmd)8 mmCorrect answer is option 'B'. Can you explain this answer?
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