Acartridge electrical heater is shaped as a cylinder of length l=200mm and outer diameter d=20 mm. under normal operating conditions the heater dissipates 2kw while submerged in a water flow that is 20℃ and provides a convection heat transfer coefficient of ℎ = 5000 w/m2k. neglecting heat transfer from the ends of the heater, determine its surface temperature . if the water flow is in advertently terminated while the heater continues to operate, the heater surface is exposed to air that is also at 20℃ but for which ℎ = 50 w/m2k. what is the corresponding surface temperature?

T(water)=50.32℃

T(air)=3052.6℃

Explanation:

Hello!

To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.

The equation is as follows!

Q = heat

h = heat transfer coefficient

Ts = surface temperature

T = fluid temperature

a = heat transfer area

The surface area of a cylinder is calculated as follows

Where

D=diameter=20mm=0.02m

L=leght=200mm)0.2m

solving

For water

Q=2Kw=2000W

h=5000W/m2K

a=0.01319m^2

Tα=20C

solving for ts

for air

Q=2Kw=2000W

h=50W/m2K

a=0.01319m^2

Tα=20C

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