An Example Of Convective Heat Transfer


Question:

Calculate the heat transferred to a mercury thermometer intially at 70oF that is placed in an oven at 400oF. The thermometer bulb has a diameter of 0.24 inches. The mean convective heat-transfer coefficient between the outside of the thermometer and the air in the oven will be taken as h = 2 Btu/(h*ft2*oF). To find the area, assume the bulb is circular.

Answer:

We first want to apply the convective heat transfer equation and decide how to use it in this particular example.

Q = -hA(Ts – T)

Where h = 2 Btu/(h*ft2*oF)

  T = 400oF = oven temperature

  Ts = 70oF = surface temperature

  A = area of a circular bulb = 4*Pi*(Diameter/2)2

  = 4P *(D/2)2 = 4*P *(0.24inches/2)2 note: P =3.14

  = 0.181 inches2

When looking at our units of the heat transfer coefficient, h, units of square feet, not inches are used. Therefore the units of area should be in square feet.

A = 0.181 inches2 * (1foot / 12 inches)2 = 0.00126 ft2

We could now plug these values into the original equation and find the heat transferred.

Q = -hA(Ts – T)

Q = - ( 2 Btu/(h*ft2*oF) )*0.00126 ft2*(70oF – 400 oF)

Q = 0.8316 Btu/h

Let’s consider why the heat turned out to be a positive value. We were focusing on the thermometer, and the thermometer gained heat when raising temperature. A gain of heat is represented by a positive value, while a heat loss is represented by a negative value.

When doing an example on radiant heat transfer, the same equation would be used, the only difference would be the value of the heat transfer coefficient, hr.


Return to Convective Heat Transfer
Return to Heat Transfer
[Introduction | Kinetics | Heat Transfer | Mass Transfer | Bibliography]

This project was funded in part by the National Science Foundation and is advised by Dr. Masel and Dr. Blowers at the University of Illinois.

© 2007 Arizona Board of Regents for The University of Arizona