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1 August, 13:49

An electric resistance heater is embedded in a long cylinder of diameter 30 mm. when water with a temperature of 25 c and velocity of 1 m/s flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of 90 c is 28 kw/m.

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  1. 1 August, 17:36
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    What is known: Long, 30mm diameter cylinder with entrenched electrical heater; power to maintain a specified surface temperature for water and air flows.

    What to look for: Convection coefficients for the water and air flow convection processes, h sub w and h sub a, correspondingly.

    Assumptions: the flow is cross-wise over cylinder which is very long in the direction normal to flow.

    Analysis: The convection heat rate from the cylinder per unit length of the cylinder has the form:

    q’ = h (πD) (Tsub s - T sub infinite)

    and solving for the heat transfer convection coefficient is

    h = q’ / (πD) (Tsub s - T sub infinite)

    Solution: substitute the numerical values for the water and air:

    Water

    h sub W = 2.8 x 10^3 W/m / π x 0.030m (90-25) degress Celsius

    = 4,570 W/m^2 K <

    Air

    h sub a = 400W/m / π x 0.030m (90-25) = 65 W/m^2 K <

    Other info: the air velocity is 10 times that of the water flow, yet h sub w = 70 x h sub a
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