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
