Newtons Law of cooling

For fluid flow over a rectangular flat plate, is it valid if h = h(x), where "x" is the length from the leading edge of the plate?

if we start with

dq = h*(T-T_amb)dA

where

dA = w*dx (w is the width of the plate)

T= T(x)

h=h(x)

q = w*Int(x,0){ h(x)( T(x) - T_amb )*dx

I seem to get pretty wacky results using this method.

I had to wiki it to understand your question.

Newtons law:
dq/dt = h.A(Tenv - T(t))

Where:
q is the thermal energy in joules
h is the heat transfer coefficient
Tenv is the temperature of the environment
T(t) is the object temperature at a given time

Are these two even equivalent, one being a rate of change of the thermal energy in terms of time, the other in terms of area..?

Originally posted by talzamir
Are these two even equivalent, one being a rate of change of the thermal energy in terms of time, the other in terms of area..?

Yeah, they are equivalent. I would have changed the variable in the time rate to "Q" to clarify, but he has the terms defined correctly