This screencast involves an internal flow problem with a constant surface temperature,
and what we have is water that's flowing at a mass flow rate of 3 kg/s through a 50 millimeter
diameter tube, so that's 0.05 meters. It's to be heated from 30 degrees C to some outlet
temperature, and that's what we're looking for. The length of our tube is 12 meters,
and the surface temperature is maintained at 110 degrees C. So the first thing we need
to do is look up the properties of water, and so the first thing is the heat capacity
which is 4181 joules per kilogram-K. Our viscosity is 547*10^-6 newton-seconds per
meter squared. Our k, and remember that it's the k of the fluid, is 0.643 watts per meter
K, and finally, our Prandtl Number is 3.56. And looking up these, or showing you these
properties probably means that we're going to have to use some kind of correlation to
find h, our heat transfer coefficient. So we are going to let delta Ti equal T of the
surface minus Tmi and our delta To is going to be T of the surface minus Tmo. Now, our
heat transfer is through convection, which is m*Cp*(delta Ti - delta To). But it's also
equal to h, which is the overall average heat coefficient for the entire tube, times the
surface area times our temperature log mean. And hopefully you remember that whenever we're
working with a constant surface temperature, the best way to use our Tm in and out is in
a logarithmic fashion. So our delta T log mean is delta To minus delta Ti over ln of
this delta To over delta T in. And the problem with this, as you can see, is we have our
Tm out involved in a number of the different terms. So the first thing we can do is take
that delta T out from the natural log term. So now we have Ts-Tmo divided by Ts-Tmi equals
exponential of minus h*As, surface area, n*Cp. And now if we want to solve for our Tmo, it's
just Ts - (Ts-Tmi) times the exponential of -h*As, n*Cp. So when we look at that equation,
we see that we have everything we need except for h. And so we're going to have to use a
correlation for internal flow to find our h. So the first thing we do is find if it's
turbulent or laminar, so our Reynolds Number is 4 times the mass flow rate over pi*d*mu,
and this equals 1.397x10^5, certainly over that 2300 that we need for laminar flow, so
it is turbulent. That means that we can use the Dittus-Boelter Correlation, where the
Nusselt Number equals 0.023Re^(4/5) and Pr^n, where n equals 0.3 for cooling and 0.4, where
we're going to be, using it for heating. So when we calculate our Nusselt Number, we come
out with 499.3, our h equals k over D times the Nusselt Number, which is 6421 W/m^2*K.
So finally, we can put it into our equation that we had before, where Tmo equals the surface
temperature, which is 110, minus 110-30 exponential of minus h, or -6421, so our surface area
is pi*d*L, or 1.885 meters, and we divide this by the mass flow rate, which is 3 kilograms
per second times that Cp, and when we finish calculating, our Tm out is going to equal
79.5 degrees C.