This is a pool boiling problem where we have ten copper tubes that are 15 millimeters in
diameter and each one is 0.5 meters long, and they're used to boil saturated water at
1 atmosphere. And if the tubes are operated at 80% of the critical heat flux, what we
want to know is what is the vapor production rate with these ten tubes. In addition, we
want to know what is the corresponding surface temperature of the tubes. So the heat
transfer rate is the surface flux times the surface area, which is the surface flux times
the number of tubes times pi D L. And this is equal to m dot h sub fg. And we should
realize that this is nucleate boiling because we are only at 80% of the critical heat flux. So what
we're looking for is this vapor production rate, or our m dot, which is our heat flux
times the number of tubes times pi D L, divided by h sub fg, or the heat of vaporization.
So the first thing we're going to need to do is find this heat flux, which is, again,
0.8 times q max. So let's do our properties for our saturated water. So our rho sub l
is 957.9 kilograms per meter cubed, that's the density of the liquid. The heat capacity
of the liquid is 4217 joules per kilogram K. The viscosity of the liquid is 279 times
10 to the minus 6 newton seconds per meter squared. The Prandtl number of the liquid
is 1.76, and the heat of vaporization is 2257 kilojoules per kilogram. And we're actually
going to change that to 2257 times ten to the third joules per kilogram since everything
else is in joules. Finally, we have the surface tension, which is 58.9 times 10 to the minus
3rd newtons per meter. And our density of the vapor, which is 0.5955 kilograms per meter
cubed. So the equation for the critical heat flux is 0.149 h sub fg rho of the vapor, the
surface temperature times gravity times the difference in densities, over the density
of the vapor squared, and this is all raised to the 1/4. And when we calculate that out,
we get 1.26 times 10 to the 6 watts per meter squared. So our q, or our heat flux, is 1.008
times 10 to the 6 watts per meter squared. And when we put that number times ten times
pi D L, and divide it by our heat of vaporization, our vapor production rate, or m dot, is 0.105
kilograms per second. In order to find the corresponding surface temperature of a tube,
we're going to need to find our difference in excess temperature. And so we're going
to use the Rohsenow correlation, which equals C sub sf h sub fg Prandtl to the n, divided
by our heat capacity of the liquid, and this is multiplied by our surface heat flux divided
by our liquid viscosity h sub fg, all raised to the 1/3, and this is multiplied by the
surface tension divided by gravity and the difference in the densities, and this whole
thing is raised to the 1/6. Now the C sub sf and our n are dependent on the solid and
liquid combination, and since this is copper and water, C sub sf is 0.0128 and our n is
equal to one. So when we calculate our delta T e, we come up with 19.2 degrees Celsius.
Therefore, our surface temperature is T sat plus delta T e, or 100 plus 19.2, or that's
going to equal 119.2 degrees Celsius.