Factoring the volume of the water inside the pipe in a heated pipe calculation

Question

I have been using a formula to calculate how much heat needs to be generated by a heated pipe in order to raise the temperature of the water flowing through the pipe by a certain amount, e.g. from 10𝑜 C to 30𝑜 C. The formula is here.

The formula is Heat generation in Watts per cubic meter = Heat Transfer rate in Watts divided by Volume in Cubic meters

𝑄𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟=𝑚𝑐𝑝(𝑇𝑜𝑢𝑡−𝑇𝑖𝑛) Where

𝑚 = flow rate in kg/s

𝑐𝑝=Heat capacity of water in J/Kg-K

𝑇𝑜𝑢𝑡 = desired temperature in C

𝑇𝑖𝑛 = starting temperature in C

𝑉=𝜋4(𝑑2𝑜𝑢𝑡−𝑑2𝑖𝑛)𝑙 where 𝑑 stands for diammeter and 𝑙 for the lenght of pipe. All in meters. As far as I can tell, the volume in this case is the volume of the pipe walls, not the volume of the water flowing through it.

The speaker does an example calculation and shows the answer in watts per cubic meter (3:56 in the video). To arrive at this figure in watts per cubic meter, we have divided the heat transfer rate by the volume of the walls of the pipe, but if my pipe contains less than a cubic meter of water as the water flows through it, should I divide this watts per cubic meter figure by the volume of water in my pipe? For example, if my pipe contains 0.1 cubic meters, would the energy needed to cause the temperature change be the heat transfer rate divided by the volume of the pipe walls, in watts per cubic meter, multiplied by 0.1 cubic meters?

Answer 1

The amount of water (and correspondingly, the internal volume of the pipe) is already incorporated, via the $\dot m$ term. This is the rate of water movement through the pipe in kg/s.

The water enters at one temperature and leaves at a higher temperature. What is heating it? Volumetric heating within the walls of the pipe. The video shows the calculation of the necessary volumetric power (in W/m³) within the pipe material to heat the continuous flow of water.

It's important to write the equations precisely to understand what's going on. The power needed to heat the water flow is $\dot m c\left(T_\mathrm{outlet}-T_\mathrm{inlet}\right)$ (verify that this has units of W). The volume of pipe material is $\frac{\pi}{4}L\left(d_\mathrm{outer}^2-d_\mathrm{inner}^2\right)$ (verify that this has units of m³). So the necessary volumetric heat generation within the pipe is $\frac{4\dot m c\left(T_\mathrm{outlet}-T_\mathrm{inlet}\right)}{\pi L\left(d_\mathrm{outer}^2-d_\mathrm{inner}^2\right)}$.

Now, you could use the interior volume of the pipe, $\frac{\pi}{4}Ld_\mathrm{inner}^2$, to calculate the number of times the water is fully replaced every second by the flow, for instance. The volumetric flow of water (in m³/s) is $\frac{\dot m}{\rho}$, where $\rho$ is the water density in kg/m³. So the rate of water replacement is $\frac{4\dot m}{\pi\rho Ld_\mathrm{inner}^2}$ in units of 1/s.

You could also use the internal area of the pipe, $\frac{\pi}{4}d_\mathrm{inner}^2$, to calculate the speed of the water in m/s. This is $\frac{4\dot m}{\pi\rho d_\mathrm{inner}^2}$.

You could also calculate the energy gained by a slug of water with volume equal to the pipe's interior, or $\frac{\pi}{4}Ld_\mathrm{inner}^2$, as the water passes through the pipe. This water has mass $\frac{\pi}{4}\rho Ld_\mathrm{inner}^2$, so the energy needed to heat it from $T_\mathrm{inlet}$ to $T_\mathrm{outlet}$ is $\frac{\pi}{4}\rho Ld_\mathrm{inner}^2 c\left(T_{inlet}-T_{outlet}\right)$ (verify that this has units of J). Noting that this occurs $\frac{4\dot m}{\pi\rho Ld_\mathrm{inner}^2}$ times per second, as calculated above, we can multiply the two terms to reconfirm that the power transfer is $\dot mc\left(T_{inlet}-T_{outlet}\right)$.

So it's essential to keep in mind which volume is relevant. I cannot think of a reason to "divide this watts per cubic meter figure by the volume of water in my pipe"; this would give watts per meters to the sixth power, which doesn't really have any meaning. Nor can I think of a reason to calculate "the heat transfer rate divided by the volume of the pipe walls, in watts per cubic meter, multiplied by 0.1 cubic meters". This would be multiplying a heat generation value per unit volume of pipe by a volume of water, which also doesn't really have any meaning.

Answer 2

The surface area for heat transfer to the fluid is $\pi DL$. The rate of heat entering the flowing fluid is $\pi DL\dot{q}$, where $\dot{q}$ is the Watts per unit area. The heat balance is then $$\dot{m}C(T_{out}-T_{in})=\pi DL\dot{q}$$where $\dot{m}$ is the mass flow rate of water and $T_{out}$ is the mixing cup average temperature of the water exiting the pipe.