Does Watt's Law apply to flowing water in a pipe?

[Jack of some trades]

Is the heat generated by flowing water in a pipe proportional to the product of the rate of flow and the pressure drop across the section of pipe in question, analogous to an electrical circuit? If so, what units would you plug into get an answer in watts?

 

[Chestermiller]

If the flow is at steady state and the pipe is insulated, then then change in specific enthalpy of the water is zero. So, $$C_p\Delta T+v(1-\alpha \bar{T})\Delta P=0$$where $\alpha$ is the volumetric coefficient of thermal expansion, $C_p$ is the specific heat capacity, v is the specific volume of the water and $\bar{T}$ in the second term is the average temperature in the pipe (that doesn't change much). So, the change in temperature from inlet to outlet is $$\Delta T=\frac{v(1-\alpha \bar{T})(-\Delta P)}{C_p}$$
If you imagine that the "heat generated" is $\dot{m}C_p\Delta T$ where $\dot{m}$ is the mass flow rate, then$$\dot{m}C_p\Delta T=\dot{m}v(1-\alpha \bar{T})(-\Delta P)=(1-\alpha \bar{T})\dot{V}(-\Delta P)$$,where $\dot{V}$ is the volumetric flow rate.

 

[tech99]

Summary:: Is the heat generated by flowing water in a pipe proportional to the product of the rate of flow and the pressure drop?

Is the heat generated by flowing water in a pipe proportional to the product of the rate of flow and the pressure drop across the section of pipe in question, analogous to an electrical circuit? If so, what units would you plug into get an answer in watts?

This seems to be a correct statement (we do not want to confuse Temperature with Heat). With the electrical analogue I think the flow rate would be Amperes and the pressure drop Volts.