# Relation between conservation of energy and conservation of mass of a fluid flow

Let us consider the following hypotheses for fluid flow:

1. Incompressible flow

That being stated, let us consider a horizontal pipe with constant diameter, thus constant cross-sectional area. According to conservation of energy, we have

$$$$\frac{P_1}{\rho g} + \frac{V_1^2}{2g} = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + h_L \tag{1} \label{1}$$$$

where the term $$h_L$$ relates all losses between point 1 and 2. Applying Eq. \ref{1} to calculate $$V_2$$, it is possible to define the following equation

$$$$\frac{V_2^2}{2g} = \frac{P_1 - P_2}{\rho g} + \frac{V_1^2}{2 g} - h_L \tag{2} \label{2}$$$$

It is also possible to calculate $$V_2$$ by applying conservation of mass

$$$$V_2 A_2 = V_1 A_1 \tag{3} \label{3}$$$$

As the pipe has the same cross-sectional area, Eq. \ref{3} can be written as $$$$V_2 = V_1 \tag{4} \label{4}$$$$

Comparing Eq. \ref{2} and \ref{4}, the unique way to Eq. \ref{2} be true is if $$$$\frac{P_1 - P_2}{\rho g} - h_L = 0 \tag{5} \label{5}$$$$

Is it Eq. \ref{5} valid, exclusively, due to hypothesis 2? As the flow is steady, it shouldn't be accelerating. Is it correct?

• Professor, you shouldn't consider conservation of energy and conservation of mass separately when dealing with fluid flow. BOTH equations have to be satisfied simultaneously, even for compressible flow. May 11, 2021 at 22:29
• @DavidWhite, yes, exactly! To satisfy conservation of mass, $V_2$ must be equal to $V_1$. Taking this information into conservation of energy, $\frac{P_1 - P_2}{\rho g} - h_L = 0$. That is my question. $\frac{P_1 - P_2}{\rho g} - h_L = 0$ due to steady flow? May 11, 2021 at 22:47

Is it Eq. 5 valid, exclusively, due to hypothesis 2? As the flow is steady, it shouldn't be accelerating. Is it correct?

Not exclusively, no. You also need $$(3)$$ and $$(4)$$.

Regarding:

$$\frac{P_1 - P_2}{\rho g} - h_L = 0\tag{5}$$

It can be re-written as:

$$P_1-P_2=\rho g h_L$$

This can then be seen as a balance of forces because $$h_L$$ is really just a mathematical device$$^{\dagger}$$, where $$\rho gh_L$$ is a pressure equal to $$P_1-P_2$$. With the forces on the fluid balancing, there can be no acceleration (Newton's 2nd Law)

$$^{\dagger}$$ I call $$h_L$$ a 'mathematical device' because the sum total of the pressure losses (between points $$1$$ and $$2$$), i.e. $$P_L$$, can be written as:

$$P_L=\rho g h_L$$

In pipe system engineering circles $$h_L$$ is often referred to as the head loss.

• Thank you for answering. One more question: if $h_L$ is larger than $\frac{P_1 - P_2}{2g}$, what this mean physically? May 11, 2021 at 21:13
• Hi. That's simply impossible. Think of the $\rho g h_L$ term as friction. Now, by analogy, imagine a block in a horizontal, rough (friction!) plane. Now we exert a horizontal force $F$. The responsive friction force $F_f$ is so that $F\geq F_f$. If $F<F_f$ then the block would start moving in the direction opposite to $\vec{F}$! Impossible!
– Gert
May 12, 2021 at 7:55
• Right, this make sense! Finally, what exactly does mean a situation which $P_1-P_2=\rho g h_L$ besides that the fluid is not accelerating? To me seems strange, because the flow loses just its potencial energy but it keeps its kinetic energy, since his velocity is the same $(V_1 = V_2)$. Does this make sense? May 12, 2021 at 11:35
• Remember that energy is lost when the flow runs past valves, bends, rough patches, constrictions, metering devices, filters etc etc, that are positioned between $1$ and $2$. The potential energy is lost to non-conserved friction energy when the flow runs past these 'obstacles'. Kinetic energy isn't affected because $v_1=v_2$ and there's no acceleration. And $v_1=v_2$ is of course the continuum equation for incompressible fluids.
– Gert
May 12, 2021 at 14:23