# -Energy Analysis

To calculate the energy head loss due to friction, we must use the frictional properties of the metal over the vertical height gains. Using the formula below, we can find the head loss in the carbon steel and ductile iron pipes. $h_{L_{1}} = f\frac{l}{D}\frac{v^{2}}{2g}$

The water flow is assumed to have a constant velocity down the pipe at all points along the pump. Water in this case is assumed to be incompressible, having a constant density. Given the volumetric flow from the data sheet to be 1000 gal/min, converted into standard units we can find the velocity in m/s. $Q = 0.631\frac{m^{3}}{s}$  and $A = \frac{\pi D^{2}}{4} = 0.636\ m^{2} \Rightarrow v = 1\frac{m}{s}$ $l = 35.7 + 2.4 + 12.6 = 70.7\ m$ $\epsilon = 0.045\ mm$, and $\frac{\epsilon}{D} = 0.00005\$ $Re=\frac{\rho VD}{\mu}= \frac{1000\times 1\times 0.9}{0.0009}=1000000$ $f = 0.013 \Rightarrow from\ moody\ diagram$ $h_{L_{1}} = 0.013\ \times \frac{70.7}{0.9}\ \times \frac{1}{2 \times 9.81} = 0.052\ m$ $l = 2.2 + 9.7 = 11.9\ m$ $\epsilon = 0.26\ mm$ , and $\frac{\epsilon}{D} = 0.00028\$

Like above, $Re=\frac{\rho VD}{\mu}= \frac{1000\times 1\times 0.9}{0.0009}=1000000$ $f = 0.0155 \Rightarrow from\ moody\ diagram$ $h_{L_{2}} = 0.0155\ \times \frac{11.9}{0.9}\ \times \frac{1}{2 \times 9.81} = 0.0104\text{\ m}$

Together, total head loss is $h_{L} = h_{L_{1}} + {h_{L}}_{2} = 0.052 + 0.0104 = 0.0624\ m$

We assume the loss due to the bends are negligible. Starting from the sump, the initial pressure comes from the water depth. Accounting for the frictional loss, we can solve for the pressure at the pipe with suspected cavitation.

Bernoulli’s Equation with datum line at pipe from sump: $\frac{p_{1}}{\gamma} + \frac{v_{1}^{2}}{2g} + h_{1} - h_{L} = \frac{p_{2}}{\gamma} + \frac{v_{2}^{2}}{2g} + h_{2}$

Because the datum line is drawn at the pipe from the sump, $h_{1} = 0 \ and\ v_{1} = 0$. The initial pressure in the sump is also defined using the simple depth relation $p_{1} = \gamma h$.

On the problem side, the final velocity in the pipe is taken to be the velocity throughout found above $v_{2} = 1\frac{m}{s}$. Because of where the datum line is drawn, we just take the difference in depth between the pipes $h_{2} = 224.1 - 220.1 = 4$. $3.7 + 0 + 0 - 0.0624 = \frac{p_{2}}{\gamma} + \frac{1}{2 \times 9.81} + 4$ $p_{2} = \ - 4055\ Pa\ \left( \text{gage} \right) = \ 97270\ Pa\ (absolute)$

This is the gage pressure at the problematic site of the piping. Convert to absolute pressure by adding 101325 Pa.

In the next two calculations, we neglect frictional head loss, so $h_{L} \Rightarrow 0$ because it is much smaller than the other values. Also, remember $v_{1} = 0$ and $h_{1} = 0$.

1.4.1 Flow rate achieved without air being entrained at the problematic site of the piping system $\frac{p_{1}}{\gamma} + \frac{v_{1}^{2}}{2g} + h_{1} = \ \frac{p_{2}}{\gamma} + \frac{v_{2}^{2}}{2g} + h_{2}$ $3.7 = \frac{p_{2}}{9810} + \frac{v^{2}}{2g} + 4$

Let $P_{2} = vapor\ pressure \rightarrow get\ v$ $v = 10.36\frac{m}{s}$

1.4.2 Maximum allowed elevation of the highest pipe point to prevent entrained air. $\frac{p_{1}}{\gamma} + {\frac{v_{1}^{2}}{2g}} + h_{1} = \frac{p_{2}}{\gamma} + \frac{v_{2}^{2}}{2g} + h_{2}$

Let $P_{2} = vapor\ pressure$ , and let $V_{2} = 1 \frac{m}{s} \Rightarrow get\ h_{2}$ $h_{2} = 3.64\ m$

The actual height from the datum line is 4 m as calculated earlier. Maximum height of the pipe should be limited to 3.64 m from the datum to avoid problems. This value was calculated setting pressure in the pipe to the lowest possible value realistic to the situation–atmospheric pressure, as if the water were sitting in open air.

Thinking further given the conclusion that the pipe is not running completely filled, we analyzed the air trapped at this problematic pipe. Assuming the pipe is airtight, and that there is no error in the opening and closing of valves, all of the air comes dissolved in the water from the source at the sump. With the help of The Engineering Toolbox and knowledge of all of the salient variables, we can solve for the mass of the air dissolved in the water per cubic meter, giving a final density of $0.9145 \frac{g}{m^{3}}$.

This is a large amount of air coming out of solution within the pipe, especially when considering the sheer volume of water travelling through the pipe at any given time. However, not every unit of volume will release the air as described in the code above since it does not take into account the change in pressure within the problematic portion of the pipe as the air becomes trapped within the system and increases the pressure in that section of the pipe. What this means is that while this process won’t create an ever-growing bubble of air within the system, it will always lead to a bubble of a certain size until equilibrium is reached.

That is of course, assuming there isn’t some kind of operator or design error within the pipe, such as a valve closing too quickly and creating a void in the flow, or bend in the same pipe that is somehow doing something similar. This assumption likely isn’t valid however, as this situation is a very real possibility and there likely is a valve that is closing too quickly and partially creating the problematic bubble.