System curve and Bernoulli equation problem

[spinoza]

Homework Statement

I have a pump pumping 50 gallons per minute of water from a tank at atmospheric pressure to another tank at atmospheric pressure. The main discharge pipe (internal diameter is 4.03 inches) is divided onto 5 smaller pipes (internal diameter is 1.05 inches). The pressure on each on the smaller pipes is 26.3 psia. But as I said before, the discharge tank is at atmospheric pressure.
I’m trying to do 2 different things:
1. Apply the Bernoulli equation to this system, specifically between points A, which is the pump suction, and points B1, B2, B3, B4 and B5 which are the 5 discharge points in order to know the work that the pump must supply to achieve this pressure on points B1-B5.
2. Draw the system curve for this situation. The system curve is basically total head vs flow rate.

Homework Equations

From what I understand the total head of a pump can be expressed as follows:

Total head = Static head + Friction loss throughout the system + pressure head (if the system has it) + velocity head (negligible)

Where:

Static Head: Static head represents the net change in height, in feet, that the pump must overcome
Pressure Head: When liquid is pumped from a vessel at one pressure to a vessel at another pressure, pressure head exists. If there is no difference between the 2 vessels, then it does not.

When applying the Bernoulli equation between the points, I get:

ηWp = (5Pb/ρ) + (g*Zb/gc) + (5(Vb^2)/(2*gc)) - (Pa/ρ) + ((Va^2)/(2*gc))

I know there are friction losses in this system due to several factors, but I decided to ignore them on this post because I don't have doubts about that.

The Attempt at a Solution

Where:
Pb is 26.3 psia
Pa is 14.7 psia
Zb is 50 ft
Vb is 0.98 ft/s
Va is 0.33 ft/s

5Pb and 5Vb because the pressure and the velocity on all 5 pipes is the same. When I solve the equation with the values given above, I get:
ηWp = 319.75 ft*lbf /lbm = 1.77 hp

On the other hand when I assume Pb to be 14.7 psia, ηWp = 1.03 hp

Is my energy balance correct?
Or, should I assume Pb to be 14.7 psia since the tank is under atmospheric pressure?

I preferred to assume Pb = 26.30 psia because this problem is very much like a shower problem where the water needs to have a certain pressure (above 14.7 psia) when it exits the shower nozzle and one needs to assume Pb as the pressure of the water exiting the nozzle even though the bathroom is under atmospheric pressure. As a mater of fact, I’ve got nozzles on all 5 pipes. Am I right about my approach to this problem?

Second question: According to the pressure head definition above, this system does not have pressure head because the pressure on both tanks is the same. Or should I assume the pressure head to be 26.30-14.7 like I did on the Bernoulli equation when drawing the system curve?

Third question: can the static head for the system curve and Zb for the Bernoulli equation be different?

 

[erobz]

The power required by the pump is given by:

$$ \dot W_p = \gamma Q h_p \tag{1} $$

$\gamma$ is the specific weight of the fluid
$Q$ is the volumetric flowrate
$h_p$ is the change in head across the pump

You have given us $\gamma$ "water", and $Q = 50 \, \rm{gpm}$, what is left is to determine $h_p$.
Using the "Energy Equation" from Fluid Mechanics ( not Bernoulli's Equation which is only valid for "inviscid flows", or in practice over short distances with of low viscosity flow), If the pump suction is at $0$ head (e.g. approximately just at the surface of a large reservoir), to determine the $h_p$ you must simultaneously solve a system of $5$ non-linear equations ( point $o$ is at the discharge inlet side of the pump) of the form:

$$ h_p = z_i + \frac{Q_i^2}{A_i^2 2g} + \frac{P_i}{\gamma} + \sum_{o \to i } h_l $$

Where:

$z_i$ is the elevation of branch $i$'s discharge relative to the pump.
$Q_i$ is the volumetric flow (flow velocity assumed uniform across discharge) through branch $i$'s discharge
$P_i$ is the pressure (gauge pressure) branch $i$'s discharge.

and another equation constraining the total discharge:

$$ Q_o = 50 \, \rm{gpm} = \sum Q_i $$

Where $Q_i$ is the volumetric discharge in each branch.

$\sum_{o \to i } h_l$ is the sum of head loss between the pump discharge and the system discharge of the branch $i$. This term is always positive in any viscous flow and is usually modeled (for turbulent flow) as $h_l = f \frac{L}{D A_i^2} \frac{Q_i^2}{2g}$ for the major head loss associated with the system "pipe friction". $f$ itself is a function of the Reynolds Number, and pipe roughness, and is commonly found graphically in the "Moody Diagram". Other forms "component friction" are referred to as minor head loss and are the result of bends, elbows, tees, etc...that are used to route the pipe. They are usually just modeled as $k \frac{Q_i^2}{A_i^2 2g}$ where $k$ will depend on $V_i$ and some other geometric factors for the type of component used.

In general, the total head loss $\sum_{o \to i} h_l$ is the algebraic sum of the major and minor head loss.

I stated that everything was in reference to branch $i$'s discharge. If you have a pressure/velocity requirement at some point in the pipe that needs to be met, just shift the equations to go between $o$ and that point. The OP has $P = 26.3 \, \rm{psia}$ at some point just before the discharge. That is fine, but the onus will be on the designer to ensure that that the downstream piping after the point of interest (which would be ignored in this case) is able to supply the required "back pressure" and velocity to meet the requirement(s) at the point of interest.

Getting back to the power calculation. After you have solved the system of non-linear equations for the branch discharges (which usually requires the techniques of linearization and iteration), you can determine the head of the pump $h_p$ by computing it for any particular branch in the system using one of the 5 equations in the system

$$ h_p = z_i + \frac{Q_i^2}{A_i^2 2g} + \frac{P_i}{\gamma} + \sum_{o \to i } h_l $$

and plug it into $(1)$

I believe this answers questions 1 and 2.

Question 3:

The static head for the system curve will generally be the minimum head required for flow to initiate in the system. This will generally be the lowest branch elevation in the system. In multi branch systems, the "system curve" is not described by a single resulting equation over the entire range of flows. If you imagine two tanks at different elevations being served by a single pump (similar piping systems), flow will initiate in the lower tank first. That will be the "static head" of the system. The flow must increase in the lower branch until the pressure required to initiate flow in the upper tank is met. From that point on the "system curve" will be a combination of both branch curves as flow further increases.