Water distribution main for several tanks

 This problem is adapted from the book Hydraulics of Pipeline Systems by BE Larock et al.

The present situation is approached in detailed to show the calculations and to explain the physics behind.

The problem

Consider the following situation for water service supply to a series of reservoir tanks in a large process facility. Water, with kinematic viscosity $\nu=$1.23786 e-5 ft$^2$/s, is to be pumped from a larger reservoir, where the fluid level is kept constant, located 110 ft below the first tank (#1) and 600 ft far (along pipe [1]), tanks 2,3 and 4 are located 90 ft above tank 1. Tank #2 is 400 ft far from tank #1 (along the pipe) while Tanks 2, 3, 4 are spaced (along the pipe) 420 ft. Next, tanks 5 to 10 are located 170 ft above tank 1.  Tanks 4-10 are spaced (along the pipe) 600 ft.

At tank 1 a flowrate 0.37 ft$^3$/s is supplied while tanks 2-4 are feeded with 0.34 ft$^3$/s and tanks 5-10 are feeded with 0.32 ft$^3$/s. In order to guarantee liquid flow from the main reservoir to tanks 1-10 a constant hydraulic grade line HGL slope 0.002 shall be kept. It has been decided that the pipeline shall assembled from commercial steel schedule 80 pipes with roughness 0.0018 in and that the pressure of discharge for tank 10 (node 11) should be, at least, 38 lb/in$^2$. Take the acceleration due to gravity as 32.1740 ft/s$^2$, of the specific weight ad 62.43 lb$_f$/ft$^3$ and make use of the Colebrook equation as well. You should estimate the following:

• flow rate for every pipe section in ft$^3$/s,
• the inner diameter of every pipe section,
• the nearest pipe nominal size corresponding to estimated inne pipe diameter,
• pressure in lb/ft$^2$ at every node,
• the value of HGL at every node,
• the linear velocity for every pipe section, and
• the head loss h$_L$ for every pipe section.

The distribution pipe is schematically represented as follows (not to scale).

Solution

First, some comments,

1. It is reasonable to consider that the flow starts traveling from the surface of the main reservoir. We can say that the hydrostatic pressures compensates the head provided by the pump at that small part.
2. By keeping a constant HGL slope we force the fluid motion from the main reservoir til node 11.
3. A HGL slope is to be define by analogy with the slope of a straight line as follows:

$HGL_{slope}=\frac{HGL_{n+1}-HGL_{n}}{L_{n+1}-L_n}$        Eq. (1)

where $n$ stands for the node number and,

$L_{n+1}-L_n$        Eq. (2)

is the distance along the pipe between the nodes $n$ and $n+1$. It may seem strange to define a HGL slope as above but it makes sense if remember thath HGL changes with the lenth of the pipe. Also, height and pressure are already included in HGL.

From a quick inspection of the pipe it is noticed that the solution starts from node 11. This is, that flow rates can be easily determined by a simple mass balance going backwards. These results are tabulated as follows in column 2 (from left to right),

Results per pipe section

Pipe section	Flow rate (ft$^3$/s)	Lentgh (ft)	Inner diameter (in)	ND (in)	Velocity (ft/s)
1	3.26	400	13.9321	13.9321	3.0793
2	2.92	420	13.3640	13.3640	2.9976
3	2.58	420	12.7534	12.7534	2.9082
4	2.24	600	12.0908	12.0908	2.8093
5	1.92	600	11.4076	11.4076	2.7051
6	1.60	600	10.6496	10.6496	2.5865
7	1.28	600	9.7909	9.7909	2.4481
8	0.96	600	8.8762	8.7862	2.2800
9	0.64	600	7.5445	7.5445	2.0615
10	0.32	600	5.8183	5.8183	1.7330

Pipe length is already given in the problem but was written in column 3 for the sake of completeness. Besides, this data will be useful to determine HGL at every node. Inner, and later nominal, diameters are to be determined by repetitive application of the energy equation:

$\frac{p_1}{\gamma}+z_1-h_L+\frac{v_1^2}{2g}=\frac{p_2}{\gamma}+z_2+\frac{v_2^2}{2g}$        Eq. (3)

However, pressure is known. All what is known is presure at node 11:

$p_{11}=38$ lb/in$^2$

$p_{11}=5472$ lb/ft$^2$

On the other hand, height for every node is given in the statement above. Then, it follows that HGL can be readily determined along the pressure from node 11 to node 1 (at the water surface in tank #1). Height for every node is shown in column 2 (from left to right) in the table below,

Results per node

Node	Height (ft)	Pressure (lb/ft$^2$)	HGL (ft)
1	0	23631.6384	378.5301
2	110	16714.3944	377.7301
3	200	11667.5532	376.8901
4	200	11615.1120	376.0501
5	200	11540.1960	374.8501
6	280	6470.8800	373.6501
7	280	6395.9640	372.4501
8	280	6321.0480	371.2501
9	280	6246.1320	370.0501
10	280	6171.2160	368.8501
11	280	5472.0000	367.6500

HGL at node 12 can be easily found by using Eq. (1) by simple isolation. Next, pressure at node 12 is also found since height of 12 is known too. In order to find the inne diameter of each pipe the energy Eq. (3) must be used so that the Reynolds number formula and Colebrook equation need to be expressed with the diamenter as an unknown. This diameter must be such that the corresponding flow rate at pipe 10 could be transported. The linear velocity can then be easily calculated since:

$Velocity=\frac{Flow}{Area}$

Finally, the nominal diameter must be found from commercial steel pipe geometrical datasheets. Further pipe sections and nodes unknowns are found by repeating the process described. Data in the previous tables were estimated in this way.

Notice that all results shown in the tables above make sense. The flow is to travel upwards and a constant HGL slope is to be kept. There is a leap in the pressure result at node 5 but it can be easily understood by considering that the fluid has to overcome a vertical discance (height) to fulfill the requirement of the HGL slope.

Any questtion? Write in the comments and I shall try to help.

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Ildebrando.