Question 1 of 6: Water-Delivery Pipe Network

Question A1: Water-Delivery Pipe Network (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Water $\rho=998.2\ \mathrm{kg/m^3}$, $\mu=0.001002\ \mathrm{Pa\cdot s}$; commercial-steel $\varepsilon=0.0457\times10^{-3}$ m; three schedule-40 lines out of node $P_1$:

Find. (a) the missing 2.5-in entries and the three friction factors and head losses; (b) tank pressure $P_1$; (c) pump shaft power; (d) the loss coefficient $k$ of the throttle valve on the 3-in branch.

[Figure not reproduced: Figure A1 — Network reconstructed from the exam schematic. Elevations are measured from the lower-reservoir surface, as in the exam schematic. The pump lifts $20\ \mathrm{m^3/h}$ to node $P_1$ at $z=20$ m; the 2.5-in branch (no adjustable fitting) drops to the 250 kPa tank at $z. See the official exam paper.]

Approach. Close the table with continuity ($v=Q/A$), the Reynolds number and a Colebrook (Fanning) friction factor; then write a mechanical-energy balance on each branch — the valve-free 2.5-in branch fixes $P_1$, the feed balance gives the pump head, and the 3-in balance is solved for the valve $k$.

1. Complete the flow table (part a). For the 2.5-in line, $D=2.5\times0.0254=0.06350$ m, $A=\tfrac{\pi}{4}D^2=0.0031670\ \mathrm{m^2}$, $v=Q/A=(10/3600)/0.0031670=0.8771$ m/s, $\mathrm{Re}=\rho vD/\mu=55\,486$ and $\varepsilon/D=0.00072$. Colebrook (Fanning) gives the three factors below; the feed mass flow is $\dot m_4=\rho Q_4=998.2(20/3600)=5.545$ kg/s.

   Quantity	4-in	3-in	2.5-in
   $v$ (m/s)	0.6852	0.6091	0.8771
   $\mathrm{Re}$	69 357	46 238	55 486
   $\varepsilon/D$	0.00045	0.00060	0.00072
   $f$ (Fanning)	0.00531	0.00579	0.00572
   $h_{\text{pipe}}+h_{\text{fit}}$ (m)	0.368	0.330 (no valve)	0.482

2. Head loss per branch. Using the loss-coefficient form $h_f=\left(4f\tfrac{L}{D}+\sum k\right)\tfrac{v^2}{2g}$ with $k=0.75$ per elbow, e.g. the 2.5-in line: $$h_{f,2.5}=\left(4(0.00572)\tfrac{30}{0.0635}+2(0.75)\right)\frac{0.8771^2}{2(9.81)}=0.482\ \mathrm{m}.$$ The equivalent-length method ($4f\tfrac{L+n\cdot30D}{D}\tfrac{v^2}{2g}$) agrees within 1–2%.
3. Tank pressure $P_1$ from the valve-free branch (part b). Take the lower-reservoir surface as datum, so $z_{P_1}=20$ m and $z_{\text{tank}}=15$ m. A mechanical-energy balance from the near-stationary $P_1$ liquid down to the 250 kPa tank, with $h_{P_2}=250\,000/(998.2\times9.81)=25.53$ m: $$\frac{P_1}{\rho g}=h_{P_2}+(z_{\text{tank}}-z_{P_1})+h_{f,2.5}=25.53+(15-20)+0.48=21.01\ \mathrm{m}$$ $$\Rightarrow\ \boxed{P_1=21.01\,\rho g\approx2.06\times10^{5}\ \mathrm{Pa}=206\ \mathrm{kPa(g)}}.$$ The line runs 5 m downhill, so gravity supplies part of the driving head and $P_1$ sits below the sum of the tank pressure and friction.
4. Pump power (part c). Balancing the reservoir surface (open, 20 m below $P_1$) to the $P_1$ liquid, the pump must supply $$h_{\text{pump}}=\frac{P_1}{\rho g}+z_{P_1}+h_{f,4}=21.01+20+0.37=41.4\ \mathrm{m},$$ so with $Q_4=0.005556\ \mathrm{m^3/s}$ and $\eta=0.65$, $$\dot W_{\text{shaft}}=\frac{\rho g Q_4 h_{\text{pump}}}{\eta}=\frac{998.2(9.81)(0.005556)(41.4)}{0.65}=\boxed{3.46\ \mathrm{kW}}.$$
5. Valve coefficient on the 3-in branch (part d). This branch discharges to the open reservoir surface at $z=40$ m, i.e. 20 m above $P_1$; the valve supplies whatever loss balances the surplus head. With $v_3^2/2g=0.01891$ m and pipe+elbow term $4f_3\tfrac{L_3}{D_3}+3(0.75)=17.45$: $$k_{\text{valve}}=\frac{P_1/\rho g+z_{P_1}-z_{\text{res}}}{v_3^2/2g}-\Bigl(4f_3\tfrac{L_3}{D_3}+3k_{\text{elb}}\Bigr)=\frac{21.01+20-40}{0.01891}-17.45=\boxed{36.1}.$$ From Table A1 this is a globe (plug-disk) valve about half-open ($k=36.0$) — a near-exact match that confirms the network reading.