# Three Reservoirs Connected by Pipes at a Common Junction

**Situation**

Three reservoirs *A*, *B*, and *C* are connected respectively with pipes 1, 2, and 3 joining at a common junction *P*. Reservoir *A* is at elevation 80 m, reservoir *B* at elevation 70 m and reservoir *C* is at elevation 60 m. The properties of each pipe are as follows:

*L*= 5000 m,

*D*= 300 mm

Pipe 2:

*L*= 4000 m,

*D*= 250 mm

Pipe 3:

*L*= 3500 m

The flow from reservoir *A* to junction *P* is 0.045 m^{3}/s and *f* for all pipes is 0.018.

- Find the elevation of the energy grade line at
*P*in m.

A. 75.512

B. 73.805

C. 72.021

D. 74.173 - Determine the flow on pipe 2 in m
^{3}/s.

A. 0.025

B. 0.031

C. 0.029

D. 0.036 - Compute the diameter appropriate for pipe 3 in mm.

A. 175

B. 170

C. 178

D. 172

**Solution**

$h_{f1} = \dfrac{0.0826(0.018)(5000)(0.045^2)}{0.3^5} = 6.195 ~ \text{m}$

$\text{Elev of } P’ = \text{Elev of } A - h_{f1} = 80 - 6.195$

$\text{Elev of } P’ = 73.805 ~ \text{m}$ ← [ B ] *answer for part 1*

The energy at *P*, denoted as *P*', is higher than reservoir *B*, thus, the flow in pipe 2 is towards *B*

$h_{f2} = \text{Elev of }P' - \text{Elev of } B = 73.805 - 70$

$h_{f2} = 3.805 ~ \text{m}$

$\dfrac{0.0826(0.018)(4000){Q_2}^2}{0.25^5} = 3.085$

$Q_2 = 0.025 ~ \text{m}^3/\text{s}$ ← [ A ] *answer for part 2*

At junction *P*, inflow = outflow:

$Q_1 = Q_2 + Q_3$

$0.045 = 0.025 + Q_3$

$Q_3 = 0.02 ~ \text{m}^3/\text{s}$

$h_{f3} = \text{Elev of } P' - \text{Elev of } C = 73.805 - 60$

$h_{f3} = 13.805 ~ \text{m}$

$\dfrac{0.0826(0.018)(3500)(0.02^2)}{{D_3}^5} = 13.805$

$D_3 = 0.172 ~ \text{m} = 172 ~ \text{mm}$ ← [ D ] *answer for part 3*

- Log in to post comments