# N°6 – Pipe selection training – A level

As we have already indicated, the sizing of closed circuits, where “we don’t want any noise” is carried out with reference to head loss charts in which the pipe head loss (j) is limited to j = 20 [daPa/m].

#### Question

Size a pipe allowing the transmission of 800 [l/h] (3.52 US gpm) of water. The value of the flow is placed on the flow rate scale (here 800 [l/h]) (3.52 US gpm); we trace a vertical line to limit of j = 20 [daPa/m]; from limit of j = 20 [daPa/m]; we descend the vertical line until it meets the first bold type curve corresponding to a piping DN.
In our case the DN 20 is above the limit of j = 20 [daPa/m].
The DN 25 is the first curve encountered when descending the vertical line.
From the intersection we draw a horizontal line which meets the pressure loss scale at 6.5 [daPa/m].
For the velocity, the corresponding curves are practically perpendicular to the DN curves. The previous intersection point is situated between the curves of 0.3 and 0.35 [m/s].  #### Question

Size a pipe allowing a flow rate of 4.5 [m³/h] (19.8 US gpm) of water.  #### Question

Size a pipe allowing the transmission of 450 [kW] at a temperature regime of 75/65 [°C] (167°F/149°F). The flow rate has first to be determined:
qv = P / (1.16 x ∆T)
= 450 / (1.16 x ((75 – 65))
= 450 / (1.16 x 10)
= 38.8 [m³/h] (171 US gpm)
Then by referring to the chart we obtain the following results: #### Question

Size a pipe allowing the transmission of 1,800 [W] at a temperature regime of 75/65 [°C] (167°F/149°F). The flow rate has first to be determined: qv = P/(1.16 x ∆T)
= 1,800/(1.16 x ((75 – 65))
= 1,800/(1.16 x 10)
= 155.2 [l/h] (0.68 US gpm)
Then, by referring to the chart , we obtain the following results: Note: radiator supply piping is generally in DN15. For steel pipes it is usual to terminate the circuits with a DN 15, even if a DN 12 would be in theory sufficient.

#### Question

Size the pipe in the installation below. DN = 100
j in [daPa/m] = 11
Total pressure loss in the pipe J in [mH2O] = 1.32 (4.3 ftH2O)

Explanation:
To begin with we must determine the flow:
qv = P / (1.16 x ∆T)
= 800 / (1.16 x ((80 – 60))
= 800 / (1.16 x 20)
= 34.5 [m³/h] (152 US gpm)
Then, referring to the chart we obtain the following results:
DN 100 and j = 11 [daPa/m], which gives:
J = j x L = 11 x 60 x 2 = 1,320 [daPa] or J = 1.32 [mH2O] (4.3 ft H20)