When a hydrant is connected and static pressure is 80 psi, a 1-1/2 inch line is flowing 100 GPM; with the line placed in service the pressure drops to 70 psi. The estimated remaining GPM available is what?

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Multiple Choice

When a hydrant is connected and static pressure is 80 psi, a 1-1/2 inch line is flowing 100 GPM; with the line placed in service the pressure drops to 70 psi. The estimated remaining GPM available is what?

Explanation:
In hose hydraulics, friction loss in a given hose length grows roughly with the square of the flow for a fixed hose size. Here, when 100 GPM is flowing, the pressure drops from 80 psi to 70 psi, so the drop is 10 psi. This gives a square-law constant: k = drop / Q^2 = 10 / (100^2) = 0.001 psi per (GPM)^2. The maximum total flow before the pressure fully falls to zero occurs when the loss equals the available static pressure, 80 psi. So Q_total = sqrt(80 / 0.001) = sqrt(80000) ≈ 283 GPM. The remaining capacity beyond the current 100 GPM is about 283 − 100 ≈ 183 GPM, which is about 200 GPM in practical terms. Therefore, the estimated remaining GPM available is about 200 GPM.

In hose hydraulics, friction loss in a given hose length grows roughly with the square of the flow for a fixed hose size. Here, when 100 GPM is flowing, the pressure drops from 80 psi to 70 psi, so the drop is 10 psi. This gives a square-law constant: k = drop / Q^2 = 10 / (100^2) = 0.001 psi per (GPM)^2.

The maximum total flow before the pressure fully falls to zero occurs when the loss equals the available static pressure, 80 psi. So Q_total = sqrt(80 / 0.001) = sqrt(80000) ≈ 283 GPM.

The remaining capacity beyond the current 100 GPM is about 283 − 100 ≈ 183 GPM, which is about 200 GPM in practical terms. Therefore, the estimated remaining GPM available is about 200 GPM.

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