Pipe Pressure Drop Calculator

About the Pipe Pressure Drop Calculator

Estimates the head loss (pressure drop) in a water pipe using the Hazen-Williams equation from flow, pipe diameter, length and roughness coefficient C.

How to use this calculator

1. Enter the pipe length, flow rate, Hazen-Williams coefficient, and pipe diameter.
2. Make sure your units match the calculator inputs, especially for length, flow, and diameter.
3. Click calculate to find the head loss, which is the pressure drop expressed as fluid head.
4. Use the result to compare pipe sizes, check system losses, or estimate the pressure available downstream.

The formula explained

The formula computes the head loss, or pressure drop, due to friction in a pipe carrying water, using the Hazen-Williams method. It shows that longer pipes and higher flow rates increase loss, while larger diameters and smoother pipes reduce it.

• \(h_f\) = head loss, or pressure drop, in the pipe
• L = pipe length
• Q = flow rate
• C = Hazen-Williams roughness coefficient
• d = inside pipe diameter

Step by step method

1. Write down the values for \(L\), \(Q\), \(C\), and \(d\) using the same unit system the calculator expects.
2. Substitute the values into \(h_f = 10.67 \frac{L Q^{1.852}}{C^{1.852} d^{4.87}}\).
3. Calculate the numerator and denominator, then divide to get \(h_f\).
4. If needed, convert head loss into pressure drop using the fluid and unit system for your problem.

Worked example

Problem. A water line is \(100\) m long, carries \(0.02\) m^3/s, has a Hazen-Williams coefficient of \(130\), and an inside diameter of \(0.10\) m. Estimate the head loss.

1. Substitute the values into \(h_f = 10.67 \frac{L Q^{1.852}}{C^{1.852} d^{4.87}}\).
2. Compute \(0.02^{1.852}\), \(130^{1.852}\), and \(0.10^{4.87}\), then combine them with \(L = 100\).
3. The result is about \(h_f \approx 4.4\) m of head loss.

Tips and common mistakes

• A common mistake is mixing units, so keep the calculator inputs consistent with the formula.
• A larger diameter lowers pressure drop a lot because \(d\) is raised to the power \(4.87\).

Frequently asked questions