English is currently available.
Chemical Engineering DISCUSSION
I am calculating the pressure drop for water at 20 °C flowing at 1 m/s through 10 m of smooth 25 mm bore pipe. Using a friction factor read from one textbook chart I get about 1.3 kPa, but a classmate using a different textbook gets about 5 kPa, and so does the simulation software.
We are using the same velocity, diameter and fluid properties. Where does a factor of four come from, which of us is right, and how should pipe roughness be included?
You have met the two definitions of the friction factor. The Darcy (or Moody) factor and the Fanning factor describe the same physics but differ by exactly four: f_Darcy = 4 × f_Fanning. Each has its own pressure drop equation:
The error comes from reading a Fanning value from a chart and putting it into the Darcy equation. To tell which chart you have, look at the laminar line: if it is labelled 64 / Re it is a Darcy chart, if 16 / Re it is Fanning. Chemical engineering texts often use Fanning; mechanical and civil texts mostly use Darcy.
Working the numbers: Re = ρ × v × D / µ = 1000 × 1 × 0.025 / 0.001 = 25,000, which is turbulent. For a smooth pipe in this range the Blasius correlation gives f_D = 0.316 / Re^0.25 = 0.316 / 12.57 = 0.0251, or f_F = 0.0063.
The velocity pressure is ρ × v² / 2 = 500 Pa and L / D = 400. So ΔP = 0.0251 × 400 × 500 = 5.0 kPa. Your classmate and the software are right. Putting 0.0063 into the Darcy form gives 1.26 kPa, which is your figure.
For real pipe, roughness raises the friction factor in turbulent flow. The Colebrook equation covers it but is implicit in f; the Swamee-Jain formula is an explicit approximation for the Darcy factor that is accurate enough for design work:
f_D = 0.25 / [log10(ε / (3.7 × D) + 5.74 / Re^0.9)]²
For commercial steel with ε = 0.045 mm in your 25 mm pipe, ε / (3.7 D) = 0.00049 and 5.74 / Re^0.9 = 0.00063. The logarithm of their sum is -2.95, so f_D = 0.25 / 8.71 = 0.029 and the pressure drop becomes 5.7 kPa, about 14 percent more than for smooth pipe.
Check the flow regime before choosing a formula. Below a Reynolds number of about 2100 the flow is laminar, f_D = 64 / Re regardless of roughness, and pressure drop is proportional to velocity. Between roughly 2100 and 4000 the flow is transitional and no correlation is dependable, so avoid designing there.
In turbulent flow pressure drop goes with velocity to a power between 1.75 and 2, so doubling the flow rate multiplies it by 3.4 to 4. It is even more sensitive to diameter: at a fixed flow rate it varies with nearly the fifth power of 1 / D. Going from 25 mm to 32 mm bore cuts the friction loss by roughly 70 percent.
Two more terms complete a real line calculation. Fittings and valves add losses of K × ρ × v² / 2 each, with K taken from tables: of the order of 0.5 to 1 for a standard elbow and much higher for a globe valve. At 1 m/s one velocity head is only 500 Pa, but in a short line with many fittings these can exceed the straight-pipe loss.
Elevation is the other. Every metre of rise costs ρ × g × h = 9.8 kPa for water, so in your example a lift of just 0.5 m would equal the entire friction loss of the 10 m run. Friction, fittings and static head are summed to get the pressure the pump must provide.
Back to Chemical Engineering community
