Internal Flows: Pipe Flow, Friction Factor and Moody Chart, Minor Losses, Pipe Systems and Pumping Power
2. Laminar pipe flow: Hagen-Poiseuille results and f = 64/Re
A force balance on a fluid cylinder of radius r gives a shear stress that is linear in r, τ = (−dp/dx) r/2, and Newton's law with the no-slip wall then gives the Poiseuille parabola. For a pipe of radius R, diameter D and pressure drop Δp over length L (horizontal):
- Centreline velocity u_max = (Δp/L)R²/(4μ) and mean velocity V = u_max/2.
- Flow rate Q = πR⁴Δp/(8μL) = πD⁴Δp/(128μL), so Δp = 32μLV/D² = 128μLQ/(πD⁴).
- Wall shear stress τ_w = (Δp/L)(R/2) = 4μV/R = 8μV/D.
- Friction factor f = 8τ_w/(ρV²) = 64μ/(ρVD) = 64/Re.
Worked example. Oil of ρ = 900 kg/m³ and μ = 0.10 Pa·s flows at V = 0.50 m/s in a 40 mm pipe. Re = 900 × 0.50 × 0.040/0.10 = 180, so the flow is laminar and f = 64/180 = 0.356. Δp per metre = f (1/D)(ρV²/2) = 0.356 × 25 × 112.5 = 1000 Pa/m. Check with Δp/L = 32μV/D² = 32 × 0.10 × 0.50/0.0016 = 1000 Pa/m. Both routes agree, which is a useful habit in the exam.
3. Turbulent flow, the Darcy-Weisbach relation and the Moody chart
The Darcy-Weisbach relation gives the major (friction) loss in any developed pipe flow, laminar or turbulent: h_f = f (L/D) V²/(2g), or Δp = f (L/D)(ρV²/2), where f is the Darcy friction factor, equal to four times the Fanning factor. A force balance on the fluid in a length L, Δp·πD²/4 = τ_w·πDL, gives τ_w = ΔpD/(4L) = (f/8)ρV², a result worth remembering because it links the pressure drop to the wall shear in every regime. Dimensional analysis shows f = φ(Re, ε/D), where ε is the roughness height; the Moody chart plots this function.
| Region | f depends on | Pressure drop at fixed pipe |
|---|---|---|
| Laminar, Re < about 2300 | Re only: f = 64/Re | ∝ V |
| Critical zone, about 2300 to 4000 | unreliable, avoid | not predictable |
| Smooth turbulent (Blasius f = 0.316 Re^(−1/4), Re < 10⁵) | Re only | ∝ V^1.75 |
| Transitionally rough (Colebrook) | Re and ε/D | between V^1.75 and V² |
| Fully rough, high Re | ε/D only (curves go flat) | ∝ V² |
The implicit Colebrook equation, 1/√f = −2 log₁₀[ε/(3.7D) + 2.51/(Re√f)], reproduces the turbulent part of the chart; its fully rough limit is 1/√f = −2 log₁₀[ε/(3.7D)]. In the exam f is normally read from the chart and given in the stem, so the work is in using it correctly. Roughness matters in turbulent flow once the roughness height exceeds the thin viscous sublayer at the wall; in laminar flow, and for a pipe that is hydraulically smooth, it does not.
Worked example. Water (ρ = 998 kg/m³) flows at V = 2.0 m/s in a 0.10 m pipe, 200 m long, with f = 0.020 from the chart. h_f = 0.020 × (200/0.10) × 2.0²/(2 × 9.81) = 8.15 m. Δp = ρgh_f = 998 × 9.81 × 8.15 = 79.8 kPa, which equals f (L/D)(ρV²/2) = 0.020 × 2000 × 1996 = 79,840 Pa. The wall shear is τ_w = (f/8)ρV² = 0.0025 × 998 × 4 = 9.98 Pa, which agrees with ΔpD/(4L) = 79,840 × 0.10/800 = 9.98 Pa.
4. Minor losses, loss coefficients and equivalent length
Entrances, exits, bends, valves, enlargements and contractions disturb the flow and dissipate energy over a short distance. These minor losses are written h_m = K V²/(2g), where the loss coefficient K depends mainly on the fitting's geometry and is found from tables or tests, and V is the velocity at the reference section the K is quoted for (take it from the stem). Standard values: a sharp-edged entrance from a large reservoir has K = 0.5, and an exit into a reservoir has K = 1.0, because all the exit kinetic energy is lost. For a sudden enlargement, momentum and energy balances give the Borda-Carnot result h_m = (V₁ − V₂)²/(2g), that is K = (1 − A₁/A₂)² on the upstream velocity V₁; the exit to a reservoir is its limit A₂ → ∞.
Equivalent length. A fitting is equivalent to a straight pipe of length L_eq that gives the same loss: f L_eq/D = K, so L_eq = KD/f. The total loss in a line is then h = (f L/D + ΣK) V²/(2g), valid when one pipe size runs throughout. Minor losses are not negligible in short lines or with throttled valves; in a very long pipeline they are small beside the friction loss.
Worked example. A sudden enlargement from D₁ = 0.05 m to D₂ = 0.10 m carries V₁ = 4 m/s. Then V₂ = V₁(A₁/A₂) = 1 m/s and h_m = (4 − 1)²/(2 × 9.81) = 0.459 m. By the coefficient, K = (1 − 0.25)² = 0.5625 and h_m = 0.5625 × 16/19.62 = 0.459 m. The pressure still rises across the enlargement because the velocity head falls by (16 − 1)/19.62 = 0.765 m, more than the 0.459 m lost.
Energy equation for a line between sections 1 and 2, steady and incompressible: p₁/(ρg) + α₁V₁²/(2g) + z₁ + h_pump = p₂/(ρg) + α₂V₂²/(2g) + z₂ + h_f + Σh_m. Here α is the kinetic-energy correction factor (2 for developed laminar flow, about 1 for turbulent flow) and h_pump is the head added by a pump. Use gauge or absolute pressures consistently, and neglect the velocity head at the free surface of a large reservoir.
5. Pipes in series and in parallel
Express the friction loss in terms of flow rate: with V = 4Q/(πD²), h_f = f (L/D) V²/(2g) = 8fLQ²/(π²gD⁵) = R Q², where R is the resistance of the pipe. Pipes in series carry the same Q and their losses add, so R_total = ΣR_i. Pipes in parallel between two junctions have the same head loss h and their flows add: Q_total = ΣQ_i, with Q_i = √(h/R_i). Parallel connection lowers the total resistance: 1/√R_eq = Σ 1/√R_i.
Worked example. Two parallel pipes with the same f and the same length L, with D₂ = 2D₁. Equal head loss means Q ∝ √(D⁵/(fL)), so Q₂/Q₁ = 2^(5/2) = 5.66 and the larger pipe carries 5.66/6.66 = 85% of the total flow. In a real network f depends on Re and hence on the flow, so branch problems are solved by guessing the flow split, computing h in each branch, and correcting until the heads agree; in the exam f is given.
6. Power needed for pumping and the system curve
A pump must supply the total head H = (p₂ − p₁)/(ρg) + (V₂² − V₁²)/(2g) + (z₂ − z₁) + h_f + Σh_m. For a line between two open reservoirs this is the static lift plus the friction and minor losses. The system curve H_sys = H_static + R_total Q² rises with Q, and the pump runs where its own head-flow curve crosses it, the operating point.
The power delivered to the fluid is P_hyd = ρgQH, and the shaft power is P_shaft = ρgQH/η for pump efficiency η; the motor's electrical input is larger still by its own efficiency. If the static head is zero, H ∝ Q² and so P ∝ Q³; if static head dominates, P ∝ Q.
Worked example. Water (ρ = 1000 kg/m³) is lifted 15 m between open tanks at Q = 0.020 m³/s through a 0.10 m pipe, L = 150 m, f = 0.025, ΣK = 5 (exit included). V = 4 × 0.020/(π × 0.10²) = 2.546 m/s and V²/2g = 0.3305 m. h_f = 0.025 × 1500 × 0.3305 = 12.39 m and Σh_m = 5 × 0.3305 = 1.65 m. H = 15 + 12.39 + 1.65 = 29.05 m. P_hyd = 1000 × 9.81 × 0.020 × 29.05 = 5.70 kW. With η = 0.65, P_shaft = 5.70/0.65 = 8.77 kW.
Key takeaways
- Fully developed flow has a constant profile, constant dp/dx and constant wall shear. Laminar L_e ≈ 0.05 Re D (constant stated in the stem); turbulent flow develops in far fewer diameters. Re ≈ 2300 separates laminar from turbulent flow in a circular pipe.
- Laminar: u_max = 2V, Δp = 32μLV/D², τ_w = 8μV/D, f = 64/Re (Darcy, no roughness effect). Check Re before using it, and never confuse it with the Fanning 16/Re.
- Darcy-Weisbach: h_f = f (L/D) V²/2g and τ_w = (f/8)ρV². On the Moody chart f = φ(Re, ε/D): Re only when laminar or smooth, ε/D only when fully rough (Δp ∝ V²).
- Minor loss h_m = K V²/2g with the velocity the K refers to; equivalent length L_eq = KD/f; sudden enlargement K = (1 − A₁/A₂)² on V₁; entrance 0.5 and exit 1.0 are the standard values.
- Series: same Q, losses add. Parallel: same head loss, flows add, Q_i = √(h/R_i), and for equal f, Q ∝ √(D⁵/L). h_f = 8fLQ²/(π²gD⁵) gives the 1/D⁵ scaling at fixed Q.
- Pump head H = static lift + h_f + Σh_m (+ pressure and velocity terms); shaft power = ρgQH/η. With negligible static head, P ∝ Q³.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
For fully developed laminar flow in a circular pipe, the Darcy friction factor is given by
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Answer: D — 64/Re, independent of the wall roughness
Force balance plus Newton's law give f = 8τ_w/(ρV²) = 64/Re, with no roughness dependence. 16/Re is the Fanning factor, a quarter of the Darcy value, and is the tempting wrong choice. 0.316/Re^(1/4) is the Blasius smooth-pipe turbulent correlation, and the roughness-dependent chart reading applies to turbulent flow.Which statement describes fully developed flow in a straight pipe of constant diameter?
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Answer: A — The velocity profile is unchanged along the pipe and the pressure gradient is constant
Once the boundary layers meet, u(r) stops changing with x, so dp/dx and the wall shear stress are constant and the pressure falls linearly. The wall shear and centreline velocity do change in the entrance region (wall shear falls, centreline speed rises), which is the tempting description. Pressure must fall for a real fluid to keep flowing against wall shear, so it is never uniform.Water flows through a long, rough pipe at a high Reynolds number, in the fully rough regime so that f is constant. If the flow speed is doubled, the pressure drop over the same length becomes
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Answer: C — 4 times, since Δp is proportional to V² when f is constant
In the fully rough regime f depends only on ε/D, so Δp = f (L/D)(ρV²/2) ∝ V² and doubling V gives 4 times. The factor 2 belongs to laminar flow, where Δp ∝ V, and a smooth turbulent pipe gives a factor of about 2^1.75 = 3.4. The factor 8 corresponds to power, which scales as Q·Δp ∝ V³.A sudden enlargement in a pipe has an area ratio A₂ = 3A₁. Using the Borda-Carnot result, the loss coefficient based on the upstream velocity V₁ (h_m = K V₁²/2g) is
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Answer: B — 0.44, equal to (1 − A₁/A₂)²
h_m = (V₁ − V₂)²/2g with V₂ = V₁/3 gives h_m = (V₁²/2g)(1 − 1/3)², so K = (2/3)² = 0.444. The value 0.67 forgets to square the bracket. K = 1 is the limit for A₂ → ∞, not for a finite ratio.Two pipes connect the same two reservoirs in parallel. Which statement is correct?
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Answer: B — The flow rates add and each pipe has the same head loss
Both pipes span the same pair of junctions, so both have the same head loss, and by continuity the total flow is the sum of the two. The flows are equal only if the pipes are identical. The pipe with the larger resistance R carries less flow, since Q = √(h/R).A pump lifts water through a pipe in which the static head is negligible and the friction factor stays constant. The volume flow rate is doubled. The shaft power required (at the same efficiency) becomes
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Answer: D — 8 times
With no static head, the head needed is the friction loss, H ∝ Q². Shaft power is ρgQH/η ∝ Q × Q² = Q³, so doubling Q multiplies power by 8. The factor 4 is the increase in head alone, and 2 would hold only for a pure static lift.In the fully rough zone at the right-hand end of the Moody chart, the friction factor f depends on
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Answer: C — the relative roughness ε/D only
At high Re the roughness elements protrude through the viscous sublayer and form drag on them sets the wall shear, so f becomes independent of Re and is a function of ε/D alone (flat curves). The Re-only dependence belongs to laminar flow and smooth pipes. f = 0.02 is only a typical value, not a rule.Oil flows laminarly at a fixed volume flow rate Q through a long pipe. The pipe is replaced by one of half the diameter (same length). The pressure drop becomes
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Answer: B — 16 times
For laminar flow Δp = 128μLQ/(πD⁴) ∝ 1/D⁴, so halving D multiplies Δp by 2⁴ = 16. The factor 32 is what a fully rough turbulent line gives (Δp ∝ 1/D⁵ at constant f), and 4 is only the rise in mean velocity (V ∝ 1/D²), while Δp = 32μLV/D² carries a further 1/D².Water flows laminarly in a pipe of fixed diameter D, with entrance length given by L_e = 0.05 Re D. If the flow rate is doubled and the flow stays laminar, the entrance length
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Answer: D — doubles, because Re is proportional to the flow rate at fixed D
At fixed D, Re = 4ρQ/(πμD) is proportional to Q, so doubling Q doubles Re and hence L_e. The faster-developing idea is wrong because in laminar flow the boundary layer grows by viscous diffusion, and a faster stream travels farther before it has spread across the pipe. D does not change in this question.Which of the following statements about fully developed pipe flow are correct?
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Answer: A — In laminar flow in a circular pipe the centreline velocity is twice the mean velocity; B — The wall shear stress is related to the pressure drop by τ_w = ΔpD/(4L) for a horizontal pipe
The parabolic profile gives u_max = 2V, and a force balance on the fluid in the pipe gives τ_w = ΔpD/(4L) in any regime. The laminar friction factor 64/Re has no roughness dependence, and the Darcy factor is four times, not one quarter of, the Fanning factor.Which of the following statements about minor losses are correct?
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Answer: A — A minor loss is written h_m = K V²/(2g), with K depending mainly on the fitting geometry; B — A fitting with loss coefficient K is equivalent to a straight pipe of length KD/f; D — For a free discharge into a large reservoir all the exit kinetic energy is lost, so K = 1
K V²/2g is the standard form, L_eq = KD/f follows from f L_eq/D = K, and an exit jet into a quiescent reservoir loses its whole velocity head. The word minor is historical: in a short line, or with a nearly closed valve, the total of the minor losses can exceed the friction loss.Which of the following statements about pipes in series and in parallel are correct?
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Answer: B — In series, the same flow rate passes through every pipe and the head losses add; D — In parallel, for equal friction factors, a branch's flow rate is proportional to √(D⁵/L)
Series pipes carry one flow and their losses add, and from h_f = 8fLQ²/(π²gD⁵) a parallel branch at fixed head carries Q ∝ √(D⁵/(fL)). Series pipes of different size have different pressure drops, and parallel branches of equal length carry equal flow only if their diameters are also equal.Oil of density 900 kg/m³ and dynamic viscosity 0.20 Pa·s flows steadily through a horizontal 20 mm diameter pipe, 10 m long, at a volume flow rate of 2.0 × 10⁻⁴ m³/s. The flow is laminar and fully developed. The pressure drop over the 10 m length, in kPa to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 101.9
V = 4Q/(πD²) = 4 × 2.0 × 10⁻⁴/(π × 0.02²) = 0.6366 m/s. Re = 900 × 0.6366 × 0.02/0.20 = 57, so the flow is laminar. Δp = 128μLQ/(πD⁴) = 128 × 0.20 × 10 × 2.0 × 10⁻⁴/(π × 1.6 × 10⁻⁷) = 101,859 Pa = 101.9 kPa. The same follows from 32μLV/D² = 32 × 0.20 × 10 × 0.6366/(4 × 10⁻⁴).Water flows at 2.0 m/s through a 50 mm diameter pipe 40 m long with Darcy friction factor f = 0.025. The line has a sharp-edged entrance (K = 0.5), two elbows (K = 0.9 each), a valve (K = 10) and a free exit into a reservoir (K = 1.0), all based on the 2.0 m/s velocity. Take g = 9.81 m/s². The total head loss, in metres to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 6.79
ΣK = 0.5 + 2 × 0.9 + 10 + 1.0 = 13.3. f L/D = 0.025 × 40/0.05 = 20. V²/2g = 4/19.62 = 0.2039 m. Total h = (20 + 13.3) × 0.2039 = 33.3 × 0.2039 = 6.79 m. Of this, 4.08 m is friction and 2.71 m is minor loss; the equivalent length of the fittings is ΣK D/f = 26.6 m.A pump lifts water (ρ = 1000 kg/m³) between two open tanks whose surface levels differ by 20 m, at a flow rate of 0.040 m³/s. The pipe is 0.15 m in diameter and 300 m long with f = 0.020, and the minor losses total ΣK = 6.0 (exit included) on the pipe velocity. The pump efficiency is 0.65. Take g = 9.81 m/s² and neglect velocity heads at the tank surfaces. The shaft power of the pump, in kW to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 19.3
V = 4 × 0.040/(π × 0.15²) = 2.264 m/s, V²/2g = 0.2611 m. h_f = 0.020 × (300/0.15) × 0.2611 = 10.45 m and Σh_m = 6.0 × 0.2611 = 1.57 m. H = 20 + 10.45 + 1.57 = 32.01 m. P_hyd = 1000 × 9.81 × 0.040 × 32.01 = 12.56 kW, and P_shaft = 12.56/0.65 = 19.3 kW. Multiplying by the efficiency instead of dividing gives 8.2 kW, and omitting the losses gives 7.8/0.65 = 12.1 kW.Water is supplied to a junction at a total flow of 0.060 m³/s and divides into two parallel pipes that rejoin downstream. Pipe 1 has D = 0.10 m and L = 100 m; pipe 2 has D = 0.20 m and L = 400 m. Both have the same friction factor, and minor losses are neglected. The flow rate in pipe 2, in litres per second to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 44.3
Equal head loss with equal f gives Q ∝ √(D⁵/L). Q₂/Q₁ = √[(0.20/0.10)⁵ × (100/400)] = √(32/4) = √8 = 2.828. Q₁ + Q₂ = 60 L/s, so Q₂ = 60 × 2.828/3.828 = 44.3 L/s and Q₁ = 15.7 L/s. Using only the diameter ratio 2^(5/2) = 5.66 would ignore the four-times longer pipe.In a horizontal circular pipe of diameter 0.080 m and length 50 m, the pressure drop under steady, fully developed flow is 12 kPa. The average wall shear stress, in Pa to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 4.8
For a horizontal pipe a force balance gives Δp·πD²/4 = τ_w·πDL, so τ_w = ΔpD/(4L) = 12,000 × 0.080/(4 × 50) = 4.8 Pa. The result holds for laminar and turbulent flow alike. Using Δp·D/L without the factor 4 gives 19.2 Pa.