What is Pipe pressure loss calculation?
Which equations Engineering CAD uses for the pressure loss of straight pipes, bends, reducers, valves and other elements, and in what order — with worked examples.
To trust a result you need to know which equations and assumptions produced it. It also lets you check that a hand calculation or another program uses the same equations and assumptions before comparing.
Pumping 10 m³/h of water at 20 °C through 100 m of 50 mm steel pipe loses about 45 kPa to friction alone — roughly 4.6 m of water head. Double the flow and the loss grows about fourfold (3.7× in fact), because it rises with roughly the square of the velocity.
Pressure changes are split into two kinds:
- Irreversible losses: pressure lost to friction and eddies (straight-pipe friction, bends, reducers, valves, other K, branches, reactors …). Each element can be switched to 0 Pa.
- Static pressure changes: pressure converted into potential or kinetic energy rather than lost (elevation, change of velocity). These remain when a loss is switched off.
p_{\mathrm{out}} = p_{\mathrm{in}} - (\text{friction} + \text{local losses}) + \text{static change}
Example: the water pipe above (300 kPa at the inlet, 100 m long) rising 10 m towards its outlet.
- Friction loss: 45.3 kPa (irreversible)
- Elevation: \rho g \Delta z = 998.4 \times 9.807 \times 10 = 97.9 kPa drop (static change)
- Outlet pressure: 300 - 45.3 - 97.9 = 156.8 kPa
With friction switched to 0 Pa the outlet is at 202.1 kPa — the 97.9 kPa for the climb remains, because the pressure spent lifting the water is not lost but turned into potential energy.
The block results and the network summary list friction, each local loss, elevation and velocity change separately. The total pressure loss is friction plus local losses (45.3 kPa here), so it differs from the inlet–outlet pressure difference (143.2 kPa) by the static change.
Unless stated otherwise, the examples use: water at 20 °C (\rho = 998.4\ \mathrm{kg/m^3}, \mu = 1.002\ \mathrm{mPa\,s}), 10 m³/h (\dot{m} = 2.773 kg/s), inner diameter D = 50 mm, roughness \varepsilon = 0.045 mm (typical of commercial steel), 300 kPa at the inlet. The numbers come from this site's calculation engine; they are illustrative inputs, not recommended or standard values.
## Straight pipe: Darcy–Weisbach
\Delta p_f = f\,\frac{L}{D}\,\frac{\rho u^2}{2}
f: Darcy friction factor, L and D: length and inner diameter, \rho: density, u = \dot{m}/(\rho A): cross-section average velocity (A = \pi D^2/4). \rho u^2/2 is the dynamic pressure.
Example: the cross-section is A = \pi \times 0.05^2/4 = 1.963 \times 10^{-3} m², the velocity u = 2.773/(998.4 \times 1.963 \times 10^{-3}) = 1.415 m/s and the dynamic pressure \rho u^2/2 = 999 Pa.
### Friction factor
The Reynolds number uses the mass flux G = \dot{m}/A, which is constant along the pipe:
Re = \frac{\rho u D}{\mu} = \frac{G D}{\mu}
- Laminar (Re < 2300): Hagen–Poiseuille, f = 64/Re
- Turbulent (Re > 4000): Colebrook (1939). f appears on both sides, so it is iterated to a relative change of 10^{-13}, starting from the Swamee–Jain approximation.
\frac{1}{\sqrt{f}} = -2\log_{10}\left(\frac{\varepsilon}{3.7D} + \frac{2.51}{Re\sqrt{f}}\right)
- Transitional (2300 \le Re \le 4000): whether the flow is laminar or turbulent cannot be predicted, so f is interpolated linearly between 64/2300 and the Colebrook value at Re = 4000, and the result carries a warning.
Example (turbulent): Re = 998.4 \times 1.415 \times 0.05 / 0.001002 = 70{,}500, so the flow is turbulent. Colebrook with relative roughness \varepsilon/D = 0.0009 gives f = 0.02269. Over 100 m:
\Delta p_f = 0.02269 \times \frac{100}{0.05} \times 999 = 45{,}300\ \mathrm{Pa} = 45.3\ \mathrm{kPa}
Example (laminar): a viscous oil (\rho = 850 kg/m³, \mu = 0.05 Pa·s) at 0.5 kg/s through 20 m of 50 mm pipe has u = 0.300 m/s and Re = 255 — laminar. f = 64/255 = 0.251 and \Delta p_f = 0.251 \times 400 \times 38.1 = 3.83 kPa. In laminar flow f does not depend on roughness and the loss is proportional to the flow (in turbulent flow, roughly to its square).
### Marching through segments
Density and viscosity change with temperature and pressure, so the pipe is divided into n_segments segments (50 by default). From the inlet onward, each segment's outlet state is iterated until it satisfies (subscript m: segment average):
p_2 = p_1 - f\,\frac{\Delta x}{D}\,\frac{\rho_m u_m^2}{2} - \rho_m g\,\Delta z - G\,(u_2 - u_1)
h_2 + \frac{u_2^2}{2} + g z_2 = h_1 + \frac{u_1^2}{2} + g z_1 + \frac{\delta Q}{\dot{m}}
The three terms on the right of the first equation are friction loss, elevation and acceleration (the velocity rise as a gas expands). The state (temperature, density) follows from (h_2, p_2): Cantera's ideal gas for gases, the water property model or the entered constant properties for liquids. Viscosity is updated per segment too, so Re and f vary along the pipe. g = 9.80665\ \mathrm{m/s^2}.
Heat exchange with the surroundings, \delta Q, uses the exact exponential form per segment, and the resulting change of density and viscosity feeds back into the pressure loss (heat loss from pipes):
\delta Q = -\dot{m} c_p (T_1 - T_\infty)\left(1 - e^{-UA/(\dot{m} c_p)}\right)
### Gas density
- auto (default): the inlet density is kept while the pressure change is at most 10 % of the inlet pressure and the Mach number at most 0.3; beyond that the compressible march is used and a note says so.
- compressible: density is always updated segment by segment.
- constant_density: refused outside the limits above.
In every case Mach 0.9 or more is refused as close to choking.
Example: air at 20 °C and 200 kPa (\rho = 2.37 kg/m³), 0.1 kg/s through a 50 mm pipe (u = 21.5 m/s, Re = 139{,}000, f = 0.0212, Mach 0.06).
- 20 m long: constant density gives \Delta p_f = 4.64 kPa, 2.3 % of the inlet pressure, so the constant density is kept.
- 100 m long: constant density would give about 23.2 kPa (11.6 %), over 10 %, so the compressible march takes over. As the pressure falls the air expands and speeds up from 21.5 to 24.6 m/s; the friction loss is 24.8 kPa (about 7 % more than at constant density) and the outlet is at 175.1 kPa. The 0.16 kPa spent on acceleration is shown separately as a static change.
## Local losses: K
\Delta p = K\,\frac{\rho u^2}{2}
u is the pipe velocity. Across a fitting h + u^2/2 is conserved, and the acceleration of an expanding gas is counted separately as a static change (loss coefficient). An element with K = 1 takes one dynamic pressure — 999 Pa in the water example.
### Bends
Enter K, or use the Crane TP-410 relation for 90° bends, K = n\,f_T: n is interpolated linearly in a table of the bend radius ratio r/D (1→20, 1.5→14, 2–3→12, 4→14, 6→17, 8→24, 10→30, 12→34, 14→38, 16→42, 20→50) and f_T is the fully rough friction factor (Colebrook with Re \to \infty):
f_T = \left[-2\log_{10}\left(\frac{\varepsilon}{3.7D}\right)\right]^{-2}
N bends lose N K \rho u^2/2. No correlation is registered for 45° bends, so their K is entered by the user.
Example: four long-radius elbows with r/D = 1.5. f_T = [-2\log_{10}(0.0009/3.7)]^{-2} = 0.0191, K = 14 \times 0.0191 = 0.268 each, 4 \times 0.268 = 1.07 in total. The loss is 1.07 \times 999 = 1.07 kPa — equivalent to L = K D / f = 1.07 \times 0.05 / 0.02269 \approx 2.4 m of straight pipe.
### Reducers and expanders
The reference velocity is the one in the smaller diameter. With \beta = small/large diameter and \theta = total included angle (Crane TP-410):
- contraction, \theta \le 45°: K = 0.8\sin(\theta/2)(1-\beta^2)
- contraction, \theta > 45°: K = 0.5(1-\beta^2)\sqrt{\sin(\theta/2)}
- enlargement, \theta \le 45°: K = 2.6\sin(\theta/2)(1-\beta^2)^2
- enlargement, \theta > 45°: K = (1-\beta^2)^2
The change of velocity with diameter, \rho(u_1^2 - u_2^2)/2, is a static change, separate from the loss.
Example: a sudden enlargement from 50 to 100 mm (\theta = 180°, \beta = 0.5) has K = (1 - 0.25)^2 = 0.5625. Referred to the small-side velocity (1.415 m/s) the loss is 0.5625 \times 999 = 0.56 kPa. Meanwhile the velocity in the large pipe falls to a quarter (0.354 m/s), so the static pressure rises by 998.4 \times (1.415^2 - 0.354^2)/2 = 0.94 kPa. On balance the outlet pressure is 0.37 kPa higher than the inlet (300.0 → 300.4 kPa): despite the loss, the pressure rises because slowing down turns kinetic energy back into pressure.
### Valves
Given by K, Kv or Cv. Kv is the water flow in m³/h at a 1 bar pressure difference (VDI/VDE 2173):
\Delta p\,[\mathrm{bar}] = \left(\frac{Q\,[\mathrm{m^3/h}]}{K_v}\right)^2 \frac{\rho}{1000\,\mathrm{kg/m^3}}
K_v = C_v \times \frac{0.2271247}{\sqrt{0.0689476}} \approx 0.865\,C_v
Without an expansion factor this is refused for a gas whose pressure difference exceeds 10 % of the pressure. With a valve opening, Kv follows the inherent flow characteristic of IEC 60534 (linear / equal percentage).
Example: 10 m³/h of water through a valve with Kv = 40 gives \Delta p = (10/40)^2 \times 0.998 = 0.0624 bar = 6.24 kPa. In Cv terms, C_v = 40/0.865 \approx 46.
## Other elements
- Splitter: each outlet loses \Delta p = K\rho u^2/2 (u: that outlet's velocity). No flow-ratio-dependent junction correlation is registered, so K is entered by the user.
- Reactors, heaters, mixers: zero (0 Pa) / fixed (given value) / k_factor (\Delta p = K\rho_{\mathrm{ref}} u_{\mathrm{ref}}^2/2 with u_{\mathrm{ref}} = \dot{m}/(\rho_{\mathrm{ref}} A_{\mathrm{ref}})), applied at the outlet at constant enthalpy. Example: for the air above (2.37 kg/m³, 0.1 kg/s) with K = 2 and A_{\mathrm{ref}} = 1.963 \times 10^{-3} m², u_{\mathrm{ref}} = 21.5 m/s and \Delta p = 2 \times 2.37 \times 21.5^2/2 = 1.10 kPa.
- Solving a split: the fraction x at which both branches end with the same pressure surplus is found by the Illinois method (an improved false position), calculating every block of both branches (flow split). The shorter, easier branch takes more of the flow.
## Worked example: a connected line
Connect the water example as a 100 m straight pipe rising 10 m → four 90° bends (r/D = 1.5) → a valve (Kv = 40). The outlet pressure of each block:
1. Straight pipe: 300.0 → 156.8 kPa (friction 45.3 kPa, elevation 97.9 kPa)
2. Bends: 156.8 → 155.7 kPa (1.07 kPa)
3. Valve: 155.7 → 149.4 kPa (6.24 kPa)
The irreversible losses add up to 45.3 + 1.07 + 6.24 = 52.7 kPa, the static change is −97.9 kPa, and the inlet–outlet difference is 150.6 kPa. In this line the climb matters most, then straight-pipe friction. To cut the loss, a larger pipe helps most: at the same flow the loss falls roughly with the fifth power of the diameter.
The Crane TP-410 coefficients (the n table for 90° bends and the reducer equations) are treated as provisional until checked against the original. Enter K directly where accuracy matters. Only single-phase flow is modeled; near-sonic flow, reverse flow and two-phase flow are not calculated. The numbers in the examples are illustrative inputs, not recommended or standard values.