Reference ID: MET-0ED4 | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
Pipe-flow energy-balance calculations are the backbone of process-plant hydraulics. They determine how much liquid will flow by gravity between two vessels, how large a pump must be to hit a target rate, or whether a given line can handle a future throughput. The balance couples elevation, pressure, velocity, and all resistances (friction plus fittings) into a single scalar equation that must be satisfied. Because the friction factor itself depends on the unknown velocity, the problem is implicit and must be solved iteratively. The sheet below gives the compact theory and the algebraic forms used in the companion Python module.
Methodology & Formulas
Control-volume energy balance (steady, incompressible, single-phase)
\[
\frac{p_1}{\rho g} + z_1 + \frac{V_1^2}{2g} + h_p = \frac{p_2}{\rho g} + z_2 + \frac{V_2^2}{2g} + h_L
\]
where \(h_p\) is the head added by a pump (zero for gravity-driven cases) and \(h_L\) is the total head loss between 1 and 2.
Reynolds number
\[
Re = \frac{4\rho Q}{\pi D\mu}
\]
Flow regime
Reynolds range
Laminar
\(Re \le 2300\)
Transition
\(2300 \lt Re \lt 4000\)
Turbulent
\(Re \ge 4000\)
Haaland explicit approximation of Colebrook friction factor
\[
\frac{1}{\sqrt{f}} = -1.8\log_{10}\left(\frac{\varepsilon/D}{3.7} + \frac{6.9}{Re}\right)
\]
The Python code iterates this expression five times starting from \(f_0=0.02\) to obtain the turbulent \(f\) to within 0.1 %.
Residual form for solver
Define the residual
\[
R(Q) = (z_1 - z_2) + h_L(Q)
\]
For gravity-driven flow with both reservoirs open to atmosphere (\(p_1=p_2=0\), \(V_1=V_2\), \(h_p=0\)) the energy balance collapses to \(R(Q)=0\). Any 1-D root-finder (e.g. Brent) can be used to obtain the unknown \(Q\).
For incompressible flow, the steady-state energy balance (first law of thermodynamics) written between two pipe sections 1 and 2 is:
ΔH + Δ(v²/2gc) + gΔz/gc = Q̇ – Ẇs
ΔH – enthalpy change (kJ kg⁻¹); for liquids, this collapses to vΔP when density is constant
Δ(v²/2gc) – kinetic energy change; kept when velocity varies (e.g., diameter change)
gΔz/gc – potential energy change; kept when elevation differs
Q̇ – heat added (positive) or lost (negative) per unit mass
Ẇs – shaft work done by the fluid (positive for turbines, negative for pumps)
In most process piping, only ΔP and friction loss terms survive; kinetic and potential changes are dropped when velocities and elevations are nearly identical.
Use the density of the fluid at flowing conditions:
hL = ΔPfriction / (ρ g)
ΔPfriction in Pa (or kPa × 1000)
ρ in kg m⁻³
g = 9.81 m s⁻²
The result, hL, is in metres of fluid and can be inserted directly into the mechanical-energy balance as a loss term.
Include Δ(v²/2gc) when the change in velocity head exceeds ~5 % of the frictional loss. Typical triggers:
sudden pipe enlargement or reduction
control valve or orifice where downstream velocity is much lower
flashing liquid where gas formation increases velocity
For long pipelines with constant diameter, the kinetic term is negligible.
Treat the pump as negative shaft work Ẇs. The hydraulic power required is:
Phyd = ṁ Ẇs = ṁ ΔPpump / ρ
ΔPpump = pressure rise across the pump (Pa)
ṁ = mass flow rate (kg s⁻¹)
Divide by pump efficiency η to obtain driver (motor) power. Insert Ẇs = –Phyd/ṁ into the energy balance; the negative sign indicates work is done on the fluid.
Worked Example: Sizing a Cooling-Water Supply Line
A small chemical plant needs to pump 25 m³ h⁻¹ of cooling water from an elevation of 0 m (pump suction sump) to an open header tank at 10 m above grade. The 50 m long, 100 mm internal diameter commercial-steel pipe contains two 90° elbows and a globe valve. Estimate the mechanical energy that must be added by the pump (i.e. the head rise required) assuming the water is at 15 °C.
Knowns
Volume flow rate, \(Q = 0.026\) m³ s⁻¹
Pipe diameter, \(D = 0.100\) m
Pipe length, \(L = 50\) m
Total minor-loss coefficient, \(K_{\text{tot}} = 2.5\)
Surface roughness, \(\varepsilon = 0.00025\) m
Elevation difference, \(z_2 - z_1 = 10\) m
Static pressures, \(p_1 = p_2 = 0\) (both tanks open to atmosphere)
Water density, \(\rho = 1000\) kg m⁻³
Water dynamic viscosity, \(\mu = 0.001\) Pa·s
Gravitational acceleration, \(g = 9.807\) m s⁻²
Step-by-step calculation
Calculate the mean velocity
\[
V = \frac{4Q}{\pi D^{2}} = \frac{4(0.026)}{\pi (0.100)^{2}} = 3.308\ \text{m s}^{-1}
\]
Compute the Reynolds number
\[
Re = \frac{\rho V D}{\mu} = \frac{1000 \times 3.308 \times 0.100}{0.001} = 3.31 \times 10^{5}
\]
Determine the friction factor (Colebrook equation for turbulent flow)
\[
\frac{1}{\sqrt{f}} = -2 \log_{10}\left( \frac{\varepsilon/D}{3.7} + \frac{2.51}{Re\sqrt{f}} \right) \rightarrow f = 0.019
\]
Evaluate the head loss due to friction
\[
h_{f} = f\frac{L}{D}\frac{V^{2}}{2g} = 0.019 \frac{50}{0.100} \frac{3.308^{2}}{2 \times 9.807} = 5.285\ \text{m}
\]
