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Energy Equation - Pressure Loss vs. Head Loss

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The total energy per mass unit in a given point in a fluid flow consists of elevation (potential) energy , velocity (kinetic) energy and pressure energy .

The Energy Equation states that energy can not disappear - the energy upstream in the fluid flow will always be equal to the energy downstream in the flow and the energy loss between the two points.

E1 = E2+ E loss (1)


E1 = energy upstream (J/kg, Btu/lb)

E2= energy downstream (J/kg, Btu/lb)

E loss = energy loss (J/kg, Btu/lb)

The energy in a specific point in the flow

E flow = E pressure + E kinetic + E potential (2)


E pressure = p / ρ = pressure energy (J/kg, Btu/lb)

E kinetic = v2/ 2 = velocity (kinetic) energy (J/kg, Btu/lb)

E potential = g h = elevation (potential) energy (J/kg, Btu/lb)

E loss = Δ p loss / ρ = major and minor energy loss in the fluid flow (J/kg, Btu/lb)

p = pressure in fluid (Pa (N/m2), psi (lb/in2))

Δ p loss = major and minor pressure loss in the fluid flow (Pa (N/m2), psi (lb/in2))

ρ = density of fluid (kg/m3, slugs/ft3 )

v = flow velocity (m/s, ft/s)

g = acceleration of gravity (m/s2, ft/s2)

h = elevation (m, ft)

Eq. 1 and 2 can be combined to express the equal energies in two different points in a stream line as

p1 / ρ + v12/ 2 + g h1 = p2/ ρ + v2 2 / 2 + g h2+ Δp loss / ρ (3)

or alternatively

p1 + ρ v12/ 2 + ρ g h1 = p2+ ρ v2 2 / 2 + ρ g h2+ Δp loss (3b)

For a horizontal steady state flow v1 = v2 and h1 = h2, - and (3b) can be simplified to:

Δ p loss = p1 - p2(3c)

The pressure loss is divided in

  • major loss due to friction and
  • minor loss due to change of velocity in bends, valves and similar

The major friction loss in a pipe or tube depends on the flow velocity, pipe or duct length, pipe or duct diameter, and a friction factor based on the roughness of the pipe or duct, and whether the flow us turbulent or laminar - the Reynolds Number of the flow. The pressure loss in a tube or duct due to friction, major loss, can be expressed as:

Δ p major_loss = λ (l / d h ) (ρ v2/ 2) (4)


Δ p major_loss = major friction pressure loss (Pa, (N/m2), lb/ft2)

λ = friction coefficient

l = length of duct or pipe (m, ft)

d h = hydraulic diameter (m, ft)

Eq. (3) is also called the D'Arcy-Weisbach Equation . (3) is valid for fully developed, steady, incompressible flow .

The minor or dynamic loss depends flow velocity, density and a coefficient for the actual component.

Δ p minor_loss = ξ ρ v2/ 2                            (5)


Δ p minor_loss = minor pressure loss (Pa (N/m2), lb/ft2)

ξ = minor loss coefficient


Head and Head Loss

The Energy equation can be expressed in terms of head and head loss by dividing each term by the specific weight of the fluid. The total head in a fluid flow in a tube or a duct can be expressed as the sum of elevation head , velocity head and pressure head .

Note ! The heads in the equations below are based on the fluid itself as a reference fluid. Read more about head here .

p1 / γ + v12/ 2 g + h1 = p2/ γ + v2 2 / 2 g + h2+ Δh loss (6)


Δ h loss = head loss (m "fluid", ft "fluid")

γ = ρ g = specific weight of fluid (N/m3, lb/ft3 )

For horizontal steady state flow v1 = v2 and p 1 = p2, - (4) can be transformed to:

h loss = h1 - h2(6a)


Δ h = p / γ = head (m "fluid", ft "fluid")

The major friction head loss in a tube or duct due to friction can be expressed as:

Δ h major_loss = λ (l / d h ) (v2/ 2 g) (7)


Δ h loss = head loss (m, ft)

The minor or dynamic head loss depends flow velocity, density and a coefficient for the actual component.

Δ p minor_loss = ξ v2/ (2 g)                            (8)

Friction Coefficient - λ

The friction coefficient depends on the flow - if it is

  • laminar,
  • transitional or
  • turbulent

and the roughness of the tube or duct.

To determine the friction coefficient we first have to determine if the flow is laminar, transitional or turbulent - then use the proper formula or diagram.

Friction Coefficient for Laminar Flow

For fully developed laminar flow the roughness of the duct or pipe can be neglected. The friction coefficient depends only the Reynolds Number - Re - and can be expressed as:

λ= 64 / Re (9)


Re = dimensionless Reynolds number

The flow is

  • laminar when Re < 2300
  • transitional when 2300 < Re < 4000
  • turbulent when Re > 4000

Friction Coefficient for Transitional Flow

If the flow is transitional - 2300 < Re < 4000 - the flow varies between laminar and turbulent flow and the friction coefficient is not possible to determine.

Friction Coefficient for Turbulent Flow

For turbulent flow the friction coefficient depends on the Reynolds Number and the roughness of the duct or pipe wall. On functional form this can be expressed as:

λ = f( Re, k / d h ) (10)


k = absolute roughness of tube or duct wall (mm, ft)

k / d h = the relative roughness - or roughness ratio

Roughness for materials are determined by experiments. Absolute roughness for some common materials are indicated in the table below

Pressure Loss due to Friction - Roughness Coefficients vs. Surface Materials
SurfaceAbsolute Roughness - k
(10-3 m) (feet)
Copper, Lead, Brass, Aluminum (new) 0.001 - 0.002 3.3 - 6.7 10-6
PVC and Plastic Pipes 0.0015 - 0.007 0.5 - 2.33 10 -5
Epoxy, Vinyl Ester and Isophthalic pipe 0.005 1.7 10 -5
Stainless steel, bead blasted 0.001 - 0.006 (0.00328 - 0.0197) 10-3
Stainless steel, turned 0.0004 - 0.006 (0.00131 - 0.0197) 10-3
Stainless steel, electropolished 0.0001 - 0.0008 (0.000328 - 0.00262) 10-3
Steel commercial pipe 0.045 - 0.09 1.5 - 3 10-4
Stretched steel 0.015 5 10 -5
Weld steel 0.045 1.5 10-4
Galvanized steel 0.15 5 10-4
Rusted steel (corrosion) 0.15 - 4 5 - 133 10-4
New cast iron 0.25 - 0.8 8 - 27 10-4
Worn cast iron 0.8 - 1.5 2.7 - 5 10-3
Rusty cast iron 1.5 - 2.5 5 - 8.3 10-3
Sheet or asphalted cast iron 0.01 - 0.015 3.33 - 5 10 -5
Smoothed cement 0.3 1 10-3
Ordinary concrete 0.3 - 1 1 - 3.33 10-3
Coarse concrete 0.3 - 5 1 - 16.7 10-3
Well planed wood 0.18 - 0.9 6 - 30 10-4
Ordinary wood 5 16.7 10-3

The friction coefficient - λ - can be calculated by the Colebrooke Equation :

1 / λ 1/2 = -2,0 log 10 [ (2,51 / (Re λ 1/2 )) + (k / d h ) / 3,72 ] (11)

Since the friction coefficient - λ - is on both sides of the equation, it must be solved by iteration. If we know the Reynolds number and the roughness - the friction coefficient - λ - in the particular flow can be calculated.

A graphical representation of the Colebrooke Equation is the Moody Diagram :

With the Moody diagram we can find the friction coefficient if we know the Reynolds Number - Re - and the

Relative Roughness Ratio - k / d h

In the diagram we can see how the friction coefficient depends on the Reynolds number for laminar flow - how the friction coefficient is undefined for transitional flow - and how the friction coefficient depends on the roughness ratio for turbulent flow.

For hydraulic smooth pipes - the roughness ratio limits zero - and the friction coefficient depends more or less on the Reynolds number only.

For a fully developed turbulent flow the friction coefficient depends on the roughness ratio only.

Example - Pressure Loss in Air Ducts

Air at 0 oC is flows in a 10 m galvanized duct - 315 mm diameter - with velocity 15 m/s .

Reynolds number can be calculated:

Re = d h v ρ / μ (12)


Re = Reynolds number

v = velocity  (m/s)

ρ = density of air (kg/m3 )

μ = dynamic or absolute viscosity ( Ns/m2)

Reynolds number calculated:

Re = (15 m/s) (315 mm) (10-3 m/mm ) (1.23 kg/m3 ) / (1.79 10 -5 Ns/m2)

= 324679 (kgm/s2)/N

= 324679 ~ Turbulent flow

Turbulent flow indicates that Colebrooks equation (9) must be used to determine the friction coefficient - λ -.

With roughness - ε - for galvanized steel 0.15 mm , the roughness ratio can be calculated:

Roughness Ratio = ε / d h

= (0.15 mm) / (315 mm)

= 4.76 10 -4

Using the graphical representation of the Colebrooks equation - the Moody Diagram - the friction coefficient - λ - can be determined to:

λ = 0.017

The major loss for the 10 m duct can be calculated with the Darcy-Weisbach Equation (3) or (6):

Δp loss = λ ( l / d h ) ( ρ v2/ 2 )

= 0.017 ((10 m) / (0.315 m)) ( (1.23 kg/m3 ) (15 m/s)2/ 2 )

= 74 Pa (N/m2)

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