# Piping Elbows - Thrust Block Forces

In a piping structure without adequately support fluid flow and internal pressure may create intolerable forces and tensions.

The resultant force - or required support force - on a thrust block - or an anchor - for a bend depends on

- the fluid mass flow, or flow velocity
- the change of flow direction
- the internal pressure

With no flow and no pressure there is no force.

### Online Pipe Bend Resulting Force Calculator

The calculator below can used to calculate the resulting force in a piping bend:

#### Metric Units

#### Imperial Units

### Resulting force due to Mass flow and Flow Velocity

The resulting force in x-direction due to mass flow and flow velocity can be expressed as:

R_{x}= m v (1 - cosβ) (1)

= ρ A v^{2}(1 - cosβ) (1b)

= ρ π (d / 2)^{2}v^{2}(1 - cosβ) (1c)

where

R_{x}= resulting force in x-direction (N)

m = mass flow (kg/s)

v = flow velocity (m/s)

β = turning bend angle (degrees)

ρ = fluid density (kg/m^{3})

d = internal pipe or bend diameter (m)

π = 3.14...

The resulting force in y-direction due to mass flow and flow velocity can be expressed as:

R_{y}= m v sinβ (2)

= ρ A v^{2}sinβ (2b)

= ρ π (d / 2)^{2}v^{2}sinβ (2c)

R_{y}= resulting force in y direction (N)

The resulting force on the bend due to force in x- and y-direction can be expressed as:

R = (R_{x}^{2}+ R_{y}^{2})^{1/2}(3)

where

R = resulting force on the bend (N)

### Example - Resulting force on a bend due to mass flow and flow velocity

The resulting force on a *45 ^{o}* bend with

- internal diameter
*102 mm = 0.102 m* - water with density
*1000 kg/m*^{3} - flow velocity
*20 m/s*

can be calculated by as

Resulting force in x-direction:

R_{x}= (1000 kg/m^{3}) π ((0.102 m) / 2)^{2}(20 m/s)^{2}(1 - cos(45))

= 957 N

Resulting force in y-direction:

R_{y}= (1000 kg/m^{3}) π ((0.102 m) / 2)^{2}(20 m/s)^{2}sin(45)

= 2311 N

Resulting force on the bend

R = (957 N)^{2}+ (2311 N)^{2})^{1/2}

= 2501 N

Note - if β is *90*^{o }the resulting forces in x- and y-directions are the same.

### Resulting force due to Static Pressure

The pressure "acting" on the end surfaces of the bend creates resulting forces in x- and y-directions.

The resulting force in x-direction can be expressed as

R_{px}= p A (1- cos β) (4)

= p π (d / 2)^{2}( 1- cos β) (4b)

where

R_{px}= resulting force due to pressure in x-direction (N)

p = gauge pressure inside pipe (Pa, N/m^{2})

The resulting force in y-direction can be expressed as

R_{py}= p π (d / 2)^{2}sinβ (5)

where

R_{py}= resulting force due to pressure in y-direction (N)

The resulting force on the bend due to force in x- and y-direction can be expressed as:

R_{p}= (R_{px}^{2}+ R_{py}^{2})^{1/2}(6)

where

R_{p}= resulting force on the bend due to static pressure (N)

### Example - Resulting force on a bend due to pressure

The resulting force on a *45 ^{o}* bend with

- internal diameter
*102 mm = 0.102 m* - pressure
*100 kPa*

can be calculated by as

Resulting force in x-direction:

R_{x}= (100 10^{3}Pa) π ((0.102 m) / 2)^{2}(1 - cos(45))

= 239 N

Resulting force in y-direction:

R_{y}= (100 10^{3}Pa) π ((0.102 m) / 2)^{2}sin(45)

= 578 N

Resulting force on the bend

R = ((239 N)^{2}+ (577 N)^{2})^{1/2}

= 625 N

## Related Topics

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Piping systems design strategies - documentation, P&ID, flow diagrams - capacities and limits.

### • Pressure Ratings of Pipes and Fittings

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