When fluid flows through a pipe bend or elbow, the fluid changes direction and hence there is a change in in momentum. As per Newtons Second Law: "The rate of change of momentum of a body is equal to the net force acting on the body". Slug forces are generated when there is 'two-phase' (gaseous and liquid phase) fluid flowing through pipes and slugs of liquid that are formed intermittently travel at relatively high velocity.

The liquid slug can cause large reaction forces at changes in direction such as elbows or tee connections. High slug forces have the potential to cause large pipe deflections or damage to the supports if they are not designed to withstand them. When the velocity of slug flow is high, the slug force is considered to act at one location at any instant of time for the piping system under consideration. The slug load lasts until the slug has traversed the elbow and then drops to near zero. The time duration of the load is calcuated as slug length divided by the fluid velocity. Let us assume a horizontal pipe bend turning through an angle θ as shown in figure below.

The following details apply:

• V = Velocity of flow
• Q = Flow Rate
• ρ = Fluid Density
• A = Cross-section area of pipe ID
• θ = Change in direction at Bend

The forces exerted on the bend by the flowing fluid are Fx and Fy respectively.

### Force in X-direction

Net force acting on the bend in x-direction (Fx)  = Rate of change of momentum in x-direction

= Mass Flow * (Final velocity in x-direction - Initial velocity in x-direction)

= Density(ρ) * Flow Rate(Q) * (Vcosθ - V) = (ρ) * (Q) * (Vcosθ - V)

Net force acting on the bend in x-direction (Fx) =  ρQV(cosθ - 1)...........(1)

### Force in Y-direction

Net force acting on the bend in x-direction (Fy)  = Rate of change of momentum in y-direction

= Mass Flow * (Final velocity in y-direction - Initial velocity in y-direction)

= Density(ρ) * Flow Rate(Q) * (Vsinθ - 0) = (ρ) * (Q) * (Vsinθ)

Net force acting on the bend in y-direction (Fy) =  ρQ(Vsinθ)...........(2)

### Resultant Force acting on the Bend

The resultant force acting on the bend (FR) = $$\sqrt[]{(F_x)^2 + (F_y)^2 }$$..........(3)

Solving equation (3) gives

The resultant force acting on the bend (FR) = ρQV$$\sqrt[]{2(1-cosθ)}$$..........(4)

Substituting Q = A*V

The resultant force acting on the bend (FR) = ρAV2$$\sqrt[]{2(1-cosθ)}$$..........(5)

For 90 elbow or bend, cosθ = 0, hence

The resultant force acting on the bend (FR) = ρAV2$$\sqrt[]{2}$$ .........(6)

For equivalent static analysis in Caesar-II, the slug load calculated by above formula shall be multiplied by a Dynamic Load Factor (DLF) of two, unless more accurate data is available.

Example Problem:

A slug of length 3.05m passes a 12" sch 40, 90° elbow with a velocity of 15m/s. The density of the fluid is 800 kg/m3. Calculate slug load acting on the elbow. Consider a DLF of 2.

Solution:

The slug load is calculated using equation (6)

Density (ρ) = 800kg/m3

ID of pipe = 323.8 - 2*10.31 = 303.18mm = 0.303m

Cross-section Area (A) = π * (0.303)2 / 4 = 0.07211 m2

Velocity (V) = 15 m/s

Length of Slug (L) = 3.05m

Resultant Force acting on the bend = (DLF) * ρAV2$$\sqrt[]{2}$$ = 2 * 800 * 0.07211 * 152 * $$\sqrt[]{2}$$ = 36707 N.

If R is the radius of elbow, then Rθ(in radian) is the length of bend.

90° = π/2 = 1.57 radians, hence the length of bend (Rθ) = 0.457 * 1.57 = 0.7175m.

It takes the slug $$\frac{Rθ}{V}$$ seconds to exit the bend and attain the maximum slug force of 36707 N. The slug force - time history is as indicated in graph above.

The slug force increases to maximum value in $$\frac{Rθ}{V}$$ seconds = $$\frac{0.7175}{15}$$ = 0.04783 sec.

The maximum slug force acts for $$\frac{L-Rθ}{V}$$ seconds = $$\frac{(3.05-0.7175)}{15}$$ = 0.1555 sec.

The slug force drops to zero in the next $$\frac{Rθ}{V}$$ seconds = $$\frac{0.7175}{15}$$ = 0.04783 sec.

Thus t1 = 0.04783s, t2 = 0.04783 + 0.1555 = 0.2033s and t3 = 0.2033 + 0.04783 = 0.2512s.