Force on a Pipe Bend

Consider a flow through a pipe bend as shown. The flow enters the bend with a speed V₁ and leaves it a speed V₂, the corresponding areas of cross section being A₁ and A₂ respectively. The velocities have components u and v in x and y directions. As the flow negotiates the bend it exerts a force upon it. This force is readily calculated by the momentum theorem.

Figure 3.30: Force on a Pipe Bend

The continuity equation yields

$$\rho~V_1~A_1~=~\rho~V_2~A_2~=~\dot{m}$$ (3.96)

Carrying out a force balance in x-direction, we have

$$p_1~A_1~-~p_2~A_2~\cos\theta~+~F_x~ =~\rho V_2~A_2~V_2~\cos\theta~-~\rho~V_1~A_1~V_1$$

$$= \dot{m}~V_2~\cos\theta~-~\dot{m}~V_1~=~\dot{m}~(V_2~\cos\theta~-V_1)$$ (3.97)

In the y-direction we have,

$$0~-~p_2~A_2~\sin\theta~+~F_y~=~\rho V_2~A_2~V_2~\sin\theta~-~0$$

$$\noalign{giving} ~p_2~A_2~\sin\theta~+~F_y~=~\dot{m}~V_2~\sin\theta~$$ (3.98)

Thus the force components acting on the bend are

$$F_x~ =~p_2~A_2~\cos\theta~-~p_1~A_1~-~\dot{m}~(V_2~\cos\theta~-~V_1)$$

$$F_y~ =-p_2~A_2~\sin\theta~+~\dot{m}~V_2~\sin\theta~$$ (3.99)

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney