Forces due to Flow Through Expanding or Reducing Pipe Bends

Let us consider, a fluid flow through an expander shown in Fig. 11.1a below. The expander is held in a vertical plane. The inlet and outlet velocities are given by V₁ and V₂ as shown in the figure. The inlet and outlet pressures are also prescribed as p₁ and p₂. The velocity and pressure at inlet and at outlet sections are assumed to be uniform. The problem is usually posed for the estimation of the force required at the expander support to hold it in position.

Fig 11.1a   Flow of a fluid through an expander

For the solution of this type of problem, a control volume is chosen to coincide with the interior of the expander as shown in Fig. 11.1a. The control volume being constituted by areas 1-2, 2-3, 3-4, and 4-1 is shown separately in Fig.11.1b.

The external forces on the fluid over areas 2-3 and 1-4 arise due to net efflux of linear momentum through the interior surface of the expander. Let these forces be F_x and F_y. Since the control volume 1234 is stationary and at a steady state, we apply Eq.(10.18d) and have for x and y components

		(11.1a)
and		(11.1b)
or,		(11.2a)
and		(11.2b)

where = mass flow rate through the expander. Analytically it can be expressed as

where A₁ and A₂ are the cross-sectional areas at inlet and outlet of the expander and the flow is considered to be incompressible.
M represents the mass of fluid contained in the expander at any instant and can be expressed as

where is the internal volume of the expander.

Thus, the forces F_x and F_y acting on the control volume (Fig. 11.1b) are exerted by the expander. According to Newton’s third law, the expander will experience the forces R_x (= − F_x) and R_y ( = − F_y) in the x and y directions respectively as shown in the free body diagram of the expander. in fig 11.1c.

Fig 11.2     Effect of atmospheric pressure on the expander

From Fig.11.2 the net x and y components of the atmospheric pressure force on the expander can be written as

The net force on the expander is therefore,

	(11.3a)
	(11.3b)

 or,

	(11.4a)
	(11.4b)

Note: At this stage that if F_x and F_y are calculated from the Eqs (11.2a) and (11.2b) with p₁ and p₂ as the gauge pressures instead of the absolute ones the net forces on the expander E_x and E_y will respectively be equal to −F_x and −F_y.