Hydraulic Fluid Power. Andrea Vacca

Hydraulic Fluid Power

alt="Schematic illustration of considering the quantities of series-parallel pipe."/>

Pipe A is in series with B and C, which are in parallel. Using the relations derived above:

p 1 minus p 3 equals left-parenthesis upper R Subscript upper A Baseline plus StartStartFraction 1 OverOver StartFraction 1 Over upper R Subscript upper B Baseline EndFraction plus StartFraction 1 Over upper R Subscript upper C Baseline EndFraction EndEndFraction right-parenthesis dot upper Q

which means

upper R Subscript series minus parallel Baseline equals upper R Subscript upper A Baseline plus StartStartFraction 1 OverOver StartFraction 1 Over upper R Subscript upper B Baseline EndFraction plus StartFraction 1 Over upper R Subscript upper C Baseline EndFraction EndEndFraction

3.8 Momentum Equation

The momentum equation is based on Newton's second law applied to a mechanical system, which states that the sum of all forces acting on the system is equal to the time rate of change of linear momentum of the system. In fluid mechanics problems, the same principle can be applied to a CV, having some bounding surfaces, CS, permeable to fluid:

(3.43)

The derivation of Eq. (3.43) can be found in basic fluid mechanics textbooks, such as [15]. The expression states that the sum of the forces acting on the CV is equal to the rate of change of momentum inside the CV (first term at the second member) added to the net rate at which momentum is leaving the CV through the CS (last term in the equation). The forces acting on CV can be of different nature: surface forces, ModifyingAbove upper F With right-arrow Subscript upper S , acting on the control surface, and body forces , acts throughout the volume. The surface forces are those related to fluid pressure and frictional effects, while the only body forces usually taken into consideration is gravity.

The above momentum equation is very useful to study the interaction between the fluid and the surrounding solid surfaces. In fluid power systems, the momentum equation is typically used to determine the force applied by the fluid in a piping system. The following problem provides a representative example.

Example 3.2 Force on an elbow

A 90° elbow is installed on a return line of a tank open to atmosphere. The areas at the elbow entrance and exit are respectively 5 and 3 cm². The return flow rate is 50 l/min. The pressure at the elbow entrance is 1 bar. Determine the overall force acting on the connecting bolts, assuming that no load is transmitted to piping upstream the elbow. Consider the frictional effects in the elbow negligible. The volume of the fluid contained in the elbow is 70 cm³.

Schematic illustration of the force applying on an elbow, showing the overall force acting on the connecting bolts.

Given:

Flow rate Q = 50 l/min; elbow entrance cross‐sectional area A₁ = 5 cm²; elbow exit area A₂ = 3 cm²; elbow turning angle α = 90°; tank pressure p₂ = p_atm = 1 bar; pressure at elbow entrance p₁ = 1 bar; volume of the fluid in the elbow V = 70 cm³; fluid density ρ = 870 kg/m³

Find:

Forces acting on the elbow, F_x, F_y;

Solution:

The force components acting on the elbow can be found by applying the momentum Eq. (3.43) on the CV that includes the elbow. The CV is conveniently chosen to include control surfaces (CS) where the fluid velocity is known. In particular, the CS includes the entrance and exit surfaces, A₁ and A₂, as well as surfaces where there is no flow (therefore with null fluid velocity).

Schematic illustration of the force components acting on the elbow that can be found by applying the momentum Eq. (3.43) on the CV.

The velocities v₁ and v₂ can be calculated from the flow rates:

v 1 equals StartFraction upper Q Over upper A 1 EndFraction equals StartFraction 50 left-bracket l slash min right-bracket Over 5 left-bracket italic c m squared right-bracket EndFraction equals 1.67 m slash s

v 2 equals StartFraction upper Q Over upper A 2 EndFraction equals StartFraction 50 left-bracket l slash min right-bracket Over 3 left-bracket italic c m squared right-bracket EndFraction equals 2.78 m slash s

The momentum Eq. (3.43) can be solved separately in both the x and y components, considering the CV shown in the above figure.

x‐component: For the calculation of the horizontal component of the total force, only the surface force term has to be considered. Neglecting frictional effects in the elbow, the only force component is given by the pressure force on the area A₁ and the force exerted by the elbow walls. Considering the elbow in atmosphere, it is convenient to use gage pressure instead of absolute pressure:

upper F Subscript upper S comma x Baseline equals p Subscript 1 g Baseline upper A 1 plus upper R Subscript x

where R_x = − F_x, meaning that the surface force seen by the fluid has equal but with opposite sign with respect to the force that the fluid exerts to the containing walls.

The x‐component of the second member of Eq. (3.43) can be written by considering the assumption of stationary conditions, which implies that the momentum does not change over time. Therefore, the contribution from the control surfaces is the only one different from zero. Indicating the scalar horizontal velocity component with v_x, the terms at the second member of the momentum equation becomes

integral Underscript italic upper C upper S Endscripts v Subscript x Baseline rho ModifyingAbove v With right-arrow dot d ModifyingAbove upper A With right-arrow

There is only one section crossed by the flow where the u velocities are not null, namely, the entrance section A₁. Therefore, the CS term above reduces to the only section A₁.

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