FLUIDS IN RELATIVE EQUILIBRIUM

2.0 FLUIDS IN RELATIVE EQUILIBRIUM

2.1 PRESSURE

Pressure is defined as a normal force exerted by a fluid per unit area. Pressure is defined as force per unit area; it has the unit of Newton per square meter (N/m²), which is called a Pascal (Pa). That is, 1 Pa = 1 N/m²

EXAMPLE: Absolute Pressure of a Vacuum Chamber

A vacuum gage connected to a chamber reads 30kPa at a location where the atmospheric pressure is 100kPa. Determine the absolute pressure in the chamber.

Analysis: The absolute pressure is easily determined.

Discussion: Note that the local value of the atmospheric pressure is used when determining the absolute pressure.

P_abs = P_atm - P_vac

        = 100 - 30

        = 70 kPa

2.2 PRESSURE MEASUREMENT

Manometer

P₂ = P₁

             P₂ = P_atm + pgh

P_atm + p1gh1 + p2gh2 + p3gh3 = P₁

A relation for the pressure difference P₁ – P₂ can be obtained by starting at point 1 with P₁, moving along the tube by adding or subtracting the rgh terms until we reach point 2, and setting the result equal to P₂

Assume that, P_A = P_B, so

P₁ + p₁g (a + h) = P₂

P₁ – P₂ = (p2- r1) ph

EXAMPLE: Measuring Pressure with a Multifluid Manometer

The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in figure above. The tank is located on a mountain at an altitude of 1200m where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if h₁ 0.1 m, h₂ 0.2 m, and h₃ 0.35 m. Take the densities of water, oil, and mercury to be 1000kg/m³, 850kg/m³, and 13,600kg/m³, respectively.

P_A = P_B

P₁ + p_water gh₂ + p_oil gh₁ – p_mercury gh₃ = P_atm

P₁ = P_atm – p_water gh₂ – p_oil gh₁ + p_mercury gh₃

P_atm + g (p_mercuryh₃ – p_water h₁ – p_oilh₂)

85.6kPa + (9.81m/s²) [(13600)(0.35m) – (1000)(0.1m) – (850)(0.2)][ [

=130kPa

2.3 HYDROSTATIC FORCES

2.3.1 ON SUBMERGED PLANE SURFACES

A plate exposed to a liquid, such as a gate valve in a dam, the wall of a liquid storage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface. On a plane surface, the hydrostatic forces form a system of parallel forces, and we often need to determine the magnitude of the force and its point of application, which is called the center of pressure. When analyzing hydrostatic forces on submerged surfaces, the atmospheric pressure can be subtracted for simplicity when it acts on both sides of the structure.

 

EXAMPLE: Determine the force and center of force on the wall if:

Fluid = gasoline = S_gasoline = 0.68

Total depth = 12 ft

Length of wall = 40 ft

Average pressure = γ (h/2)

= (0.68) (62.4) (12/2) = 254.6

Force on wall, F = pA

= (254.6) (12) (40) = 122204 lb

Center of pressure = 12/3 from bottom

       = 4 ft from bottom

Submerged Rectangular Plate

The resultant hydrostatic force on the upper surface is equal to the average pressure, which is the pressure at the midpoint of the surface, times the surface area A. That is,

F_R = P_cA

     = (P₀ + pg (s + b/2) sinØ) ab

                

EXAMPLE: Hydrostatic Force Acting on the Door of a Submerged Car

A heavy car plunges into a lake during an accident and lands at the Bottom of the lake on its wheels as shown. The door is 1.2 m high and 1 m wide, and the top edge of the door is 8 m below the free surface of the water. Determine the hydrostatic force on the door and the location of the pressure center, and discuss if the driver can open the door.

Analysis: The average pressure on the door is the pressure value at the centroid (midpoint) of the door and is determined to be. Then the resultant hydrostatic force on the door becomes

P_ave = P_C = rgh_C = rg (s + b/2)

        = (1000 kg/m3) (9.81 ms2) (8+ 1.2/2 m) (1 kN/1000 kgms²)

        = 84.4 kNm²

Then the resultant hydrostatic force on the door becomes

FR = P_aveA = (84.4 kN/m2) (1 m x1.2 m) = 101.3kN

The pressure center is directly under the midpoint of the door, assume P₀ = 0

Y_p = s+ b/ 2 +b2/ (12(s+ b))

      = 8 +1.2/2 + 1.22/ (12(8 + 1.2)) = 8.61 m

2.3.2 ON SUBMERGED CURVED SURFACES

·         For a submerged curved surface, the determination of the resultant hydrostatic force is more involved since it typically requires the integration of the pressure forces that change direction along the curved surface.

·         The pressure at the centroid of the portion of the surface in fluid i and Ai is the area of the plate in that fluid. The line of action of this equivalent force can be determined from the requirement that the moment of the equivalent force about any point is equal to the sum of the moments of the individual forces about the same point.

Noted: The concept of the pressure prism in this case is not much help either because of the complicated shapes involved.

Pc,i = Po + ri ghc,i

           

Example :

The height of a water reservoir is controlled by a cylindrical gate hinged to the reservoir. The hydrostatic force on the cylinder and the weight of the cylinder per ft. length are to be determined.

Assume the density of water, p = 62.4 lbm/ft³

Then the magnitude and direction of the hydrostatic force acting on the cylindrical surface become

Noted: When the water level is 15-ft high, the gate opens and the reaction force at the bottom of the cylinder becomes zero. Then the forces other than those at the hinge acting on the cylinder are its weight, acting through the center, and the hydrostatic force exerted by water. Taking a moment about the point A where the hinge is and equating it to zero gives,

 

2.4 BUOYANCY & STABILITY

What is buoyancy?

The force that tends to lift the body and is denoted by Fb.

Why buoyant force occur?

The increase of pressure in a fluid with depth.

 

 

The difference between these two forces is a net upward force, which is the buoyant force, 

Fb   = F bottom – F top

         = rfg(s +h) A - rfgsA

      = rfghA

                  = rfgV

Example :

A crane is used to lower weights into the sea (density 1025 kg/m3) for an underwater construction project as shown. Determine the tension in the rope of the crane due to a rectangular 0.4-m x 0.4-m x 3-m concrete block (density = 2300 kg/m3) when it is (a) suspended in the air and (b) completely immersed in water.

                                                        

 

Analysis: 

(a) Consider the free-body diagram of the concrete block. The forces acting on the concrete block in air are its weight and the upward pull action (tension) by the rope. These two forces must balance each other, and thus the tension in the rope must be equal to the weight of the block

V = (0.4 m)(0.4 m)(3 m) =   0.48 m³

FT,air = W =r concrete gV

(b) When the block is immersed in water, there is the additional force of buoyancy acting upward. The force balance in this case gives : Fb = rfgV

Therefore, F T ,water = W – FB = 10.8 -4.8 = 6.0 kN

Stability of Immersed and Floating Bodies

·         Analogy of “ball on the floor” can explain the fundamental concepts of stability and instability.

                    I.            Stable since any small disturbance (someone moves the ball to the right or left) generates a restoring force (due to gravity) that returns it to its initial position

                  II.            Neutrally stable because if someone moves the ball to the right or left, it would stay put at its new location. It has no tendency to move back to its original location, nor does it continue to move away.

                III.            Unstable when in a situation which is the ball may be at rest at the moment, but any disturbance, even an infinitesimal one, causes the ball to roll off the hill—it does not return to its original position; rather it diverges from it.