PITOT TUBE

Dynamics of Fluid Flow

The lower end, which is bent 90^o is directed against flow direction as shown. The liquid rises up in the tube due to conversion of kinetic energy into pressure energy. Due to the stagnation pressure

developed at section 2, the liquid rises in the vertical bent pipe by a height h above the surrounding surface.

Applying the Bernoulli’s equation between the two sections 1 and 2, we have,

The static pressure heads at sections 1 and 2 respectively are given by

(

H h

)

Where C_v is the coefficient of velocity which is the ratio of the actual velocity of flow to the theoretical velocity.

Velocity of flow in a pipe line

The manometer reading depends on the difference of pressure between the hydrostatic pressure head shown by piezometer or tube perpendicular to the flow direction and stagnation pressure shown by pipe placed parallel to the flow direction. The pressure head difference is computed in the same way as that of venturimeter. If x is the manometer reading, then depending on the manometer whether it is differential manometer or inverted U-tube manometer, the differential head h is respectively computed as





 



 −

= 1

o m

S x S

h or 

 



 −

o m

S x S

h 1

Where Sm and So is the specific gravities of the manometric fluid and pipe fluid respectively. The velocity is computed as Hence in general v= C_v 2gh

A submarine moves horizontally in a sea and has its axis 15 m below the surface of water.

A pitot-tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of U-tube manometer containing mercury which reads 170 mm. Find the

x x

speed of the sub-marine knowing that the specific gravity of sea water as 1.026 with respect to fresh water.

Solution:

x = 0.17 m, S_o = 1.026, Assume S_m = 13.6 and C_v = 1

We know that 1 2.0834

026 . 1

6 . 17 13 . 0

1 =



 



 −

=



 



 −

o m

S x S

h m

455 . 6 0834 . 2 10 2 2

V= C_v gh= × × = m/s = 23.24 kmph (Ans)

A pitot-tube is inserted in a pipe of 300 mm diameter. The static pressure is 100 mm of mercury (vacuum). The stagnation pressure at the centre of pipe recorded by the pitot-tube is 10 kPa. Calculate the rate of flow of water through the pipe, if the mean velocity of flow is 0.85 times the central velocity. Take C_v = 0.98.

Solution:

D = 0.3 m, Static pressure head = 100 mm Hg (vacuum) = - 0.1 x 13.6 = -1.36 m of water Stagnation pressure = 10 kPa = 10 x 10 ³/ (1000 x 10) = 1 m.

assume g = 10 m/s², ρ = 1000 kg/m³ and SHg= 13.6

h = Stagnation pressure – Static pressure = 1 – (-1.36) = 2.36 m of water.

87 . 6 36 . 2 10 2 2

V= C_v gh = × × = m/s.

Mean velocity = V = 0.85 x 6.87 = 5.8395 m/s.

From discharge continuity equation,

Q = AV = 0.3 5.8395 0.4128 4

2× = × × =

× π

π D V m³/s (Ans).

Momentum equation

It is based on the law of conservation of momentum or on momentum principle, which states that the net force acting on a fluid mass is equal to the rate of change of momentum of flow in that direction. If the force acting on a mass of fluid m is F_x along x direction, the net force along the direction is given by Newton’s second law of motion as

Fx = m ax

Where ax is the acceleration produced due to the force Fx along the same direction.

But dt a_x = du

Hence

( )

dt mu d dt mdu ma

F_x = _x= = (as m is constant for incompressible flow)

The above equation is called momentum principle. The same equation can also be written as F_xdt=d

( )

mu which is known as impulse momentum principle and can be stated as

“The impulse of a force acting on a fluid of mass m in a short interval of time dt along any direction is given by the rate of change of momentum d(mu) along the same direction.

Force exerted by a Flowing fluid on a pipe Bend

Consider a flow occurring in a pipe bend which is changing its cross sectional area along the bend as shown in the Fig. Let θ be the angle of bend and Fx and F_y be the force exerted by the fluid on the bend along the x and y directions respectively. The force exerted by the bend on the mass of fluid is --F_x and --F_y. The other forces acting on the mass of fluid are hydrostatic pressure forces at the two sections 1 and 2 p₁A₁along the flow direction and p₂A₂ against the flow direction respectively. From the momentum equation, the net force acting on the fluid mass along x direction is given by the rate of change of momentum in x direction.

F_y

V₂Sinθ

V₂Cosθ 1

θ V2

p1A1

p2A2

θ p2A2

p₂A₂Cos θ p₂A₂Sin θ

i.e. p1A1 – p2A2 cos θ - Fx = (Mass of fluid per second) (change in velocity) = ρ Q (Final velocity – initial velocity) along x

= ρ Q (V2 cos θ – V1)

Fx = ρ Q (V1 – V2 cos θ) + p1A1 – p2A2 cos θ …(01) Similarly the momentum equation in y direction gives

0 – p2A2 sin θ - Fy = ρ Q (V2 Sin θ – 0)

Fy = ρ Q (– V2 Sin θ ) – p2A2 sin θ

…(02)

The resultant force F acting on the bend is given by F = F_x² +F_y² and the angle made by it with x axis is given by

x y

= F α tan Problems

A 45^o degree bend is connected in a pipe line, the diameters at the inlet and outlet of the bend being 600 mm and 300 mm respectively. Find the force exerted by water on the bend if intensity of pressure at inlet to bend is 88.29 kPa and rate of flow of water is 600 lps.

Solution:

θ = 45⁰, D1 = 0.6 m, D2 = 0.3 m p1 = 88.29 kPa, Q = 0.6 m³/s

Assume g = 10 m/s² and ρ = 1000 kg/m³

2827 . 4 0

6 . 0 4

2 2 1

1 =π D =π× =

A m²

07068 . 4 0

3 . 0 4

2 2

2 =π D =π× =

A m²

45^o V2

p1A1

p2A2

From discharge continuity equation, we have Q = A V 122

Applying Bernoulli’s equation between Sections 1 and 2, we get

α with horizontal.

July/Aug 2005

6 b) Water flows up a reducing bend of weight 80 kN place in a vertical plane. For the bend, the inlet diameter is 2 m, outlet diameter is 1.3 m, angle of deflection is 120^o and vertical height (distance between the inlet and the outlet) is 3 m. If the discharge is 8.5 m³/s, pressure at the inlet is 280 kPa and the head loss is half the kinetic head at the exit, determine the force on the bend (12)

Applying Modified Bernoulli’s equation between the two sections of the bend shown in Fig. we get

Forces acting on the bend in x and y direction respectively are Fx = ρ Q [V1 – V2 cos θ ] + p1A1 – p2A2 cos θ = 1,048,423.63 N

α with horizontal.

60^o

Hydrostatics

In this chapter, only forces acting on fluid or acted up on the fluid at rest are considered. The forces considered are pressure forces and gravity forces.

Pressure or Pressure intensity (p): It is the Fluid pressure force per unit area of application. Mathematically,

P= p . Units are Pascal or N/m².

Total Pressure (P): This is that force exerted by the fluid on the contact surface (of the submerged surfaces), when the fluid comes in contact with the surface always acting normal to the contact surface. Units are N.

Centre of Pressure: It is defined as the point of application of the total pressure on the contact surface.

The submerged surface may be either plane or curved. In case of plane surface, it may be vertical, horizontal or inclined. Hence, the above four cases may be studied for obtaining the total pressure and centre of pressure.

Horizontal Plane surface submerged in liquid

Consider a horizontal surface immersed in a static fluid as shown in Fig.

As all the points on the plane are at equal depth from the free surface of the liquid, the pressure intensity will be equal on the entire surface and given by p = ρg y , where y is the depth

of the fluid surface

Let A = Area of the immersed surface The total pressure force acting on the immersed surface is P

P = p x Area of the surface = ρg yA P = ρgAy

Where y is the centroidal distance immersed surface from the free surface of the liquid and h is the centre of pressure.

y P h

Vertical Plane surface submerged in liquid

Consider a vertical plane surface of some arbitrary shape immersed in a liquid of mass density ρ as shown in Fig.

Let

A = Total area of the surface

y = Depth of Centroid of the surface from the free surface G = Centroid of the immersed surface

C = Centre of pressure

h = Depth of centre of pressure

Consider a rectangular strip of breadth b and depth dy at a depth y from the free surface.

Total Pressure:

The pressure intensity at a depth y acting on the strip is p = ρgy

Total pressure force on the strip = dP = (ρgy)dA

∴ The Total pressure force on the entire area is given by integrating the above expression over the entire area

P = ∫ dP = ∫ (ρgy)dA= ρg∫ y dA ..(01)

But ∫ y dA is the Moment of the entire area about the free surface of the liquid given by

∫ y dA = A y

Substituting in Eq. 01, we get

P = ρg Ay= γ A y ..(02)

Where γ is the specific weight of water.

G P C

y h ^y

dy b

For water, ρ=1000 kg/m³ and g = 10 m/s². The force will be expressed in Newtons (N).

Centre of Pressure:

This is computed on the principle of Theorem of moments.

The moment of the pressure force about the free surface is given by

M = P x h = ρg A y h ...(03)

Similarly, the moment of the differential pressure force about the free surface is given by

dM = dPxy = (ρgy)dAxy = ρgy²dA

The moment of the pressure force about the free surface is given by integrating the above expression over the entire area A

M = ∫ ρgy²dA=ρg∫y²dA

But ∫y²dA is the Moment of Inertial of the entire submerged area about the free surface given by I_o

Hence M = ρgIo …(04)

From Eqs. 03 and 04, we get ρg A y h = ρgIo

y A h= I^o

But from parallel axis theorem of Moment of inertia, the moment of inertia of the given area about any axis is sum of the moment of inertia of the area about its centroidal axis and product of area and square of the distance between the two axes. Mathematically

Io = Ig + Ay²

Substituting we get y

A y I h= + ^g

Where y is the centroidal depth and h is the centre of pressure.

Inclined Plane surface submerged in liquid

Consider a vertical plane surface of some arbitrary shape immersed in a liquid of mass density ρ, in such a way that the plane is making an angle θ with the free surface as shown in Fig.

Let

A = Total area of the surface

y = Depth of Centroid of the surface from the free surface

y = Depth of Centroid of the surface from the free surface along the plane ∗

G = Centroid of the immersed surface C = Centre of pressure

h = Depth of centre of pressure from the free surface

h = Depth of centre of pressure from the free surface along the plane ∗

Consider a rectangular strip of breadth b and depth dy at a depth y from the free surface.

From the triangle, we have

∗

∗ = ∗ =

h h y

y y θ y

sin …(1)

G C P

h ^y

dy y*

O O

θ y

h y

y *

Total Pressure:

The pressure intensity at a depth y acting normal to the plane on the strip is p = ρgy

Total pressure force on the strip = dP = (ρgy)dA

∴ The Total pressure force on the entire area is given by integrating the above expression over the entire area along the plane

P = ∫ dP = ∫ (ρgy)dA= ρg∫ y dA

But y and dA are on different planes and hence substituting for y from Eq.

1, we get

P= ρg∫ y* sin θ dA = ρg sin θ ∫ y* dA …(2)

But ∫ y* dA is the Moment of the entire area about the free surface of the liquid given by

∫ y* dA = A y^∗sinθ = A y Substituting in Eq. 2, we get

P = ρg A y = γ A y … (3)

Where γ is the specific weight of water.

Centre of Pressure:

This is computed on the principle of Theorem of moments.

The moment of the pressure force about the free surface is given by

M = P x h = ^∗ ρg A y (h/sinθ) ...(4)

Similarly, the moment of the differential pressure force about the free surface is given by

dM = dPxy* = (ρgy)dAxy* = ρg(y* sinθ) y*dA=ρg y*²sinθ dA

The moment of the pressure force about the free surface is given by integrating the above expression over the entire area A

M = ∫ ρg y*²sinθ dA =ρ g sinθ ∫y*²dA

But ∫y*²dA is the Moment of Inertial of the entire submerged area about the free surface given by I_o

Hence M = ρgIosinθ …(5)

From Eqs. 4 and 5, we get ρg A y (h/sinθ) = ρgIosinθ

y A

h= I^o sin²θ

Io = Ig + Ay∗²= Ig + θ

2 2

sin y A

Substituting we get

y A y I

h= + ^g sin²θ

Where y is the centroidal depth, h is the centre of pressure and θ is the angle of inclination of the plane with the free surface.

From the above equation it is seen that h is always either equal to or greater than y , which means that the in general for a plane surface immersed in a liquid, the Centre of pressure always lies below the centroid.

Curved surface submerged in liquid

Consider a curved surface AB immersed in a liquid of mass density ρ as shown in Fig. Let dA be a differential area located on the curved surface at a depth y from the free surface such that it ma kes an angle θ with the horizontal.

θ dA

dA sinθ

dA cosθ C

O A

θ θ dF dFx

dFy

dA y

The pressure intensity p acting on the differential area dA is given by p = ρ g y

The total differential pressure force dF acting normal to dA is given by dF = p x Area = ρ g y dA

As can be seen from the Fig. this dF goes on changing its inclination and is not acting constantly at a particular angle. Hence, let us resolve dF into horizontal and vertical directions as dFx and dFy as shown in Fig. where dFx = dF sin θ

= ρ g y dA sin θ dF_y = dF cos θ = ρ g y dA cos θ

dF_x acts on the vertically projected area of dA = dA sin θ = dAy and dF_y acts on the horizontally projected area of dA = dA cos θ = dAx.

Hence

dFx = ρ g y dAy and dFy = ρ g y dAx

Integrating we get

∫ dFx = F_x = ∫ρ g y dAy = ρ g ∫ y dAy = ρ gA_yy

Where Ay is the projected area of the curved surface on vertical plane equivalent to the vertical plane surface and y is the centroidal distance of the projected area from the free surface.

Similarly, its location can be obtained by y

In document VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS FLUID MECHANICS Kinematics of Fluid Flow (Page 26-39)