Fluid Flow measurement

Revision

First Law of Thermodynamics

Types of energy

3 ways that energy transferred to a system.

General Mechanical Energy Equation

3 types of non-flow work

Head form of Mechanical Energy Eqn

Bernoulli’s Equation

Application: Nozzle and diffuser

MZA@UTPChemEFluidMech

Fdm

dW

2

ΔVzg

ρ

ΔP n.f

2

ggdm

dW

2g

ΔVΔz

ρg

ΔP n.f

2 F

02

Vzg

ρ

P 2

Toricelli’s equation


Fluid Flow Measurement

Pitot tube

Pitot static tube

Venturi meter

Orifice meter

02

Vzg

ρ

P 2

Chapter 3 Week 4


• A useful concept associated with Bernoulli’s equation deals with the stagnation and dynamic pressures.

• These pressures arise from the conversion of kinetic energy in a flowing fluid into a “pressure rise” as the fluid is brought to rest.

• Bernoulli’s equation can be rewritten as:

02

ρΔVρgΔzΔP

2


• Static pressure.

– One could measure static pressure in a flowing fluid by drilling a hole on a flat surface, and attach a piezometer tube.

02

ρΔVρgΔzΔP

2


• Hydrostatic pressure.

– not actually a pressure but does represent the change in pressure possible due to potential energy variations of the fluid as a result of elevation changes.

02

ρΔVρgΔzΔP

2


• Dynamic pressure.

– Its interpretation can be seen in by considering the pressure at the end of a small tube inserted into the flow and pointing upstream.

02

ρΔVρgΔzΔP

2


• Dynamic pressure.

– After the initial transient motion has died out, the liquid will fill the tube to a height of H as shown.

– The fluid in the tube, including that at its tip,(2),will be stationary

02

ρΔVρgΔzΔP

2


• Applying Bernoulli’s equation between (1) and (2):

• Or, the sum of static and dynamic pressures is equal to stagnation pressure.

• V2 = 0 Stagnation point

2

ρVPP

2

112


• Stagnation point:

– Exist in any solid body placed into a flowing fluid.

– For symmetrical objects stagnation point is clearly at the tip or front of the object.

– For nonsymmetrical objects the location of the stagnation point is not always obvious.


• Knowledge of the values of the static and stagnation pressures in a fluid implies that the fluid speed can be calculated.

• This is the principle on which the Pitot-static tube is based.


1) Static holes at appropriate distance from the tip of the tube.

2) Open tip of the tube.

3) Towards high pressure side.

4) Towards low pressure side


P

V1

P2

V 1

• Instead of measuring static or stagnation pressures separately, it is customary to measure their difference with a transducer.


Example

• Air is flowing in a duct. The pressure-difference gage attached to the pitot-static tube indicates a difference of 200 Pa. what is the air velocity? (air = 1.20 kg/m3)

• Solution:

s

m 28.81

2.1

2002

P2V 1


• The primary disadvantage of Pitot tube is that it must be aligned with the flow direction, which is unknown.

• For yaw angles greater than 5o, there are substantial errors in both the static and stagnation pressures measurements.


• Because of the slow response of the fluid filled tubes leading to the pressure sensors, it is not useful for unsteady flow measurements.

• The Pitot-static tube is useful in liquid and gases; for gases a compressibility correction is needed if the stream Mach number is high.

• It is not suitable for low-velocity measurement in gases because of the small pressure differences developed.


Venturi meter

• Consist of:

– Truncated cone with decreasing cross sectional area

– Short cylindrical section (throat)

– Truncated cone with increasing cross sectional area

Perpendicular to the flow


• From Bernoulli’s equation:

……………..(1)

• Friction is negligible then F is dropped

• From continuity equation, inserting into (1)

02

VV

ρ

PP2

1

2

212

A

A1

ρ

PP2

V

2

1

2

2

21

2


Example

• The venturi meter has water flowing through it. The pressure difference P1 – P2 is 1 psi. The diameter at point 1 is 1 ft, and at point 2 is 0.5 ft. Determine the velocity through this meter.


Solution

• Assumption: Water incompressible fluid, density constant

( = 62.4 Ibm/m3)

s

ft12.7

1

0.51

sIbf

ftIbm 32.2

ft

Ibm62.4

ft

in 144

in

Ibf1 2

A

A1

ρ

PP2

V

2

2

2

3

2

2

2

2

1

2

2

21

2


• V2 that is obtained is the ideal V2

• In venturi, actual V2 is less than ideal

V2

• A correction factor is needed

A

A1

PP2

V

2

1

2

2

21

2


• Correction factor Discharge Coefficient, Cv


• The actual velocity, V2

• A reliable flow-measuring device

• Cause less pressure loss (F is small)

• Used for particularly large-volume liquid and gas flow

A

A1

PP 2

CV

2

1

2

1

2

2

21

vactual2,


Orifice meter

• Venturi is expensive and complex to construct.

• To measure flow rate in small pipeline orifice meter

• Orifice meter cheap, easy to construct


• Consist of flat orifice plate with circular hole drilled in it.

• Vena contracta contraction of flow occurs

• Cause actual outlet velocity less than ideal outlet velocity

Vena contracta


• Correction factor Discharge Coefficient, Cv


• From Bernoulli’s equation:

A

A1

PP 2

CV

2

1

2

1

2

2

21

vactual2,


Example

The flow rate of methanol at20C (=788.4 kg/m3 and=5.85710-4 kg/ms) through4-cm-diamter pipe is to bemeasured with a 3-cm-diameter orifice meterequipped with a mercury(=13,600 kg/m3) manometeracross the orifice place (seefigure). If the differential of themanometer is read to be 11cm, determine the flow rate ofmethanol through the pipe andthe average flow velocity.

Fluid Flow measurement

Documents

Transcript of Fluid Flow measurement