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

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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

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Fluid Flow Measurement

Pitot tube

Pitot static tube

Venturi meter

Orifice meter

02

Vzg

ρ

P 2

Chapter 3 Week 4

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• 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

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• 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

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• 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

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• 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

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• 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

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• 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

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• 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.

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• 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.

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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

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P

V1

P2

V 1

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

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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

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• 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.

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• 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.

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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

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• 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

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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.

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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

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• 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

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• Correction factor Discharge Coefficient, Cv

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• 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,

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Orifice meter

• Venturi is expensive and complex to construct.

• To measure flow rate in small pipeline orifice meter

• Orifice meter cheap, easy to construct

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• 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

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• Correction factor Discharge Coefficient, Cv

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• From Bernoulli’s equation:

A

A1

PP 2

CV

2

1

2

1

2

2

21

vactual2,

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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.