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Home , Fluid mechanics

PART II

• Some applications of Fluid Mechanics –

• Pressure = F/A • Units: Newton's per square meter, Nm-2, kgm-1 s-2 • The same unit is also known as a Pascal, Pa, i.e. 1Pa = 1 Nm-2)

• English units: lbf/sqft, or inches of H2O • Also frequently used is the alternative SI unit the bar, where 1 bar = 105 Nm-2 • Dimensions: M L-1 T-2

Fluid Mechanics – Pressure Fluid Mechanics – Specific Gravity

• Gauge pressure: • (r): mass per unit volume. -3 -3 -3 pgauge = _ gh • Units are M L , (slug ft , kg m ) • Absolute Pressure: • Specific weight (SW): wt per unit volume.

pabsolute = _ gh + patmospheric • Units are F L-3, (lbf ft-3, N m-3) • Head (h) is the vertical height of fluid for • sw = rg constant gravity (g): • Specific gravity (s): ratio of a fluid’s density to the density of h = p/ _ g water at 4° C • When pressure is quoted in head, density (_) must also be given. s = r/rw -3 -3 • rw = 1.94 slug ft , 1000 kg m

1 Continuity and Fluid Mechanics – Continuity and

Conservation of Matter • Flow through a pipe: . • . Conservation. of mass for steady state (no storage) says • (m ) = Mass of fluid flowing through a m in = m out -1 control surface per unity time (kg s ) . . • Volume flow rate, or Q = volume of fluid flowing m m through a control surface per unit time (m3 s-1) • Mean flow (V ): m _1A1Vm1 = _ 2A2Vm2 • For incompressible , density does not changes (_ = _ ) V = Q/A 1 2 m so A1Vm1 = A2Vm2 = Q

Fluid Mechanics – Bernoulli’s equation

• The equation of continuity states that for an Y incompressible fluid flowing in a tube of varying 2 A V cross-sectional area (A), the mass flow rate is the 2 2 same everywhere in the tube: . . m1= m2 _ 1A1V1 = _ 2A2V2 _1A1V1= _2 A2V2 • Generally, the density stays constant and then it's Y For simply the flow rate (Av) that is constant. 1 A1 V1 A1V1= A2V2

Assume steady flow, V parallel to streamlines & no

2 Bernoulli Equation – energy Bernoulli Equation – work • Consider energy terms for steady flow: • Consider work done on the system is x distance • We write terms for KE and PE at each point • We write terms for force in terms of Pressure and area

Y Y 2 2 Wi = FiVi dt =PiViAi dt A2 V2 A2 V2 E = KE + PE i i i Note ViAi dt = mi/_i 1 2 E1 = 2 m&1V1 + gm&1 y1 W1 = P1m&1 / ρ1 2 Y 1 Y 1 E2 = 2 m&2V2 + gm&2 y2 1 W2 = −P2m&2 / ρ2 A1 V1 A1 V1

As the fluid moves, work is being done by the external Now we set up an energy balance on the system. to keep the flow moving. For steady flow, the work requires that the change in done must equal the change in mechanical energy. energy equals the work done on the system.

Bernoulli equation- energy balance Forms of the Bernoulli equation

Energy accumulation = _Energy – Total work • Most common forms: 1 2 1 2 0 = (E2-E1) – (W1+W2) i.e. no accumulation at steady state P1 + 2 ρV1 = P2 + 2 ρV2 + gρΔh Or W +W = E -E Subs terms gives: 1 2 2 1 PS1 + PV1 = PS 2 + PV 2 + ΔPht & & P1m1 P2m2 1 2 1 2 − = ( 2 m&V + gm& y ) − ( 2 m&V + gm&y ) The above forms assume no losses within the volume… ρ ρ 2 2 2 2 1 1 1 1 1 2 If losses occur we can write: Pm& P m& 1 1 1 m&V 2 gm&y 2 2 1 m&V 2 gm& y + 2 1 1 + 1 1 = + 2 2 2 + 2 2 PS1 + PV1 = PS 2 + PV 2 + losses + ΔPht ρ1 ρ2 And if we can ignore changes in height: For incompressible steady flow m&1 = m&2 and ρ1 = ρ2 = ρ

1 2 1 2 PS1 + PV1 = PS 2 + PV 2 + losses Key eqn P1 + 2 ρV1 + gρy1 = P2 + 2 ρV2 + gρy2

3 Application of Bernoulli Equation Bernoulli Equation for a venturi

developed the most important equation in fluid • A venturi measures flow rate in a duct using a pressure in 1738. this equation assumes constant density, difference. Starting with the Bernoulli eqn from before: irrotational flow, and velocity is derived from velocity potential:

PS1 + PV1 = PS 2 + PV 2 + losses + ΔPht

• Because there is no change in height and a well designed venturi will have small losses (<~2%) We can simplify this to:

PS1 + PS 2 = PV 2 + PV1 or − ΔPS = ΔPV

• Applying the continuity condition (incompressible flow) to get: 2(P − P ) V S1 S 2 1 = 2  A2  ρ1− 2   A1 

Venturi Meter Duct

Pv P P2 1 Ps 2(PS1 − PS 2 ) V C • PT=PS + PV  i.e. an energy balance 1 = e 2  A2  • – potential energy ρ1− 2  • Velocity pressure – KE A1   • Total pressure – total energy • Recall KE is _ mV2 so KE term is proportional to V2 • Discharge Coefficient C corrects for losses = f(R ) 2 e e • At NTP, PV = (V/4005) for V in ft/min, PV in inches H2O – Or V = 4005√PV {at non std conditions, V = 4005√(PV/d)}

4 Pitot tube Pressure, & flow rates The static and Pitot tube are often combined into the one-piece Pitot- static tube. • Units of pressure = “wg Total pressure port – The height of a column that would be

Static pressure port supported by that pressure – 406”wg = 1 atm – Traditional units not used since small pressure differentials are involved

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Static pressure measurement VP Measurement

upstream downstream upstream downstream SUCTION SIDE, INLET PRESSURE SIDE, OUTLET SP<0, TP<0, VP>0 SP>0, TP>0, VP>0 SUCTION SIDE, INLET PRESSURE SIDE, OUTLET SP<0, TP<0, VP>0 SP>0, TP>0, VP>0 F F A A N N

TP - SP = (SP + VP) = VP

5 Velocity & VP Density correction

• Velocity pressure can also be determined by • Density of standard air = 0.075 lb/ft3 measuring velocity at a given point and using – Air density affected by: moisture, & the following relationship: altitude above sea level – Roughly, density corrections are needed, when: • Moisture exceeds 0.02 lbs water/lb of air V = 4005√VP • Air temp outside of 40 – 100F range V = velocity in fpm • Altitude exceeds +1000 ft relative to sea level VP = velocity pressure in “wg

Density correction Pressures in LEV

• Total pressure (TP) 2 V  Density  VP   actual – sum of static pressure & velocity pressure =    3   4005   0.075lb / ft  – - upstream of fan – + downstream of fan

0.075lb 530F Bar.pressure Density = x x actual ft 3 (460 + t)F 29.92 TP = SP + VP Densityactual = air density in lb/ft3 T = temperature in oF Bar. Pressure = pressure in inches of mercury

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