PART II. Fluid Mechanics Pressure. Fluid Mechanics Pressure. Fluid Mechanics Specific Gravity. Some applications of fluid mechanics

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ART II Some applications of fluid mechanics Fluid Mechanics ressure ressure = F/A Units: Newton's per square meter, Nm -, kgm - s - The same unit is also known as a ascal, a, i.e. a = Nm - ) English units: lbf/sqft, or inches of H O Also frequently used is the alternative SI unit the bar, where bar = 0 5 Nm - Dimensions: M L - T - Fluid Mechanics ressure Gauge pressure: p gauge = _ gh Absolute ressure: p absolute = _ gh + p atmospheric Head (h) is the vertical height of fluid for constant gravity (g): h = p/ _ g When pressure is quoted in head, density (_) must also be given. Density (r): mass per unit volume. Units are M L -3, (slug ft -3, kg m -3 ) Specific weight (SW): wt per unit volume. Units are F L -3, (lbf ft -3, N m -3 ) sw = rg Fluid Mechanics Specific Gravity Specific gravity (s): ratio of a fluid s density to the density of water at 4 C s = r/r w r w =.94 slug ft -3, 000 kg m -3

Fluid Mechanics Continuity and Conservation of Matter Mass flow rate ( ṁ) = Mass of fluid flowing through a control surface per unity time (kg s - ) Volume flow rate, or Q = volume of fluid flowing through a control surface per unit time (m 3 s - ) Mean flow velocity (V m ): V m = Q/A Continuity and Conservation of Mass Flow through a pipe: Conservation of mass for steady state (no storage) says ṁin = ṁout ṁ _ A V m = _ A V m For incompressible fluids, density does not changes (_ = _ ) so A V m = A V m = Q ṁ Fluid Mechanics Continuity Equation Bernoulli s equation The equation of continuity states that for an incompressible fluid flowing in a tube of varying cross-sectional area (A), the mass flow rate is the same everywhere in the tube: _ A V = _ A V Generally, the density stays constant and then it's simply the flow rate (Av) that is constant. Y Y A V A V ṁ = ṁ _ A V = _ A V For incompressible flow A V = A V Assume steady flow, V parallel to streamlines & no viscosity

Bernoulli Equation energy Consider energy terms for steady flow: We write terms for KE and E at each point Bernoulli Equation work Consider work done on the system is Force x distance We write terms for force in terms of ressure and area Y A V Y A V W i = F i V i dt = i V i A i dt Y A V E E E i = KE i + E i & + & = mv gm y = m& V y Y A V Note V i A i dt = m i /_ i = m& / ρ W W = m& / ρ As the fluid moves, work is being done by the external forces to keep the flow moving. For steady flow, the work done must equal the change in mechanical energy. Now we set up an energy balance on the system. Conservation of energy requires that the change in energy equals the work done on the system. Bernoulli equation- energy balance Energy accumulation = _Energy Total work 0 = (E -E ) (W +W ) i.e. no accumulation at steady state Or W +W = E -E Subs terms gives: m& m& = m& V y) ( m& V ρ ρ m& m& + m& y V y = + m& V ρ ρ ( y For incompressible steady flow + ρ V + gρy = + ρv + gρy m & = m& and ρ = ρ = ρ ) Forms of the Bernoulli equation Most common forms: + ρv = + ρv + gρδh S If losses occur we can write: S + Δ The above forms assume no losses within the volume And if we can ignore changes in height: ht + losses + Δ S + losses Key eqn ht 3

Application of Bernoulli Equation Daniel Bernoulli developed the most important equation in fluid hydraulics in 738. this equation assumes constant density, irrotational flow, and velocity is derived from velocity potential: Bernoulli Equation for a venturi A venturi measures flow rate in a duct using a pressure difference. Starting with the Bernoulli eqn from before: S + losses + Δ Because there is no change in height and a well designed venturi will have small losses (<~%) We can simplify this to: S + S = V or Δ Applying the continuity condition (incompressible flow) to get: ( S S ) V = ρ A A S ht = Δ V Venturi Meter Duct pressures V = C ( S S ) ρ A A Discharge Coefficient C e corrects for losses = f(r e ) e s v T = S + V i.e. an energy balance Static pressure potential energy Velocity pressure KE Total pressure total energy Recall KE is _ mv so KE term is proportional to V At NT, V = (V/4005) for V in ft/min, V in inches H O Or V = 4005 V {at non std conditions, V = 4005 ( V /d)} 4

itot tube The static and itot tube are often combined into the one-piece itotstatic tube. Total pressure port Static pressure port ressure, velocities & flow rates Units of pressure = wg The height of a liquid column that would be supported by that pressure 406 wg = atm Traditional units not used since small pressure differentials are involved 5. Static pressure measurement V Measurement upstream SUCTION SIDE, INLET S<0, T<0, V>0 F A N downstream RESSURE SIDE, OUTLET S>0, T>0, V>0 upstream SUCTION SIDE, INLET S<0, T<0, V>0 F A N downstream RESSURE SIDE, OUTLET S>0, T>0, V>0 T - S = (S + V) = V 5

Velocity & V Velocity pressure can also be determined by measuring velocity at a given point and using the following relationship: V = 4005 V V = velocity in fpm V = velocity pressure in wg Density correction Density of standard air = 0.075 lb/ft3 Air density affected by: moisture, temperature & altitude above sea level Roughly, density corrections are needed, when: Moisture exceeds 0.0 lbs water/lb of air Air temp outside of 40 00F range Altitude exceeds +000 ft relative to sea level V Density correction actual = 3 V Density 4005 0.075lb / ft Density 0.075lb 530F Bar. pressure = x x actual 3 ft (460 + t) F 9.9 Density actual = air density in lb/ft3 T = temperature in o F Bar. ressure = pressure in inches of mercury Total pressure (T) ressures in LEV sum of static pressure & velocity pressure - upstream of fan + downstream of fan T = S + V 6

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