Chater 7: Energy Princile By Dr Ali Jawarneh Hashemite University
Outline In this chater we will: Derive and analyse the Energy equation. Analyse the flow and shaft work. Derive the equation for steady flow. Derive the equation for incomressible, steady flow in a ie. Discuss some alications on the energy equation: Sudden Exansion Discharge of ie into a reservoir Introduce the concet of energy and hydraulic grade lines.
First Law of Thermodynamics st law of thermodynamics for a given system: Where: ΔE E = Q W Q: Heat transferred to the system. W: Work done by the system on the surroundings. E: the energy of a system E = E u E k E E u : Internal Energy (atoms, chemical, electrical, ) E k : kinetic energy E : otential energy 3
In terms of rate of energy: Q ( ) de dt = Q W system W ( ) W Q and W are ath function A P C B olume 4
7.: Derivation Energy Equation Control olume equation db dt sys d = bρ d bρda dt cv Let: B sys = E sys = E u E k E, then b=e=ue k e Where (e) is the energy er unit mass and e k = /, and e =gz cs de dt sys = d dt cv e ρ d e ρ. da cs 5
Recall the st law of thermodynamics Q& W& = d Q& W& = dt de sys d dt dt = Q W cv e ρ d e ρ. da cs d Q& W& = ( gz u ) ρd ( gz u ) ρ. d dt ( ek e u) ρ d ( ek e u) ρ. A cv cv cs A Work can be divided into: - Shaft work (through turbine or um). - Flow work (due to ressure). cs = s W W W Shaft work rate flow work rate f 6
Flow work: Work done by ressure forces as the system moves through sace W & f = A W& f = A W f = ρ cs ρ da 7
Substitute the shaft and flow work back in the energy equation: d Q& W& s = ( gz u) ρd ( gz u ) ρ. d dt ρ A cv But by definition the enthaly h is: cs h = u ρ then d Q& W& s = ( gz u) ρd ( gz h) ρ. d dt A cv cs 8
7.: Simlified Forms of the Energy Equation For steady flow: Q& W& s = ( g z h) ρ. da For steady and uniform roerties: cs Q& W& s = ( g z h) ρ A. cs 9
Examle: A turbine receives steam at.8 Ma, 500 o C (h=3470 kj/kg) at a velocity of 5 m/s. The steam exits at an enthaly 0f 630 kj/kg with a velocity of 70 m/s. The steam flows through at a rate of kg/s, and the turbine develos 830 kw. Calculate the heat transfer from the turbine. Neglect the otential energy due to the elevation difference. 0
Solution: Q& W& s = ( g z h) ρ. da cs Q W ( h )( m ) ( h )( m ) s = 5 70 Q 830 = ( 3470 )( ) ( 630 )( ) 000 000 Q = 7. 6 kw
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Suction cavitation Suction cavitation occurs when the um suction is under a low ressure/high vacuum condition where the liquid turns into a vaor at the eye of the um imeller. This vaor is carried over to the discharge side of the um where it no longer sees vacuum and is comressed back into a liquid by the discharge ressure. This imloding action occurs violently and attacks the face of the imeller. An imeller that has been oerating under a suction cavitation condition can have large chunks of material removed from its face or very small bits of material removed causing the imeller to look songe like. Both cases will cause remature failure of the um often due to bearing failure. Suction cavitation is often identified by a sound like gravel or marbles in the um casing. 5
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Energy equation for steady flow of an incomressible fluid in a ie A more general form of the equation: Q& W& s = ( g z u ) ρ. A ρ cs Re-writing the equation: 3 3 ρ ρ ρ ρ Q& W& s ( g z u ) da da ( g z u ) da da ρ = ρ A A A A 8
cs ρ da = ρ A = m α = A cs [ ] 3 da Integrating the velocity distribution: & & Q W m& ( g z ) & u m ( g z u s α = α ρ ρ ( Where α and α are kinetic energy correction factors: α= Uniform flow α= Laminar flow α=.05 Turbulent flow (aroximated as α=) ) 9
Re-arranging: Re arranging: ) ( u z g u z g W Q m s α ρ α ρ = & & & The shaft work can be written as: m ρ ρ Writing in head form: t s W W W & & & = Writing in head form: Q u u z W z W t & & & = α α mg g g z mg g z mg & & & = α γ α γ 0
The equation is most commonly known in the following form: γ g γ z α h = z α g h t h L Where: Thermal energy Mechanical energy α = Laminar flow α = Turbulent flow h h h t L W& W& = = mg & γq W& & t W = = t mg & γq u u Q& = g mg & Pum Head Turbine Head Head Loss Note: The bar over is usually omitted
Pums and turbines lose energy due to: - mechanical friction - viscous dissiation 3- leakage η fluid shaft shaft W fluid W γqh = = Pum effeciency wt W shaft W shaft ηt = = Turbine effeciency γ Qh W t
Examle: In this system, d=6 in., D= in., Δz =6 ft, and Δz = ft. The discharge of water in the system is 0 cfs. Is the machine a um or a turbine? What are the ressures at oint A and B? Neglect head losses. Assume α=. 3
0 Solution: Q 0 = B = = = 50. 95 A π ( 6 / ) 4 Energy equation between & and assume the machine is um: 8 0 0 0 0 α z h = α z ht h h. L = 3 γ g γ g 0 Therefore a machine is um 3. ft/s ft/s ft Energy between B & : γ B g 6 0 g B αb zb h = α z 6.4 Ibf/ft 3 g γ 0 0 0 0 B = 374. 4 sfg( Ibf / ft ) =. 6 sig ( Ibf / in h t h L ) 4
Continuity: ft/s. A A A A A B B A B B A A 74 = = = Energy between A & : A 0 0 0 0 0 8 L t A A A A h h z g α γ h z g α γ = sig. sfg A 58 = 35 = 8 5
Examle: Water is flowing at a rate of 0.5 m 3 /s, and it is assumed that h L = /g from the reservoir to the gage, where is the velocity in the 30-cm ie. What ower must the um suly? 6
Solution: Q 0. 5 30 = = = = = 3. 54 m/s A π 30 0. 3 4 ρd 000 3. 54 0. 3 Re = = = 80687 Turb. α μ. 3 0 3 = ρ and μ From Table A.5 00 000 /g 0 0 6 0 0 γ g g α z h = α z W γ h = 6. m = Qγh = 0. 5 980 6. = 39. 5 kw h t h L Note: the minor losses are negligible 7
7.3: Alication of the Energy, Momentum, and Continuity Princiles in Combination Abrut Exansion Considering the sudden exansion from a smaller to a larger ie: The flow searates from the boundary as shown. The one-dimensional energy equation between sections and : 8
h L g z g z = α γ α γ The momentum equation for fluid in th l i the larger ie: = m m F s & & sin A A L A A A ρ ρ α γ sin A A L A A A ρ ρ α γ = ) ( ga A g z z = γ γ 9 ga g γ γ
Using this equation along with the energy equation and the continuity equation: ( h ) L = g 30
Discharge into a Reservoir When a ie discharges into a reservoir, =0: h L = g The energy is dissiated i d by viscous action of the liquid. 3
Examle: This abrut exansion is to be used to dissiate the high-energy flow of water in the 5-ft diameter enstock. a- What ower (in horseower) is lost through the exansion b- If the ressure at section is 5 sig, what is the ressure at section? c- What force is needed to hold the exansion in lace? sig: Pounds er Square Inch 3
Solution: A = = 6. 5 A A ft/s ( ) ( 5 6. 5 ) h L = = = 5. 46 g 3. ft h = 550 Ibf.ft/s h = 745. 7 W Q = A = 490. 9 ft 3 /s a- Qγ hl 490. 9 6. 4 5. 46 P = horseower = = = 304 550 550 (h) ρd. 94 5 5 Re5 = = = 888784 Turb. α = 5 μ. 73 0 ρd. 94 6. 5 0 Re0 = = = 44439 Turb. α = 5 μ. 73 0 ρ and μ from Table A.5 33
b- γ 0 0 α z h = α z ht g γ g h L 5 44 5 = 6. 4 3. 6. 4 = 946. 6 sfg = 6. 57 6. 5 3. sig 5. 46 c- A A Fx = m m π m = ρ A =. 94 5 5 = 95. 8 4 = m m m F x = 49 Ibf 34
Examle: For this sihon the elevations at A, B, C, and D are 30 m, 3 m, 7 m, and 6 m, resectively. The head loss between the inlet and oint B is ¾ of the velocity head, and the head loss in the ie itself between oint B and the end of the ie is ¼ of the velocity head. For these conditions, what is the discharge and what is the ressure at oint B? The diameter=30 cm. 35
Solution: h L, = 4 g 3 4 g = g gg Abrut Energy fro A to B Note: the minor losses are negligible excet the abrut 36
EXAMPLE: Water discharges to the atmoshere from a large faucet (control valve) as shown in the figure with a ressure at the faucet inlet is 50 kpa, gauge. The faucet is held stationary at oint A. Let α s are equal to one, the total head loss in the faucet is 0 m and the inlet/outlet diameters are 5 & cm, resectively. Neglect the weight of faucet and the water inside it. d = 0.35 m and d =0.5m. What are the inlet and outlet velocities? What are the comonents of the force to hold the faucet stationary? What is the torque (moment) necessary to kee the faucet from twisting? 37
SOLUTION: m = m ρ A = ρ A A = A = 0. 64 50000 γ 3 0 0 0 0 0 α z h = α z g γ g 4.7 m / s h t h = M 0 =.9 m / s = ρ A = m 59.8 kg / s L M 0 P A Fx = m Fx = 8. 4 kn to left F z = m( ) = 5. 95 kn to down 38
M = ( r v ) ρ. da M o P A ( d ) = ( r( k ) v( i) )( m ) ( r( i) v( k ) )( m ) = m( r v r v ) r r = d = d = 0.35 = 0.5 m m M =. 765 kn m o 39
7.4: Concets of the Hydraulic and Energy Grade Lines z α h = z α ht hl γ g γ g z γ Piezometric head z γ Total head g HGL=Hydraulic grade line= line to describe the iezometric head EGL= Energy grade line= line to describe the total head 40
Hydraulic & Energy Grade Lines Recalling the energy equation between the surface of the reservoir and the downstream section: z = z α γ g hl 4
Hydraulic & Energy Grade Lines The height of the /γ Z line is called the hydraulic grade line (HGL). The height of the total energy line is called the energy grade line (EGL). HGL and EGL are useful means of sotting velocity and ressure distributions in the system. 4
Hydraulic & Energy Grade Lines EGL is ositioned above HGL by α /g. If velocity is zero, then HGL and EGL coincide. If there is an increase in the diameter of the ie, this means that the distance between EGL and HGL should decrease. If there is a decrease in the diameter of the ie, this means that the distance between EGL and HGL should increase. 43
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Hydraulic & Energy Grade Lines 45
Hydraulic & Energy Grade Lines Head loss in a ie means that EGL sloes downward in the direction of the flow. (The same alies for HGL) The sloe at which EGL dros is steeer for ies with smaller diameters. For steady flow in a ie with uniform hysical characteristics, HGL and EGL will be arallel (having the same sloe). 46
Hydraulic & Energy Grade Lines 47
Hydraulic & Energy Grade Lines A um causes an abrut rise in HGL and EGL. A turbine causes an abrut dro in HGL and EGL. When HGL coincides with the ie, this means that the ressure is equal to zero. When HGL dros below the ie, this means negative ressure. 48
Hydraulic & Energy Grade Lines 49
Hydraulic & Energy Grade Lines 50
Hydraulic & Energy Grade Lines 5
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Examle: What horseower must be sulied to the water to um 3.0 cfs at 68 o F from the lower to the uer reservoir? Assume that the head loss in the ies is given by h L =0.05(L/D)( /g), where L is the length of the ie in feet and D is the ie diameter in feet. Sketch the HGL and the EGL. 53
Solution: Q 3 = = = 8. A π 59 ( 8 / ) 4 ft/s h = h h = 0. 05 (L/D)( / g) L,total L,ie L,abrut g 8. 59 h L,total = 0. 05( 3000/( 8/ ))( 8. 59 / 3. ) = 78. 5 3. γ 0 90 0 g γ g α z h = α z h = 8. 5 ft 40 h t 0 h L ft P = horseower = Q γ h 550 = 3 6. 4 8. 5 = 43. 7 Note: the minor losses are negligible excet the abrut h 54
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