Microsoft PowerPoint - HydL_Ch3_Energy [호환 모드]

Size: px

Start display at page:

Download "Microsoft PowerPoint - HydL_Ch3_Energy [호환 모드]"

쳥아 간
5 years ago
Views:

1 3. Energy & Power in Hydraulic Systems Hydraulic Energy & Power Efficiency Pascal s Law Hydraulic Jack Air-to-Hydraulic Pressure Booster Conservation of Energy Law Continuity Equation Hydraulic Power Potential Energy & Kinetic Energy Bernoulli s Equation Torricelli s Theorem The Siphon -1-

2 Pressure & Flow -2-

3 3.1 Hydraulic Energy & Power Energy: the ability to perform work Input ME Lost HE = Output ME Power: the rate of doing work or expending energy Hydraulic System Is not a source of energy Energy transfer system Is much more versatile to transmit power variable speed, reversibility, overload protection, ti high-power-to-ratio, h ti immunity it to damage under stalled conditions -3-

4 Source of Power -4-

5 Mechanical Power Transmission -5-

6 Hydraulic Power Transmission -6-

7 3.2 Review of Mechanics Newton s 1 st Law 관성의법칙 : An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Newton s 2 nd Law 가속도의법칙 : The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. Newton s 3 rd Law 작용반작용의법칙 : For every action, there is an equal and opposite reaction. -7-

8 Mechanics Linear Motion v = s/t F = ma W = Fs P = W/t / = Fs/t / = Fv Rotational Motion ω = θ/t T = Fr W = Tθ P = Tω Torque horsepower, Brake horsepower(bhp) Efficiency η = output power/input power X 100 (%) 동력손실 : 유압유의누설, 유체내부의마찰, 기계적마찰 -8-

9 3.3 Pascal s Law Pascal s law 유체의압력은관에대하여직각으로작용한다. 각점의압력은모든방향으로같다. 밀폐된용기속의유체의일부분에가해진압력은동시에각부에같은세기를가지고전달된다. 기계시스템의레버에의한힘의증폭은유공압시스템에서파스칼의법칙에의한힘의증폭과유사하다. P = F 1 /A 1 = F 2 /A 2-9-

10 3.4 Hydraulic Jack -10-

11 Hydraulic Multiplication -11-

12 Simple Hydraulic Jack & Mechanical Lever -12-

13 Analysis of Simple Hydraulic Jack Pascal s Law p 1 = p 2 F 1 /A 1 = F 2 /A 2 F 2 /F 1 = A 2 /A 1 V 1 = V 2 A 1 S 1 = A 2 S 2 S /S A /A S 2 /S 1 = A 1 /A 2 F 2 /F 1 = S 1 /S 2 F 1 S 1 = F 2 S 2 Conservation of energy 다른동력시스템과마찬가지로유압시스템은동력을생성할수없다. -13-

14 Hydraulic Jack 19C 영국의 J.Bramah -14-

15 Air-to-Hydraulic Pressure Booster 공간절약, 운영및유지관리비용절감압력비 = 출력오일압력 / 입력공기압력 = 공기피스톤넓이 / 유압피스톤넓이 100psi의공기압으로 30,000psi000psi 까지의유압으로바꾸어줄수있다. Pressure ratio = 1000psi / 100psi = 10in 2 / 1in 2 =

16 3.5 Conservation of Energy Conservation of Energy 에너지 = 힘 X 움직인거리 : [Nm] [J] 입력에너지와출력에너지는같다. F 1 s 1 = F 2 s 2 시스템의총에너지는항상같다. 즉, 에너지는생성되지도소멸되지도않는다. Energy = Potential Energy + Kinetic Energy Elevation Potential Energy (EPE) EPE = WZ Pressure Potential Energy (PPE) PPE = WP/γ Kinetic Energy (KE) KE = ½ W/g v 2 E T = WZ + WP/γ + ½ W/g v 2 = constant -16-

17 3.6 Continuity Equation 정상유동의관로에서모든단면의무게유량 (weight flow rate) 은같다. w 1 = w 2 γ 1 A 1 v 1 = γ 2 A 2 v 2 Q 1 = A 1 v 1 = A 2 v 2 = Q 2 v 1 /v 2 = A 2 /A 1 = (D 2 /D 1 ) 2-17-

18 Continuity Equation Integral Relation for a Control Volume Conservation of mass dm dt d = 0 = ( ρdv ) + ρ( V ˆ) r n da dt CV.. CS.. sys Derivation of the Continuity Equation for Lumped Fluid Circuit ρ V = m = constant VΔ ρ + ρδ V = Δ V V Δ ρ = ρ 0 V βe = ΔV ΔP = ρ ΔP Δρ -18-

19 Generalized Flow-Continuity Equation Generalized Flow-Continuity Equation Mass Flow Rate Continuity dm dt d = dt sys ( ρdv ) CV.. d ( ρivi) d ( ρovo) = d ( ρv) = V dρ + ρ dv dt dt dt dt dt ρ Δ ρ = ΔP ΔV 0 β e Generalized Flow-Continuity i Equation Q i dρ ρ dp = dt β dt V dp dv Q = + o (for β dt dt Lumped model, ρ =ρ =ρ) i o ) e e -19-

20 3.7 Hydraulic Power Hydraulic horsepower (HHP; 유압마력 ) 유체가액츄에이터에전달하는마력 HHP[kW] = p[pa]q[m 3 /s]/10 3 = p[kgf/cm 2 ]Q[l/min]/612 ]/612 Output horsepower (OHP; 출력마력 ) 액츄에이터가부하에전달하는마력 출력마력은마찰손실이나누설손실에의하여항상유압마력보다작다. -20-

21 Hydraulic Cylinder Example Questions How do we determine how large a piston diameter is required for the cylinder? What is the pump flow rate required to drive the cylinder through its stroke in a specified time? How much hydraulic horsepower does the fluid deliver to the cylinder? Answers pa = F load A = F load /p V D = AS Q = V D /t = AS/t = Av E = FS = pas P = E/t = pas/t = pav = pq HHP[kW] = p[pa]q[m 3 /s]/10 3 = p[kgf/cmg 2 ]Q[l/min]/

22 Conversion of Power Mechanical power = force x linear velocity = torque x angular velocity Electrical power = voltage x electric current Hydraulic power = pressure x volume flow rate -22-

23 3.8 Bernoulli s Equation Z : elevation head p/γ : pressure head v 2 /2g : velocity head 관로의마찰손실을무시할수있다면 1 지점과 2 지점에서 Wlb 의유체가갖는총에너지는같다. WZ 1 + Wp 1 /γ + Wv 12 /2g = WZ 2 + Wp 2 /γ + Wv 22 /2g : [Nm] Z 1 + p 1 /γ + v 12 /2g = Z 2 + p 2 /γ + v 22 /2g : [m] -23-

24 Bernoulli s Equation Bernoulli s Equation Incompressible, steady state에서유선을따라 Euler s equation을적분 2 p V z constant along streamline γ + 2g + = Basic differential momentum equation for an infinitesimal element 2 dv ρ g p + μ V = ρ dt Navier-Stokes Equation Newtonian Fluid의경우전단력이속도구배에비례 dv u u i j ρg p+ τij = ρ τij μ = + dt xj x i Euler s Equation Inviscid Flow의경우점성력을무시 dv ρ g p = ρ dt -24-

25 Energy Equation 유체 1lb의덩어리가 1지점에서갖고있는총에너지에서펌프에의해가해지는에너지를더하고, 유압모터에의해제거되는에너지를빼고, 마찰손실에의한에너지를빼면, 이유체가 2 지점에도달했을때의총에너지와같다. Z 1 + p 1 /γ + v 12 /2g + H p H m H L = Z 2 + p 2 /γ + v 22 /2g : [m] H L : head loss H p : pump head H m : motor head p=γh, HHP[kW]=p[Pa]Q[m 3 /s]/10 3 H p [m] = 1000 HHP[kW]/Q[m 3 /s] 로부터 -25-

26 Venturi Application A 1 v 1 = A 2 v 2 p /γ + v 12 /2g = p 2 /γ + v 22 /2g p 1 - p 2 = γ/2g(v 22 - v 12 ) > 0-26-

27 3.9 Torricelli s Theorem 이상적인유체의자유분사 (free jet) 속도는중력가속도와수두의곱의 2배에제곱근을취한것과같다. Z / 2 H H / p 1 /γ + v 12 /2g + H p H m H L = Z 2 + p 2 /γ + v 22 /2g h = v 22 /2g v 2 = sqrt(2gh) 이상유체가아니면구멍으로밀어내는순수한압력수두가작아져서분사속도가줄어들며, 또한실제의분사속도는유체의점도에영향을받는다. v 2 = sqrt(2g(h-h L L)) -27-

28 Operating Principle of Siphon Conditions The elevation of the free end must be lower than the elevation of the liquid surface inside the container The fluid must initially be forced to flow up from the container into the center portion of the U-tube. This is normally done by temporarily providing a suction pressure at the free end of the siphon. Z 1 + p 1 /γ + v 12 /2g + H p H m H L = Z 2 + p 2 /γ + v 22 /2g Z H L = Z v 22 /2g v 2 = sqrt(2g(z 1 -Z 2 -H L )) = sqrt(2g(h-h L )) -28-

29 The Siphon: Toilet -29-

30 Report Text Problems Due date: 2 주후 -30-

Microsoft PowerPoint - 7-Work and Energy.ppt

Microsoft PowerPoint - 7-Work and Energy.ppt Chapter 7. Work and Energy 일과운동에너지 One of the most important concepts in physics Alternative approach to mechanics Many applications beyond mechanics Thermodynamics (movement of heat) Quantum mechanics...