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Dec. 18, 2013 Dr. Celik Energy in Steady Flow Motion of a real fluid, Review A P 1 Energy in Steady Flow General Energy Equation, steady flow, incompressible fluid Review Change in the internal energy per unit weight of a fluid (e.g. due to friction, hL=hf) coincides with change in temperature: 2 Energy in Steady Flow Power considerations in fluid flow: Review P = γhQ Power in BG units Power in SI units Horsepower = P = γhQ/550 Kilowatts = P = γhQ/1000 ( [Q] = cfs, [h] = ft, [γ] = pcf ) ( [Q] = m3/s, [h] = m, [γ] = N/m3 ) P : power put into flow by a pump, then h = hpump Pump efficiency, ηpump = (γhpumpQ ) / (power input) P : power extracted from flow by a turbine, then h = hturbine Turbine efficiency, ηturbine = (power output) / (γhturbineQ ) P : power lost because of friction, then h = hL 3 Energy in Steady Flow Energy and Power considerations in fluid flow, Stories Why are electricity rates cheaper at night in some places? Because at night (low demand) the generated energy http://www.energymanagertraining.com will be otherwise wasted. How should I eliminate cavitation in a pipe flow? How can I transfer fermented beer (homebrew) over to another vessel in a way that does not expose it to air or other contaminants? You can use a simple siphon. http://www.planetware.com/ You can use a control valve. http://www.planetware.com/ http://www.valvedirectory.com or reduce the operating temperature. 4 Energy in Steady Flow Power considerations in fluid flow Story There are approximately 95,000 miles nationwide of refined products pipelines It has been estimated that industrial pumping systems account for nearly 25% of industrial electrical energy demand in the United States. 5 Energy in Steady Flow Definition of Hydraulic Grade Line and Energy Line 6 Energy in Steady Flow Definition of Hydraulic Grade Line and Energy Line EL (also called Energy Grade Line, EGL) is positioned above the HGL by an amount equal to velocity head (V2/2g) Real fluid 7 Energy in Steady Flow Tips for Drawing HGL & EL 1) As the velocity goes to zero, the HGL and the EL approach each other. Thus, in a reservoir, they are identical and lie on the surface. 2) The EL and HGL slope downward in the direction of the flow due to the head loss in the pipe (exceptions: pump, diffuser). For steady flow in a pipe, the slope will be constant (ΔhL/ΔL = const.) 3) A jump occurs in the HGL and the EL whenever energy is added to the fluid as occurs with a pump, and a drop occurs if energy is extracted from the flow, as in the presence of a turbine. 8 Energy in Steady Flow Tips for Drawing HGL & EL 4) When a flow passage changes diameter, the distance between the EL and HGL will change, because velocity changes. In addition, for real fluids, the slope of the EL will change because the head loss per length will be larger in a conduit with larger flow velocity. 5) At points where the HGL passes through the centerline of the pipe, the pressure is zero. If the pipe lies above the HGL, there is a vacuum in the pipe, a potential location for cavitation, also a condition that is often avoided, if possible, in the design of piping systems; an exception would be in the design of a siphon. 9 Energy in Steady Flow Tips for Drawing HGL & EL 6) When a fluid discharges with velocity V from the end of the pipe into a tank reservoir that is so large that the velocity within it is negligible, the entire kinetic energy of the flow dissipates. This situation is called the loss of head at submerged discharge. 10 Energy in Steady Flow Problem 2 In the pipe: hL = 1.6V2/2g patm = 90 kPa For the liquid in the suction pipe, V = 1.8 m/s What is the maximum allowable value of z if the liquid were a) Water at 20° C b) Gasoline with pv = 49 kPa abs with a specific weight of 8 kN/m3 11 Energy in Steady Flow Problem 2 , Solution 1) Apply energy equation between A and B. When maximum z is considered, to prevent cavitation, pb abs = p min = pv zmax = (90-pv)/γ – 0.264 2) For water at 20° C, γ = 9.789 kN/m3 and pv = 2.34 kN/m2 abs zmax = 8.69 m 3) For gasoline with pv = 49 kPa abs and γ =8 kN/m3 zmax = 4.86 m 12 Energy in Steady Flow 2 Problem 1 40 m 1 L = 5000 m Water 10 m pump Q D = 0.5 m 5m hL = 0.01(L/D)V2/2g Q = 3 m3/s What is the power (hp) supplied to the flow? Draw the HGL and EL for the system. 13 Energy in Steady Flow Problem 3 A piping system with a black box shows a large EGL change at the box (steady flow-uniform diameter). What is the flow direction? What could be in the “black box”? A pump, or a turbine? 14 Energy in Steady Flow Problem 4 The EGL and HGL are as shown for a certain system. Find: (a) Direction of flow. (b) Whether there is a reservoir. (c) Whether the diameter at E is uniform or variable. (d) Whether there is a pump. (e) Sketch a physical set up that could exist between C and D. 15 Energy in Steady Flow Problem 5 Two tanks are connected by a uniformly tapered pipe. Draw the HGL and EGL. 16 Momentum and Forces in a Fluid Momentum principle is important in flow problems where we need to determine the forces. Examples: forces on vanes, pipe bends, the thrust produced by a rocket or turbojet, torque produced by a hydraulic turbine Such forces occur whenever the velocity of a stream of fluid changes in direction and/or magnitude Newton’s 2nd law: http://www.flowtite.com/ d (mV ) S F dt The sum of external forces on a body of fluid or system is equal to the rate of change of linear momentum of that body (of the fluid or system) Change in momentum must be in the same direction as the force (acting on the fluid) By the law of action and reaction, the fluid exerts an equal and opposite on the body (boundary, wall) that is producing the change (in velocity) 17 Momentum and Forces in a Fluid Momentum Principle Newton’s 2nd law: d (mV ) F dt The sum of external forces on a body of fluid or system is equal to the rate of change of linear momentum of that body (of the fluid or system). Change in momentum must be in the same direction as the force 0 for steady flow out d (mV ) S d (mV )CV d (mV )CV d (mV )inCV F dt dt dt dt out d (mV ) S d (mV )CV d (mV )inCV F dt dt dt For steady flow, the net force on the fluid mass is equal to the net rate of outflow of momentum across the control surface 18 Momentum and Forces in a Fluid Momentum Principle d (mV ) F dt out CV d (mV ) dt in CV Control surface is normal to the velocity at sections 1 and 2 Velocity is constant across the control surface d (mV )1 d (m1 ) V1 m V1 1Q1V1 dt dt F Q V 2 2 2 2 Vin V1 d (mV ) 2 2Q2V2 dt 1Q1V1 2Q2 1Q1 Q m F QV Vout V2 for steady flow QV1 Q(V2 V1 ) Q(V ) 19 Momentum and Forces in a Fluid Momentum Principle , Force and Velocity Considerations F QV 2 QV1 Q(V2 V1 ) Q(V ) Same direction! Summation of all forces acting on the fluid mass including: gravity forces (!), shear forces, pressure forces exerted by surrounding fluid and by solid boundary 20 y Momentum and Forces in a Fluid Momentum Principle, Scalar Components x F Q(V ) Q(V 2 V1 ) ΔV Vectorial Scalar: F x Q(Vx ) Q(V2 x V1x ) ΔV ΔVx Fy Q(Vy ) Q(V2 y V1y ) ΔV Force components applied on the fluid. The force of same magnitude but opposite direction is the one applied by the fluid on its surroundings ΔVy 21 Momentum and Forces in a Fluid Momentum Principle, Comments 1) 1 F QV out QVin 3 F (Q5V5 Q3V3 ) (Q1V1 Q2V2 Q4V4 ) 22 Momentum and Forces in a Fluid Momentum Principle, Comments 2) Draw a suitable system of axes. It is important to establish positive x & y directions for vectorial relationships. 3) Draw a suitable control volume(CV). The CV should include all the changes that the velocity experiences (i.e. magnitude and direction) 4) For the summation of forces, all the external forces applied on the CV should be considered. If we have Fx, then the x component of all the forces should be included. Gravity is the force applied in the vertical direction 5) The result we find from the momentum principle is a force applied upon the CV. Most often we are asked to determine the forces applied by the fluid. This is equal in magnitude but opposite in direction compared to the former force. 6) We don’t need to know the energy losses or we don’t care how complicated the flow field is when dealing with the momentum. We only need to know the conditions at the end sections of our CV (and gravity forces). 7) The direction of the forces in x and y axes applied by the fluid are not known 23 before hand. Momentum and Forces in a Fluid Pressure Conduits 24 Momentum and Forces in a Fluid Momentum Principle, Problem The flow turns by 90° in a horizontal plane before it discharges into the open. Determine the forces, Fx and Fy applied by the flow upon the pipe. ρ = 1000 kg/m3 2 D2 = 0.5 m Q = 1.5 m3/s Forces applied upon the pipe: 1 p1 = 30 kPa F ’x= 49,987.3 N D1 = 1 m F’y = 11,459 N 25 Momentum and Forces in Fluid Flow Example: Force applied on a spillway Flow 2.1 m Spillway 0.6 m Find the force applied by the flow on the spillway (assume that the width normal to the page is 1 m, ρ = 1000 kg/m3) 26 Momentum and Forces in Fluid Flow Example: Force applied on a spillway CV FH1 2.1 m Ff FH2 0.6 m 1 2 x 27 Momentum and Forces in Fluid Flow Free Jet Gage pressures in the jet are “0” 28 Momentum and Forces in Fluid Flow Choice of alternative control volumes (Example: Vertical reducing section) Neglect losses but include gravity, determine the force (F) on the contraction. V12 p2 V22 h 2g 2g p1 CV 1 p2 V2 V12 V22 h 2g p2 F z W V1 p1 h p1 F z W p1 A1 p2 A2 F Q(V2 V1 ) 29 Momentum and Forces in Fluid Flow Choice of alternative control volumes (Vertical reducing section) CV 2 p’2 V2 F V12 V22 ( h h' ) 2g p '2 h’ z W’ z p1 V1 Same as CV 1 F F p '2 h F p1 p2 h' W ' p1 A1 p'2 A2 F Q(V2 V1 ) W ' W A2 h' z (W A2 h' ) p1 A1 ( p2 h' ) A2 F Q(V2 V1 ) z W p1 A1 p2 A2 F Q(V2 V1 ) 30 Momentum and Forces in Fluid Flow Example: Water jet V3 x 0.4 m (diameter) V1=20 m/s V2=20 m/s V4 (Horizontal plane) A water jet is aimed at the hole in the wall; 25% of the water escapes through the hole, and the remaining water is turned 90° to the water jet. Find the force on the wall by the jet. ρ = 1000 kg/m3 Discussion: What If this was a vertical plane? (Vyin =0, Vyout=V3 and V4) 31