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MECH1300 Basic Principles of Hydraulics Topics • • • • • • • Pascal’s Law Transmission of Force Properties of Hydraulic Fluids Liquid Flow Static Head Pressure Power Hydraulic Systems Chapter 2 MECH1300 Basic Principles of Hydraulics Hydraulic Systems usually operate from 500 to 5000 lbs per square inch. They are precise with speed and positioning. Smooth transitions. Have more power. MECH1300 Pascal’s Law Pressure is the intensity of a force, it’s created when a force from one object acts over an area of another object. Pascal’s Law – The pressure exerted on a confined fluid at rest is transmitted undiminished in all directions and acts at right angles to the containing surfaces. Pressure transmits force and energy, not the fluid. Most industrial hydraulic systems use pressurized fluids and are considered hydrostatic MECH1300 Pressure 𝐹𝑜𝑟𝑐𝑒 𝑁 𝑙𝑏𝑠 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = = 2= 2 𝐴𝑟𝑒𝑎 𝑚 𝑖𝑛 𝐹 𝑝= 𝐴 One kPa is equal to 1000 Pascals, and 6.9 kPa equals 1 psi 1 bar = 100KPascal 𝑙𝑏𝑠 𝑖𝑛2 is abbreviated as psi (U.S. Units) 𝑁 𝑚2 𝑖𝑠 𝑎𝑏𝑏𝑟𝑒𝑣𝑖𝑎𝑡𝑒𝑑 𝑎𝑠 𝑎 𝑝𝑎𝑠𝑐𝑎𝑙𝑠 (𝑃𝐴) (S.I. Units) 𝑝= 5 = 2 𝑝𝑎𝑠𝑐𝑎𝑙𝑠 2.5 𝑝= 5 = 5 𝑝𝑎𝑠𝑐𝑎𝑙𝑠 1 MECH1300 Pressure Ex: What is the amount of pressure in the given containier? 10 𝑝= = 100 𝑝𝑎𝑠𝑐𝑎𝑙𝑠 0.1 MECH1300 Pressure Ex: The piston has a diameter of 1.5in. We don’t want to exceed a pressure of 500psi. What is the maximum force this system can withstand? 𝜋 ∙ 𝐷2 3.14159 ∗ (1.5)2 𝐴= = = 1.767𝑖𝑛2 4 4 𝐹 = 𝑝 ∙ 𝐴 = 500 𝑙𝑏𝑠 ∙ 1.767𝑖𝑛2 = 884 𝑙𝑏𝑠 2 𝑖𝑛 MECH1300 Pressure Ex: The piston is required to support a force of 10kN (10,000N). We do not want to exceed a pressure of 70 bar. What size cylinder is required? 𝑁 70 𝑏𝑎𝑟 = 7,000,000 𝑃𝑎 = 7,000,000 2 𝑚 𝐴= 𝐹 𝑝 = 10,000𝑁 𝑁 7,000,000 𝑚 Solve A = 𝐷= 4∙𝐴 𝜋 = 0.001429𝑚2 𝜋∙𝑑2 4 for D = 4∙(0.001429) 3.14159 = 0.04265m MECH1300 Pressure Ex: The piston has a diameter of 40mm. We don’t want to exceed a pressure of 3500 kPa. What is the maximum force this system can withstand? 𝑁 3500𝑘𝑃𝐴 = 3,500,000 𝑃𝑎 = 3,500,000 2 𝑚 𝜋 ∙ 𝐷2 3.142 ∙ 0.04𝑚 𝐴= = 4 4 2 = 0.001257𝑚2 𝑁 𝐹 = 𝑝 ∙ 𝐴 = 3500000 2 ∙ 0.001257𝑚2 = 4399𝑁 𝑚 MECH1300 Multiplying Force 𝑊𝑜𝑟𝑘 = 𝐹𝑜𝑟𝑐𝑒 ∙ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑊 = 𝐹 ∙ 𝑑 𝐹𝑜𝑢𝑡 𝐴𝑟𝑒𝑎𝑜𝑢𝑡 = ∙ 𝐹𝑖𝑛 𝐴𝑟𝑒𝑎𝑖𝑛 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑖𝑛 = 𝐴𝑟𝑒𝑎𝑜𝑢𝑡 ∙ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑜𝑢𝑡 𝐴𝑟𝑒𝑎𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦𝑖𝑛 = 𝐴𝑟𝑒𝑎𝑜𝑢𝑡 ∙ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦𝑜𝑢𝑡 𝐴𝑟𝑒𝑎𝑖𝑛 This is a closed system – Fluid must remain constant. Whatever amount of fluid flows out of the input piston must flow into the output piston. MECH1300 Multiplying Force Ex: If we have an input cylinder with a diameter of 1 in and an output cylinder with a diameter of 2.5 in. A force of 250 lbs is applied to the input cylinder. What is the output force? How far would we need to move the input cylinder to move the output cylinder 1 in? Input piston area 𝜋 ∙ 𝐷2 3.142 ∙ 1𝑖𝑛 2 𝐴= = = 0.7854𝑖𝑛2 4 4 Output piston area 𝜋 ∙ 𝐷2 3.142 ∙ 2.5𝑖𝑛 2 𝐴= = = 4.909𝑖𝑛2 4 4 Output Force 𝐴𝑟𝑒𝑎𝑜𝑢𝑡 4.909𝑖𝑛2 𝐹𝑜𝑢𝑡 = ∙ 𝐹𝑖𝑛 = ∙ 250𝑙𝑏𝑠 = 1563𝑙𝑏𝑠 𝐴𝑟𝑒𝑎𝑖𝑛 0.7854𝑖𝑛2 Input distance 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑖𝑛 𝐴𝑟𝑒𝑎𝑜𝑢𝑡 4.909𝑖𝑛2 = ∙ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑜𝑢𝑡 = ∙ 1𝑖𝑛 = 6.25𝑖𝑛 𝐴𝑟𝑒𝑎𝑖𝑛 0.7854𝑖𝑛2 MECH1300 Basic Properties of Hydraulic Fluids Mass – How much matter an object contains. (Slugs, Kg) Weight – The measure of how much force an objects mass creates due to gravity. (lbs, N) Gravity (Earth) = 32.2 𝑓𝑡 𝑠𝑠 = 9.81 𝑚 𝑠2 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑎𝑠𝑠 ∙ 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 EX: An object weighs 500 lbs. What is its mass? 𝑤 500𝑙𝑏𝑠 𝑙𝑏 ∙ 𝑠 2 𝑚= = = 15.53 = 15.53 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡 𝑔 𝑓𝑡 32.2 2 𝑠 EX: An object weighs 1000N. What is its mass? 𝑤 1000 𝑁 𝑁 ∙ 𝑠2 𝑚= = = 101.94 = 101.94 𝐾𝑔 𝑔 9.81 𝑚 𝑚 𝑠2 𝑤 = 𝑚𝑔 MECH1300 Basic Properties of Hydraulic Fluids Density – A substance’s mass per unit volume. (𝑠𝑙𝑢𝑔𝑠 𝑓𝑡 3 , 𝐾𝑔 𝑚𝑎𝑠𝑠 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑚3 ) 𝑚 𝜌= 𝑉 𝑤𝑒𝑖𝑔ℎ𝑡 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 = 𝑤 𝛾= 𝑉 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑙𝑢𝑖𝑑 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑤𝑎𝑡𝑒𝑟 𝛾𝑤𝑎𝑡𝑒𝑟 = 62.4 𝑙𝑏𝑠 𝑠𝑔𝑥 = 𝛾𝑥 𝛾𝑤𝑎𝑡𝑒𝑟 𝑓𝑡 3 Specific gravity does not have units associated with it. Specific weight of Petroleum based oils is 56 𝑙𝑏𝑠 𝑓𝑡 3 𝑜𝑟 8800 𝑁 𝑚3 MECH1300 Basic Properties of Hydraulic Fluids EX: Calculate the specific gravity of oil. 𝑠𝑔𝑜𝑖𝑙 = 𝛾𝑜𝑖𝑙 𝛾𝑤𝑎𝑡𝑒𝑟 = 56 𝑙𝑏𝑠 62.4 𝑙𝑏𝑠 𝑓𝑡 3 𝑓𝑡 3 = 0.9 MECH1300 Basic Properties of Hydraulic Fluids Viscosity – The measure of a fluid’s thickness or resistance to flow. 𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝜇 = 𝐹∙𝑦 𝑣∙𝐴 𝐹 = 𝑓𝑜𝑟𝑐𝑒 𝑙𝑏𝑠, 𝑁 𝑦 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 (𝑓𝑡, 𝑚) 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑓𝑡 𝑠 , 𝑚 𝑠) 𝐴 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑣𝑖𝑛𝑔 𝑝𝑙𝑎𝑡𝑒 (𝑓𝑡 2 , 𝑚2 ) (wetted area) 𝑙𝑏∙𝑠 𝑁∙𝑠 , 𝑓𝑡 2 𝑚2 MECH1300 Basic Properties of Hydraulic Fluids EX: The top plate shown has a wetted area of 0.5𝑓𝑡 2 and is moving at a velocity of 10 𝑓𝑡 𝑠. The force required to maintain this speed is found to be 5 lbs. The fluid film thickness is 0.05 in. What is the viscosity of the fluid? 𝜇= 𝐹∙𝑦 𝑣∙𝐴 0.05 𝑖𝑛 ∙ 𝜇= 1 𝑓𝑡 12 𝑖𝑛 = 0.00417 𝑓𝑡 5 𝑙𝑏𝑠∙(0.00417𝑓𝑡) 10𝑓𝑡 𝑠∙ 0.5𝑓𝑡 2 = 0.00417 𝑙𝑏∙𝑠 𝑓𝑡 2 MECH1300 Basic Properties of Hydraulic Fluids Kinematic viscosity – "momentum diffusivity“- The dynamic viscosity divided by the density. 𝑣= 𝜇 𝜌 𝑙𝑏∙𝑠 Kinematic viscosity of oil = 0.00417 2 𝑓𝑡 1.74 𝑙𝑏∙𝑠2 𝑓𝑡4 = 0.00240 𝑓𝑡 2 𝑁∙𝑠 = 0.0833 2 𝑚 𝑁∙𝑠2 897 4 𝑚 = 0.0000929 𝑚 𝑠 2 𝑠 MECH1300 Basic Properties of Hydraulic Fluids Viscosity is determined using a Saybolt viscometer. It measures the time for a 60-milliliter sample of oil to drain through a standard orifice, resulting in saybolt seconds universal (SSU or SUS). Viscosity decreases with increasing temperature. Viscosity Index (V.I.) - a scale from 1-100. The higher the number, the more stable the fluids viscosity with varying temperature. If viscosity is too high, then it becomes more difficult for fluid to flow through a system. If viscosity is too low, it creates excessive wear because the oil is not thick enough to provide good lubrication. It can also cause leakage. MECH1300 Bulk Modulus The Bulk Modulus of a liquid is a measure of it’s stiffness or incompressibility. 𝐵= 𝐵𝑢𝑙𝑘 𝑆𝑡𝑟𝑒𝑠𝑠 𝐵𝑢𝑙𝑘 𝑆𝑡𝑟𝑎𝑖𝑛 = 𝐹𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎 ∆𝑉 𝑉 = −∆𝑝 ∆𝑉 𝑉 𝐵 = 𝑏𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 (𝑝𝑠𝑖, 𝑃𝑎) ∆𝑝 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑝𝑠𝑖, 𝑃𝑎) ∆𝑉 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑖𝑛3 , 𝑚3 ) 𝑇ℎ𝑒 ∆𝑉 𝑉 𝑡𝑒𝑟𝑚 𝑖𝑠 𝑎 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 MECH1300 Bulk Modulus EX: In a system using hydraulic oil (B = 250,000 psi), the pressure is 4000 psi. By what percentage is the oil being compressed relative to the unpressurized state (𝑝 = 0 𝑝𝑠𝑖)? Calculate the change in pressure: ∆𝑝 = 4000 𝑝𝑠𝑖 − 0 𝑝𝑠𝑖 = 4000 𝑝𝑠𝑖 Calculate the proportional change in volume by solving for ∆𝑉 𝑉 −∆𝑝 −4000 𝑝𝑠𝑖 ∆𝑉 𝑉 = = = −0.016 𝐵 250,000 𝑝𝑠𝑖 Multiply by 100 to get the % change −0.016 × 100 = −1.6% At 4000 psi the oil is only 1.6% of its original volume. This justifies that oil is relatively incompressible. MECH1300 Liquid flow Flow Rate – The volume of fluid that moves through a system in a given period of time. It determines the speed at which the output device will operate. Flow velocity – Is the distance the fluid travels in a given period of time. 𝑄 =𝑣∙𝐴 𝑖𝑛3 𝑚3 𝑄 = 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 , 𝑚𝑖𝑛 𝑚𝑖𝑛 𝑖𝑛 𝑚 𝑣 = 𝑓𝑙𝑜𝑤 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 , 𝑚𝑖𝑛 𝑚𝑖𝑛 2 2 𝐴 = 𝑎𝑟𝑒𝑎 𝑖𝑛 , 𝑚 MECH1300 Liquid flow 𝑖𝑛 500 𝑚𝑖𝑛 EX: A fluid flows at a velocity of through a conduit with an ID (inner diameter) of 3 in. Determine the flow rate. Calculate the conduit area: 𝜋 ∙ 𝐷2 3.142 ∙ 3 𝑖𝑛 𝐴= = 4 4 2 = 7.07 𝑖𝑛2 Calculate the flow rate: 3 𝑖𝑛 𝑖𝑛 𝑄 = 𝑣 ∙ 𝐴 = 500 ∙ 7.07𝑖𝑛2 = 3534.75 𝑚𝑖𝑛 𝑚𝑖𝑛 Convert to gpm: 𝑖𝑛3 1 𝑔𝑎𝑙 𝑔𝑎𝑙 3534.75 ∙ = 15.30 𝑚𝑖𝑛 231𝑖𝑛3 𝑚𝑖𝑛 MECH1300 Liquid flow 𝑚 50 𝑚𝑖𝑛 EX: A fluid flows at a velocity of through a conduit with an ID (inner diameter) of 20 mm. Determine the flow rate. Calculate the conduit area: 𝜋 ∙ 𝐷2 3.142 ∙ 0.020 𝑚 𝐴= = 4 4 2 = 0.000314 𝑚2 Calculate the flow rate: 3 𝑚 𝑚 𝑄 = 𝑣 ∙ 𝐴 = 50 ∙ 0.000314𝑚2 = 0.0157 𝑚𝑖𝑛 𝑚𝑖𝑛 Convert to lpm: 𝑚3 1000 𝑙 𝑙 0.0157 ∙ = 15.71 𝑚𝑖𝑛 1𝑚3 𝑚𝑖𝑛 MECH1300 Liquid flow EX: A fluid flows at a rate of 20 𝑙𝑝𝑚 through a conduit with an ID (inner diameter) of 30 mm. Determine the flow velocity in m/s. Calculate the conduit area: 𝜋 ∙ 𝐷2 3.142 ∙ 0.030 𝑚 𝐴= = 4 4 Convert to 𝑚 3 2 = 0.000707 𝑚2 𝑚𝑖𝑛: 𝑙 1𝑚3 𝑚3 20 ∙ = 0.02 𝑚𝑖𝑛 1000 𝑙 𝑚𝑖𝑛 Calculate the flow velocity: 𝑚3 0.02 𝑚𝑖𝑛 𝑄 𝑚 𝑣= = = 28.29 𝐴 0.000707𝑚2 𝑚𝑖𝑛 Convert to m/s: 𝑚 1 𝑚𝑖𝑛 𝑚 28.29 ∙ = 0.4715 𝑚𝑖𝑛 60 𝑠 𝑠 MECH1300 Continuity Equation Steady Flow – The volume of fluid that flows through one section of a system must flow through any other section. This occurs if a pump is producing a constant flow rate so that we can assume that the fluid is incompressible. MECH1300 Continuity Equation 𝑄1 = 𝑄2 𝑣1 ∙ 𝐴1 = 𝑣2 ∙ 𝐴2 This is a system with a steady flow rate, with a reduction in area. The reduction in area will correspond to an increase in flow velocity. MECH1300 Continuity Equation Ex: A fluid flows at a velocity of 100 in/min at point 1. Its diameter at point 1 is 3 in and the diameter at point 2 is 2 in. Determine the flow velocity at point 2. Determine the flow rate in gpm. Calculate the conduit areas: 𝜋 ∙ 𝐷2 3.142 ∙ 3 𝑖𝑛 2 𝐴1 = = = 7.07 𝑖𝑛2 4 4 2 𝜋∙𝐷 3.142 ∙ 2 𝑖𝑛 2 𝐴2 = = = 3.142 𝑖𝑛2 4 4 Calculate the flow velocity: 𝐴1 𝑖𝑛 7.07𝑖𝑛2 𝑖𝑛 𝑣2 = 𝑣1 ∙ = 100 ∙ = 207.21 𝐴2 𝑚𝑖𝑛 3.142𝑖𝑛2 𝑚𝑖𝑛 Calculate the flow rate: 3 𝑖𝑛 𝑖𝑛 𝑄 = 𝑣1 ∙ 𝐴1 = 100 ∙ 7.07𝑖𝑛2 = 707 𝑚𝑖𝑛 𝑚𝑖𝑛 Convert to gpm: 𝑖𝑛3 1 𝑔𝑎𝑙 𝑔𝑎𝑙 707 ∙ = 3.06 𝑚𝑖𝑛 231𝑖𝑛3 𝑚𝑖𝑛 MECH1300 Bernoulli’s Equation Bernoulli’s equation describes the total energy of an incompressible fluid. 1. Potential Energy (due to elevation and gravity) = 𝑤 ∙ ℎ 𝑝 2. Pressure Energy = w ∙ 𝛾 3. Kinetic Energy (due to velocity) = 𝑤 𝑤 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑙𝑏𝑠, 𝑁 ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 (𝑖𝑛, 𝑚) 𝑝 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒(𝑙𝑏𝑠 𝑖𝑛2 , 𝑁 𝑚2 ) 𝛾 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡(𝑙𝑏𝑠 𝑖𝑛3 , 𝑁 𝑚3 ) 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑖𝑛 𝑠 , 𝑚 𝑠) 𝑔 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 396.4 𝑖𝑛 𝑠2 = 9.81 𝑚 𝑠2 𝑣2 ∙ 2∙𝑔 MECH1300 Bernoulli’s Equation Potential Energy 1+Pressure Energy 1+Kinetic Energy 1 = Potential Energy 2+Pressure Energy 2+Kinetic Energy 2 𝑤 ∙ ℎ1 + 𝑤 𝑝 ℎ1 + 1 𝛾 + 𝑝1 ∙ 𝛾 𝑣1 2 2∙𝑔 +𝑤 = 𝑣1 2 ∙ 2∙𝑔 𝑝 ℎ2 + 2 𝛾 = 𝑤 ∙ ℎ2 + 𝑤 ∙ 𝑝2 𝛾 +𝑤∙ 𝑣2 2 2∙𝑔 𝑣2 2 + 2∙𝑔 It tells us that an increase in energy in any one of the three areas must be balanced by a reduction in one or more of the other two areas of an equal amount. Elevation Head + Pressure Head + Velocity Head = total head Head is a measure of energy a fluid posses per pound. MECH1300 Bernoulli’s Equation 3 EX: A fluid 𝛾 = 0.0324 𝑙𝑏𝑠 𝑖𝑛3 flows at a constant rate of 150 𝑖𝑛 𝑠. The areas at point 1 and point 2 are equal. The pressure at point 1 is 100 psi and h=200 in. Determine the pressure at point 2. 𝑝1 𝑣1 2 𝑝2 𝑣2 2 ℎ1 + + = ℎ2 + + 𝛾 2∙𝑔 𝛾 2∙𝑔 Since the two areas are equal and the flow rate is constant, we can ignore the velocity terms because they cancel out, so the equation reduces to: 𝑝1 𝑝2 ℎ1 + = ℎ2 + 𝛾 𝛾 Solve for 𝑝2 𝑝2 = 𝑝1 − 𝛾 ℎ2 − ℎ1 Substitute known values 𝑙𝑏𝑠 𝑙𝑏𝑠 𝑙𝑏𝑠 𝑝2 = 100 2 − 0.0324 3 200𝑖𝑛 = 93.5 2 𝑖𝑛 𝑖𝑛 𝑖𝑛 MECH1300 Bernoulli’s Equation 3 EX: A fluid 𝛾 = 8000 𝑁 𝑚3 flows at a constant flow rate of 0.004 𝑚 𝑠. The areas are 𝐴1 = 0.003𝑚2 and 𝐴2 = 0.002𝑚2 . If the pressure at point 1 is 500kPa, determine the pressure at point 2. Calculate the flow velocities: 𝑣1 = 𝑣2 = 𝑄 𝐴1 𝑄 𝐴2 𝑚3 = = 0.004 𝑠 0.003𝑚2 𝑚3 𝑠 0.002𝑚2 0.004 = 1.33 =2 𝑚 𝑠 𝑚 𝑠 Because ℎ1 = ℎ2 the elevation terms drop out. Solve for 𝑝2 : 𝛾 𝑝2 = 𝑝1 + 𝑣1 2 − 𝑣2 2 2∙𝑔 Reducing the size of the conduit causes and increase in velocity and a decrease in pressure. Substitute known values: 𝑁 𝑁 3 𝑚 𝑝2 = 500,000 2 + 𝑚 𝑚 2 ∙ 9.81 2 𝑠 8000 𝑚 1.33 𝑠 2 𝑚 − 2 𝑠 2 𝑁 = 99090 2 = 99.09𝑘𝑃𝑎 𝑚 MECH1300 Bernoulli’s Equation Venturi – reducing a conduits size to intentionally reduce pressure. The reduced diameter section is called the throat. The heights are equal and the elevation terms drop out. Bernoulli’s equation becomes: 𝑝1 𝛾 𝑣1 2 + 2∙𝑔 = 𝑝2 𝛾 + 𝑣2 2 2∙𝑔 Then solve for 𝑣2 𝐴 using the continuity equation plug in 𝑣1 = 𝑣2 ∙ 𝐴2 1 𝑣2 = 2 ∙ 𝑔 ∙ 𝑝1 − 𝑝2 𝐴 𝛾 ∙ 1 − 𝐴2 1 MECH1300 Torricelli’s Theorem A special case of Bernoulli’s equation that applies to a liquid draining from a tank. In this case the pressure at points 1 and 2 are both 0 psi and the velocity at point 1 is zero because it is going to be very small compared to the velocity at point 2. Bernoulli’s equation is reduced to: ℎ1 = ℎ2 + 𝑣2 2 2∙𝑔 We solve for 𝑣2 𝑣2 = 2 ∙ 𝑔 ℎ1 − ℎ2 However, ℎ1 − ℎ2 is simply just h, so 𝑣2 = 2∙𝑔∙ℎ MECH1300 Torricelli’s Theorem EX: The tank is fill with liquid to a height of 4 feet, determine the velocity at the outlet. 𝑣2 = 2∙𝑔∙ℎ 𝑣2 = 2 ∙ 32.2 𝑓𝑡 𝑠2 ∙ 4𝑓𝑡 = 16.04 𝑓𝑡 𝑠 EX: The tank is fill with liquid to a height of 2 meters, determine the velocity at the outlet. 𝑣2 = 2∙𝑔∙ℎ 𝑣2 = 2 ∙ 9.81 𝑠2 ∙ 2𝑚 = 6.26 𝑚 𝑚 𝑠 MECH1300 Laminar and Turbulent Flow Laminar Flow – A thin layer of fluid adheres to the conduit wall, as you move away from the conduit wall, the velocity increases at each layer until it is at it’s maximum in the center of the conduit. If the average velocity increase the flow can transition from Laminar to Turbulent Turbulent Flow – Fluid flows in a chaotic manner with parcels of fluid flowing in all directions, as a type of “churning” action MECH1300 Laminar and Turbulent Flow We can predict the type of flow by calculating the Reynolds number. 𝑅𝑒 = 𝑣∙𝐷 v 𝑅𝑒 = 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑛𝑜 𝑢𝑛𝑖𝑡𝑠 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑡 𝑠2 , 𝑚 𝑠2 𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑓𝑡, 𝑚 𝑓𝑡 2 𝑚2 v = 𝑘𝑖𝑛𝑒𝑚𝑎𝑡𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑠, 𝑠 Laminar flow 𝑅𝑒 < 2000 2000 < 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛𝑎𝑙 < 4000 Laminar Turbulent Flow 4000 < 𝑅𝑒 Type of flow is difficult to predict in the transitional range MECH1300 Laminar and Turbulent Flow 2 EX: A fluid with a kinematic viscosity of 0.0045 𝑓𝑡 𝑠 is flowing through a 3-in diameter conduit at a flow rate of 400 gpm. Determine if the flow will be laminar or turbulent. 3 Convert gpm to 𝑖𝑛 𝑠: 𝑔𝑎𝑙 231𝑖𝑛3 1 𝑚𝑖𝑛 𝑖𝑛3 𝑄 = 400 ∙ ∙ = 1540 𝑚𝑖𝑛 1 𝑔𝑎𝑙 60 𝑠 𝑠 Calculate the area: 𝜋 ∙ 𝐷2 𝜋 ∙ 3𝑖𝑛 𝐴= = 4 4 2 = 7.069𝑖𝑛2 Calculate the flow velocity: 𝑖𝑛3 𝑄 1540 𝑠 𝑖𝑛 𝑣= = = 217.87 𝐴 7.069𝑖𝑛 𝑠 𝑠 Convert to 𝑓𝑡 𝑠: 𝑖𝑛 1 𝑓𝑡 𝑓𝑡 𝑣 = 217.87 ∙ = 18.16 𝑠 12 𝑖𝑛 𝑠 Convert diameter into feet: 1 𝑓𝑡 𝐷 = 3𝑖𝑛 ∙ = 0.25 𝑓𝑡 12 𝑖𝑛 Calculate 𝑅𝑒 : 𝑓𝑡 𝑣 ∙ 𝐷 18.16 𝑠 ∙ 0.25 𝑓𝑡 𝑅𝑒 = = = 1008.89 𝑓𝑡 2 v 0.0045 𝑠 Laminar MECH1300 Static Head Pressure Static Head Pressure – The deeper that you submerge an object underwater it is subject to pressure. The pressure is do to the weight of the water that is above it, pressing down on it. Its magnitude is determined by: 𝑝𝐻 = 𝛾 ∙ ℎ 𝑝𝐻 = ℎ𝑒𝑎𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝛾 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 (𝑖𝑛, 𝑚) 𝑙𝑏𝑠 𝑙𝑏𝑠 𝑖𝑛3 𝑖𝑛3 ,𝑁 ,𝑁 𝑚3 𝑚3 MECH1300 Static Head Pressure EX: A tank is fill with oil with a specific weight of 8000 𝑁 at the bottom of the tank? 𝑚3 to a depth of 8 m. What is the pressure 𝑁 𝑁 𝑝𝐻 = 𝛾 ∙ ℎ = 8000 𝑚3 ∙ 8𝑚 = 64000 2 = 64𝑘𝑃𝑎 𝑚 MECH1300 Pressure loss • When there is a flowing fluid, pressure drops as the fluid flows through the system due to energy loss (due to viscosity of the fluid resisting flow). • When flow stops, pressure equalizes again. • Lost energy is transformed into heat, raising the temperature of the fluid. • As temperature rises, viscosity goes down. • It’s ability to lubricate weakens, causing wear and tear to components MECH1300 Power Review 𝑤 𝑡 = 𝐻𝑃 = 𝐹∙𝑣 550 𝑃= 1 ℎ𝑝 = 550 𝑓𝑡∙𝑙𝑏𝑠 𝑠 𝐹∙𝑑 𝑡 𝑣= 𝑑 𝑡 so 𝑃 = 𝐹 ∙ 𝑣 HP means power is in the horsepower 𝑘𝑊 = 𝐹∙𝑣 1000 EX: A 20,000 N load must be moved at a velocity of 2 m/s. How much power is required? 𝑘𝑊 = 20000𝑁 ∙ 2 1000 𝑚 𝑠 = 40𝑘𝑊 MECH1300 Power In Hydraulics Power in terms of pressure 𝑤 𝐹∙𝑑 𝑑 𝑃= = 𝑣= so 𝑃 = 𝐹 ∙ 𝑣 and 𝐹 = 𝑝 ∙ 𝐴 𝑡 𝑡 𝑃 =𝑝∙𝐴∙𝑣 𝑡 𝑝𝑜𝑤𝑒𝑟 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∙ 𝑎𝑟𝑒𝑎 ∙ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑄 = 𝐴 ∙ 𝑣 𝑃 =𝑝∙𝑄 𝐻𝑃𝐻 = 𝑝∙𝑄 1714 𝑘𝑊𝐻 = 𝑝∙𝑄 60000 MECH1300 Power In Hydraulics EX: Oil is flowing through a hydraulic system at 15 gpm and 2700 psi. What is the hydraulic power of this system? 𝑝 ∙ 𝑄 2700 𝑝𝑠𝑖 ∙ 10 𝑔𝑝𝑚 𝐻𝑃𝐻 = = = 15.75 ℎ𝑝 1714 1714 MECH1300 Hydraulic Systems Three main segments: 1. Power supply – supplies flow to the system. The pump takes the mechanical power from the prime mover (motor or engine) and converts it to fluid power 2. Control – Controls the direction of the fluid with valves, allowing a cylinder to extend or retract. 3. Output – Output force is controller by the pressure of the fluid.