Denunciar

Compartilhar

Seguir

•0 gostou•1,513 visualizações

•0 gostou•1,513 visualizações

Denunciar

Compartilhar

Baixar para ler offline

TBA

Seguir

- 1. KV Keith Vaugh BEng (AERO) MEng CENTRIFUGAL PUMPS
- 2. KV PUMP A pump is a hydraulic machine that converts mechanical energy into hydraulic energy in the form of Pressure Energy When the conversion of mechanical energy into hydraulic energy is done by means of a centrifugal force, then the pump is referred to as a Centrifugal Pump
- 3. KV MAIN PARTS ① ② ③ ④ 1. Impeller 2. Volute or Casing 3. Suction Section 4. Delivery Section At the eye of the impeller there is a negative pressure causing the suction from the sump
- 4. KV PRINCIPLE Rise of pressure head is directly proportional to the tangential velocity
- 5. KV therefore AREA ⬆︎ increases VELOCITY ⬆︎ increases VELOCITY ⬇︎ decreases AREA ⬇︎ decreases 1︎⃣ 2︎⃣ 1︎⃣ decreasing tangential velocity 2︎⃣ increasing area The continuity equation states
- 6. KV So how does this decrease in Kinetic Energy increase the Pressure Head? According to Bernoulli’s equation: Pressure Energy Kinetic Energy Potential Energy + + = Constant In the case of a centrifugal pump, we can assume that the suction inlet and discharge outlet are on the same level, therefore the potential energy in Bernoulli’s equation can be ignored Pressure Energy Kinetic Energy + = Constant So as the Pressure Energy increases, the Kinetic Energy decreases or if the Kinetic Energy increases, the Pressure Energy decreases
- 7. KV During a laboratory test on a centrifugal pump water at 20 ºC is pumped from a sump to a reservoir. The flow rate of water passing through the pump was measured at 15 m3/h and the torque applied to the pump shaft was measured to be 6.4 N-m. Other measurements taken during the test are tabulated in Table 2. Assume the pump operates at 3000rpm. EXAMPLE 32 Parameter Inlet Section Outlet Section Gauge pressure, p (kPa) 86 ? Elevation above datum, z (m) 2.0 10 Velocity, U (ms-1) 2.0 4.6 Construct a detailed diagram and label the components in such a system, and identify the key assumptions that can be made. If the pump and electric motor efficiencies are 75% and 85% respectively, determine; (a) the electric power required, (b) the gauge pressure at the outlet.
- 8. KV
- 9. KV ASSUMPTIONS: • Flow is steady • Irreversible losses are negligible GOVERNING EQUATIONS 𝑃𝑜𝑤𝑒𝑟 = 𝜌 × 𝑔 × 𝑉 · × 𝐻 𝑘𝑔 𝑚3 × 𝑚 𝑠2 × 𝑚3 𝑠 × 𝑚 𝑘𝑔 𝑚3 × 𝑚5 𝑠3 𝑘𝑔 × 𝑚2 𝑠3 𝑘𝑔 × 𝑚 𝑠2 × 𝑚 𝑠 𝑁 × 𝑚 𝑠 = 𝐽 𝑠 = 𝑊𝑎𝑡𝑡𝑠
- 11. KV Manipulating to obtain the discharge pressure (pressure at the outlet) Solving for the the electric power required to develop the head From the governing equations:
- 12. KV The flow system used to test a centrifugal pump at a nominal speed of 1750 rpm is shown in the figure. The liquid water at 27 °C and the suction and discharge pipe diameters are 150 mm. Data measured during the test are given in the table. The electric motor is supplied at 460 V, 3-phase, and has a power factor of 0.875 and a constant efficiency of 90%. EXAMPLE 33 Zd
- 13. KV Calculate the net head delivered and the pump efficiency at a volumetric flow rate of 227m3/s. Plot the pump head, power input and efficiency as functions of the volumetric flow rate. EXAMPLE 33 Table 1: Test results - Centrifugal pump Volumetric Flow Rate (m3/s) Suction Pressure (kPa-gauge) Discharge Pressure (kPa-gauge) Motor Current (amp) 0 -25 377 18.0 114 -29 324 25.1 182 -32 277 30.0 227 -39 230 32.6 250 -43 207 34.1 273 -46 179 35.4 318 -53 114 39.0 341 -58 69 40.9
- 14. KV GOVERNING EQUATIONS: SOLUTION: GIVEN: Pump test flow system and data shown ASSUMPTIONS: • Steady and incompressible flow • Uniform flow at each section • U1 = U2 • Correct all heads to same elevation FIND: • Pump head and efficiency at Q = 227 m3/s • Pump head, power input, and efficiency as a function of volumetric flow rate • Plot the results
- 15. KV ANALYSIS: Since U1 = U2, the pump head is where the discharge and suction pressures, corrected to the same elevation, are designated P2 and P1, respectively Correct measured static pressures to the pump centreline
- 16. KV Calculate the pump head Compute the hydraulic power delivered to the fluid
- 17. KV Calculate the motor power output (the mechanical power input to the pump) from the electrical information The corresponding pump efficiency is Perform these calculations for each of the other points, and plot the results against the Volumetric Flow Rate:
- 18. KV Keith Vaugh BEng (AERO) MEng IMPELLER CALCULATIONS BEng Mechanical Engineering
- 19. KV A fan has a bladed rotor of 0.3 m outside diameter, 0.13 m inside diameter and runs at 1725 rpm. The width of each rotor blade is 0.025 m from blade inlet to outlet. The volumetric flow rate is steady at 0.11m3/s and the absolute velocity of the air at the blade inlet U1 is purely radical. The blade discharge angle is 30 deg measured with respect to the tangential direction at the outside diameter of the rotor. Determine • A reasonable blade inlet angle (measured with respect to the tangential direction at the inside diameter of the rotor)? • The power required to run the fan. EXAMPLE 1 - FAN
- 20. KV
- 21. KV SOLUTION: A fan turbine is driven at a specified angular speed by fluid entering along the axis of rotation. The available power at the rotor shaft and an appropriate inlet angles at the blades is to be determined. ASSUMPTIONS: The flow is steady Irreversible losses are negligible PROPERTIES: Assume the density of the fluid to be 1.23 kg/m3 ANALYSIS: The stationary and non-deforming control volume shown in the sketch is used. To determine a reasonable blade inlet angle we assume that the blade should be tangent to the relative velocity at the inlet. The inlet velocity triangle can be sketched as
- 22. KV ω Runner blade u1→ Absolute velocity vector. Velocity of the flow at the inlet. v1→ Velocity of blade at inlet i.e. 𝑣1 = 𝜔𝑟1 𝝑1→ Angle of flow at inlet; the angle between the velocity vector of the blade and the direction of the flow at inlet 𝝑1 v1
- 23. KV β1 ω Runner blade ur1 𝝑1 ur1→ The relative velocity vector of the flow at the inlet uf1→ The velocity of flow in the radial direction uw1→ The component of the velocity of the flow in horizontal/tangential direction β1→ Blade angle at inlet. Angle between the relative velocity vector and the direction of motion of the blade at the inlet 𝝑1 u1→ Absolute velocity vector. Velocity of the flow at the inlet. v1→ Velocity of blade at inlet i.e. 𝑣1 = 𝜔𝑟1 𝝑1→ Angle of flow at inlet; the angle between the velocity vector of the blade and the direction of the flow at inlet v1
- 25. KV β1 ω Runner blade ur1 𝝑1 𝝑1 v1 uf2 u2→ Absolute velocity vector. Velocity of the flow at the outlet. v2→ Velocity of blade at inlet i.e. 𝑣2 = 𝜔𝑟2 ur2→ The relative velocity vector of the flow at the outlet uf2→ The velocity of flow in the radial direction uw2→ The component of the velocity of the flow in horizontal/tangential direction 𝝑2→ Angle of flow at outlet; the angle between the velocity vector of the blade and the direction of the flow at outlet β2→ Blade angle at outlet. Angle between the relative velocity vector and the direction of motion of the blade at the outlet
- 26. KV ϑ Opposite Adjacent 𝑠𝑖𝑛𝜃 = 𝑂𝑃𝑃 𝐻𝑌𝑃 𝑐𝑜𝑠𝜃 = 𝐴𝐷𝐽 𝐻𝑌𝑃 𝑡𝑎𝑛𝜃 = 𝑂𝑃𝑃 𝐴𝐷𝐽 Also remember that 𝑡𝑎𝑛𝜃 = 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑐𝑜𝑡𝜃 = 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 𝑠𝑒𝑐𝜃 = 1 𝑐𝑜𝑠𝜃 𝑐𝑠𝑐𝜃 = 1 𝑠𝑖𝑛𝜃 Recall from Trigonometry in Mathematics
- 27. KV uf1 uf1 u1 uw1 ur1 uf1 u1 uf1 ur1 Velocity triangles Use trigonometry to solve for each vectors uw1 Two triangles within uw1 uw1 uw1 - v1 v1 - uw1 NOTE - These triangles use different notation - from Renewable module
- 28. KV (a) With the velocity triangle, we conclude that Now 𝛽1 = 𝑡𝑎𝑛−1 𝑢1 𝑣1 eq 1 and 𝑢1 = 𝑉 · 𝐴1 = 𝑉 · 2𝜋𝑟1ℎ1 𝑢1 = 0.11𝑚3 /𝑠 2𝜋 × 0.065𝑚 × 0.025𝑚 = 10.8𝑚/𝑠 𝑉1 = 𝑟1𝜔 𝑉1 = 2𝜋𝑟𝑎𝑑/𝑟𝑒𝑣 × 0.065𝑚 × 1725𝑟𝑒𝑣/𝑚𝑖𝑛 60𝑠/𝑚𝑖𝑛 = 11.7𝑚/𝑠 u1 v1 β1 𝝑1 r1 ur1
- 29. KV Therefore from equation 1 𝛽1 = 𝑡𝑎𝑛−1 10.8𝑚/𝑠 11.7𝑚/𝑠 = 43∘ (b) The power required can be obtained from; 𝑊 · 𝑠ℎ𝑎𝑓𝑡 = 𝑚 · 2𝑣2𝑢𝑤2 eq 2 The mass flow rate maybe obtained as follows 𝑚 · 2 = 𝜌𝑉 · = 1.23𝑘𝑔/𝑚3 × 0.11𝑚3 /𝑠 = 0.1353𝑘𝑔/𝑠
- 30. KV and 𝑣2 = 𝑟2𝜔 𝑣2 = 2𝜋𝑟𝑎𝑑/𝑟𝑒𝑣 × 0.15𝑚 × 1725𝑟𝑒𝑣/𝑚𝑖𝑛 60𝑠/𝑚𝑖𝑛 = 27.1𝑚/𝑠 The value of Uθ2︎ maybe obtained by considering the velocity triangle for the flow leaving the rotor at section 2. The relative velocity at the rotor exit is considered to be tangent to the blade there. The rotor exit flow velocity triangle is sketched as r2
- 31. KV therefore 𝑢𝛽2 = 𝑢𝑓2 𝑡𝑎𝑛30∘ = 𝑉 · 2𝜋𝑟2ℎ2 𝑡𝑎𝑛30∘ = 0.11𝑚2/𝑠 2𝜋 × 0.15𝑚 × 0.025𝑚 𝑡𝑎𝑛30∘ = 8.1𝑚/𝑠 Now 𝑢𝑤2 = 𝑣2 − 𝑢𝛽2 and 𝑢𝑤2 = 27.1𝑚/𝑠 − 8.1𝑚/𝑠 = 19𝑚/𝑠 and from equation 2 𝑊 · 𝑠ℎ𝑎𝑓𝑡 = 0.1353𝑘𝑔/𝑠 × 27.1𝑚/𝑠 × 19𝑚/𝑠 = 69.7𝐽/𝑠
- 32. KV TURBINE CALCULATIONS Keith Vaugh BEng (AERO) MEng BEng Mechanical Engineering & Renewable Energy
- 33. KV ω Runner blade β1 u1→ Absolute velocity vector. Velocity of the flow at the inlet. It’s direction is governed by the guide vane angle α v1→ Velocity of blade at inlet ur1→ The relative velocity vector of the jet at the inlet uf1→ The velocity of flow in the radial direction uw1→The component of the velocity of the jet u1 in horizontal/tangential direction α1→ Guide vane angle, or the angle between the velocity vector of the blade and the direction of the jet β1→ Blade angle at inlet. Angle between the relative velocity vector and the direction of motion of the blade at the inlet u0 u0
- 34. KV ω Runner blade u0 u1→ Absolute velocity vector. Velocity of the flow at the inlet. It’s direction is governed by the guide vane angle α v1→ Velocity of blade at inlet ur1→ The relative velocity vector of the jet at the inlet uf1→ The velocity of flow in the radial direction uw1→The component of the velocity of the jet u1 in horizontal/tangential direction α1→ Guide vane angle, or the angle between the velocity vector of the blade and the direction of the jet β1→ Blade angle at outlet. Angle between the relative velocity vector and the direction of motion of the blade at the inlet
- 35. KV ω Runner blade u0 u2→ Absolute velocity vector water leaving the runner. It’s direction is governed by the guide vane angle α and if it is in the radial direction, will be equal to uf2 v2→ Velocity of blade at exit ur2→ The relative velocity vector of the water leaving the impeller uf2→ The velocity of flow in the radial direction uw2→The component of the velocity for exit u2 in horizontal/tangential direction α2→ The angle between the velocity vector of the blade and the direction of the u2 exiting β2→ Blade angle at outlet, Angle between the relative velocity vector and the direction of motion of the blade at the outlet uf2 β2 𝜶2 u0
- 36. KV uf1 uf1 u1 uw1 ur1 uf1 u1 uf1 ur1 Velocity triangles Use trigonometry to solve for each vectors uw1 Two triangles within uw1 uw1 uw1 - v1 v1 - uw1
- 37. KV ϑ Opposite Adjacent 𝑠𝑖𝑛𝜃 = 𝑂𝑃𝑃 𝐻𝑌𝑃 𝑐𝑜𝑠𝜃 = 𝐴𝐷𝐽 𝐻𝑌𝑃 𝑡𝑎𝑛𝜃 = 𝑂𝑃𝑃 𝐴𝐷𝐽 Also remember that 𝑡𝑎𝑛𝜃 = 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑐𝑜𝑡𝜃 = 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 𝑠𝑒𝑐𝜃 = 1 𝑐𝑜𝑠𝜃 𝑐𝑠𝑐𝜃 = 1 𝑠𝑖𝑛𝜃 Recall from Trigonometry in Mathematics
- 38. KV A water turbine wheel rotates at the rate of 50 rpm in the direction shown in figure 1. The inner radius r2 of the blade row is 0.6 m and the outer radius r1 is 1.2 m. The absolute velocity vector at the turbine rotor entrance makes an angle of 20° with the tangential direction. The inlet blade angle is 60° relative to the tangential direction. The blade output angle is 120°. The flow rate is 0.57m3/s. For the flow tangent to the rotor blade surface at inlet and outlet, determine an appropriate constant blade height, b, and the corresponding power available at the rotor shaft. EXAMPLE 1 - TURBINE
- 39. KV
- 40. KV SOLUTION: A water turbine is driven at a specified angular speed by water entering it at a specified rate and angle. The available power at the rotor shaft and an appropriate height for the blades is to be determined. ASSUMPTIONS: The flow is steady Irreversible losses are negligible PROPERTIES: Assume the density of water to be 1 kg/L = 1000 kg/m3
- 41. KV The use of the “+” or “-” with V2Uθ,2 depends on whether Uθ,2 is opposite to or in the same direction as V2 respectively. To determine the value of UR,1 use the velocity triangle at section (1), thus: ANALYSIS: equation (1) The shaft power is obtained with the moment of momentum power equation equation (2)
- 42. KV However equation (3) With the velocity triangle we get
- 43. KV Therefore and b can be calculated
- 44. KV For Uθ,1 we use the velocity triangle at section (1) to obtain For the blade velocities in equation (2)
- 45. KV For Uθ,2 we construct the section (2) velocity triangle sketch below and it is noted that equ (4) From conservation of mass
- 46. KV So with equation 4 we obtain Finally from equation 2 we obtain
- 47. KV A Francis radial flow hydroturbine has the following dimensions, where locations 2 is the inlet and 1 is the outlet; r1 = 2.0 m, r2 = 1.3 m, b1 = 0.85 m and b2 = 2.1 m. The runner blade angles are β1 = 66° and β2 = 18.5° at the turbine inlet and outlet respectively. The runner rotates a ṅ = 100 rpm. The volume flow rate at the design conditions is 80.0 m3/s. Irreversible losses are neglected in this preliminary analysis. Calculate the angle α1 through which the wicket gates should turn the flow, where, α1 is measured from the radial direction at the runner inlet. Calculate the swirl angle α2, where α2 is measured from the radial direction at the runner outlet. Does this turbine have a forward or reverse swirl? Predict the power output (MW) and the required net head (m). EXAMPLE 2 - TURBINE
- 48. KV
- 49. KV SOLUTION A Francis turbine is driven at a specified angular speed by water entering it at a specified rate and angle.We are to calculate runner blade angles, required net head, and power output for a Francis turbine. ASSUMPTIONS: • The flow is steady. • The fluid is water at 20°C • The blades are infinitesimally thin. • The flow is everywhere tangent to the runner blades. • We neglect irreversible losses through the turbine. PROPERTIES: Assume the density of water to be 998 kg/m3
- 50. KV ω Runner blade β1 u1→ Absolute velocity vector. Velocity of the flow at the inlet. It’s direction is governed by the guide vane angle α v1→ Velocity of blade at inlet ur1→ The relative velocity vector of the jet at the inlet uf1→ The velocity of flow in the radial direction uw1→The component of the velocity of the jet u1 in horizontal/tangential direction α1→ Guide vane angle, or the angle between the velocity vector of the blade and the direction of the jet β1→ Blade angle at inlet. Angle between the relative velocity vector and the direction of motion of the blade at the inlet u0 u0
- 51. KV Analysis The angular velocity is 𝜔 = 2𝜋𝑛 60 = 10.4719𝑟𝑎𝑑/𝑠. Solve for the normal component of velocity at the inlet, 𝑢𝑓1 = 𝑉 · 2𝜋𝑟1𝑏1 = 80𝑚3/𝑠 2𝜋 2.0𝑚 0.85𝑚 = 7.4896𝑚/𝑠 (1) The tangential velocity component of the absolute velocity at the inlet is obtained from trigonometry to be 𝑢𝑤1 = 𝜔𝑟1 − 𝑢𝑓1 𝑡𝑎𝑛𝛽1 = 10.4719𝑟𝑎𝑑/𝑠 × 2𝑚 − 7.4896𝑚/𝑠 𝑡𝑎𝑛66∘ = 17.609𝑚/𝑠 (2) From these two components of u1 in the absolute coordinate system, we calculate the angel α1 through which the wicker gates should turn the flow, 𝛼1 = 𝑡𝑎𝑛−1 𝑢𝑓1 𝑢𝑤1 = 𝑡𝑎𝑛−1 7.4896𝑚/𝑠 17.609𝑚/𝑠 = 23.04∘ When measured from the radial direction - 90∘ − 23.04∘ = 66.96∘ (3)
- 52. KV ω Runner blade u0 u2→ Absolute velocity vector water leaving the runner. If it is in the radial direction, will be equal to uf2 v2→ Velocity of blade at exit ur2→ The relative velocity vector of the water leaving the impeller uf2→ The velocity of flow in the radial direction uw2→The component of the velocity for exit u2 in horizontal/tangential direction α2→ The angle between the velocity vector of the blade and the direction of the u2 exiting β2→ Blade angle at outlet, Angle between the relative velocity vector and the direction of motion of the blade at the outlet uf2 β2 𝜶2 u0
- 53. KV Analysis The angular velocity is 𝜔 = 2𝜋𝑛 60 = 10.4719𝑟𝑎𝑑/𝑠. Solve for the normal component of velocity at the inlet, 𝑢𝑓2 = 𝑉 · 2𝜋𝑟2𝑏2 = 80𝑚3/𝑠 2𝜋 1.3𝑚 2.1𝑚 = 4.6639𝑚/𝑠 (1) The tangential velocity component of the absolute velocity at the inlet is obtained from trigonometry to be 𝑢𝑤2 = 𝜔𝑟2 − 𝑢𝑓2 𝑡𝑎𝑛𝛽2 = 10.4719𝑟𝑎𝑑/𝑠 × 1.3𝑚 − 4.6639𝑚/𝑠 𝑡𝑎𝑛18.5∘ = −0.325𝑚/𝑠 (2) From these two components of u1 in the absolute coordinate system, we calculate the angel α1 through which the wicker gates should turn the flow, 𝛼2 = 𝑡𝑎𝑛−1 𝑢𝑓2 𝑢𝑤2 = 𝑡𝑎𝑛−1 −0.325𝑚/𝑠 4.6639𝑚/𝑠 = −3.986∘ Since α2 is negative, this turbine operates with a small amount of reverse swirl.
- 54. KV Using equations 2 and 4, the shaft output power is estimated from the euler turbo machinery equation 𝜌𝑉 · 𝑣1𝑢𝑤1 − 𝑣2𝑢𝑤2 = 𝜌𝜔𝑉 · 𝑟1𝑢𝑤1 − 𝑟2𝑢𝑤2 𝜌𝜔𝑉 · 𝑟1𝑢𝑤1 − 𝑟2𝑢𝑤2 = 998𝑘𝑔/𝑚3 × 10.472𝑟𝑎𝑑/𝑠 × 80𝑚3 /𝑠 2𝑚 × 17.609𝑚/𝑠 − 1.3𝑚 × −0.325𝑚/𝑠 𝜌𝜔𝑉 · 𝑟1𝑢𝑤1 − 𝑟2𝑢𝑤2 = 297.9𝑀𝑊 the required net head assuming that the turbine efficiency is 100% since we have irreversibility’s 𝐻 = 𝑊 · 𝜌𝑔𝑉 · = 297.9𝑀𝑊 998𝑘𝑔/𝑚3 × 9.81𝑚/𝑠2 × 80𝑚3/𝑠 = 38.05𝑚 Since the required net head is less than the gross net heat, the design is feasible
- 55. KV 𝑃𝑜𝑤𝑒𝑟 = 𝜌 × 𝑔 × 𝑉 · × 𝐻 𝑘𝑔 𝑚3 × 𝑚 𝑠2 × 𝑚3 𝑠 × 𝑚 𝑘𝑔 𝑚3 × 𝑚5 𝑠3 𝑘𝑔 × 𝑚2 𝑠3 𝑘𝑔 × 𝑚 𝑠2 × 𝑚 𝑠 𝑁 × 𝑚 𝑠 = 𝐽 𝑠 = 𝑊𝑎𝑡𝑡𝑠 Examine the Units
- 56. KV Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of useful mechanical power to the water. The free surface of the upper reservoir is 45 m higher than the surface of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine the irreversible head loss of the system and the lost mechanical power during this process. EXAMPLE 1
- 57. KV ANALYSIS: The mass flow rate of water through the system is SOLUTION: Water is pumped from a lower reservoir to a higher one. The head loss and power loss associated with this process are to be determined. ASSUMPTIONS The flow is steady and incompressible. The elevation difference between the reservoirs is constant. PROPERTIES: We take the density of water to be 1000 kg/m3 .
- 58. KV We choose points 1 and 2 at the free surfaces of the lower and upper reservoirs, respectively, and take the surface of the lower reservoir as the reference level (z1 = 0). Both points are open to the atmosphere (P1 = P2 = Patm) and the velocities at both locations are negligible (V1 = V2 = 0). Then the energy equation for steady incompressible flow for a control volume between 1 and 2 reduces to
- 59. KV Substituting the lost mechanical power and head loss are calculated as Noting that the entire mechanical losses are due to frictional losses in piping and thus, the irreversible head loss is determined to be
- 60. KV DISCUSSION: The 6.76 kW of power is used to overcome the friction in the piping system. Note that the pump could raise the water an additional 23 m if there were no irreversible head losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 20 kW of power from the water.
- 61. KV Water in a partially filled large tank is to be supplied to the roof top, which is 8 m above the water level in the tank, through a 2.5 cm internal diameter pipe by maintaining a constant air pressure of 300 kPa (gauge) in the tank. If the head loss in the piping is 2 m of water, determine the discharge rate of the supply of water to the roof top. EXAMPLE 8
- 62. KV ANALYSIS: We take point 1 at the free surface of the tank, and point 2 at the exit of the discharge pipe. Noting that the fluid velocity at the free surface of the tank is very low (V1≅ 0) and water discharges into the atmosphere (and thus P2 = Patm), the energy equation written in the head form simplifies to SOLUTION: Water from a pressurised tank is supplied to a roof top. The discharge rate of water from the tank is to be determined. ASSUMPTIONS: The flow is steady and incompressible. The effect of the kinetic energy correction factor is negligible and thus α2 = 1 (we examine the effect of this approximation in the discussion). PROPERTIES: We take the density of water to be 1000 kg/m3.
- 63. KV Manipulating Solving for U2 and substituting the discharge velocity is determined to
- 64. KV DISCUSSION: This is the discharge rate that will prevail at the beginning. The mean flow velocity will decrease as the water level in the tank decreases. If we assume that the flow in the hose at the discharge is fully developed and turbulent, α2 ≈ 1.05, then the results change to V2 = 19.610 m/s ≈ 19.6 m/s, and the Volflowrate = 0.0096263 m3/s 9.63 L/s , a decrease (as expected since we are accounting for more losses) of about 2.4%. Then the initial rate of discharge of water becomes
- 65. KV Water is pumped from an underground reservoir by a 78% efficient 5 kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. (Assuming an irreversible head of 4m) Determined • The maximum flow rate of the water • The pressure difference across the pump Assume the elevation difference better the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible. EXAMPLE 9 & 11
- 66. KV ANALYSIS: (a) The pump-motor draws 5 kW of power, and is 78% efficient. Then the useful mechanical (shaft) power it delivers to the fluid is SOLUTION: Underground water is pumped to a pool at a given elevation. The maximum flow rate and pressures at the inlet and outlet of the pump are to be determined. ASSUMPTIONS: •The flow is steady and incompressible. •The elevation difference between the inlet and the outlet of the pump is negligible. •We assume the frictional effects in piping to be negligible since the maximum flow rate is to be determined, •The effect of the kinetic energy correction factors is negligible, α = 1. PROPERTIES: We take the density of water to be 1000 kg/m3.
- 67. KV ANALYSIS: (a) The pump-motor draws 5 kW of power, and is 78% efficient. Then the useful mechanical (shaft) power it delivers to the fluid is We take point 1 at the free surface of underground water, which is also taken as the reference level (z1 = 0), and point 2 at the free surface of the pool. Also, both 1 and 2 are open to the atmosphere (P1 = P2 = Patm), the velocities are negligible at both points (V1 ≅ V2 ≅ 0), and frictional losses in piping are disregarded. Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to
- 68. KV In the absence of a turbine, and Thus, Then the mass and volume flow rates of water become (NOTE - Solution for 11 in Blue)
- 69. KV ANALYSIS: (b) We take point 3 and 4 at the inlet and the exit of the pump respectively, were the flow velocities are We take the pump as the control volume. Noting that z3 = z4 the energy equation for this control volume reduces to
- 70. KV Substituting, DISCUSSION: In an actual system, the flow rate of water will be less because of friction in the pipes. Also, the effect of flow velocities on the pressure change across the pump is negligible in this case (under 2%) and can be ignored.
- 71. KV A utility company is selling electric power for €0.08/kWh at night and is willing to pay €0.195/kWh for power produced during the day. To take advantage of this opportunity an individual is considering building a large reservoir 40 m above the level of a lake. Availing of the cheap electricity rates at night, this individual intends to pump water from the lake to reservoir. During the day, the water will be allowed to flow back through the pump-motor, which will act as a turbine- generator (i.e. the pump-motor operating in reverse), thereby producing electricity. Initial analysis shows that a water flow rate of 2 m3⋅s-1 can be used in either direction, and the irreversible head loss of the piping system is 4 m. The combined pump-motor and turbine-generator efficiencies are expected to be 75% each. Assuming the system operates for 10 hours each in pump and turbine modes during a typical day, determine the potential revenue this pump- turbine system can generate per year. EXAMPLE 10
- 72. KV ANALYSIS: Choose points 1 and 2 at the free surfaces of the lake and the reservoir, and take the surface of the lake as the datum. Both points are open to the atmosphere (P1 = P2 = Patm); velocities at both are negligible (V1 = V2 = 0). SOLUTION: Water is pumped to a reservoir at a given elevation during the night to avail of night rate electricity. During the day the water flows back through the pump operating in reverse and generates electricity. Required to determine if this a cost making venture. ASSUMPTIONS: • The flow in each direction is steady and incompressible • The elevation difference between the lake and the reservoir can be taken to be constant • The elevation change of reservoir during charging and discharging is disregarded • Unit prices remain constant • The system operates every day of the year for 10 hours in each mode PROPERTIES: We take the density of water to be 1000 kg/m3.
- 73. KV Then the energy equation in terms of heads for steady incompressible flow through a control volume between these two points that includes the pump (or the turbine) and the pipes reduces to; The pump and turbine power corresponding to there heads are;
- 74. KV Then the power cost of the pump, the revenue generated by the turbine, and the net income (revenue minus cost) per year become; It appears that this pump-turbine system has a potential annual income of about €41,135.5 per annum. A decision on such a system will depend on the initial cost of the system, its life, the operating and maintenance costs, the interest rate, and the length of the contract period.
- 75. KV VELOCITY TRIANGLES Radii of the impeller - ri Absolute velocity vector - U Tangential velocity vector - V Relative velocity vector - Ur Angles between the relative velocity vector and tangential velocity α and β Angles between the relative velocity vector and tangential velocity θ and Φ
- 76. KV

- Note Suction Section is made up of a strainer or screen, a suction value, a pressure gauge as well as the suction pipe Discharge Section is made up pressure gauge, delivery gauge, the delivery pipe You also have the Eye of the Impeller
- The water enters through the eye of the impeller from the suction pipe as the impeller is rotated. The water gains velocity as it passes through the impeller - this velocity is know as the tangental velocity. There is also a rise in pressure in the impeller casing or volute. The outlet pressure head is directly proportional to the tangential velocity. As the impeller rotates, water is sucked in through the inlet an rotated outward as shown. The kinetic energy from the rotation is converted into pressure energy
- The continuity equation states that the product for the area and the velocity is constant. 1︎⃣ As the tangential velocity decreases the Kinetic Energy decreases because KE = ½mU² 2︎⃣ As the area increases, the velocity decreases by the continuity equation
- As the impeller rotates the velocity is decreasing as the Area in the casing/Volute is increasing and consequently the Kinetic Energy is decreasing causing the Pressure Energy to increase As the water under the pressure head leaves the pump, it tases through a non return delivery valve.
- The velocity triangle shown is obtained as follows: The fluid enters through an imaginery surface at r1 and the inlet velocity triangle is constructed by drawing the vector representing the absolute velocity u1 at an angle 𝝑1 to the tangent of this surface.. The tangential velocity of the impeller V1 is then subtracted from it vectorially in order to obtain the relative velocity vector Ur1 of the fluid with respect to the impeller blade at the radius R1. In the basic velocity triangle the absolute velocity vector is resolved into two components one in the radial direction called the velocity of flow uf1 and the other vector is perpendicular to it and known as the tangential velocity vector or sometimes called velocity of whirl uw1. These two components are useful in the analysis and therefore are always shown as part of the velocity triangle.
- The velocity triangle shown is obtained as follows: The fluid enters through an imaginery surface at r1 and the inlet velocity triangle is constructed by drawing the vector representing the absolute velocity u1 at an angle 𝝑1 to the tangent of this surface.. The tangential velocity of the impeller V1 is then subtracted from it vectorially in order to obtain the relative velocity vector Ur1 of the fluid with respect to the impeller blade at the radius R1. In the basic velocity triangle the absolute velocity vector is resolved into two components one in the radial direction called the velocity of flow uf1 and the other vector is perpendicular to it and known as the tangential velocity vector or sometimes called velocity of whirl uw1. These two components are useful in the analysis and therefore are always shown as part of the velocity triangle.
- The velocity triangle shown is obtained as follows: The fluid enters through an imaginery surface at r1 and the inlet velocity triangle is constructed by drawing the vector representing the absolute velocity u1 at an angle 𝝑1 to the tangent of this surface.. The tangential velocity of the impeller V1 is then subtracted from it vectorially in order to obtain the relative velocity vector Ur1 of the fluid with respect to the impeller blade at the radius R1. In the basic velocity triangle the absolute velocity vector is resolved into two components one in the radial direction called the velocity of flow uf1 and the other vector is perpendicular to it and known as the tangential velocity vector or sometimes called velocity of whirl uw1. These two components are useful in the analysis and therefore are always shown as part of the velocity triangle.
- Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinϑ, therefore u0A0 = u1 A1 sinϑ
- Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinϑ, therefore u0A0 = u1 A1 sinϑ
- A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, rΩ. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges. The sectional view of the airfoil illustrates, the velocity vectors, the angle Φ which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this.
- Consider an inward flow Francis turbine. The total head available is H and the fluid velocity entering is u0. The velocity leaving the guide vanes is u1 and is related to u0 by the continuity equation; u0A0 = uf1A1 However, uf1= u1sinϑ, therefore u0A0 = u1 A1 sinϑ
- α and θ are associated with the inlet β and φ are associated with the outlet