Centrifugal Pump: - Theory [PDF]

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Centrifugal Pump • A machine which imparts energy to a liquid causing the liquid to flow or rise to a higher level or both. • Demour’s centrifugal pump – 1730



• Theory – conservation of angular momentum – conversion of kinetic energy to potential energy



• Pump components ❑ Rotating element – impeller - takes the power (mechanical) from the rotating shaft and accelerates the fluid.



❑ Enclosing the rotating element and sealing the pressurized liquid inside – casing or housing – acts as a diffuser and transforms high fluid velocity (kinetic energy) into pressure.



Centrifugal Pump



Schematic diagram of basic elements of centrifugal pump



Main Parts of Centrifugal Pumps 1. Impeller which is the rotating part of the centrifugal pump. It consists of a series of backwards curved vanes (blades). The impeller is driven by a shaft which is connected to the shaft of a prime mover, e.g., an electric motor, an IC engine, etc.



2. Casing Which is an air-tight passage surrounding the impeller designed to direct the liquid to the impeller and lead it away Volute casing. It is of spiral type in which the area of the flow increases gradually.



Main Parts of Centrifugal Pumps 3.



Suction Pipe



The conduit connecting the sump (reservoir) up to the centerline of the pump impeller.



4.



Delivery Pipe



The conduit connecting the pump impeller centerline up to the delivery point.



5.



Impeller Shaft



The shaft which carries the mechanical power from the engine/motor drive to the impeller.



6.



Engine/Motor Driver



Supplies mechanical power to the rotating impeller shaft. It can be mounted directly on the pump, above it, or adjacent to it.



Effect of Impeller Diameter and Speed Broad range of applicable flows and heads. Higher discharges and heads can be achieved by increasing the rotational speed or the diameter of the impeller although with corresponding change in power input. 2 3



𝑄∝ 𝑁



𝐻∝𝑁



𝑃∝𝑁



𝑄 ∝ 𝐷3



𝐻 ∝ 𝐷2



𝑃 ∝ 𝐷5



▪ Discharge varies directly with the change in speed or as cube of the change in impeller diameter ▪ Head varies as square of the change in speed or impeller diameter. ▪ Power input varies as the cube of the change in speed or fifth power to the impeller diameter. For instance, the pump speed is doubled: ▪ Discharge will double. ▪ Head will increase by a factor of 22 = 4. ▪ Power input will increase by a factor of 23 = 8.



Application of Centrifugal Pumps Centrifugal pumps (radial-flow pumps) are the most used pumps for hydraulic purposes where high discharge with relatively low head is required such as irrigation, cooling tower of central A/C system/power plants, etc. Shrouded Impeller is used for very Thin (less viscous) clear liquid, e.g. water.



Semi-enclosed Impeller is used for moderately thick (viscous) liquid, e.g. thick oils.



Fully-open Impeller is used for very thick (highly viscous), stingy liquid, e.g., dredging fluid (mud), paper stock, etc.



Types of Centrifugal Pump Impellers



(a) Volute (casing) pump



(b) Double-volute (casing) pump



Types of Centrifugal Pumps (casing)



(c) Turbine (casing with guide vanes) pump



Euler Equation for Centrifugal Pumps Using subscripts 1 for inlet and 2 for outlet quantities: 𝐷1 and 𝐷2 =Diameters at inlet of outlet of the impeller 𝑁 =Rotational Speed of the impeller in rpm



Outlet Velocity Triangle



(𝑢2 −𝑣𝑤2 )



𝑣𝑤2



𝑉1 and 𝑉2 =Absolute velocities at inlet and outlet 𝑢1 =



𝜋𝐷1 𝑁 60



and 𝑢2 =



𝜋𝐷2 𝑁 60



=Tangential velocities at



inner and outer periphery of the impeller 𝑣𝑟1 and 𝑣𝑟2 =Relative velocities at inlet and outlet 𝛼1 and 𝛼2 =Direction of absolute velocities at inlet and outlet. 𝛼 is the angle made by the absolute velocity vector 𝑉 with the positive direction of the peripheral velocity 𝑢. 𝛽1 and 𝛽2 =Vane (blade) angle at inlet and outlet made by the relative velocity vector 𝑣𝑟 with the negative direction of the peripheral velocity 𝑢. 𝑣𝑤1 and 𝑣𝑤 2 =Whirl velocities at inlet and outlet (Tangential components of 𝑉1 and 𝑉2 respectively).



𝑣𝑓1 and 𝑣𝑓 =Flow velocities at inlet and outlet (Radial components of 2 𝑉1 and 𝑉2 respectively) .



𝑣𝑤1 Inlet Velocity Triangle



Euler Equation for Centrifugal Pumps Euler Head, 𝑯𝒆 : [Theoretical head transferred to liquid by the impeller or work done by the impeller per kg of liquid] From Euler’s Pump Equation, Euler Head, 𝑯𝒆 :



Outlet Velocity Triangle



(𝑢2 −𝑣𝑤2 )



𝑣𝑤2



𝑽𝟐𝟐 − 𝑽𝟐𝟏 𝒗𝟐𝒓𝟏 − 𝒗𝟐𝒓𝟐 𝒖𝟐𝟐 − 𝒖𝟐𝟏 𝑯𝒆 = + + 𝟐𝒈 𝟐𝒈 𝟐𝒈 Where: 𝑉22 −𝑉12 2𝑔



represents the increase in absolute kinetic energy of fluid;



𝑢22 −𝑢12 2𝑔



represents the increase in static pressure due to centrifugal action;



2 −𝑣 2 𝑣𝑟1 𝑟2 2𝑔



represents the change in the kinetic



energy due to retardation of flow.



𝑣𝑤1 Inlet Velocity Triangle



Euler Equation for Centrifugal Pumps 𝑽𝟐𝟐 − 𝑽𝟐𝟏 𝒗𝟐𝒓𝟏 − 𝒗𝟐𝒓𝟐 𝒖𝟐𝟐 − 𝒖𝟐𝟏 𝑯𝒆 = + + 𝟐𝒈 𝟐𝒈 𝟐𝒈 From the inlet and outlet velocity triangles, we have:



Outlet Velocity Triangle



(𝑢2 −𝑣𝑤2 )



𝑣𝑤2



2 𝑣𝑟1 = 𝑢12 + 𝑉12 − 2𝑢1 𝑉1 𝑐𝑜𝑠𝛼1 2 𝑣𝑟2 = 𝑢22 + 𝑉22 − 2𝑢2 𝑉2 𝑐𝑜𝑠𝛼2



Using these relations in 𝐻𝑒 equation and rearranging, we get:



1 𝐻𝑒 = 𝑢2 𝑉2 𝑐𝑜𝑠𝛼2 − 𝑢1 𝑉1 𝑐𝑜𝑠𝛼1 𝑔 𝟏 ⇒ 𝑯𝒆 = 𝒖 𝒗 − 𝒖𝟏 𝒗𝒘𝟏 𝒈 𝟐 𝒘𝟐 𝑣𝑤1 Inlet Velocity Triangle



Euler Equation for Centrifugal Pumps Assuming Radial Entry at inlet for a centrifugal pump,



Euler Head (Theoretical),



𝟏 𝑯𝒆 = 𝒖𝟐 𝒗𝒘𝟐 𝒈 Actual Head Developed by the impeller,



Manometric Head (Measured), 𝑯𝒎



Manometric (Hydraulic) Efficiency,



𝜼𝒎



𝑯𝒎 = 𝑯𝒆



Inlet Velocity Triangle (radial entry, 𝜶𝟏 = 𝟗𝟎°)



𝑉1 = 𝑉𝑓1



𝑣𝑟1



𝛽1



𝛼1



𝑢1



Minimum Starting Speed of a Centrifugal Pump At the time of start, the fluid velocities are zero and the only head that is operating



is the centrifugal head



𝑢22 −𝑢12 2𝑔



. This centrifugal force must overcome the



manometric head for the fluid to move, i.e., Again,



𝑢22 − 𝑢12 ≥ 𝐻𝑚 2𝑔 Now, 𝑢1 =



𝜋𝐷1 𝑁 60



𝑎𝑛𝑑 𝑢2 =



𝜋𝐷2 𝑁 60



So that, equating



𝑢22 − 𝑢12 = 𝐻𝑚 2𝑔 𝑵𝒎𝒊𝒏



𝑯𝒎 = 𝟖𝟒. 𝟓𝟗𝟔 𝑫𝟐𝟐 − 𝑫𝟐𝟏



𝑢22 − 𝑢12 ≥ 𝜂𝑚 𝐻𝑒 2𝑔 Equating



𝑢22 − 𝑢12 𝑢2 𝑣𝑤2 = 𝜂𝑚 2𝑔 𝑔



𝑵𝒎𝒊𝒏



𝟏𝟐𝟎𝜼𝒎 = 𝝅



𝒗𝒘𝟐 𝑫𝟐 𝑫𝟐𝟐 − 𝑫𝟐𝟏



Minimum Impeller Diameter at a Given Speed Usually the impeller outer diameter is designed as twice the inner (inlet) diameter, i.e., 𝑫𝟐 = 𝟐𝑫𝟏 . Using this relationship, the minimum diameter of the impeller for fluid to move,



𝑢22 − 𝑢12 ≥ 𝐻𝑚 2𝑔



Here, 𝑢1 =



𝜋𝐷1 𝑁 60



=



𝜋𝐷2 𝑁 120



𝑎𝑛𝑑 𝑢2 =



𝜋𝐷2 𝑁 60



So that, equating



𝑢22 − 𝑢12 = 𝐻𝑚 2𝑔



𝜋𝐷2 𝑁 60



2



𝜋𝐷2 𝑁 − 120



2



= 2𝑔𝐻𝑚



𝑯𝒎 This equation is used in practical situations to 𝑫 𝟐 𝒎𝒊𝒏 = 𝟗𝟕. 𝟔𝟖 design impeller for liquid pumping at a given 𝑵 speed.



Priming of Centrifugal Pump When the pump casing and the suction conduit are completely filled with water, as the impeller rotates, the pressure at the pump suction side becomes lower than the atmospheric pressure. Due to this difference in pressure head between the water surface of the sump and the inlet of the pump, the atmospheric pressure pushes the water from the sump to the pump casing. However, an impeller running in air would produce only a small head. This cannot create the necessary differential head of water between the sump and the pump inlet as the density of air is much less than that of water. Consequently, the pump does not do its work of pumping of water. Further, dry running of the pump may damage several parts of the pump. This is, therefore, necessary to ensure that the pump casing, impeller, suction pipe and the portion of the pump delivery pipe up to the delivery valve are always filled with water before the start of the pump. Filling is done by pouring water into the funnel or priming-cup provided for this purpose. An air vent in the casing is provided for the air to escape. This air vent must be closed after filling. This filling process is called the “priming” of the pump. Most centrifugal pumps are not self-priming, so they always need priming. However, a one-way valve, called foot valve is used at the entrance of the suction pipe, which keeps the suction conduit filled-up even when the pump is stopped. Thus when the pump is restarted, it does not need priming.



Why pump pressure is expressed in liquid head (liquid height in meters or feet)? A pump throws liquid in the opposite direction of a free falling body from a certain height. A free falling body starts from zero velocity at a certain height and then reached maximum velocity at the ground. On the other hand, a pump throws liquid at a maximum velocity at its impeller exit to reach zero velocity at a certain height. The height of liquid travel from impeller exit of a pump with impeller diameter of 500 mm running at 1800 rpm, can be determined as follows: 𝜋𝐷 𝑁



𝜋×0.5×1800



Peripheral velocity of the Impeller at exit point u2 = 602 = = 47.12 𝑚/s, at this 60 velocity the liquid will be thrown by the impeller at the impeller exit. Therefore, the height of liquid travel is given by: 0 = 𝑢22 − 2𝑔ℎ ⇒ ℎ = 𝑢22 /2𝑔 ⇒ ℎ = 113 𝑚 It can be observed that no liquid property enters the equation, maximum velocity of the liquid depends only on the impeller size and speed.



Thus a pump always produce the same head regardless of the type of liquid being pumped. However, pressure will increase or decrease in direct proportion to a liquid’s specific gravity (𝑝 = 𝜌𝑔𝐻) and power input to the pump will also vary directly with liquid’s specific gravity. Thus a centrifugal pump can develop the same 100 m of head when pumping water, brine or kerosene. The resulting pressures, however, will vary. Therefore, the pump pressures are expressed in liquid height in meters or feet (liquid head).



Example Problems Problem–1: A centrifugal pump impeller runs at 950 rpm (𝑁). Its external and internal diameters are 500 mm (𝐷2 )and 250 mm (𝐷1 ) respectively. The vanes are set back at an angle of 35o (𝛽2 ) to the outer rim. If the radial velocity of water through the impeller is maintained constant at 2 m/s (𝑉𝑓1 = 𝑉𝑓2 ), find the angle of the vanes at inlet (𝜷𝟏 ), the velocity and direction of water at outlet (𝑽𝟐 , 𝜶𝟐 ) and the work done by the impeller per kg of water (𝑯𝒆 ). Solution: Assuming radial entry of the fluid, i.e., 𝑉1 = 𝑉𝑓1 ▪ 𝑢1 =



𝜋𝐷1 𝑁 60



▪ 𝑡𝑎𝑛𝜷𝟏 = ▪ 𝑢2 =



▪ 𝑡𝑎𝑛𝛽2 =



=



𝑉𝑓1 𝑢1



𝜋𝐷2 𝑁 60



𝜋×0.25×950 60



=



2 12.44



𝑉𝑓2 𝑢2 −𝑣𝑤2



= 12.44 𝑚/𝑠



𝑢2



𝛼2



= 0.16 ⇒ 𝜷𝟏 = 𝟗. 𝟏𝟑°



𝜋×0.5×950 60



=



𝑣𝑤2



𝑉2



𝑣𝑟2



= 24.87 𝑚/𝑠



⇒ tan 35° =



2 24.87−𝑣𝑤2



⇒ 𝑣𝑤2 = 22.1 𝑚/𝑠 ▪ 𝑽𝟐 =



2 𝑉𝑓2



+



▪ 𝜶𝟐 = tan−1 ▪ 𝑯𝒆 =



𝑢2 𝑣𝑤2 𝑔



2 𝑣𝑤2



=



𝑉𝑓2



= tan−1



𝑣𝑤2



=



22



+



24.87×22.1 9.81



22.12 2 22.1



= 𝟐𝟐. 𝟐 𝒎/𝒔 = 𝟓. 𝟏𝟕°



= 𝟓𝟔 𝒌𝒈 𝒎/𝒔



𝑣𝑟1



𝑉1 = 𝑉𝑓1



𝛽1



𝑢1



𝛽2



Inlet



Outlet



Example Problems Problem–2: A centrifugal pump is used to lift water at a rate of 0.085 m3/s (𝑄). The outer diameter of the impeller is 325 mm (𝐷2 ) and the breadth of the wheel at outlet is 15 mm (𝑏2 ). The manometric head of the pump is 38 m (𝐻𝑚 ) and manometric efficiency is 85% (𝜂𝑚 ). If the pump runs at 1500 rpm (𝑁), determine the blade angle at exit (𝜷𝟐 ). Solution: Assuming radial entry of the fluid, i.e., 𝑉1 = 𝑉𝑓1



▪ 𝑢2 =



𝜋𝐷2 𝑁 60



▪ 𝜂𝑚 =



𝐻𝑚 𝐻𝑒



=



=



𝜋×0.325×1500 60



𝐻𝑚 𝑢2 𝑣𝑤2 𝑔



= 25.52 𝑚/𝑠



⇒ 0.85 =



𝑣𝑤2



38



𝑢2



𝛼2



25.52×𝑣𝑤2 9.81



𝑉2



⇒ 𝑣𝑤2 = 17.19 𝑚/𝑠



𝑣𝑟2



▪ 𝑄 = 𝜋𝐷2 𝑏2 𝑉𝑓2 ⇒ 0.085 = 𝜋 × 0.325 × 0.015 × 𝑉𝑓2 ⇒



𝑉𝑓2 = 5.55 𝑚/𝑠



𝑣𝑟1 ▪ 𝑡𝑎𝑛𝜷𝟐 = 𝑢 ⇒ 𝜷𝟐 =



𝑉𝑓2



2 −𝑣𝑤2 tan−1



𝑉1 = 𝑉𝑓1



𝛽1 5.55 25.52−17.19



= 𝟑𝟑. 𝟔𝟔°



𝑢1



𝛽2



Inlet



Outlet



Example Problems Problem–3: The inner and outer diameters of the impeller of a centrifugal pump are 300 mm (𝐷1 ) and 600 mm (𝐷2 ) respectively. The constant velocity of flow is 2.2 m/s (𝑉𝑓1 = 𝑉𝑓2 ) and the vanes are curved backward at an angle of 45o at the exit (𝛽2 ) . If the manometric efficiency is 75% (𝜂𝑚 ), find the minimum starting speed (𝑵𝒎𝒊𝒏 ) of the pump. Solution: Assuming radial entry of the fluid, i.e., 𝑉1 = 𝑉𝑓1



▪ 𝑢2 =



𝜋𝐷2 𝑁 60



=



▪ 𝑡𝑎𝑛𝛽2 = 𝑢



𝜋×0.6×𝑁𝑚𝑖𝑛 60



𝑉𝑓2



2 −𝑣𝑤2



= 0.0314159𝑁𝑚𝑖𝑛



⇒ tan 45° =



𝑣𝑤2



𝑢2



𝛼2



2.2 0.0314159𝑁𝑚𝑖𝑛 −𝑣𝑤2



𝑉2



𝑣𝑟2



⇒ 𝑣𝑤2 = (0.031416𝑁𝑚𝑖𝑛 − 2.2) 𝑚/𝑠



▪ 𝑵𝒎𝒊𝒏 = =



120𝜂𝑚 𝜋



𝑣𝑤2 𝐷2 𝐷22 −𝐷12



120 × 0.75 𝜋



0.031416𝑵𝒎𝒊𝒏 − 2.2 × 0.6 0.62 − 0.32



⇒ 𝑵𝒎𝒊𝒏 = 63.662 0.031416𝑵𝒎𝒊𝒏 − 2.2 ⇒ 𝑵𝒎𝒊𝒏 = 𝟏𝟒𝟎 𝒓𝒑𝒎



𝑣𝑟1



𝑉1 = 𝑉𝑓1



𝛽1



𝑢1



𝛽2



Inlet



Outlet



Example Problems Problem–4: A centrifugal pump delivers 0.20 m3/s water against a head of 26 m while running at 950 rpm. The constant velocity of flow is 2.9 m/s and the vanes are curved backward at an angle of 30o. If the manometric efficiency is 77%, find the diameter and the width (or breadth) of the impeller at outlet. Solution: Assuming radial entry of the fluid, i.e., 𝑉1 = 𝑉𝑓1



▪ 𝑄 = 𝜋𝑫𝟐 𝒃𝟐 𝑉𝑓2 ⇒ 0.20 = 𝜋 × 𝐷2 × 𝑏2 × 2.9 ⇒ 𝑫𝟐 𝒃𝟐 = 0.02195



▪ 𝑢2 =



𝜋𝐷2 𝑁 60



𝜋×𝐷2 ×950 60



=



▪ 𝑡𝑎𝑛𝛽2 = 𝑢



𝑉𝑓2



2 −𝑣𝑤2



𝑣𝑤2



= 49.74𝑫𝟐



⇒ tan 30° =



𝑢2



𝛼2



𝑉2



2.9 49.74𝑫𝟐 −𝑣𝑤2



𝑣𝑟2



⇒ 𝑣𝑤2 = 49.74𝑫𝟐 − 5.023



▪ 𝜂𝑚 =



𝐻𝑚 𝐻𝑒



𝑔𝐻𝑚 2 𝑣𝑤2



=𝑢



9.81 × 26 49.74𝑫𝟐 49.74𝑫𝟐 − 5.023 ⇒ 𝑫2𝟐 − 0.134𝑫𝟐 − 0.101 = 0 ∴ 𝑫𝟐 = 0.3918 𝑚 = 𝟑𝟗𝟐 𝒎𝒎 0.02195 And 𝒃𝟐 = = 0.056 𝑚 = 𝟓𝟔 𝒎𝒎 ⇒ 0.77 =



𝐷2



𝑣𝑟1



𝑉1 = 𝑉𝑓1



𝛽1



𝑢1



𝛽2



Inlet



Outlet



Example Problems Problem–5: A centrifugal pump impeller has an outside diameter of 200 mm (𝐷2 ) and it rotates at 2900 rpm (𝑁). Determine the head generated (𝑯𝒎 ) if the vanes are curved backward at an angle of 25o to the outer rim (𝛽2 ) and the velocity of flow throughout the wheel is constant at 3 m/s (𝑉𝑓1 = 𝑉𝑓2 ). Assume hydraulic efficiency of 75% (𝜂𝑚 ). Determine also the power in kW (𝑷) required to run the impeller if the breadth of the wheel at outlet is 15 mm (𝑏2 ). Neglect the effect of vane thickness and mechanical friction and leakage losses. Solution: Assuming radial entry of the fluid, i.e., 𝑉1 = 𝑉𝑓1



▪ 𝑢2 =



𝜋𝐷2 𝑁 60



=



▪ 𝑡𝑎𝑛𝛽2 = 𝑢



𝜋×0.2×2900 60



𝑉𝑓2



2 −𝑣𝑤2



𝑣𝑤2



= 30.37 𝑚/𝑠



𝑢2



𝛼2



3



⇒ tan 25° = 30.37−𝑣



𝑉2



𝑤2



𝑣𝑟2



⇒ 𝑣𝑤2 = 23.94 𝑚/𝑠 ▪ 𝑯𝒎 = 𝜂𝑚 𝐻𝑒 = 𝜂𝑚



𝑢2 𝑣𝑤2 𝑔



30.37 × 23.94 ⇒ 𝑯𝒎 = 0.75 × = 𝟓𝟓. 𝟔 𝒎 9.81



▪ 𝑷 = 𝛾𝑄𝐻𝑚 /𝜂𝑚 = 𝛾(𝜋𝐷2 𝑏2 𝑉𝑓2 )𝐻𝑚 /𝜂𝑚 ⇒ 𝑷 = 9810 × 𝜋 × 0.2 × 0.015 × 3 × 55.6/0.75 ⇒ 𝑷 = 𝟐𝟎. 𝟓𝟔 𝐤𝐖



𝑣𝑟1



𝑉1 = 𝑉𝑓1



𝛽1



𝑢1



𝛽2



Inlet



Outlet



Expressions for Manometric Head Applying Bernoulli’s Equation between points 1 and 4 𝑝𝑎 𝑝𝑎 𝑉𝑑2 + 0 + 0 + 𝐻𝑚 = + 𝐻𝑠 + 𝐻𝑑 + (𝐻𝑓𝑠 +𝐻𝑓𝑑 ) + 𝛾 𝛾 2𝑔 𝑉𝑑2 ⇒ 𝐻𝑚 = 𝐻𝑠 + 𝐻𝑑 + (𝐻𝑓𝑠 +𝐻𝑓𝑑 ) + 2𝑔 2 𝑉𝑑 ⇒ 𝐻𝑚 = 𝐻𝑠𝑡𝑎𝑡 + Σ𝐿𝑜𝑠𝑠𝑒𝑠 + 2𝑔 ⇒ 𝐻𝑚 = 𝐻𝑠𝑡𝑎𝑡 + Σ𝐿𝑜𝑠𝑠𝑒𝑠 Manometric head of a pump is the gross head that must be provided by the impeller for the liquid to flow from the sump to the delivery point. Applying Bernoulli’s Equation between points 2 and 3 𝑝3 𝑉𝑑2 𝑝2 𝑉𝑠2 𝐻𝑚 = + + 𝑍3 − + + 𝑍2 𝛾 2𝑔 𝛾 2𝑔 𝑝3 𝑝2 𝑉𝑑2 𝑉𝑠2 𝐻𝑚 = − + − 𝛾 𝛾 2𝑔 2𝑔 𝐻𝑚 =



𝑝3 𝑝2 − 𝛾 𝛾



Typical Pump Setup



Ideal Increase in Pressure Head in the Impeller Energy of a liquid at inlet + Energy added by the impeller = Energy of liquid at outlet



𝑣𝑤2



𝑝1 𝑉12 𝑢2 𝑣𝑤2 𝑝2 𝑉22 + + 𝑍1 + = + + 𝑍2 𝛾 2𝑔 𝑔 𝛾 2𝑔 Increase in piezometric head at the pump



𝑢2



𝛼2



𝑉2



𝑣𝑟2



𝑝2 𝑝1 𝑉12 𝑉22 𝑢2 𝑣𝑤2 + 𝑍2 − + 𝑍1 = Δ𝐻𝑝 = − + 𝛾 𝛾 2𝑔 2𝑔 𝑔 𝑝2 𝑝1 𝑉12 𝑉22 𝑢2 𝑣𝑤2 − = 𝐻𝑚 = Δ𝐻𝑝 = − + 𝛾 𝛾 2𝑔 2𝑔 𝑔 Δ𝑝𝑖 = 𝐻𝑚 & 𝐻𝑚 = Δ𝐻𝑝 = 𝐻𝑒 − Δ𝐾𝐸 𝛾 Δ𝑝𝑖 1 Δ𝐻𝑝 = = 𝑉12 − 𝑉22 + 2𝑢2 𝑣𝑤2 𝛾 2𝑔



𝑣𝑟1



𝑉1 = 𝑉𝑓1



𝛽1



𝑢1



𝛽2



Inlet



Outlet



Ideal Increase in Pressure Head in the Impeller ▪ 𝑡𝑎𝑛𝛽2 = 𝑢



𝑉𝑓2



𝑣𝑤2



2 −𝑣𝑤2



⇒ 𝑣𝑤2



𝑢2



𝛼2



𝑉𝑓2 = 𝑢2 − = 𝑢2 − 𝑉𝑓2 𝑐𝑜𝑡𝛽2 tan𝛽2



𝑉2



𝛽2



𝑣𝑟2



2 2 ▪ 𝑉22 = 𝑣𝑤2 + 𝑉𝑓2



2 2 = 𝑢22 + 𝑉𝑓2 cot 2 𝛽2 − 2𝑢2 𝑉𝑓2 𝑐𝑜𝑡𝛽2 + 𝑉𝑓2 2 = 𝑉𝑓2 (1 + cot 2 𝛽2 ) + 𝑢22 − 2𝑢2 𝑉𝑓2 𝑐𝑜𝑡𝛽2 2 = 𝑉𝑓2 cosec 2 𝛽2 + 𝑢22 − 2𝑢2 𝑉𝑓2 𝑐𝑜𝑡𝛽2



▪ Δ𝐻𝑝 =



Δ𝑝𝑖 𝛾



1



= 2𝑔 𝑉12 − 𝑉22 + 2𝑢2 𝑣𝑤2



𝑣𝑟1



𝑉1 = 𝑉𝑓1



𝛽1



𝑢1



Inlet



Δ𝑝𝑖 1 2 2 = 𝑉𝑓1 − 𝑉𝑓2 cosec 2 𝛽2 − 𝑢22 + 2𝑢2 𝑉𝑓2 𝑐𝑜𝑡𝛽2 + 2𝑢2 (𝑢2 −𝑉𝑓2 𝑐𝑜𝑡𝛽2 ) 𝛾 2𝑔



∴



Δ𝑝𝑖 1 2 2 = 𝑉𝑓1 − 𝑉𝑓2 cosec 2 𝛽2 + 𝑢22 𝛾 2𝑔



Outlet



Head Losses in Centrifugal Pumps Mechanical losses are the frictional losses in bearings, glands, packages, etc. and the disc friction between the impeller and the liquid which fills the clearance space between the impeller and the casing. Some leakage loss also take place between impeller and casing, at mechanical seals, glands, etc.



Hydraulic losses are due to: ▪ Circulatory flow at the passages of the impeller and independent of the discharge.



▪ Fluid friction at the flow passage: this loss depends on the fluid contact area and the roughness of the surface and hence equal to 𝐾1 𝑄2 where 𝐾1 is a coefficient. ▪ Shock losses at the entrance to impeller: this loss occurs due to improper entry angle of the flow with respect to the blade angle. At design condition, this loss is practically zero and increases at reduced or increased flow from normal values.



Head Losses in Centrifugal Pumps To account for various losses, several efficiencies are defined.



Volumetric efficiency, 𝜼𝒗 =



𝑸 𝑸+𝑸𝑳



where 𝑄 is Discharge reaching the pump outlet, 𝑄𝐿 is the leakage flow which does not reach the pump outlet and 𝑄 + 𝑄𝐿 is Discharge entering the eye of the impeller.



Manometric (hydraulic) efficiency, 𝜼𝒎 =



𝑯𝒎 𝑯𝒆



=



𝒈𝑯𝒎 𝒖𝟐 𝒗𝒘𝟐



=



𝑯𝒎 𝑯𝒎 +𝑯𝒇𝒍



where 𝐻𝑓𝑙 is hydraulic losses in the impeller and the casing. Mechanical efficiency, 𝜼𝒎𝒆𝒄𝒉 =



𝜸 𝑸+𝑸𝑳 𝑯𝒆 𝑷



or 𝜼𝒎𝒆𝒄𝒉 = 𝑯



𝑯𝒆 𝒆 +𝑯𝒎𝒆𝒄𝒉



where 𝑃 is the mechanical power input to the impeller shaft by the prime mover and 𝐻𝑚𝑒𝑐ℎ is the mechanical head losses. Overall efficiency, 𝜼𝒐 =



𝜸𝑸𝑯𝒎 𝑷



or



𝜼𝒐 = 𝜼𝒗 𝜼𝒎 𝜼𝒎𝒆𝒄𝒉 =



𝑸 𝑸+𝑸𝑳



×



𝑯𝒎 𝑯𝒆



×



𝜸 𝑸+𝑸𝑳 𝑯𝒆 𝑷



Effect of Outlet Blade Angle



Please note here 𝑽𝒖𝟐 𝐢𝐬 𝐮𝐬𝐞𝐝 𝐢𝐧 𝐩𝐥𝐚𝐜𝐞 𝐨𝐟 𝒗𝒘𝟐



Effect of Outlet Blade Angle



Please note here 𝑽𝒖𝟐 𝐢𝐬 𝐮𝐬𝐞𝐝 𝐢𝐧 𝐩𝐥𝐚𝐜𝐞 𝐨𝐟 𝒗𝒘𝟐



Theoretical Head-Discharge Relationship of a Pump



Please note here 𝑽𝒖𝟐 𝐢𝐬 𝐮𝐬𝐞𝐝 𝐢𝐧 𝐩𝐥𝐚𝐜𝐞 𝐨𝐟 𝒗𝒘𝟐



Theoretical Head-Discharge Relationship of a Pump Please note here 𝑽𝒖𝟐 𝐢𝐬 𝐮𝐬𝐞𝐝 𝐢𝐧 𝐩𝐥𝐚𝐜𝐞 𝐨𝐟 𝒗𝒘𝟐



Theoretical Head-Discharge Relationship of a Pump



Head-Discharge (He-Q) relationship for ideal pumps



Head-Discharge (He-Q) relationship for actual pumps



From the ideal 𝐻 − 𝑄 curve on the left, it can be inferred that forward curved vanes produce higher head at higher discharge. However, both forward-curved and radial vane pumps result in poor efficiency. Forward-curved vane produce larger absolute velocities that require very efficient diffusers to convert the exit kinetic energy into pressure energy. So the energy losses are high. Further, an instability called pump surging occurs in forward-curved vanes. Therefore, in actual practice, backward-curved vanes in the range of 20° – 40° are of common use.



In actual 𝐻 − 𝑄 curve shown on the right the head decreases with increase in discharge for all types of vanes due to hydraulic losses present in real-world applications.



Typical Characteristics of Actual Centrifugal Pumps Constant



Diameter = Constant



Head-Discharge (𝑯 − 𝑸) characteristics



▪ From the 𝐻 − 𝑄 curve, it can be observed that for the same pump (same impeller/diameter) as the speed increases, the head also increases (𝐻 ∝ 𝑁 2 ) at the same discharge, or the discharge increases (𝑄 ∝ 𝑁) at the same head. The higher the pump speed, the higher will be the head or discharge. ▪ From the 𝜂 − 𝑄 curve, it can be observed that for the same pump (same impeller/diameter) as the speed increases, both the maximum efficiency and the maximum discharge increase. The higher the speed, the higher will be the efficiency and discharge.



▪ However, the 𝑃 − 𝑄 curve shows that for a given discharge, higher speed needs higher power input.



Variation of Efficiency with discharge



Diameter = Constant



Variation of Power with discharge



Main Characteristics Curve of Actual Centrifugal Pumps



Head (Duty Point)



𝜼𝒎𝒂𝒙 (𝑩𝑬𝑷)



Discharge (Duty Point)



Pump manufacturers provide information on the performance of their pumps in the form of curves, commonly called pump characteristic curve (or simply pump curve). In pump curves the following information may be given: (i) the discharge on the x-axis, (ii) the head on the left y-axis, (iii) the pump efficiency as a percentage on the right (or left) y-axis, (iv) the pump power input on the left (or right) y-axis, (v) the NPSH of the pump on the y-axis (vi) the speed of the pump one the y-axis. The discharge-head (𝑸, 𝑯) values corresponding to BEP (𝜼𝒎𝒂𝒙 ) is called the ‘Duty Point’ of the pump.



Similarity Ratios (Affinity Laws) for Centrifugal Pumps 𝑷 𝝆𝑵𝟑 𝑫𝟓



Similarity law:



=𝒇



𝒈𝑯 𝑸 𝝆𝑵𝑫𝟐 , , 𝝁 𝑵𝟐 𝑫𝟐 𝑵𝑫𝟑



⇒



ς1 = 𝑓( ς2, ς3, ς4 ) ς𝟏 = ς𝟐 = ς𝟑 = ς𝟒 =



𝑃 𝜌𝑁3 𝐷5 𝐴 𝐻 𝑁 2 𝐷2



𝐴



𝑄 𝑁𝐷3 𝐴 𝜌𝑁𝐷2 𝜇



=



=



=



𝑃 𝜌𝑁3 𝐷5 𝐵 𝐻 𝑁2 𝐷2



𝑄 𝑁𝐷3 𝐵



𝐵



= 𝐶𝑝 = 𝑃𝑜𝑤𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡



= 𝐶𝐻 = 𝐻𝑒𝑎𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (dropping ‘𝑔’ term)



= 𝐶𝑄 = 𝐹𝑙𝑜𝑤 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡



= 𝑅𝑒 = 𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑐𝑎𝑛 𝑏𝑒 𝑖𝑔𝑛𝑜𝑟𝑒𝑑)



𝑭𝒐𝒓 𝒈𝒆𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒂𝒔 𝒘𝒆𝒍𝒍 𝒂𝒔 𝒅𝒚𝒏𝒂𝒎𝒊𝒄 𝒔𝒊𝒎𝒊𝒍𝒂𝒓𝒊𝒕𝒚 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒑𝒖𝒎𝒑𝒔 𝑨 𝒂𝒏𝒅 𝑩, 𝒂𝒍𝒍 𝒕𝒉𝒓𝒆𝒆 ς′𝒔 𝒎𝒖𝒔𝒕 𝒃𝒆 𝒆𝒒𝒖𝒂𝒍



Again,



ς3 1/2 ς2 3/4



𝑁 𝑄



= 𝐻 3/4 = 𝑁𝑠 is called specific speed . For dynamic similarity



between two pumps or model and its prototype, the value of Ns should be same.



Specific Speed Specific Speed, 𝑵𝒔



=



𝑵 𝑸 𝟑/𝟒



𝑯𝒎



▪ ▪



Specific Speed of a pump is defined as the speed of an imaginary pump which will produce the same amount of discharge under unit head. This is a numerical engineering tool for the selection of the type of the pump for installation*.



Note that The values of 𝑵, 𝑸 𝒂𝒏𝒅 𝑯𝒎 are taken at Best Efficiency Point (BEP) There is another expression for non-dimensional specific speed (also called shape factor or shape number) 𝑵 𝑸 This does not indicate the size, rather indicated the shape or type of the fluid machinery. 𝑺𝒒 = 𝒈𝑯𝒎 𝟑/𝟒 Please note the usage of units while calculating Specific Speed. Different units will result in different values for the same pump. 𝑁 𝑄 𝑁 𝑄 Ns = Specific Speed 𝑁𝑠 = 3/4 𝑁 = 𝑐 𝑠 𝐻 3/4



𝐻



N = Speed in rpm Q = Discharge in US gallons per minute H = Head in ft [1 US gallon = 3.785 Liters]



In US Customary Unit (FPS) Reciprocating Pump : 50 to 500 Centrifugal Pump: 500 to 10000 Radial Flow Pump : 500 to 4000 Mixed Flow Pump : 2000 to 8000 Axial Flow/Propeller Pump: 7000 to 20000



N = Speed in rpm Q = Discharge in m3/hr H = Head in m 𝒄 = 𝟎. 𝟖𝟔𝟏 𝑖𝑓 𝑄 𝑖𝑛 𝑚3 /ℎ𝑟, 𝒄 = 𝟔. 𝟔𝟕 𝑖𝑓 𝑄 𝑖𝑛 𝑚3 /𝑚𝑖𝑛 and 𝒄 = 𝟓𝟏. 𝟔𝟔 𝑖𝑓 𝑄 𝑖𝑛 𝑚3 /𝑠 If the factor 𝑐 is omitted, the calculated specific speeds will be different than the US customary values, for example, specific speed of Centrifugal Pump: 10 to 220 if Q is in m3/s The factor 𝑐 (might be omitted) is used to compare the value obtained with that obtained from US customary unit which are prevalent in usage among practicing professionals.



*For instance, in an application, if the required discharge and head are known, the prime mover (motor or engine) rpm is also known, then using these values 𝑁𝑠 can be calculated and a particular type of pump suitable in this 𝑁𝑠 range can be selected for installation.



Cavitation of Pumps and NPSH • • • •



In general, cavitation occurs when the liquid pressure at a given location is reduced to the vapor pressure of the liquid. For a piping system that includes a pump, cavitation occurs when the absolute pressure at the inlet falls below the vapor pressure of the water. This phenomenon may occur at the inlet to a pump and on the impeller blades, particularly if the pump is mounted above the level in the suction reservoir. Under this condition, vapor bubbles form (water starts to boil) at the impeller inlet and when these bubbles are carried into a zone of higher pressure, they collapse abruptly and hit the vanes of the impeller (near the tips of the impeller vanes). causing: ➢ ➢ ➢ ➢



•



Damage to the pump (pump impeller) Violent vibrations (and noise). Reduce pump capacity. Reduce pump efficiency



To avoid cavitation, the pressure head at the inlet should not fall below a certain minimum which is influenced by the further reduction in pressure within the pump impeller. The parameter used for the determination of cavitation is called ‘Net Positive Suction Head (NPSH)’.



Net Positive Suction Head (NPSH) NPSH is the difference between the total head at the pump inlet and the water vapor pressure head 𝐻𝑣 = 2.5 𝑚 𝑤𝑎𝑡𝑒𝑟 , i.e., 𝑵𝑷𝑺𝑯 = 𝑯𝒑𝒊 +



𝑽𝟐𝒔 𝟐𝒈



− 𝑯𝒗



the datum is taken through the centerline of the pump impeller inlet (eye). There are two values of NPSH of interest. The first is the required NPSH, denoted (NPSH)R , that must be maintained or exceeded so that cavitation will not occur and usually determined experimentally and provided by the manufacturer.



The second value for NPSH of concern is the available NPSH, denoted (NPSH)A , which represents the head that actually occurs for the particular piping system. This value can be determined experimentally, or calculated if the system parameters are known. For proper pump operation (no cavitation) :



(NPSH)A > (NPSH)R



NPSHavailable (at the installation site) > NPSHrequired (for pump) As stated above, NPSHrequired is usually given for a particular pump by the manufacturer for its installation without cavitation. NPSHavailable is calculated at the installation site.



Calculation of NPSH 𝑽𝟐𝒔 𝑵𝑷𝑺𝑯 = 𝑯𝒑𝒊 + − 𝑯𝒗 𝟐𝒈 Applying the Bernoulli’s equation between point (1) and (2), datum at pump center line 𝑉𝑠2 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝐿 = 𝐻𝑝𝑖 + 2𝑔 ∴ 𝑵𝑷𝑺𝑯𝑨 = 𝑯𝒂𝒕𝒎 − 𝑯𝒔 − 𝑯𝑳 − 𝑯𝒗



datum h𝐻s𝑠



𝐻𝑎𝑡𝑚 𝑯𝑳 = 𝑯𝒇 + 𝜮𝑯𝒎𝒊𝒏𝒐𝒓 𝒇𝒍𝒗𝟐 𝒗𝟐 = + 𝜮𝒌 𝟐𝒈𝒅 𝟐𝒈



A cavitation parameter called “Thoma’s cavitation number” is defined as: 𝑁𝑃𝑆𝐻𝐴 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝐿 − 𝐻𝑣 𝜎= = 𝐻𝑚 𝐻𝑚 A critical value 𝜎𝑐 (for no cavitation, 𝜎 ≥ 𝜎𝑐 )defined as 3% criteria, corresponds to critical value of NPSH. This critical value of NPSH is known as NPSHR. 𝑁𝑃𝑆𝐻𝑅 𝜎𝑐 = ⇒ 𝑁𝑃𝑆𝐻𝑅 = 𝜎𝑐 𝐻𝑚 𝐻𝑚 Thus, NPSHR is defined as the excess absolute head above 𝐻𝑣 , required by the pump to obtain satisfactory pumping head (i.e., no more than 3% reduction in head or efficiency at constant flow) and to prevent cavitation. It is determined by the pump manufacturer through tests.



Multiple-Pump Operation To install a pumping station that can be effectively operated over a large range of fluctuations in both discharge and pressure head, it may be advantageous to install several identical pumps at the station in parallel or in series operation. Pumps in Parallel: Pumping stations frequently contain several (two or more) pumps in a parallel arrangement. ▪ Any number of the pumps can be operated simultaneously. ▪ The discharge is increased but the pressure head remains the same as with a single pump. ▪ A common feature of sewage pumping stations where the inflow rate varies during the day.



Multiple-Pump Operation Pumps in Series: ▪ Increases the pressure head keeping the discharge approximately the same as that of a single pump. ▪ Basis of multistage pumps; the discharge from the first pump (or 1st stage) is delivered to the inlet of the second pump (or 2nd stage), and so on. ▪ The same discharge passes through each pump receiving a pressure boost in doing so.. ▪ Note that, however, all pumps in a series system must be operated simultaneously



Multi-stage Pump ▪



▪



▪



Similar to series arrangement of identical pumps; series of impellers mounted on a single compact shaft instead to save space. the discharge from the 1st stage impeller is delivered to the inlet of the 2nd stage impeller and so on. Usually even numbers of impellers are used, the inlets of one-half of the impellers are facing opposite to the inlets of the other half of the impellers to produce zero axial thrust on the shaft which is known as ‘opposed mounting’. Used when high heads are required but both impeller speed and size limitations are prohibitive.



Multi-stage pump with four stages ▪ ▪ ▪



The performance analysis is done per stage basis as for a single-stage pump. The specific speed is calculated based on manometric head per stage. The total manometric head is calculated by multiplying manometric head per stage with the number of stages.



▪ ▪



Applications include Boiler feed, deep-well pumping, water supply to very high-rise buildings, etc. Submersible pumps, Deep-well turbine pumps are some of the examples.



Multi-stage Centrifugal Pumps



Submersible Pump



Deep-well Turbine Pump



Other Types of Pumps



Propeller pump (axialflow/rotodynamic type) arranged in vertical operation



Screw pump (positive displacement type) - a screw pumps works across one pair of mating threads as shown in the figure. The liquid is similarly carried across all the pairs of mating threads.



Jet Pump - uses a jet, often of steam, to create a low pressure. This low pressure sucks in fluid and propels it into a higher pressure region. Liquid is trapped between threads at the suction end…



Selection of a Centrifugal Pump ▪



The system (demand) curve of a particular hydraulic system is calculated based on its design. It shows how much head is required to produce a certain quantity of discharge. point



Selection of a Centrifugal Pump



Selection of Pump Type The approximate ranges of application of each type of pumps are shown in the following Figure. ▪



The centrifugal pumps (radial flow) occupy a substantial part of the space.



▪



The regions of appropriate choice of jet pumps (mixed flow) and propeller pumps (axial flow) are also indicated by the thick border.



▪



The positive displacement pumps have a niche space in the high-head-lowdischarge category.



Selection of A Pump



Example Problems Problem–6: A centrifugal pump delivers 0.055 m3/s (𝑄) of water to a total height of 16 m (𝐻𝑠𝑡𝑎𝑡 ). The diameter of the pipe is 150 mm (𝑑) and it is 22 m (𝑙) long. If the overall efficiency is 75% (𝜂𝑜 ) , calculate the power (𝑷) required to drive the pump. Take 𝑓 = 0.05 for the pipe. Solution: Water velocity in the pipe, 𝑉



▪



𝑉2 𝐻𝑚 = 𝐻𝑠𝑡𝑎𝑡 + 𝐻𝑓 + 2𝑔 𝑓𝑙𝑉 2 𝑉2 ⇒ 𝐻𝑚 = 𝐻𝑠𝑡𝑎𝑡 + 2𝑔𝑑 + 2𝑔 0.05×22×3.112 ⇒ 𝐻𝑚 = 16 + 2×9.81×0.15



▪ 𝜂𝑜 =



𝛾𝑄𝐻𝑚 𝑃



⇒ 𝑷=



=



4𝑄 𝜋𝐷2



=



4×0.055 𝜋×0.152



3.112



+ 2×9.81 = 20.11 𝑚



𝛾𝑄𝐻𝑚 9810 × 0.055 × 20.11 = = 𝟏𝟒. 𝟒𝟕 𝒌𝑾 𝜂𝑜 0.75



= 3.11 𝑚/𝑠



Example Problems Problem–7: A centrifugal pump delivers 10 l/s (𝑄) of water at 1500 rpm (𝑁). The internal and external diameters of the impeller are 125 mm (𝐷1 ) and 250 mm (𝐷2 ) respectively. Width of the impeller at the inlet is 13 mm(𝑏1 ) and at the outlet 7 mm (𝑏2 ). Vanes are curved backward at an angle of 30o at the outlet (𝛽2 ). If water enters radially at the inlet (𝑉1 = 𝑉𝑓1 ), find the pressure rise in the impeller (𝚫𝒑/𝜸). Solution:



▪ 𝑄 = 𝜋𝐷1 𝑏1 𝑉𝑓1 ⇒ 10 × 10−3 = 𝜋 × 0.125 × 0.013 × 𝑉𝑓1 ⇒ 𝑉𝑓1 = 1.958 𝑚/𝑠



▪ 𝑄 = 𝜋𝐷2 𝑏2 𝑉𝑓2 ⇒ 10 × 10−3 = 𝜋 × 0.25 × 0.007 × 𝑉𝑓2 ⇒ 𝑉𝑓2 = 1.819 𝑚/𝑠



▪ 𝑢2 = ▪



𝜟𝒑 𝜸



𝜋𝐷2 𝑁 60



=



𝜋×0.25×1500 60



= 19.63 𝑚/𝑠



1



2 2 = 2𝑔 𝑉𝑓1 − 𝑉𝑓2 𝑐𝑜𝑠𝑒𝑐 2 𝛽2 + 𝑢22



𝜟𝒑 1 1.8192 2 ∴ = 1.958 − ° + 19.632 = 𝟏𝟗. 𝟏𝟔 𝒎 2 𝜸 2 × 9.81 𝑠𝑖𝑛 30



Example Problem Problem–8:



In US customary units, 𝑵𝒔 = 51.66 × 42.4 = 𝟐𝟏𝟗𝟎



Note: For radial flow pumps, 𝑁𝑠 = 500 𝑡𝑜 4000



Example Problem Problem–9: A centrifugal pump delivers 10 liter/s at 900 rpm against a head of 20 m. What head will be developed when the pump runs at 600 rpm? What will be the quantity of water delivered at that head?



Solution:



𝐻 𝑁 2 𝐷2 1



ς𝟐 =



=



𝐻 𝑁2 𝐷2 2



For the same pump, 𝑫𝟏 = 𝑫𝟐 = 𝑫



ς𝟐 = ς𝟑 =



𝐻 𝑁2 1



𝐻 𝑁2 2



=



𝑄 𝑁𝐷3 1



=



20 9002



⇒



𝐻



= 60022 ⇒



𝑯𝟐 = 𝟖. 𝟖𝟗 𝒎



𝑄 𝑁𝐷3 2



Again for the same pump, 𝑫𝟏 = 𝑫𝟐 = 𝑫



ς𝟑 =



𝑄 𝑁 1



𝑄 𝑁 2



=



⇒



10 900



𝑄



2 = 600



⇒



𝑸𝟐 = 𝟔. 𝟔𝟕 𝒍𝒊𝒕𝒆𝒓/𝒔



𝐀𝐥𝐭𝐞𝐫𝐧𝐚𝐭𝐢𝐯𝐞: 𝒖𝒔𝒊𝒏𝒈 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒔𝒑𝒆𝒆𝒅 𝑁√𝑄 𝐻 3/4 1



⇒



900×√10 203/4



=



=



𝑁√𝑄 𝐻 3/4 2



600×√(𝑄2 ) 8.893/4



= 𝑁𝑠 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑠𝑝𝑒𝑒𝑑



⇒ 𝑄2 =



900 2 600



×



8.89 3/2 20



= 6.67 𝑙𝑖𝑡𝑒𝑟/𝑠



Example Problems Problem–10: The speed of two geometrically similar centrifugal pumps is 1000 rpm. The outside diameter of the impeller of the first pump is 360 mm. It delivers 27 l/s of water against a head of 17 m. If the flow rate of second pump is half of the first pump, find the diameter of the impeller and head for the second pump. Solution: 𝑄 𝑁𝐷3 1



ς𝟑 =



𝑄 𝑁𝐷3 2



=



For the two pumps, 𝑵𝟏 = 𝑵𝟐 = 𝟏𝟎𝟎𝟎 rpm



ς𝟑 = Again, from



ς𝟐 = ς𝟐 =



𝑄 = 𝐷3 1 𝐻 𝑁 2 𝐷2 1 𝐻 𝐷2 1



=



𝑄 ⇒ 𝐷3 2 𝐻 = 𝑁2 𝐷2 2 𝐻 𝐷2 2



⇒



𝑄 0.363



17 .362



=



𝑄/2 𝐷3



⇒



𝐻



𝑫𝟐 = 𝟎. 𝟐𝟖𝟓𝟕 𝒎 = 𝟐𝟖𝟔 𝒎𝒎



2 = .2857 2 ⇒



𝑯𝟐 = 𝟏𝟎. 𝟕 𝒎



𝐀𝐥𝐭𝐞𝐫𝐧𝐚𝐭𝐢𝐯𝐞: 𝒖𝒔𝒊𝒏𝒈 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒔𝒑𝒆𝒆𝒅 𝑁√𝑄 𝐻 3/4 1



=



𝑁√𝑄 𝐻 3/4 2



= 𝑁𝑠 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑠𝑝𝑒𝑒𝑑 1000 27×10−3 = 1014 173/4 1000 27×10−3 /2 51.66 × 𝐻 3/4



For the first pump, 𝑁𝑠 = 51.66 × For the second pump, 1014 =



⇒ 𝐻2 = 10.7 𝑚



Example Problems Problem–11: The NPSHR of a centrifugal pump is given by the manufacturer as 7.5 m abs. The pump is employed to pump water at 0.3 m3/s from a sump whose water level is 2.05 m below the pump inlet. The atmospheric pressure at the site is 97 kPa abs and the vapor pressure at the relevant temperature is 2.35 kPa abs. Total head loss in the suction pipe is estimated to be 0.95 m. Determine the NPSHA and comment on the suitability of the installation against the cavitation. Solution: 𝑁𝑃𝑆𝐻𝐴 = 𝐻𝑎𝑡𝑚 − 𝐻𝑠 − 𝐻𝐿 − 𝐻𝑣 97 × 103 2.35 × 103 ⇒ 𝑁𝑃𝑆𝐻𝐴 = − 2.05 − 0.95 − = 6.65 𝑚 𝑎𝑏𝑠 9810 9810 Since 𝑵𝑷𝑺𝑯𝑨 (𝟔. 𝟔𝟓 𝒎) < 𝑵𝑷𝑺𝑯𝑹 (7.5 m), the pump will have cavitation problem. Problem–12: A pump has a critical cavitation constant = 0.12. This pump has to be installed in a well with a pipe of 10 m length and 200 mm diameter. There are an elbow (ke=1) and a valve (kv= 4.5) in the system. The flow is 0.035 m3/s and the total dynamic head Hm = 25 m. The atmospheric pressure is 9.7 m water abs and vapor pressure = 2.0 m water abs. Calculate the maximum suction height for the pump to run without cavitation. [f = 0.02] 𝑁𝑃𝑆𝐻𝑅 = 𝜎𝑐 𝐻𝑚 = 0.12 × 25 = 3 𝑚 = 𝑁𝑃𝑆𝐻𝐴|𝑚𝑖𝑛 Solution: 4𝑄 4 × 0.035 ⇒ 𝑉𝑠 = = = 1.11 𝑚/𝑠 𝜋𝑑2 𝜋 × 0.22 𝑓𝑙𝑉𝑠2 𝑉𝑠2 𝑉𝑠2 𝐻𝐿 = 𝐻𝑓 + 𝐻𝑚𝑖𝑛𝑜𝑟 = + 𝑘𝑒 + 𝑘𝑣 = 0.063 + 0.063 + 0.283 = 0.409 𝑚 2𝑔𝑑 2𝑔 2𝑔 𝑁𝑃𝑆𝐻𝐴|𝑚𝑖𝑛 = 𝐻𝑎𝑡𝑚 − 𝑯𝒔|𝒎𝒂𝒙 − 𝐻𝐿 − 𝐻𝑣 ⇒ 3 = 9.7 − 𝑯𝒔|𝒎𝒂𝒙 − 0.409 − 2.0 ∴ 𝑯𝒔|𝒎𝒂𝒙 = 𝟒. 𝟑 𝒎



Exercise Problems Problem–1: A centrifugal pump delivers 120 l/s of water against a head of 25 m while running at 1500 rpm. The outside diameter of impeller is 300 mm and the breadth of the impeller at exit is 50 mm. If the manometric efficiency is 80%, find the blade angle at outlet. Water enters the impeller radially at the inlet. [Ans. 13.6o] Problem–2: The outer and inner diameters of the impeller of a centrifugal pump are 500 mm and 245 mm respectively. The vanes are curved backward at an angle of 40o. The constant velocity of flow is 2.2 m/s at both inlet and outlet. The manometric head of the pump is 9.5 m. If the pump speed is 500 rpm, find its manometric efficiency and the vane angle at inlet. [Ans. 68%, 18.9o] Problem–3: A centrifugal pump delivers 0.20 m3/s of water to a head of 35 m at a speed of 1500 rpm. The outer diameter and width of the impeller at the outlet are 300 mm and 50 mm respectively. (i) If the manometric efficiency is 0.75, calculate the blade angle at the outlet, (ii) If the impeller diameter at the inlet is 150 mm, calculate the blade angle at inlet. [Ans. (i) 45.75o, (ii) 19.8o] Problem–4: A two-stage centrifugal pump delivers 110 l/s of water at 1200 rpm. The outer diameter and width of the impeller at the outlet are 450 mm and 25 mm respectively. The blades are curved backward at an angle of 30o at the outlet. Due to the thickness of the blades, 10% of the exit area is blocked. If the manometric efficiency is 86% and the overall efficiency is 80%, find the head developed by the pump and power required to drive the pump. [Ans. 55.22 m, 148.97 kW] [Hint: 𝑄 = 𝜋𝐷2 𝑏2 1 − 0.10 𝑉𝑓2 ] Problem–5: The outside and inside diameters of the impeller of a centrifugal pump are 550 mm and 275 mm respectively. The inlet vane angle is 30o and the outlet vane angle is 45o. The constant velocity of flow is 3 m/s. Find the speed of the pump, work done per unit of water and the pressure rise in the impeller. [Ans. 361.14 rpm, 7.85 Nm, 5.06 m water] Problem–6: Two geometrically similar pumps A and B have the same speed of 1500 rpm. Pump A has a diameter of 0.35 m and discharge of 36 L/s against a head of 25 m. Pump B gives a discharge of 18 L/s. Estimate the total head and impeller diameter of pump B. [Ans. 15.86 m, 278 mm]



Exercise Problems Problem–7: A centrifugal pump delivers 60 l/s of salt water (sp. gr. 1.2) or petrol (sp. gr. 0.71) against a pressure of 410 kPa. The overall efficiency of the pump is 66%. Prove that the same power is required to drive the pump for both the liquids. [Ans. 37.27 kW] Problem–8: The scale ratio of the model and prototype of a centrifugal pump is 1/4. The prototype delivers 1550 l/s of water at 550 mm against a head of 31 m and absorbs 750 kW. If the model works against a head of 11 m, find the speed, discharge and power required by the model. [Ans. 1310.5 rpm, 61.55 l/s, 9.91 kW] Problem–9: This is required to pump 1.3 m3/s of water to a total head of 45 m. How many pumps of specific speed 2070 (US Customary unit) and running at 1450 rpm would be needed when connected in parallel? The dynamic head in the system can be neglected. 1450× 𝑄



[Ans. 6] [Hints: 𝑁𝑠 = 2070 = 51.66 , ⇒ 𝑄 = 0.23 𝑚3 /𝑠(for one pump), 450.75 So Number of pumps required = 1.3 / 0.23 ≈ 6 pumps]



Problem–10: A discharge of 0.4 m3/s of water is needed to be pumped to a total head of 240 m. How many pumps connected in series and each having a specific speed of 1810 (US Customary unit) and running at a speed of 1500 rpm would be needed for the job? The dynamic head in the system can be neglected. [ Ans. 3] [Hints: 𝑁𝑠 = 1810 = 51.66



1500× 0.4 𝐻 0.75



, ⇒ 𝐻 = 81.3 𝑚 (for one pump),



So Number of pumps required = 240 / 81.3 ≈ 3 pumps]



Problem–11: A four-stage centrifugal pump delivering 0.76 m3/s of water at 1000 rpm against a manometric head of 66 m. Vanes are curved backward at an angle of 60o at the outlet. The ratio of the velocity of flow and the peripheral velocity at outlet is 0.25. If the hydraulic loss is 1/3 of the exit kinetic energy per unit weight of water, find the outside diameter of each impeller and the manometric efficiency. [Ans. 286 mm, 84.4%] Hints: Manometric head per stage, 𝐻𝑚 =



66



1 𝑉22



4



3 2𝑔



, Hydraulic loss, 𝐻ℎ𝑦𝑑𝑟𝑜_𝑙𝑜𝑠𝑠 =



= 36.99𝐷22



Exercise Problems Problem–12: A three-stage centrifugal pump delivers 65 l/s at 950 rpm. The outside diameter and outside width of each impeller are 380 mm and 26 mm respectively. The vanes are curved backward at an angle of 45o at the exit. If the manometric efficiency is 86%, find the manometric head developed by the pump. [Ans. 83.55 m water] [Hint: 𝐻𝑚 = 3𝐻𝑚│𝑝𝑒𝑟 𝑠𝑡𝑎𝑔𝑒 ] Problem–13: The external and internal diameters of an impeller of a centrifugal pump are 450 mm and 225 mm respectively. The pump delivers 200 l/s water at 1250 rpm. The outside and inside widths of the impeller are 70 mm and 150 mm. The vanes are curved backward at an angle of 30o at exit. If the manometric efficiency is 82%, find the Euler head and manometric head. [Ans. 77.90 m, 63.88 m]



Problem–14: A single stage centrifugal pump delivers 0.5 m3/s of water at 2000 rpm against a head of 32 m. The outside diameter of impeller is 250 mm. A geometrically similar multistage pump is required to deliver 0.75 m3/s of water at 1500 rpm against a head of 220 m. Find out the number of impellers of the multistage pump and the outside diameter of each of the impellers of the same pump. [Ans. 8, 700 mm] Problem–15: A centrifugal pump delivers 130 l/s of water at 1050 rpm. The outside diameter of impeller is 300 mm and it is 65 mm wide at exit. The blade angle at outlet is 30o. If the manometric efficiency is 86%, find the specific speed of the pump. [Ans. 60.39]



Problem–16: The scale ratio of the model and prototype of a centrifugal pump is 0.5. The outside diameter of the impeller of the model is 150 mm. The model supplies 0.045 m3/s of water at 7000 rpm against a head of 42 m. If the efficiency of the model and prototype is same, find the discharge, head and speed of the model. Find also the specific speed of the pump. [Ans. 0.09 m3/s, 10.5 m, 1750 rpm, 88.74 rpm]