CE 6403 APPLIED HYDRAULIC ENGINEERING UNIT - V PUMPS Centrifugal pups - Miniu speed to start the pup - NPSH - Cavitations in pups Operating characteristics - Multistage pups - Reciprocating pups - Negative slip - Flow separation conditions - Air vessels, indicator diagras and its variations - Savings in work done Rotary pups: Gear pup. BY. MR.KUMAR.G M.E. STRUCTURAL ENGINEERING ASSISTANT PROFESSOR DEPARTMENT OF CIVIL ENGINEERING SRI VENKATESWARA COLLEGE OF ENGINEERING SRIPERUMBUDUR 60 117
CENTRIFUGAL PUMP
1.The ipeller of a centrifugal pup has external and internal diaeters of 500 and 50 respectively. Width of outlet 50 and running of 100 rp. It works against a head of 48. The velocity of flow through the ipeller is constant and equal to 3 /sec. The vanes are set back at an angle of 40 0 at outlet. Deterine (i) Inlet vane angle, (ii) Work done by the ipeller on water per second and (iii) Manoetric efficiency. Given data External Diaeter ( D ) 500 0.50 Internal Diaeter ( D 1 ) 50 0.5 Width of outlet (B ) 50 0.05 Speed ( N ) 100 rp Head ( H ) 48 Velocity of flow V f1 V f 3 / sec Vane angle of outlet ( ϕ ) 40 0 (i) Find out Inlet vane angle ( ϴ )? (ii) Find out Work done by the ipeller on water per second? (iii) Manoetric efficiency ( ƞ an )?
Solution Tangential velocity at inlet u 1 ( π x D 1 x N ) / 60 u 1 ( π x 0.5 x 100 ) / 60 u 1 15.708 / sec Fro inlet Velocity Triangle tanθ ( V f1 / u 1 ) ( 3 / 15.708 ) tanθ 0.191 ( i ) Inlet vane angle ( ϴ ) tan -1 ( 0.191 ) 10.813 ϴ 10 0 48 Discharge ( ) π x D x B x V f ( ) π x 0.50 x 0.05 x 3 ( ) 0.36 ³ / sec Tangential velocity at outlet u ( π x D x N ) / 60 u ( π x 0.50 x 100 ) / 60 u 31.416 / sec
Fro outlet Velocity Triangle tan ϕ { V f / ( u - V w ) } tan 40 0 { 3 / (31.416 - V w ) } V w (31.416 3.575) V w 7.841 /sec ( ii ) Work done by the Ipeller per second W g V w u ρ g g V w u 1000 9.81 0.36 9.81 7.841 06418.074 N 31.416 / sec
( iii ) Manoetric Efficiency η an g V w H u η an 7 9.835. 81 48. 416 31 η an 0. 53849 η an 53. 849 % Answer * Inlet vane angle ( ϴ ) 10 0 48 *Work done by the Ipeller per second 06418.074N / sec η *Manoetric Efficiency 53. 849 % an
.The ipeller of a centrifugal pup has an external diaeter of 450 and internal diaeter of 00 and it runs at of 1440 rp. Assuing a constant radial flow through the ipeller of.5 /sec and that the vanes at exit are set back at an angle of 5 0. Deterine (i) Inlet vane angle (ii) The angle, absolute velocity of water at exit akes with the tangent and (iii) The work done per N of water. Given data External Diaeter ( D ) 450 0.45 Internal Diaeter ( D 1 ) 00 0.0 Speed ( N ) 1440 rp Velocity of flow V f1 V f.5 / sec Vane angle of outlet ( ϕ ) 5 0 (i) Find out Inlet vane angle ( ϴ )? (ii) The angle, absolute velocity of water at exit akes with the tangent? (iii) The work done per N of water? Solution Tangential velocity at inlet u 1 ( π x D 1 x N ) / 60 u 1 ( π x 0.0 x 1440 ) / 60 u 1 15.080 / sec
Fro inlet Velocity Triangle tanθ ( V f1 / u 1 ) (.5 / 15.080 ) tanθ 0.166 ( i ) Inlet vane angle ( ϴ ) tan -1 ( 0.166 ) 9.45 ϴ 9 0 5 Tangential velocity at outlet u ( π x D x N ) / 60 u ( π x 0.45 x 1440 ) / 60 u 33.99 / sec Fro outlet Velocity Triangle tan ϕ { V f / ( u - V w ) } tan 5 0 {.5 / (33.99 - V w ) } V w ( 33.99 5.361) V w 8.568 /sec
( ii ) tan ᵦ ( V f / V w ) (.5 / 8.568 ) ( 0.088 ) ᵦ tan -1 ( 0.088 ) ᵦ 5.09 ᵦ 5 0 1 Absolute velocity [ ] [ ] V V + V V w f [ 8.568 ] + [. ] 5 V 8.677 ( iii ) Work done by the Ipeller per N of water / sec 9 1 V g w 1 8. 568 33. 99. 81 98. 806 N u
Answer * Inlet vane angle ( ϴ ) 9 0 5 * Angle ᵦ 5 0 1 * Absolute velocity V 8.677 / sec *Work done by the Ipeller per N of water 98. 806 N 3. A centrifugal pup delivers water against a net head of 14.5 etres and a design speed of 1000 rp. The vanes are curved back to an angle of 30 0 with the periphery. The ipeller diaeter is 300 and outlet width 50. Deterine the discharge of the pup. If Manoetric efficiency is 95 %. Given data Net Head ( H ) 14.5 Speed ( N ) 1000 rp Vane angle of outlet ( ϕ ) 30 0 Ipeller diaeter eans the diaeter of the ipeller at outlet External Diaeter ( D ) 300 0.30 Width of outlet (B ) 50 0.05 Manoetric efficiency η 95 % 0. 95 an
Solution Tangential velocity at outlet u ( π x D x N ) / 60 u ( π x 0.30 x 1000 ) / 60 u 15.708 / sec Manoetric Efficiency η an g V w H u V w g η an H u V w 9. 81 14. 5 0. 95 15. 708 V w 9. 53 / sec Fro outlet Velocity Triangle tan ϕ { V f / ( u - V w ) } tan 30 0 { V f / (15.708 9.53 ) } V f tan 30 0 (15.708 9.53) V f 3.566 /sec
Discharge ( ) π x D x B x V f ( ) π x 0.30 x 0.05 x 3.566 ( ) 0.168 ³ / sec Answer Discharge ( ) 0.168 ³ / sec 4. A centrifugal pup having outer diaeter equal to two ties the inner diaeter and running at 100 rp works against a total head of 3. The velocity of flow through the ipeller is constant and equal to 3 / sec. The vanes are set back at an angle of 30 0 at the outlet. If the outer diaeter of the ipeller is 600 and width of outlet is 50. Deterine (i) Vane angle at inlet, (ii) Work done per second by ipeller and (iii) Manoetric efficiency. Given data Speed ( N ) 100 rp Total Head ( H ) 3 Velocity of flow V f1 V f 3 / sec Vane angle of outlet ( ϕ ) 30 0 External Diaeter ( D ) 600 0.60 Internal Diaeter ( D 1 ) 600 / 0.30 Width of outlet (B ) 50 0.05
(i) Find out Vane angle at inlet? (ii) Find out Work done per second by ipeller? (iii) Find out Manoetric efficiency? Solution Tangential velocity at inlet u 1 ( π x D 1 x N ) / 60 u 1 ( π x 0.30 x 100 ) / 60 u 1 18.850 / sec Fro inlet Velocity Triangle tanθ ( V f1 / u 1 ) ( 3 / 18.850 ) tanθ 0.159 ( i ) Inlet vane angle ( ϴ ) tan -1 ( 0.159 ) 9.034 ϴ 9 0 Discharge ( ) π x D x B x V f ( ) π x 0.60 x 0.05 x 3 ( ) 0.83 ³ / sec
Tangential velocity at outlet u ( π x D x N ) / 60 u ( π x 0.60 x 100 ) / 60 u 37.699 / sec Fro outlet Velocity Triangle tan ϕ { V f / ( u - V w ) } tan 30 0 { 3 / ( 37.699 - V w ) } V w ( 37.699 5.196 ) V w 3.503 /sec ( ii ) Work done by the Ipeller per second W g V w u ρ g g V w u 1000 9.81 9.81 0.83 3.503 37.669 346.493 10 3 N / sec
( iii ) Manoetric Efficiency η an g V w H u η an 3 9. 503. 81 3 37. 669 η an 0. 5640 η an 5. 640 % Answer * Inlet vane angle ( ϴ ) 9 0 *Work done by the Ipeller per second 346.493 10 3 N / sec *Manoetric Efficiency η an 5. 640 %
5. Find the power required to drive a centrifugal pup which delivers 0.04 ³ / sec of water to a height of 0 through a 15 c diaeter pipe and 100 long. The overall efficiency of the pup is 70% and co-efficient of friction f 0.15 in the forula h 4 flv f d Given data Discharge ( ) 0.04 ³ / sec Height ( H s ) h s + h d 0 Diaeter of pipe D s D d 15 c 0.15 Length L s + L d L 100 Overall efficiency ( ƞ o ) 70% 0.70 Co efficient of friction ( f ) 0.15 Find out Power? Solution g Velocity of water in pipe, V s V d V ( Discharge / Area of pipe ) (0.04 / ( π x 0.15² / 4 )) V.64 / sec
Frictional head loss in pipe, h f s + h f d 4 d flv g h f + h s f d 4 0.15 100.64 0.15 9.81 h f s + h f d 307.. 54 943 h f + h f 104. 499 s d Manoetric head as H [ h + h ] + [ h + h ] s d f s f d + V d g H [ 0 ] + [ 104. 499 ] +. 64 9. 81 H 0 + 104.499 + H 14. 760 0.61
Shaft Power is the power required to drive the centrifugal pup η o W 1000 H S. P S. P W 1000 H η o S. P ρ g 1000 H η o S. P 1000 9.81 0.04 14 1000 0.70.760 S. P 69. 937 kw Answer * Shaft Power S. P 69. 937 kw
6. A single acting reciprocating pup, running at 60 rp delivers 0.53 ³ of water per inute. The diaeter of the piston is 00 and stroke length 300. The suction and delivery heads are 4 and 1 respectively. Deterine ( i ) Theoretical discharge ( ii ) Co efficient of discharge ( iii ) Percentage slip of the pup, and ( iv ) Power required to run the pup. Given data Speed of the pup ( N ) 60 rp Actual discharge ( act ) 0.53 ³ / inute ( 0.53 / 60 ) 8.833 x 10-3 ³ / sec Diaeter of the piston ( D ) 00 0. Stroke length ( L ) 300 0.3 Suction head ( h s ) 4 Delivery head ( h d ) 1 Solution Area (π D² / 4) (π x 0.² ) / 4 0.031416 ² ( i ) Theoretical discharge A L the 60 N the 0.031416 0.3 60 60
the 9.45 10 3 3 / sec ( ii ) Co efficient of discharge cd act the cd cd 8.833 9.45 10 10 0.937 3 3 ( iii ) Percentage slip of the pup Percentage Slip the the act 100 9.45 3 10 8.833 3 9.45 10 10 3 100 Percentage Slip 6.81 %
( iv ) Power required to run the pup W A L N P h s + h d 60 [ ] ρ g A L N P h s + h d 1000 60 [ ] 1000 9.81 0.031416 P 1000 60 P 1. 479 kw 0.3 60 [ 4+ 1] Answer ( i ) Theoretical discharge ( ii ) Co efficient of discharge ( iii ) Percentage slip of the pup the 9.45 10 3 3 / sec cd 0.937 Percentage Slip 6.81 % ( iv ) Power required to run the pup P 1. 479 kw
7. A single acting reciprocating pup, running at 50 rp delivers 0.01 ³ / sec of water. The diaeter of the piston is 00 and stroke length 400. Deterine the theoretical discharge of the pup, Co efficient of discharge and Slip and the Percentage slip of the pup. Given data Speed of the pup ( N ) 50 rp Actual discharge ( act ) 0.01 ³ / sec Diaeter of the piston ( D ) 00 0. Stroke length ( L ) 400 0.4 Solution Area (π D² / 4) (π x 0.² ) / 4 0.031416 ² ( i ) Theoretical discharge A L the 60 N the the 0.031416 0.4 60 0.01047 3 / 50 sec
( ii ) Co efficient of discharge cd act the ( iii ) Slip Slip Slip cd cd 0 the 0.01047 0. 01. 01047 0.955 act 0.01 Slip 0.00047 3 / sec ( iv ) Percentage slip of the pup Percentage Slip the the act 100 0. 01047 0 0. 01047 Percentage Slip. 01 100 4.489 %
Answer ( i ) Theoretical discharge ( ii ) Co efficient of discharge the cd 0.01047 0.955 3 / sec ( iii ) Slip Slip 0.00047 3 / sec ( iv ) Percentage slip of the pup Percentage Slip 4.489 % 8. A single acting reciprocating pup discharges 4.5 litres per second with cylinder bore diaeter 00 and its stroke length 300. The pup runs at 350 rp and lifts water through a height of 5. The delivery pipe is 30 long and 100 in diaeter. Find the theoretical discharge and the theoretical power required to run the pup. Also deterine the percentage slip. Given data Actual discharge ( act ) 4.5 litres / sec 4.5 x 10-3 3 / sec Diaeter of cylinder ( D ) 00 0. Stroke length ( L ) 300 0.3 Speed of the pup ( N ) 350 rp height ( H ) 5
Length of delivery pipe ( l d ) 30 Suction head ( h s ) 4 Diaeter of delivery pipe ( d d ) 100 0.1 Solution Area (π D² / 4) (π x 0.² ) / 4 0.031416 ² ( i ) Theoretical discharge A L N the 60 the 0.031416 0.3 350 60 the 0. 055 3 / sec ( ii ) Theoretical Power P t ρ g 1000 the H P t 1000 9.81 0.055 5 1000 P t 13. 489 kw
( iii ) Percentage slip of the pup Percentage Slip the the act 100 0. 055 4. 5 0. 055 Percentage Slip 10 3 100 91.818% Answer ( i ) Theoretical discharge ( ii ) Theoretical Power ( iv ) Percentage slip of the pup the 0. 055 P t 13. 489 kw 3 / sec Percentage Slip 91.818 % 9. A double acting reciprocating pup running at 40 rp is discharging 1.0 ³ of water per inute. The pup has a stroke of 400. The diaeter of the piston is 00. The delivery and suction head are 0 and 5 respectively. Find the slip of the pup and power required to drive the pup.
Given data Speed of the pup ( N ) 40 rp Actual discharge ( act ) 1.0 3 / inute 1.0 / 60 0.01666 3 / sec Stroke length ( L ) 400 0.4 Diaeter of the piston ( D ) 00 0. Delivery head ( h d ) 0 Suction head ( h s ) 5 Solution Area (π D² / 4) (π x 0.² ) / 4 0.031416 ² ( i ) Theoretical discharge for double acting pup A L N the 60 the 0.031416 0.4 40 60 the 0.01675 3 / sec ( ii ) Slip Slip Slip Slip the 0.01675 0.00009 act 0.01666 3 / sec
( iii ) Power required to drive the double - acting pup W A L N P s + 60 [ h h ] ρ g A L N P h s + h d 1000 60 d [ ] 1000 9.81 0.031416 P 1000 60 P 4. 109 kw 0.4 40 [ 5+ 0] Answer ( i ) Theoretical discharge for double acting pup the 0.01675 3 / sec ( ii ) Slip Slip 0.00009 3 / sec ( iii ) Power required to drive the double - acting pup P 4. 109 kw