Centrifugal pumps
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- Centrifugal pumps use impellers to impart velocity and pressure to fluids. Impellers can have single or multiple stages. - Affinity laws describe how flow rate, head, speed, and power relate as variables change. Centrifugal pump performance can be predicted using these scaling relationships. - Specific speed is used to classify pumps based on geometric similarity. Euler's and Bernoulli's equations relate velocities, pressures, and heads in centrifugal pumps. - Stepanoff's method uses empirical coefficients to design centrifugal pumps based on design parameters like flow rate, head, speed, and impeller dimensions. Blade thickness affects internal velocities.
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Centrifugal pumps
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- 5. Cross section of high speed water injection pump Source: www.framo.no
- 6. Water injection unit 4 MW Source: www.framo.no
- 7. Specific speed that is used to classify pumps nq is the specific speed for a unit machine that is geometric similar to a machine with the head Hq = 1 m and flow rate Q = 1 m3 /s 43q H Q nn ⋅= qs n55,51n ⋅=
- 9. Affinity laws 2 1 2 1 n n Q Q = 2 2 1 2 2 1 2 1 n n u u H H = = 3 2 1 2 1 n n P P = Assumptions: Geometrical similarity Velocity triangles are the same
- 10. Exercise sm1,11 1000 1100 Q n n Q 3 1 1 2 2 =⋅=⋅= m121100 1000 1100 H n n H 2 1 2 1 2 2 =⋅ =⋅ = kW164123 1000 1100 P n n P 3 1 3 1 2 2 =⋅ =⋅ = • Find the flow rate, head and power for a centrifugal pump that has increased its speed • Given data: ηh = 80 % P1 = 123 kW n1 = 1000 rpm H1 = 100 m n2 = 1100 rpm Q1 = 1 m3 /s
- 11. Exercise • Find the flow rate, head and power for a centrifugal pump impeller that has reduced its diameter • Given data: ηh = 80 % P1 = 123 kW D1 = 0,5 m H1 = 100 m D2 = 0,45 m Q1 = 1 m3 /s sm9,01 5,0 45,0 Q D D Q n n D D cBD cBD Q Q 3 1 1 2 2 2 1 2 1 2m22 1m11 2 1 =⋅=⋅= ⇓ == ⋅⋅⋅Π ⋅⋅⋅Π = m81100 5,0 45,0 H D D H 2 1 2 1 2 2 =⋅ =⋅ = kW90123 5,0 45,0 P D D P 3 1 3 1 2 2 =⋅ =⋅ =
- 14. Slip angle Reduced cu2 Slip angle Slip Best efficiency point Friction loss Impulse loss
- 15. Power ω⋅= MP Where: M = torque [Nm] ω = angular velocity [rad/s] ( ) ( ) t 1u12u2 111222 HgQ cucuQ coscrcoscrQP ⋅⋅⋅ρ= ω⋅⋅−⋅⋅⋅ρ= ω⋅α⋅⋅−α⋅⋅⋅⋅ρ=
- 16. g cucu H 1u12u2 t ⋅−⋅ = In order to get a better understanding of the different velocities that represent the head we rewrite the Euler’s pump equation 1u1 2 1 2 1111 2 1 2 1 2 1 cu2uccoscu2ucw ⋅⋅−+=α⋅⋅⋅−+= 2u2 2 2 2 2222 2 2 2 2 2 2 cu2uccoscu2ucw ⋅⋅−+=α⋅⋅⋅−+= g2 ww g2 cc g2 uu H 2 1 2 2 2 1 2 2 2 1 2 2 t ⋅ − − ⋅ − + ⋅ − =
- 17. Euler’s pump equation g cucu H 1u12u2 t ⋅−⋅ = g2 ww g2 cc g2 uu H 2 1 2 2 2 1 2 2 2 1 2 2 t ⋅ − − ⋅ − + ⋅ − = = ⋅ − g2 uu 2 1 2 2 Pressure head due to change of peripheral velocity = ⋅ − g2 cc 2 1 2 2 = ⋅ − g2 ww 2 1 2 2 Pressure head due to change of absolute velocity Pressure head due to change of relative velocity
- 18. Rothalpy Using the Bernoulli’s equation upstream and downstream a pump one can express the theoretical head: 1 2 2 2 t z g2 c g p z g2 c g p H + ⋅ + ⋅ρ − + ⋅ + ⋅ρ = g2 ww g2 cc g2 uu H 2 1 2 2 2 1 2 2 2 1 2 2 t ⋅ − − ⋅ − + ⋅ − = The theoretical head can also be expressed as: Setting these two expression for the theoretical head together we can rewrite the equation: g2 u g2 w g p g2 u g2 w g p 2 1 2 11 2 2 2 22 ⋅ − ⋅ + ⋅ρ = ⋅ − ⋅ + ⋅ρ
- 19. Rothalpy The rothalpy can be written as: ( ) ttancons g2 r g2 w g p I 22 = ⋅ ⋅ω − ⋅ + ⋅ρ = This equation is called the Bernoulli’s equation for incompressible flow in a rotating coordinate system, or the rothalpy equation.
- 20. Stepanoff We will show how a centrifugal pump is designed using Stepanoff’s empirical coefficients. Example: H = 100 m Q = 0,5 m3 /s n = 1000 rpm β2 = 22,5 o
- 21. 4,22 100 5,0 1000 H Q nn 4343q =⋅=⋅= 1153n55,51n qs =⋅= Specific speed: This is a radial pump
- 22. 0,1Ku = sm3,44Hg2Ku Hg2 u K u2 2 u =⋅⋅⋅=⇒ ⋅⋅ = srad7,104 60 n2 = ⋅Π⋅ =ω m85,0 2u D 2 D u 2 2 2 2 = ω ⋅ =⇒⋅ω= We choose: m17,0D5,0D 1hub =⋅=
- 23. 11,0K 2m = sm87,4Hg2Kc Hg2 c K 2m2m 2m 2m =⋅⋅⋅=⇒ ⋅⋅ = m038,0 cD Q d dD Q A Q c 2m2 2 22 2m = ⋅⋅Π = ⇓ ⋅⋅Π == u2 c2 w2 cu2 cm2
- 24. Thickness of the blade Until now, we have not considered the thickness of the blade. The meridonial velocity will change because of this thickness. ( ) ( ) m039,0 cszD Q d dszD Q A Q c 2mu2 2 2u2 2m = ⋅⋅−⋅Π = ⇓ ⋅⋅−⋅Π == We choose: s2 = 0,005 m z = 5 m013,0 5,22sin 005,0 sin s s o 2 2 u == β =
- 25. 145,0K 1m = sm4,6Hg2Kc Hg2 c K 1m1m 1m 1m =⋅⋅⋅=⇒ ⋅⋅ = u1 w1 c1= cm1
- 26. 405,0 D D 2 1 = m34,0D405,0D405,0 D D 21 2 1 =⋅=⇒= m09,0 cD Q d dD Q A Q c 1mm1 1 1m11 1m = ⋅⋅Π =⇒ ⋅⋅Π == We choose: Dhub m17,0D5,0D 1hub =⋅= m27,0 2 DD D 2 hub 2 1 m1 = + = Without thickness
- 27. Thickness of the blade at the inlet m015,0 8,19sin 005,0 sin s s o 1 1 1u == β = u1 w1 Cm1=6,4 m/s sm8,17 2 34,0 7,104 2 D u 1 1 =⋅=⋅ω= β1 o 1 1m 1 8,19 8,17 4,6 tana u c tana = = =β ( ) m10,0 cszD Q d 1m1um1 1 = ⋅⋅−⋅Π =
- 29. u2=44,3 m/s c2 w2 cm2=4,87m/s cu2 sm3,21 3,44 81,996,0100 u gH c g cu H 2 h 2u h 2u2 = ⋅⋅ = ⋅η⋅ =⇒ ⋅η ⋅ = o 2u2 2m 2 9,11 3,213,44 87,4 tana cu c tana' = − = − =β ooo 2slipslip 6,109,115,22' =−=β−β=β