# radial impeller for centrifugal blower design

Question asked by MECH Design in #Coffee Room on Apr 12, 2009

MECH Design · Apr 12, 2009

Rank E2 - BEGINNER

i Want to design a centrifugal blower's radial impeller for 10000 cfm output at 200 mm water column pressure. i want to know the calculation for the output in relation to the blade width, inner and outer diameter and rpm. Thanks
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#Coffee Room

gohm · Apr 12, 2009

Rank A3 - PRO

What style of radial impeller? Blade lean, curved/flat? There are several forms of software and books that cover this area along with CFM.

To save a lot of typing, check out this page for some info:

International Water Power and Dam Construction

Ok, now for some typing 😀:

AFFINITY LAWS FOR CENTRIFUGAL PUMPS

The following laws have been established ;

FLOWRATE [Q] ¥ SPEED [N].

HEAD [H] ¥ SPEED SQUARED [N^2]

POWER [P] ¥ SPEED CUBED [N^3]

It follows from the affinity laws that the head and quantity can be altered by altering the speed or by altering the impeller diameter. Hence Q1 X N2 = N1 X Q2 or

Q1 X D2 = D1 X Q2 or N1 X D2 = D1 X N2

also , H1 X [N2]^2 = H2 X [N1]^2

and H1 x [D2]^2 = H2 X [ D1]^2

OR H1 X [ Q2]^2 = H2 X [Q1]^2

However care must be taken in the practical use of the laws where the impeller eye diameter cannot be overlooked.

If an impeller is cut or its speed changed the efficiency alters.

Efficiency change with speed

1 - E / 1 - e = [n / N ]^ 0.2

Where n and e = new speed and new efficiency and N and E are the original speed and efficiency values. For approximation purposes the efficiency change could be determined in this manner. Reduce the efficiency of the test curve known point by half of the percentage reduction in diameter. Thus 12 % reduction in diameter causes approximately 6 % reduction in efficiency value .

The flow through the impeller [rotor] can be expressed as follows

p2/r + 0.5 [v2]2 = p1/r + 0.5 [v1]2 + Wi - Li

where Wi = work input per unit mass flow

Li = mechanical energy loss per unit mass flow due to friction and eddying.

u1 and u2 are the tangential velocities

v1 and v2 are the absolute inlet and exit impeller velocities each of which can be split into two velocity components [whirl[w] and flow [f]].

[vr] = relative velocity of the fluid.

[Wi] = [w2u2-w1u1] = work input per unit mass flow

hence [p2-p1] /r = [w2u2 - 0.5[v2]2] - [[w1u1 - 0.5[v1]2] – [Li]

this equation can be expressed in terms of [f2]

thus,v2 .v2 = w2.w2 + f2.f2 = w2.w2 + vr2.vr2 - [u2- w2].[u2 -w2]

= vr2.vr2 - u2.u2 + 2w2u2

therefore, w2u2-[0.5[v2.]2] = 0.5[[u2]2-[vr2]2]

hence, [p2-p1]/r = 0.5[u2.u2-u1.u1] - 0.5[vr2.vr2-vr1.vr1] – [Li]

if there is no flow AS WHEN the delivery valve is closed then

[p2-p1]/r = 0.5W 2[ [r2]2-[r1]2 ]

where W = angular velocity = [ur]

THIS is the usual expression for a forced vortex.

The flow through the diffuser or volute casing chamber is expressed as follows

[p3/r ] +0.5[v3]2 = [p2/r ] +0.5[v2]2 – [Ld]

where suffix 3 refers to the outlet from the diffuser or volute and [Li] is the mechanical energy loss from the diffuser per unit mass flow.

The total mechanical energy equation may then be expressed as

[p3/r ] + 0.5[v3.v3] = [p1/r ] + 0.5 [v1.v1] + [Wi]- [Li+Ld]

THE ACTUAL GAIN OF HEAD = H

= [ [p3/r g] + 0.5[v3.v3/g]] - [ [p1/r g] + 0.5 [v1.v1/2g]

= [ Hi] – [H]

where [Hi] = theoretical gain of total head = [w2u2-w1u1]/g

if there is no pre-rotation then Hi = w2u2/g as w1 = 0

and hi = total internal loss of head due to friction and eddying

and = [Li + Ld]/g

The hydraulic efficiency = H / Hi or gw / Wi

the mechanical efficiency = Wi / work input to pump

To save a lot of typing, check out this page for some info:

International Water Power and Dam Construction

Ok, now for some typing 😀:

AFFINITY LAWS FOR CENTRIFUGAL PUMPS

The following laws have been established ;

FLOWRATE [Q] ¥ SPEED [N].

HEAD [H] ¥ SPEED SQUARED [N^2]

POWER [P] ¥ SPEED CUBED [N^3]

It follows from the affinity laws that the head and quantity can be altered by altering the speed or by altering the impeller diameter. Hence Q1 X N2 = N1 X Q2 or

Q1 X D2 = D1 X Q2 or N1 X D2 = D1 X N2

also , H1 X [N2]^2 = H2 X [N1]^2

and H1 x [D2]^2 = H2 X [ D1]^2

OR H1 X [ Q2]^2 = H2 X [Q1]^2

However care must be taken in the practical use of the laws where the impeller eye diameter cannot be overlooked.

If an impeller is cut or its speed changed the efficiency alters.

Efficiency change with speed

1 - E / 1 - e = [n / N ]^ 0.2

Where n and e = new speed and new efficiency and N and E are the original speed and efficiency values. For approximation purposes the efficiency change could be determined in this manner. Reduce the efficiency of the test curve known point by half of the percentage reduction in diameter. Thus 12 % reduction in diameter causes approximately 6 % reduction in efficiency value .

*THE ENERGY EQUATION FOR A CENTRIFUGAL PUMP*The flow through the impeller [rotor] can be expressed as follows

p2/r + 0.5 [v2]2 = p1/r + 0.5 [v1]2 + Wi - Li

where Wi = work input per unit mass flow

Li = mechanical energy loss per unit mass flow due to friction and eddying.

u1 and u2 are the tangential velocities

v1 and v2 are the absolute inlet and exit impeller velocities each of which can be split into two velocity components [whirl[w] and flow [f]].

[vr] = relative velocity of the fluid.

[Wi] = [w2u2-w1u1] = work input per unit mass flow

hence [p2-p1] /r = [w2u2 - 0.5[v2]2] - [[w1u1 - 0.5[v1]2] – [Li]

this equation can be expressed in terms of [f2]

thus,v2 .v2 = w2.w2 + f2.f2 = w2.w2 + vr2.vr2 - [u2- w2].[u2 -w2]

= vr2.vr2 - u2.u2 + 2w2u2

therefore, w2u2-[0.5[v2.]2] = 0.5[[u2]2-[vr2]2]

hence, [p2-p1]/r = 0.5[u2.u2-u1.u1] - 0.5[vr2.vr2-vr1.vr1] – [Li]

if there is no flow AS WHEN the delivery valve is closed then

[p2-p1]/r = 0.5W 2[ [r2]2-[r1]2 ]

where W = angular velocity = [ur]

THIS is the usual expression for a forced vortex.

The flow through the diffuser or volute casing chamber is expressed as follows

[p3/r ] +0.5[v3]2 = [p2/r ] +0.5[v2]2 – [Ld]

where suffix 3 refers to the outlet from the diffuser or volute and [Li] is the mechanical energy loss from the diffuser per unit mass flow.

The total mechanical energy equation may then be expressed as

[p3/r ] + 0.5[v3.v3] = [p1/r ] + 0.5 [v1.v1] + [Wi]- [Li+Ld]

THE ACTUAL GAIN OF HEAD = H

= [ [p3/r g] + 0.5[v3.v3/g]] - [ [p1/r g] + 0.5 [v1.v1/2g]

= [ Hi] – [H]

where [Hi] = theoretical gain of total head = [w2u2-w1u1]/g

if there is no pre-rotation then Hi = w2u2/g as w1 = 0

and hi = total internal loss of head due to friction and eddying

and = [Li + Ld]/g

The hydraulic efficiency = H / Hi or gw / Wi

the mechanical efficiency = Wi / work input to pump

MECH Design · Apr 13, 2009

Rank E2 - BEGINNER

. It is a flat impeller bladeMECH Designi Want to design a centrifugal Air blower's radial impeller for 10000 cfm output at 200 mm water column pressure. i want to know the calculation for the output in relation to the blade width, inner and outer diameter and rpm. Thanks

urvi patel · Sep 4, 2009

Rank E1 - BEGINNER

give me equetion & calculation formulaMECH Designi Want to design a centrifugal blower's radial impeller for 10000 cfm output at 200 mm water column pressure. i want to know the calculation for the output in relation to the blade width, inner and outer diameter and rpm. Thanks

urvi patel · Sep 4, 2009

Rank E1 - BEGINNER

i want centrifugal blower design for air system

urvi patel · Sep 4, 2009

Rank E1 - BEGINNER

calculation formula for cfm & pressure calculationurvi pateli want centrifugal blower design for air system

design of impeller & housing of air blower

urvi patel · Sep 4, 2009

Rank E1 - BEGINNER

calculation of cfm & pressureurvi patelcalculation formula for cfm & pressure calculation

design of impeller & housing of air blower

impeller diameter

width

blade size

blade quantity

kuppuswamy anantharaman · May 1, 2016

Rank E1 - BEGINNER

good one.