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Affinity law on open type impeller 1
- Thread starter Wkcoo
- Start date
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1
- #2
Normally yes, provided impeller / volute / wear plate clearance is correct.
Bear in mind, even for enclosed impellers, trimming dia. is a prediction only - if an exact Q or H is required it is best to trim less and then check performance, if over performance trim a little more.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
Bear in mind, even for enclosed impellers, trimming dia. is a prediction only - if an exact Q or H is required it is best to trim less and then check performance, if over performance trim a little more.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
electricpete
Electrical
As a practical matter I don't know much about the limitations of the affinity laws.
As a theoretical matter, the affinity laws can be derived from dimensional analysis under the assumption of turbulent flow. I would think that applies in most cases except when flow through tight clearances (like recirculation thru wear rings and other tight clearances) is significant (this type of flow includes a lot of friction forces which violate the turbulent assumption)
Start with a general functional relationship: DP = f {Q, rho, D, N, visc, L1/L2, L1/L3 etc}
Choose rho, D, N as our “base” variables which will be used to non-dimensionalize the other variables:
[ul]
[li]D: m:[/li]
[li]N: 1/sec[/li]
[li]rho: kg/m^3[/li]
[/ul]
Manipulate DP on LHS to a dimensionless ratio of base variables
[ul]
[li]DP: (kg*m/sec^2)/m^2 = kg/(m*sec^2)[/li]
[li]DP/rho = m^2/sec^2 (same as g*H, but g is a physical constant which doesn't belong in dimensional analysis)[/li]
[li]DP/(rho*N^2*D^2)… unitless. This will be the dependent variable on LHS of the dimensionless equation.[/li]
[/ul]
Manipulate Q on RHS to a dimensionless ratio of base variables:
[ul]
[li]Q : m^3/sec[/li]
[li]Q/(D^3*N): unitless[/li]
[li]Q/(D^3*N) will be the indep variable on the RHS of the dimensionless equation[/li]
[/ul]
Two more assumptions:
[ul]
[li]Assumption 1 - Compare only geometrically similar pump designs so L1/L2, L1/L3 are constant (and of course dimensionless) => these can dimensionless constants can be dropped from the RHS as independent variables (they will be absorbed into the function definition).[/li]
[li]Assumption 2 - Assume turblent flow. The viscosity term would end up non-dimensionalizing to something like a Reynold’s number which will be a constant friction factor IF highly turbulent flow. As long as we have only turbulent with inertia forces much larger than viscuous forces, then we can neglect viscous forces and drop viscosity from the RHS as a variable (the dimensionless constant friction factor will be absorbed into the function definition)[/li]
[/ul]
Under these last two assumptions (similarity and turbulent flow), we can write the following functional relationship:
DP/(rho*N^2*D^2) = f{ Q/(D^3*N) }
which gives rise to the affinity laws
[ul]
[li]DP/rho~N^2, Q~N, and for constant rho Fluid power = DP*Q ~ N^3[/li]
[li]DP/rho~D^2, Q~D^3[/li]
[/ul]
=====================================
(2B)+(2B)' ?
As a theoretical matter, the affinity laws can be derived from dimensional analysis under the assumption of turbulent flow. I would think that applies in most cases except when flow through tight clearances (like recirculation thru wear rings and other tight clearances) is significant (this type of flow includes a lot of friction forces which violate the turbulent assumption)
Start with a general functional relationship: DP = f {Q, rho, D, N, visc, L1/L2, L1/L3 etc}
Choose rho, D, N as our “base” variables which will be used to non-dimensionalize the other variables:
[ul]
[li]D: m:[/li]
[li]N: 1/sec[/li]
[li]rho: kg/m^3[/li]
[/ul]
Manipulate DP on LHS to a dimensionless ratio of base variables
[ul]
[li]DP: (kg*m/sec^2)/m^2 = kg/(m*sec^2)[/li]
[li]DP/rho = m^2/sec^2 (same as g*H, but g is a physical constant which doesn't belong in dimensional analysis)[/li]
[li]DP/(rho*N^2*D^2)… unitless. This will be the dependent variable on LHS of the dimensionless equation.[/li]
[/ul]
Manipulate Q on RHS to a dimensionless ratio of base variables:
[ul]
[li]Q : m^3/sec[/li]
[li]Q/(D^3*N): unitless[/li]
[li]Q/(D^3*N) will be the indep variable on the RHS of the dimensionless equation[/li]
[/ul]
Two more assumptions:
[ul]
[li]Assumption 1 - Compare only geometrically similar pump designs so L1/L2, L1/L3 are constant (and of course dimensionless) => these can dimensionless constants can be dropped from the RHS as independent variables (they will be absorbed into the function definition).[/li]
[li]Assumption 2 - Assume turblent flow. The viscosity term would end up non-dimensionalizing to something like a Reynold’s number which will be a constant friction factor IF highly turbulent flow. As long as we have only turbulent with inertia forces much larger than viscuous forces, then we can neglect viscous forces and drop viscosity from the RHS as a variable (the dimensionless constant friction factor will be absorbed into the function definition)[/li]
[/ul]
Under these last two assumptions (similarity and turbulent flow), we can write the following functional relationship:
DP/(rho*N^2*D^2) = f{ Q/(D^3*N) }
which gives rise to the affinity laws
[ul]
[li]DP/rho~N^2, Q~N, and for constant rho Fluid power = DP*Q ~ N^3[/li]
[li]DP/rho~D^2, Q~D^3[/li]
[/ul]
=====================================
(2B)+(2B)' ?
goutam_freelance
Mechanical
To apply affinity laws to a series of pump models you need to compare apple to apple, in the sense that the models must be geometrically similar. So to be accurate you must compare one closed impeller model with another closed impeller model only, not with open impeller model as strictly speaking they are geometrically dissimilar. Similarly for open impeller models. The cross comparison may bring inaccuracies.
Engineers, think what we have done to the environment !
Engineers, think what we have done to the environment !
goutam freelance.
The OP was asking about the validity of affinity laws being applied to open or semi open impellers with the inference being are they as accurate as the results for closed impellers.
Yes they are.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
The OP was asking about the validity of affinity laws being applied to open or semi open impellers with the inference being are they as accurate as the results for closed impellers.
Yes they are.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
goutam_freelance
Mechanical
Yes it can be applied to any set of pump models provided they are geometrically similar and pumping similar fluids( e.g Newtonian/Non Newtonian).
Engineers, think what we have done to the environment !
Engineers, think what we have done to the environment !
electricpete:
Interesting discussion to arrive at:
1. Q varies at the ratio of diameter / speed
2. H varies as the ratio^2 of diameter / speed
3. P varies as the ratio^3 of diameter / speed.
With reasonable accuracy for minor changes in diameter /speed, for major changes of dia. or speed, the manufacturer should be consulted.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
Interesting discussion to arrive at:
1. Q varies at the ratio of diameter / speed
2. H varies as the ratio^2 of diameter / speed
3. P varies as the ratio^3 of diameter / speed.
With reasonable accuracy for minor changes in diameter /speed, for major changes of dia. or speed, the manufacturer should be consulted.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
goutam_freelance
Mechanical
@artisi,
Please edit your post as Q prop to dia x speed etc.
Engineers, think what we have done to the environment !
Please edit your post as Q prop to dia x speed etc.
Engineers, think what we have done to the environment !
goutam_freelance
Mechanical
OK
To be more general
1. Q proportional to (diameter x speed)
2. H proportional to (diameter x speed)^2
3. P proportional to (diameter x speed)^3
Engineers, think what we have done to the environment !
To be more general
1. Q proportional to (diameter x speed)
2. H proportional to (diameter x speed)^2
3. P proportional to (diameter x speed)^3
Engineers, think what we have done to the environment !
Sure if you want to change speed and diameter at the same time, but in 40+ years, it's never been a requirement.
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)
goutam_freelance
Mechanical
You have the flexibility to apply the law keeping either speed or diameter as constant. Changing both of them has been used by me for estimation and I believe used by pump designers too. Refer Pump Handbook by Karassik.
Engineers, think what we have done to the environment !
Engineers, think what we have done to the environment !
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