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Air flow (eg in a duct) has a dynamic pressure, and when the air flow goes through e.g. a 90°-bend the bend has a dynamic-loss-coefficient.
The dynamic-loss-coefficient is used to calculate the total-pressure-loss by the formula:
Total Pressure loss = C * Velocity Pressure
(Velocity pressure is a different name for dynamic pressure. Both Total Pressure loss and velocity pressure are in Pascal, while dynamic-loss-coefficient C doesn't have a unit.)

For most bends/structures C is less than 1.
What I don't understand is what a C greater than 1 actually represents/means in the real world?
(Does it mean the air turns backwards?)

(Because if you subtract the total pressure loss from the dynamic pressure it would be negative.)

(Since this also has practical relevance there are several organizations empirically determining/collecting these values (e.g. ASHRAE and SMACNA), and in most cases C<1, but there are quite a few cases where C>1. E.g. an example of an fitting by SMACNA and an example by ASHRAE. (Edit: Heres an image of the definitions and formulas used by SMACNA.)
But the examples are just for illustration and don't matter in themselves, i'm just interested in what happens to the airflow when C>1.)

Regards

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1 Answer 1

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The $C$ coefficient simply quantifies the pressure loss of a restriction as a scale multiple or fraction of the dynamic pressure $\frac 1 2 \rho v^2$. But the dynamic pressure is usually much less than the total pressure of the flow.

For example, air flowing at 10 m/s at a static pressure of 10 atm has a dynamic pressure of

$$ \frac 1 2 \left( \frac{10}{1} 1.22 ~\rm{kg/m^3} \right)(10 ~ \rm{m/s})^2 = 610 ~\rm{Pa ~ or} ~0.006 ~\rm{atm} $$

which is far less than the static pressure of 10 atm. So your flow could lose many times the dynamic pressure going over restrictions and still have plenty of pressure left at the end. There is nothing special about the loss coefficient $C$ being greater than one.

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  • $\begingroup$ Thanks a lot for your answer. How is it determined(/is there a very rough approximation or some principle) how much of the lost total pressure is static-pressure and how much of it is dynamic pressure? (1/3) $\endgroup$
    – user84635
    Sep 27, 2023 at 1:09
  • $\begingroup$ Like say first you have just the duct and it ends straight. And then you have the same duct, but you add a bend/fitting as its end. There is now some total-pressure-reduction (as calculated by multiplying C times velocity-pressure). This total-pressure-reduction needs to be equal to the sum of static-pressure-change and dynamic-pressure-change, right? (2/3) $\endgroup$
    – user84635
    Sep 27, 2023 at 1:10
  • $\begingroup$ But how much of the total-pressure-reduction is "archived" by reducing the dynamic-pressure and how much by reducing/changing the static pressure? Is there a very rough rule? (Presumably with the same fan blowing the air through it, but now the bend adding some air-resistance the air is going at least a little slower through the same diameter, so there is likely at least some dynamic-pressure reduction.) (3/3) $\endgroup$
    – user84635
    Sep 27, 2023 at 1:10
  • $\begingroup$ The pressure losses are entirely in the static pressure, because the flow rate and therefore average velocity must be the same everywhere along the pipe, or it would violation continuity (mass conservation). In a constant area set of piping, if you had different avg velocity (=(flow rate)/area) at different stations of the pipe, that would mean either fluid is accumulating in an area, or voids are opening up, which cannot happen $\endgroup$
    – RC_23
    Sep 27, 2023 at 3:37
  • $\begingroup$ For a system with variable cross sectional area, it's a little different because there is trade off between static and dynamic pressure. E.g. if the diameter expands, the velocity (and thus dyn pressure) will decrease, while the static pressure correspondingly increases. But I would still basically consider losses as being "deducted from" the static pressure. Total amount of expected losses in the system are what determines the pressure your pump or compressor driving the flow must operate at to achieve the required flow $\endgroup$
    – RC_23
    Sep 27, 2023 at 3:42

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