A question of efficiency

[rfellows00@gmail.com]

I been reading here for a few weeks. All else held constant and
comparing apples to apples...If your goal is to have maximum wind speed
over a controlled cubic area within a cylinder and your biggest
constraint is cost of electricity consumed, would you "push" the air
through the controlled area from a blower at the entrance of the
cylinder OR "pull" the air through from a blower at the exit.
Thank you in advance. Ronnie

[Richard Saam]

rfellows00@gmail.com wrote:
> I been reading here for a few weeks. All else held constant and
> comparing apples to apples...If your goal is to have maximum wind speed
> over a controlled cubic area within a cylinder and your biggest
> constraint is cost of electricity consumed, would you "push" the air
> through the controlled area from a blower at the entrance of the
> cylinder OR "pull" the air through from a blower at the exit.
> Thank you in advance. Ronnie
>
Due to air compressibility effects

- Pull condition
maximum wind speed in a controlled cubic area within a cylinder

- Push condition
maximum wind power in a controlled cubic area within a cylinder

Richard

[Uncle Al]

rfellows00@gmail.com wrote:
>
> I been reading here for a few weeks. All else held constant and
> comparing apples to apples...If your goal is to have maximum wind speed
> over a controlled cubic area within a cylinder and your biggest
> constraint is cost of electricity consumed, would you "push" the air
> through the controlled area from a blower at the entrance of the
> cylinder OR "pull" the air through from a blower at the exit.
> Thank you in advance. Ronnie

There is no limit to how hard you can push. You only get 14.7 psi of
pull. How does efficiency vary with pressure in either direction for
a given modality of transport?

If you push you can have a laminar flow nozzle.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz3.pdf

[Cl.Massé]

<rfellows00@gmail.com> a écrit dans le message de news:
1154456389.881776.123120@s13g2000cwa.googlegroups. com

> I been reading here for a few weeks. All else held constant and
> comparing apples to apples...If your goal is to have maximum wind speed
> over a controlled cubic area within a cylinder and your biggest
> constraint is cost of electricity consumed, would you "push" the air
> through the controlled area from a blower at the entrance of the
> cylinder OR "pull" the air through from a blower at the exit.
> Thank you in advance. Ronnie

Both. You're welcome.

--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.

[Richard Saam]

rfellows00@gmail.com wrote:

> I been reading here for a few weeks. All else held constant and
> comparing apples to apples...If your goal is to have maximum wind speed
> over a controlled cubic area within a cylinder and your biggest
> constraint is cost of electricity consumed, would you "push" the air
> through the controlled area from a blower at the entrance of the
> cylinder OR "pull" the air through from a blower at the exit.
> Thank you in advance. Ronnie
>
I thought there would be more discussion on this post:

Analysis:

System A 'pulling'

P1 P3 P1
---------------------------------
/\ ^ <--
/ \ | <--
\ / | <--
|| |D QA = Q <-- vA
/ \ | <--
\ / | <--
\/ v <--
---------------------------------
<- L ->



System B 'pushing'

P1 P3 P1
---------------------------------
^ <-- /\
| <-- / \
| <-- \ /
QB = Q |D <-- vB ||
| <-- / \
| <-- \ /
v <-- \/
---------------------------------
<- L ->


Governing equations:

1. vA is average velocity in System A 'pulling' volume (V)

2. vB is average velocity in System B 'pushing' volume (V)

2. P2 - P1 = rhoB (L / D) vB^2 / 2

3. P1 - P3 = rhoA (L / D) vA^2 / 2

4. V = (pi / 4) D^2 L (V is Volume where VA = VB)

5. P V = n R T (air compressiblity
or expandibility)
or

P = (n / V) R T = (rho / M) R T

6. Power = P Q (Q is flowrate and QA = QB)

Therefore:

rho = (P1 + P2)/2 M / (R T)

and

P2 - P1 = (P1 + P2)/2 M / (R T) (L / D) vB^2 / 2
= (P1 + P2) k vB^2
and

P1 - P3 = (P1 + P3)/2 M / (R T) (L / D) vB^2 / 2
= (P1 + P3) k vA^2

rearrange:

k vB^2 = (P2 - P1)/ (P1 + P2)

k vA^2 = (P1 - P3)/ (P1 + P3)

As defined:

P2 > P1 > P3

let for example:

k = 1 P1 = 10 P2 = 11 P3 = 9

then:

vB^2 = (11 - 10)/ (10 + 11) = 1/21

vA^2 = (10 - 9)/ (10 + 9) = 1/19

therefore:

vA > vB

Due to air expandibility
Condition A 'pulling' results in faster velocity vA
than air compressiblity
Condition B 'pushing' resulting in slower velocity vB

The power is the same for A and B
Power = (P2 - P1) Q = (P1 - P3) Q

but 'power / mass' in volume 'V'
is just the opposite of 'power / velocity'.

power/(rhoA V) > power/(rhoB V)

Richard Saam