Effect of Speed of Revolution

In the previous sections, the centrifugal and axial fans were investigated for a constant rotational speed.

In this section, the rotational velocity is directly proportional to the rota­tional velocity n according to the equation u = irDn. The impeller blade an­gles remain the same regardless of the rotational velocity of the impeller. Hence, the inlet and exit velocity triangles have the same form. The axial ve­locity of an axial fan changes directly proportionally to the circumference ve­locity u. This is also valid for the radial velocity at the outer circumference of a radial impeller fan. These velocities are directly proportional to the fan flow volume; hence,

‘hit’ll 4vi ni’ where

«1 is the rotational velocity of impeller qvl is the volume flow at rotational velocity n1 n2 is another rotational velocity of the impeller qv2 is the volume flow at rotational velocity n2

As the velocity triangles’ shape remains the same,

M21 _ ^fll _ c«21 _ cu 11 _ Ј (9113)

U22 U2 cu22 cu 12

In Eq. (9.113), subscript u indicates the velocity to the tangential velocity component, the first subscript 2 indicates impeller exit, the first subscript 1 in­dicates impeller inlet, the second subscript 2 stands for n2, and the second sub­script 1 stands for «]. The proportionality constant k is

K = ^. (9.114)

N2

Even though the rotational velocity changes, the flow is still parallel to the blades, and the hydraulic efficiency remains the same regardless of rotational speed.

(9.112)

/ n

Aptotl Pi ^U22^cu22 ~ k-U2k-cu 12 Normally, with fans pj = p2. Then

V V

Aptot2 _ Pi U22cu22 ~ U2cu2 _ P2 X _ Pi ‘ ’ " Pi

Pi

(9.1151

N2 —

(.9.116)

Afitot2 =

Aptotl

If the rotational velocity change is within reasonable limits and the me­chanical efficiency does not change, Eqs. (9.76), (9.112) and (9.115) give the shaft power relation as

T9, h ;

La2 = Pl Pa, Pi

Example 3

The fan is tested at an air pressure of 102.9 kPa, temperature of 10 °C, and a rotational speed of 970 rev min-1. Under these conditions the volume flow is 0.7 m3 s-1, total pressure difference is 250 kPa, and shaft power is 250 kW. If the operating conditions change to handle an air temperature of 16 °C and pressure of 100 kPa and the efficiency remains unchanged, calculate under the new operating conditions the volume flow, total pressure difference, and shaft power.

Solution

Operating conditions 1: pl = 102.9 kPa and Tj = 10 °C = 283 K, Operating conditions 2: p2 = 100 kPa and T2 = 16 °C = 289 K.

Using ideal gas law,

Pi P 2

— /ll*?- 10,2.9.;, IQ;,-28.964 . , 2<~ kCTm RTi 8314.31-283 ‘-^/Kem

= PjM = 100 • 103 • 28.9864 = < 2Q5 . —

J? T, 8314 31-283 Kgm

500 970

Using «j = 970 rev min qvl = 0.73 m3 s_1, APtotl = 250 Pa, Pal = 250 W, and n2 = 500 rev min-1, Eq. (9.112) gives

3 -1

• 0.7 = 0.361 ms

Equation (9.115) gives the total pressure difference as

1.205

500 970

A p

• 250 = 63.2 Pa,

Totl

1.267 And Eq. (9.117) gives the shaft power as

V V

C «2 N V

‘p — 1-205 lal 1.267

500 970

P — Pl P°2 ~ Pi

The total pressure is 500 Pa, and volume flow is 2.4 m3 s~l for a fan work ­ing at standard air conditions (pressure 1 bar and temperature 20 °C), The speed of revolution is 14 rev s-1, and fan efficiency is 60%.

A. Calculate the total pressure, speed of revolution, and power if the volume flow is increased to 3.5 m3 s_1. The fan efficiency is assumed to remain constant.

B. Calculate the total pressure and power at the first volume flow if the temperature is increased to 82 °C and the speed of revolution remains constant.

Solution.

A. Denote qvl — 2.4 m3 s-1, ApX0ll = 500 Pa, nx — 14 rev s-i, and qv} = 3.5 m3 s-1. From Eq. (9.112) we get

«7 = —Ki = ^4 • 14 = 20 rev s-1.

2.4

" 4Vi 1

From Eq. (9.116) we get

20 14

• 500 = 1020 Pa.

F «?

Apmtl

V V

Hptotl — From Eq. (9.76) we get

3.5

5950 W.

■ 1020 = 0.6

B. At standard conditions we have p1 = 1 bar and T1 = 20 °C = 293 K. In the new situation we have pj = p2 = 1 bar and T2 = 82 °C = 355 K. From ideal gas law we have px = plM/RTi and = p7M/RT-> and because px =p2 of we get

Si =., h Pi ^2

From Eq. (9.115) we then get, for n1 —

Аp tot2 = P2 Apte Pi

Aptot2 = = Y2APto« = Ьf • 500 = 413 Pa’

From Eq. (9.76) we get

Pa2 = = 2.4; 413 = 16J2 w

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