# Centrifugal Fan

In this type of fan, the gas flows initially in an axial direction toward the impeller. After this, in a part of the impeller blade, the gas flow becomes radial.

To force the air to flow through the impeller blades of a centrifugal fan, a tangential force is needed. According to the momentum law this force is

F, = |(Mc2l + mc1() = qm(cZt-cu). (9.59)

Here cu is the tangential component of the air velocity at the inlet to the impeller blades, and c2t is the same at the exit. qm is the mass flow of air through the impeller.

Mechanical power is generally equal to force times velocity, so we get from Eq. (9.59)

Pu = uicu) ■ (9-60) * *

Mj is the velocity of impeller at the inlet to the blades, and u2 is the velocity of impeller at the outer edge.

Generally the velocity u can be calculated from

U = ttDh (9.61)

Where D is the diameter at the point in consideration and n is the speed of revolution of the impeller.

Consider the thermodynamic process in the fan (Fig. 9.33). As the fan is a stationary flow system, consideration is directed to the total enthalpy change. As the suction openings are often at the same, or almost the same level, the potential energy change can be neglected.

Two cases are studied:

• Adiabatic reversible (isentropic) process

• Adiabatic irreversible process in which entropy is generated

In both cases the mass flow, mechanical work, and density are constant.

The flow is denoted by subscripts 1 before the fan and 2 after the fan. If there

Is no entropy generation, subscript 2 is replaced by subscript 2s.

The mass conservation equation gives

Qm = pc^A , = pc2A2 = pc2sA2, (9.62) * *

Where

A j is the cross-sectional area of the fan entrance and A2 is the cross-sectional area of the fan exit.

Considering the air density to be constant yields c2 = c2s, and using

T ds = dh-vdp, (9.63)* *

The total enthalpy difference for the isentropic case can be written as

^2s. ~ ^I 9 ^2 ~~ 2^1* *

1 1

= V(Pls~P) + 9(c2-cl)

Z (9.64)

1 1 9 2 = ~(p2s~p l) + 2^C2_Ј:l)- |

In an adiabatic process, an entropy change occurs, giving

(2* *

T ds > 0. (9.6.5 j

I

Thus

1 / .2 .2

~ p(p2~Pl)+ = p(p2~Pl)+ |

Tds + 2(c2~ci) (9.66)* *

Tds + ^(cl-c]).* *

As the mass flow and power requirements are the same in both cases,

^ ~ A^tot, s ^ ^^tot, s A^tot„

Comparing Eqs. (9.64) and (9.66) and relating to Eq. (9.65), the following is obtained:

P2s > p2 !9-67)

In the isentropic case, the static pressure increases more than for the irreversible case, with the fan operating under the same power input and mass flow.

The total isentropic enthalpy difference can be expressed in terms of the total isentropic pressure difference by

APtot, s = pA^toM = APs + p(C2-Cl) . (9.68)

Denoting the total pressure by

Ptox = P + p^ (9,69)* *

Gives the total pressure difference as

Aptot = Ap + |p(Ac2). (9.70)

In Eq. (9.67), p is the static pressure, and pc2 is the dynamic pressure.* *

Using Eq. (9.65) gives

APtot, s > APtot (9.71)

For a fan in which the opening suction pressure, mass flow, and power are the same. Equations (9.60), (9.68), and (9.70) give

Isentropic case: u1clu-u-lc]tl = — ptms. (9.72)* *

P* *

1

Adiabatic irreversible case: M2c2w~wic1m > “ &P tot — (9.73)

The isentropic or hydraulic efficiency of the impeller is

Vs = T^- < 1- (9.74)

Tot, 5

Qm(U2C2u-U1CUt) = IT^r (9-75) R *ts |

Using Eqs. (9.60), (9.72), and (9.74) gives

Pu = qm(U2C2u-UCu) =* *

R r,

Where tjs = 1 if the flow through the impeller is isentropic.

For isentropic or nonisentropic flow, the fan shaft power is greater than the power required for the flow through the impeller.

A proportion of the shaft power is used to overcome the bearing friction. This is allowed by using the mechanical efficiency. Thus, the required axial power is

P — Ll — — Qi/ Aptot (9* *

" Vm VsVm V ’ »" ‘* *

Where

Rj is the total efficiency of the fan and

Qv = — is the air volume flow through the fan.

By using the vector equation

1 — ____ ____ __ _____ __ 2 1 __

Iv~ = tv ■ tv = (c — u) ■ (c — u) — c +u~-2u-c 1.9 77)* *

— c1 + u2 — 2uc2 the velocity expression of Eq. (9.73) can be written as

U2C2u~U1Clu = hc-c) + hul-u) + hw-w22)* *

1 L L (9.78)

= l(Pls-Pl) + j(c2-cl),* *

From which

&ps = Pis — Pi = JP(»I — u) + p(wi — w) • <9-79)** **

The centrifugal force produced increases the impeller static pressure [the first term of the right-hand side of Eq. (9.79)] and reduces the relative velocity w. To increase the static pressure by a change in the relative velocity, the following relationship is necessary:

Wl > IV2 .* *

The mass flow through the blade passage is constant; hence, > iv2, and the flow across the surface is greater for the leaving edge than the incoming edge. The cross-sectional area of the passage increases in the direction of flow.

Equation (9.79) shows that the static pressure increases with an increase in the tangential speed and the distance between the incoming and leaving edges of the impeller blade. This will not influence the nature of the flow process in the impeller.

Up to this stage, only the fan characteristics have been considered, without investigating the influence of different impeller blade shapes. Consider the flow at the edge of the impeller blade. Normally, for cost reasons, leading devices are not installed in front of the fan propeller, resulting in radial gas flow into the propeller, with the tangential velocity component clM = 0.

The velocity triangles of the incoming and leaving flow at the blade edges of a centrifugal fan are shown in Fig. 9.34. The flow to the blade is in a radial direction.

FIGURE 9.34 Velocity triangles at the blade edges of a backward-blade centrifugal fan.2 |

From Fig. 9.34, it can be seen that at the incoming edge of a backward — curved blade, the relative velocity is in the blade direction when clu — 0. The blade leaving edge may be

1. Backward curved (02 < 90°), Fig. 9.34

2. Straight (02 = 90°), Fig. 9.35, or

3. Forward curved (02 > 90°), Fig. 9.36.

FIGURE 9.35 Straight-blade centrifugal fen impeller. The blade is backward curved at the entrance.2

FIGURE 9.36 Centrifugal fen impeller, where the blade is forward curved at the leaving edge and backward curved at incoming edge.5 |

R = — ЈЈ-, (9.80)** **

^Ptot* *

By which the static pressure increase is related to the total pressure increase. The reaction ratio can be determined for different blade shapes for the isen — tropic case as

R = (9.81)

Tot, s

When cu< = 0 and rj5 = 1, Eq. (9.75) gives

&P tot, s = Pu2c2u — (9.82)* *

Using Eq. (9.79) gives

R _u-u + w-w l9 g3,* *

S lu2c2u ‘ ‘ ‘ ‘ ‘

The radial velocity is approximately constant; thus,

C] = Cr — c2r. (.9.84)* *

At the entrance edge of the blade, the velocity triangle is a right triangle and

Clu = 0, so Eq. (9.77) gives

C = w-u. (9.85)* *

Apply Eq. (9.77) to solve for the reaction ratio in isentropic flow:

R = u~w2 + C1 = lu2c2u~ cj + cr* *

‘ 2u2c1u ^ 2u2c2tt (9.86)* *

_ 2U2c2u ~ c2u = J _ 1 . Ј2m 2 ^2

In which the result c2 — clr — c2u was used.

For backward-curved blade j32 < 90° from Fig. 9.34, it is seen that C2U/u2 < 1- The reaction ratio must be less than 1. In straight-curved blade j82 = 90° of Fig. 9.35, c2u/u2 = 1 and the reaction ratio is rs — 1. In forward-curved blade j82 > 90° of Fig. 9.36, c2u/u2 > 1 and the reaction ratio is always less than |.

The reaction ratio of a straight-blade impeller is poor when compared with a backward-curved impeller. The radial (paddle) fan has the advantage that when the air contains much dust, the straight blades remain cleaner than curved blades.

The reaction ratio of forward-curved impeller blades is the lowest. This type is used when a small static pressure increase is sufficient to transport the air. An example of such type is the drum fan, with short blades. A large suction opening is required, which allows large volumes of air to be handled. With backward-curved blades, a better reaction ratio is achieved, but with a reduction in total pressure increase. This is because u2 is only slightly larger than uu and for that reason, clu will be small. By using forward-curved blades, clu increases, resulting in a larger total pressure differential.

The above discussion assumes that the gas flows parallel to the blades. This is not the case, as the propeller rotation causes the air to rotate between

The blades, against the impeller rotational direction at the same angular velocity. Due to this, the velocity clu decreases to c’2u; hence,

C>2u ~ C2u ~ ew (9.87) * *

Where

W is the impeller angular velocity

E is the effective radius of the vortex at the outer edge of the impeller (Fig. 9.37)

Figure 9.37 gives

Sin /3-» ~ —(9.88) " 7tDj/Z

Where

(9.89) |

VD2 sin /32

; is the number of blades. The fan slip factor is

TtD2 sin j32 2u2

Ij. u.. C~ 2z ‘ P2* *

Ciu ciu (9.90)

— ] 77 S’n $2 u2* *

Z c2u* *

In which co = 2u2/Di is used.

In Eq. (9.90), clu is the tangential component of the absolute velocity at the exit if the flow is exactly in the blade direction. Since the slip factor is less than 1, the total pressure increase will decrease according to Eq. (9.72) for the same impeller and isentropic flow.

KD7/2 |

The reaction ratio is improved, as shown in the following expression:

Rs = 1 — = 1 -(T—. {9.91}

S 2 w2 2 «2

The mass flow through the impeller is constant. For constant flow density and volume flow, the volume flow is

Q„ = 7i-D{b]Clr = 7rD262c2r (9.92)

Where

B1 is the blade width at the entrance (Fig. 9.38)

B2 is the blade width at the exit (Fig. 9.38)

Clr is the absolute radial velocity component at the entrance

Clr is the absolute radial velocity component at the exit

Because c. Xu = 0, we have c[r = c,.

Example I* *

The blades of the impeller in a centrifugal fan are backward curved. At the blade entrance ^ = 22°, and at the blade exit /32 = 50°. The outer and inner diameters of the impeller are D2 = 0.8 m and = 0.4 m, respectively. The width of the blade at the entrance is bx = 22 cm and at the exit is

B2 =12 cm. The impeller has 15 blades. The impeller rotational speed is n = 960 rev min-1 = 16 s_1. Calculate the volume flow, total pressure increase, static pressure increase, and reaction ratio. Take cUl = 0, and air density is 1.22 kg m~3.

Solution. The velocities of the circumferences are

U2 = 7rD2n = it ■ 0.8 • 16 = 40.2 m s_1

Ux = TrDxn = 7r ■ 0.4 • 16 = 20.1 m s^1

The velocity triangle at the entrance, taking into consideration that c1h = 0, is shown in Fig. 9.39a.

T** **

The triangle gives

CXr = cx = «i tan = 20.1 • tan 22° = 8.12 in s_1 = Jc + u = a/8.122 + 20.12 = 21.68 m s_1.

Equation (9.92) gives

= Dibi________________ C2r D, fe, ‘ "lr 0.8 • 0.12 |

~ ’ . r = 0-40 • 0-22 = 7 44 — l

Ct r A O A 1 -1 m S •

Using

2 _ 2 2 c2 c2u + c2r

Gives

C2» = Jcj — c2_r = J34.S2 — 7.442 = 34.0 m s“1.

For z = 15, Eq. (9.90) gives

„ 77- sin 02 M2 ■> tt * sin50° 40.2 no^A

From which c2u = oc2u = 27.55 m s_l, which is the tangential component of the absolute velocity parallel to the circumference velocity component when the blade number is taken into consideration. The flow process is still assumed isentropic. The velocity triangle at the exit is shown in Fig. 9.39b.

Using the velocity triangle gives

C2 = jcl + cjr = J27.552 + 7A42 = 28.5 m s’1. c2 and w2 are now the velocities when the slip factor is a = 0.81.

Using c1 = 8.12 m s_1, = 21.68 m s_1, and u^ —20.1 m s_1, the total

Pressure increase is

. 1,2222 2 2,

&Pwt, s = 2.P(ci — cj + m2 — Mi + wi — wi)* *

= 0.5 • 1.22(28.52 — 8.122 + 40.22 — 20.12 + 21.682 — 14.72)

= 1349 Pa,

And the static pressure increase is

A 1 1 1 1 2 x

Aps = jp(u2 w2)* *

= 0.5 • 1.22(40.22 — 20.12 + 21.682 — 14.72) = 894 Pa.

The reaction ratio is

= 0.663. |

_ &Ps _ 894

5 APtots 1349 For an isentropic process and clu = 0, Eq. (9.75) gives

AfW = pu2c2u — (9.93)

Using Fig. 9.33 gives

C2r* *

From which

C2u ~ u2~c2r COt0? • (9.95)

Figure 9.35 gives

——— = cot(180° — 0.) = —cot09, (9.96)

C2r ‘ “

From which

C2u = u2 “ c2r cot02. (9.97)

For all blade shape exits, we can write

°2u ~ u2 ~ c2r cot 02 when 0 < 02 < 180°. (9.98)

Using Eq. (9.92), the total pressure rise is expressed by

(9.99)

Equation (9.99) shows the total pressure increase as a function of volume flow for given impeller dimensions and rotational speed at fixed u2. The blade angle 02 at the exit is taken as parameter. This function is a linear one. When 02 > 90°, Aptot increases; for 02 = 90°, Apwcs remains constant; and for 02 < 90°, Aptot decreases for other parameters given in Eq. (9.97).

The dependency of total pressure increase on volume flow is demonstrated in Fig. 9.40.

Apn* *

Li; »

"

Yov*^

Straight blades, fS2 = 90°

•"«V

% /9 "…….. -■* *

^ T%o

——————————————————————————— ► <iv* *

FIGURE 9.40 Total pressure increase versus volume flow for different exit blade angles in isen- tropic flow. The flow follows the blade shape exactly.

Figure 9.40 is valid only for a single rotational speed. In addition, it is assumed that in the velocity triangle at the leaving edge of the impeller, the velocity component w2 has the same direction as the angle /32. The volume flow qv changes when c2r changes, which means, in the case of constant u2 and constant direction of w2, that the absolute velocity c2 changes. When clt changes, so does c]r, and the direction of w1 changes because ux is constant. This means that in Fig. 9.40 only for one volume flow is the direction of the velocity wi the same as the angle j8j for the impeller blade. For other values of qv this is not the case. We still think that the flow process between impeller blades is isentropic, but it is likely that in reality the deviation from an isentropic process is larger the more the direction of wx deviates from j8j. This will change the curves in Fig. 9.40 so that we get lower increases in the total pressure.

If the flow is isentropic but does not follow the blade shape exactly, then the slip factor is smaller than 1, and this will directly affect what is shown in Fig. 9.40. Denoting Apt’oM the total pressure rise when the flow does not follow the blade shape and calculating the slip factor from Eq. (9.90) give

Aptot, s — ^Ptous — Pulclu — Pu2c2u* *

, TT sin j87 u1

= pu2clu(l — a) = pu2c2u————- ;—(9.100)* *

Z Clu ‘ ’* *

2 7T sin/37 = pUl.

The total pressure decreases by the same quantity regardless of volume flow; the change is smaller for larger blade number z.

After the impeller the flow is flowing into the fan casing. The purpose of the casing is to collect the flow coming from the impeller and take the flow into the fan exit duct. Air flows to the casing everywhere from the exit edge of the impeller. For that reason the casing has a spiral shape (Fig. .9.41).

FIGURE 9.41 Centrifugal fan impeller and spiral casing. |

If there is no large duct after the fan which needs a large static pressure, then there is no need for spiral casing. The air then leaves steadily from the impeller circumference, and the air is not collected into one point. At the exit side in ventilating systems of buildings, the need for static pressure is small if the leaving air is directly sent to the atmosphere.

Real fan total pressure difference is smaller for the same volume flow than that of an isentropic, theoretical fan. This is a result of the fan losses. These arise from the entropy generation in adiabatic systems. We investigate the losses separately, i. e., entropy generation in the impeller and casing.

At the design point the volume flow is such that the relative velocity is parallel to the blade at the inlet of the impeller. The impeller losses at the design point are

• Friction losses between the flow and blades

• Leakage losses when the gas is transported from pressure side to suction side. Since the phenomenon is equalizing pressure in an adiabatic system, this will increase the entropy in the system.

• Friction in bearings, gasket, and impeller side walls. These losses are taken into consideration by the mechanical efficiency r m.

The casing losses are due to friction and mixing. The frictional losses and their reasons are the same as those in the impeller channel. The mixing losses develop because the velocity of the impeller exit is not the same at every point as in the spiral casing; it is an average velocity. Mixing of two different flow velocities leads to acceleration and deceleration and pressure difference, whose equalizing increases the entropy.

Outside the design point the direction of the relative velocity is not parallel to the blade, and shock losses are generated.

The theory of losses is very complex. The effect of mixing on the total pressure has been investigated.4 The calculated results qualitatively match the measured results. Friction in impeller channel and casing decreases the total pressure.

In practice, the fan characteristic curve—i. e., the total pressure difference dependency on the volume flow—is determined experimentally. The measured results are then for the impeller and casing together. Since the losses are greater outside the design point, the fan efficiency is high at the design point.

FIGURE 9.42 Typical performance curve of centrifugal fan, &Ptot-q„ chart. |

Figure 9.42 shows the typical characteristic curve of a centrifugal fan, where the blades are backward curved. The figure also shows the theoretical characteristic curve when the slip factor is 1 and when it is smaller than I. Characteristic curves for a real fan are closer to the isentropic one at the design point. At this point the efficiency is maximum.

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