Idealized Blower Performance A centrifugal blower rotates at n = 1750 rpm (183.3 rad/s). Air enters the impeller normal to the blades (α1 = 0°) and exits at an angle of 40° from radial (α2 = 40°) as sketched in Fig. 14–38. The inlet radius is r1 = 4.0 cm, and the inlet blade width b1 = 5.2 cm.

Question

Idealized Blower Performance

A centrifugal blower rotates at [latex] \dot{n}=1750 [/latex] rpm (183.3 rad/s). Air enters the impeller normal to the blades [latex]\left(\alpha_{1}=0^{\circ}\right) [/latex] and exits at an angle of 40° from radial [latex]\left(\alpha_{2}=40^{\circ}\right) [/latex] as sketched in Fig. 14–38. The inlet radius is [latex] r_{1}=4.0 [/latex] cm, and the inlet blade width [latex] b_{1}= [/latex] 5.2 cm. The outlet radius is [latex] r_{2}= [/latex] 8.0 cm, and the outlet blade width [latex] b_{2}=[/latex] 2.3 cm. The volume flow rate is 0.13 m³/s. Assuming 100 percent efficiency, calculate the net head produced by this blower in equivalent millimeters of water column height. Also calculate the required brake horsepower in watts.

Accepted Answer

We are to calculate the brake horsepower and net head of an idealized blower at a given volume flow rate and rotation rate.
Assumptions 1 The flow is steady in the mean. 2 There are no leaks in the gaps between rotor blades and blower casing. 3 The air is incompressible. 4 The efficiency of the blower is 100 percent (no irreversible losses).
Properties We take the density of air to be [latex] ho_{ ext {air }}=1.20 kg / m ^{3} [/latex].
Analysis Since the volume flow rate (capacity) is given, we calculate the normal velocity components at the inlet and the outlet using Eq. 14–12,

[latex] V_{1, n}=\frac{\dot{V}}{2 \pi r_{1} b_{1}}=\frac{0.13 m ^{3} / s }{2 \pi(0.040 m )(0.052 m )}=9.947 m / s [/latex] (1)

Volume flow rate: [latex] \dot{V}=2 \pi r_{1} b_{1} V_{1, n}=2 \pi r_{2} b_{2} V_{2, n} [/latex] (14–12)

[latex] V_{1}=V_{1, n} ext {, and } V_{1, t}=0 ext {, since } \alpha_{1}=0^{\circ} . ext { Similarly, } V_{2, n}=11.24 m / s ext {, } [/latex] and

[latex] V_{2, t}=V_{2, n} an \alpha_{2}=(11.24 m / s ) an \left(40^{\circ} ight)=9.435 m / s [/latex] (2)

Now we use Eq. 14–17 to predict the net head,

[latex] H=\frac{\omega}{g}(r_{2} V_{2, t}-r_{1} \underbrace{\cancel{V_{1, t}}}_{0})=\frac{183.3 rad / s }{9.81 m / s ^{2}}(0.080 m )(9.435 m / s )=14.1 m [/latex] (3)

Net head: [latex] H=\frac{1}{g}\left(\omega r_{2} V_{2, t}-\omega r_{1} V_{1, t} ight) [/latex] (14–17)

Note that the net head of Eq. 3 is in meters of air, the pumped fluid. To convert to pressure in units of equivalent millimeters of water column, we multiply by the ratio of air density to water density,

[latex] \begin{aligned} H_{ ext {water column }} &=H \frac{ ho_{ ext {air }}}{ ho_{ ext {water }}} \ &=(14.1 m ) \frac{1.20 kg / m ^{3}}{998 kg / m ^{3}}\left(\frac{1000 mm }{1 m } ight)= 1 7 . 0 ~ m m ~ o f ~ w a t e r ~ \end{aligned} [/latex] (4)

Finally, we use Eq. 14–16 to predict the required brake horsepower,

[latex]\begin{aligned}bhp &= ho g \dot{ V } H=\left(1.20 kg / m ^{3} ight)\left(9.81 m / s ^{2} ight)\left(0.13 m ^{3} / s ight)(14.1 m )\left(\frac{ W \cdot s }{ kg \cdot m / s ^{2}} ight) \&= 2 1 . 6 W\end{aligned}[/latex] (5)

[latex] bhp =\omega T _{ ext {shaft }}= ho \omega \dot{V}\left(r_{2} V_{2, t}-r_{1} V_{1, t} ight)=\dot{W}_{ ext {water horsepower }}= ho g \dot{V} H [/latex] (14–16)

Discussion Note the unit conversion in Eq. 5 from kilograms, meters, and seconds to watts; this conversion turns out to be useful in many turbomachinery calculations. The actual net head delivered to the air will be lower than that predicted by Eq. 3 due to inefficiencies. Similarly, actual brake horsepower will be higher than that predicted by Eq. 5 due to inefficiencies in the blower, friction on the shaft, etc.

Question 14.5: Idealized Blower Performance A centrifugal blower rotates at...

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