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

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 \rho_{\text {air }}=1.20 kg / m ^{3} .
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,

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                   (1)

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

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

V_{2, t}=V_{2, n} \tan \alpha_{2}=(11.24 m / s ) \tan \left(40^{\circ}\right)=9.435 m / s                   (2)

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

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                     (3)

Net head:          H=\frac{1}{g}\left(\omega r_{2} V_{2, t}-\omega r_{1} V_{1, t}\right)                     (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,

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

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

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

bhp =\omega T _{\text {shaft }}=\rho \omega \dot{V}\left(r_{2} V_{2, t}-r_{1} V_{1, t}\right)=\dot{W}_{\text {water horsepower }}=\rho g \dot{V} H                   (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.