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• Sarah Harun

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QUESTION 13-5: BEST CROSS SECTION OF AN OPEN CHANNEL Water is to be transported at a rate of 2 m 3/s in uniform flow in open channel whose surfaces are asphalt lined. The bottom slope is 0.001. Determine the dimensions of the best cross section if the shape of the channel is

a) Rectangular

b) Trapezoidal

SOLUTION: Assumptions: 1. The flow is steady and uniform. 2. The bottom slope is constant. 3. The roughness of the wetted surface of the channel and thus the friction coefficient are constant. Properties The Manning coefficient for an open channel with asphalt lining is n = 0.016

Analysis

(a) The best cross section for a rectangular channel occurs when the flow height is half the channel width, y = b/2. Then the cross sectional area, perimeter, and hydraulic radius of the channel are A b b p=b +2 y =2 b R h= C = 2 P 4 2

A c =by =

Substituting into the Manning equation

(

a 2 /3 1 /2 V´ = A c Rh S 0 n

b=

2 n V´ 42 /3 a √ S0

3 /8

)

¿

(

2 ( 0.016 ) ( 2 m3 / s ) 4 2 /3 1

( 1m√ 3 /s ) √ 0.001

3 /8

)

Which gives b = 1.84 m. Therefore, Ac = 1.70 m2 , p = 3.68 m, and the dimensions of the best rectangular channel are b = 1.84 m

and

y = 0.92 m

(b) The best cross section of trapezoidal channel occurs when the trapezoidal angle is 60° and flow height is

y=b √ 3 /2 . Then,

A c = y ( b+ cos θ )=0.5 √ 3 b2 ( 1+ cos 60 ° )=0.75 √ 3 b2 y 3 p=3 b Rh = = √ b 2 4 Substituting into the Manning equation, a V´ = A c R2h /3 S 10 /2 n

(

b=

3 /8

(0.016) ( 2 m3 / s ) 0.75 √ 3 ( √3 /4 )

2 /3

( 1 m1/3 /s ) √ 0.001

)

Which yields b = 1.12 m. Therefore, Ac = 1.64 m2, p = 3.37 m, and the dimensions of the best trapezoidal channel are b= 1.12 m

y = 0.973 m

and

θ = 60°

QUESTION 14-5 IDEALIZED BLOWER PERFORMANCE

A centrifugal blower rotates at

n´ =1750rpm ( 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. The outlet radius is r2 = 8.0 cm, and the outlet blade width b 2 = 2.3 cm. The volume flow rate is 0.13 m 3/s. For the idealized case, i.e., 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.

SOLUTION 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 flow is incompressible. 4. The efficiency of the blower is 100 percent (no irreversible losses). Properties: Take the density of air is to be ρair = 1.20 kg/m3.

Analysis: Since the volumes flow rate (capacity)is given, calculate the normal velocity components at the inlet using Eq. 14-12.

V 1,n=

V´ 0.13m 3 / s = =9.947 m/s 2 π r 1 b1 2 π (0.040 m)(0.052 m)

V 1=V 1,n ∧V 1,t =0, since α 1=0 ° . Similarly ,V 2, n=11.24

m ,∧¿ s

V 2,t =V 2,n tan α 2=(11.24 m/ s) tan ( 40° )=9.435 m/s

Now we use Equ. 14-17 to predict the net head,

¿

183.3 rad / s ( 0.080 m ) ( 9.435 m/s )=14.1 m 9.81m/ s2

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, H water column =H

ρair ρ water ¿ ( 14.1m )

1.20 kg /m3 1000 mm =17.0mm of water 1m 998 kg /m3

(

)

Finally, use Eq.14-16 to predict the required brake horsepower, bhp=ρg V´ H=(1.20 kg /m3 )(9.81 m/s 2)(0.13 m3 /s) ( 14.1 m) = 21.6 W

( kgW.m/. s s ) 2