Chapter 10

# Chapter 10 - Problem 10.1 Given Data on a centrifugal pump...

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U 2 r 2 ω = and V t2 U 2 V rb2 cos β 2 () = U 2 = From the outlet geometry V t1 0 = For an axial inlet The governing equation (directly derived from the Euler turbomachine equation) is β 2 90 deg = V rb2 5 m s = ω 3500 rpm = η 67 % = W in 5kW = Q 0.01 m 3 s = ρ 999 kg m 3 = The given or available data is Solution Find: Estimate basic dimensions Given: Data on a centrifugal pump Problem 10.1

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b 2 6.37mm = b 2 Q 2 π r 2 V n2 = Hence V n2 Q 2 π r 2 b 2 = From continuity V n2 5 m s = V n2 V rb2 sin β 2 () = Also r 2 50mm = r 2 W m m rate ω 2 = Hence m rate 9.99 kg s = m rate ρ Q = and W m 3.35kW = W m η W in = with W m U 2 2 m rate = r 2 2 ω 2 m rate = Hence, in Eq. 10.2b

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V t2 W m U 2 ρ Q = Hence V t1 0 = For an axial inlet The governing equation (derived directly from the Euler turbomachine equation) is D 300 mm = b 2 10 mm = ω 1750 rpm = η 75 % = W in 45 kW = Q5 0 L s = ρ 999 kg m 3 = The given or available data is Solution Find: Estimate exit angle of impeller blades Given: Data on a centrifugal pump Problem 10.12

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β 61.3deg = β atan V n2 U 2 V t2 = or tan β () V n2 U 2 V t2 = With the exit velocities determined, β can be determined from exit geometry V n2 5.31 m s = From continuity V n2 Q π D b 2 = V t2 24.6 m s = V t2 W m U 2 ρ Q = Hence W m 33.8kW = W m η W in = and U 2 27.5 m s = U 2 D 2 ω = We have

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T out T shaft Q R VU () 1 cos θ = or Hence T shaft RV U cos θ  ρ Q = m flow ρ Q = and V t2 ( ) cos θ = V t1 = U 1 U 2 = U = r 1 r 2 = R = In terms of the notation of Example Problem 10.5, for a stationary CV The governing equation is the Euler turbomachine equation Solution Problem 10.15

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The power is W out ω T out Q R ⋅ω VU () 1 cos θ = W out ρ Q U 1 cos θ = These results are identical to those of Example Problem 10.5. The proof that maximum power is when U = V /2 is hence also the same and will not be repeated here.
The governing equations (derived directly from the Euler turbomachine equation) are β 2 45 deg = b 2 7.5 mm = r 2 150 mm = β 1 25 deg = b 1 10 mm = r 1 90 mm = η 70 % = ω 1200 rpm = ρ 999 kg m 3 = The given or available data is Solution Find: Flow rate for zero inlet tangential velocity; outlet flow angle; power; head developed Given: Data on a centrifugal pump Problem 10.16

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V t1 0 m s = U 2 18.8 m s = U 2 r 2 ω = U 1 11.3 m s = U 1 r 1 ω = The power, head and absolute angle α 10.2b, 10.2c, and 1 above Q 0.0298 m 3 s = Q 29.8 L s = Q2 π r 1 2 b 1 ⋅ω tan β 1 () = r 1 ω Q 2 π r 1 b 1 tan β 1 0 = Hence V n1 Q 2 π r 1 b 1 = and from continuity V t1 0 = U 1 V rb1 cos β 1 = r 1 ω V n1 sin β 1 cos β 1 = From geometry (1) α 2 atan V t2 V n2 = We also have from geometry
(This last result can also be obtained from Eq. 10.8a W h ρ Q g H p = ) H p 19.7m = H p η H = W h 5.75kW = W h η W m =

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## This note was uploaded on 01/24/2012 for the course MECHE 343 taught by Professor Professor during the Spring '11 term at Carnegie Mellon.

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Chapter 10 - Problem 10.1 Given Data on a centrifugal pump...

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