# ch10 - Problem 10.1[2 Problem 10.2 Given Geometry of...

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Problem 10.1 [2]

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Problem 10.2 [2] Given: Geometry of centrifugal pump Find: Estimate discharge for axial entry; Head Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1.94 slug ft 3 = r 1 4in = r 2 7.5 in = b 1 1.5 in = b 2 1.5 in = ω 1500 rpm = β 1 30 deg = β 2 20 deg = From continuity V n Q 2 π r b = V rb sin β () = V rb V n sin β = From geometry V t UV rb cos β = U V n sin β cos β = U Q 2 π r b cot β = For an axial entry V t1 0 = so U 1 Q 2 π r 1 b 1 cot β 1 0 = Using given data U 1 ω r 1 = U 1 52.4 ft s = Hence Q2 π r 1 b 1 U 1 tan β 1 = Q 7.91 ft 3 s = Q 3552 gpm = To find the power we need U 2 , V t2 , and m rate The mass flow rate is m rate ρ Q = m rate 15.4 slug s = U 2 ω r 2 = U 2 98.2 ft s = V t2 U 2 Q 2 π r 2 b 2 cot β 2 = V t2 53.9 ft s = Hence W m U 2 V t2 U 1 V t1 m rate = W m 81212 ft lbf s = W m 148 hp = The head is H W m m rate g = H 164 ft =
Problem 10.3 [2] Given: Data on centrifugal pump Find: Estimate basic dimensions Solution: Basic equations: (Eq. 10.2b, directly derived from the Euler turbomachine equation) The given or available data is ρ 1.94 slug ft 3 = Q 150 gpm = Q 0.334 ft 3 s = W in 6.75 hp = η 67 % = ω 3500 rpm = V rb2 17.5 ft s = β 2 90 deg = For an axial inlet V t1 0 = From the outlet geometry V t2 U 2 V rb2 cos β 2 () = U 2 = and U 2 r 2 ω = Hence, in Eq. 10.2b W m U 2 2 m rate = r 2 2 ω 2 m rate = with W m η W in = W m 4.52hp = and m rate ρ Q = m rate 0.648 slug s = Hence r 2 W m m rate ω 2 = r 2 0.169ft = r 2 2.03in = Also V n2 V rb2 sin β 2 = V n2 17.5 ft s = From continuity V n2 Q 2 π r 2 b 2 = Hence b 2 Q 2 π r 2 V n2 = b 2 0.0180ft = b 2 0.216in =

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Problem 10.4 [2]
Problem 10.5 [2] Given: Geometry of centrifugal pump Find: Theoretical head; Power input for given flow rate Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1.94 slug ft 3 = r 1 3in = r 2 9.75 in = b 1 1.5 in = b 2 1.125 in = ω 1250 rpm = β 1 60 deg = β 2 70 deg = Q 1500 gpm = Q 3.34 ft 3 s = From continuity V n Q 2 π r b = V rb sin β () = V rb V n sin β = From geometry V t UV rb cos β = U V n sin β cos β = U Q 2 π r b cot β = Using given data U 1 ω r 1 = U 1 32.7 ft s = U 2 ω r 2 = U 2 106.4 ft s = V t1 U 1 Q 2 π r 1 b 1 cot β 1 = V t1 22.9 ft s = V t2 U 2 Q 2 π r 2 b 2 cot β 2 = V t2 104 ft s = The mass flow rate is m rate ρ Q = m rate 6.48 slug s = Hence W m U 2 V t2 U 1 V t1 m rate = W m 66728 ft lbf s = W m 121hp = The head is H W m m rate g = H 320ft =

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Problem 10.6 [2] Given: Geometry of centrifugal pump Find: Theoretical head; Power input for given flow rate Solution: Basic equations: (Eq. 10.2b) (Eq. 10.2c) The given or available data is ρ 1.94 slug ft 3 = r 1 15 in = r 2 45 in = b 1 4.75 in = b 2 3.25 in = ω 575 rpm = β 1 40 deg = β 2 60 deg = Q 80000 gpm = Q 178 ft 3 s = From continuity V n Q 2 π r b = V rb sin β () = V rb V n sin β = From geometry V t UV rb cos β
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## This note was uploaded on 07/10/2011 for the course CHE 144 taught by Professor Tuzla during the Spring '11 term at Lehigh University .

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ch10 - Problem 10.1[2 Problem 10.2 Given Geometry of...

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