Explanation Given information: The inlet radius of rotor is 2 m , the outlet radius is 1.3 m , width flow at inlet is 0.885 m and width of flow at outlet is 2.1 m , the runner blade angle at inlet is 71.4 ° and at outlet is 15.3 ° , speed of rotor is 160 rpm and volume flow rate is 80 m 3 / s . Figure-(1) represents the cross-section of turbine. Figure-(1) Write the expression for angular velocity. ω = 2 π n ˙ 60 ....(I) Here, speed of rotor is n ˙ . Write the expression for normal component of velocity at inlet. V 2,n = V ˙ 2 π r 2 b 2 ....(II) Here, volume flow rate is V ˙ , radius of rotor at inlet is r 2 and width of flow at inlet is b 2 . Write the expression for tangential component of velocity at inlet. V 2 ,t = ω r 2 − V 2 , n tan β 2 ....(III) Here, the runner blade angle at inlet is β 2 . Write the expression for normal component of velocity at outlet. V 1,n = V ˙ 2 π r 1 b 1 ....(IV) Here, radius of rotor at outlet is r 1 and width of flow at outlet is b 1 . Write the expression for tangential component of velocity at outlet. V 1,t = ω r 1 − V 1 , n tan β 1 ....(V) Here, the runner blade angle at outlet is β 1 . Write the expression for angle through which gates should turn the flow. α 2 = tan − 1 ( V 2 , t V 2 , n ) ....(VI) Write the expression for swirl angle. α 1 = tan − 1 ( V 1 , t V 1 , n ) ....(VII) Write the expression for power produced. W ˙ shaft = ρ ω V ˙ ( r 2 V 2 , t − r 1 V 1 , t ) ....(VIII) Here, density of water is ρ . Write the expression for net head. H = W ˙ shaft ρ g V ˙ ....(IX) Here, acceleration due to gravity is g . Calculation: Substitute 160 rpm for n ˙ in Equation (I). ω = 2 π ( 160 rpm ) 60 = 2 π ( 160 rpm ) 60 × 1 rad / s 1 rpm = 16.755 rad / s Substitute 80 m 3 / s for V ˙ , 2 m for r 2 , and 0.85 m for b 2 in Equation (II). V 2,n = 80 m 3 / s 2 π ( 2 m ) ( 0.85 m ) = 80 m 3 / s 3.4 π m 2 = 7.49 m / s Substitute 16.755 rad / s for ω , 2 m for r 2 , 7.49 m / s for V 2,n , 71.4 ° for β 2 in Equation (III). V 2 ,t = ( 12.56 rad / s ) ( 2 m ) − 7.49 m / s tan ( 71.4 ° ) = 25.12 m / s − 2.51056 m / s = 30.98947 m / s Substitute 80 m 3 / s for V ˙ , 1.3 m for r 1 , and 2.1 m for b 1 in Equation (IV). V 1,n = 80 m 3 / s 2 π ( 1.3 m ) ( 2.1 m ) = 80 m 3 / s 5.46 π m 2 = 4.663 m / s Substitute 16.755 rad / s for ω , 1.3 m for r 1 , 7.49 m / s for V 2,n , 15

4th Edition

ISBN: 2810022150991

Author: CENGEL

Publisher: MCG

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Applied Fluid Mechanics

A fluid is a substance that is continuously deformed by the application of shear stress. The rate of deformation of a fluid depends on its viscosity. The fluid tries to flow when it interacts with some other system.

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Chapter 14, Problem 89P

To determine

The angle through which wickets gates should turn the flow.

The swirl angle α1.

The power output of turbine.

The net required head.

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Chapter 14, Problem 88P

Chapter 14, Problem 90CP

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Chapter 14 Solutions

FLUID MECHANICS FUNDAMENTALS+APPS

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The angle through which wickets gates should turn the flow. The swirl angle α 1 . The power output of turbine. The net required head.

Solution Summary: The author explains the angle through which wickets gates should turn the flow. The net required head is 45.4249°.

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The angle through which wickets gates should turn the flow.

The swirl angle α1.

The power output of turbine.

The net required head.

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Chapter 14 Solutions

The angle through which wickets gates should turn the flow. The swirl angle α 1 . The power output of turbine. The net required head.

Solution Summary: The author explains the angle through which wickets gates should turn the flow. The net required head is 45.4249°.

Concept explainers

Videos

The angle through which wickets gates should turn the flow. The swirl angle α1. The power output of turbine. The net required head.

Want to see the full answer?

Chapter 14 Solutions

The angle through which wickets gates should turn the flow.

The swirl angle α1.

The power output of turbine.

The net required head.