Explanation Given information: The radius of the turbine at outlet is 6.6 ft , the radius of the turbine at inlet is 4 .4 ft , the blade angle at inlet is 82 ° , the blade angle at outlet is 46 ° , the rotational speed of the turbine is 120 rpm , the breadth at inlet is 2.6 ft and the breadth of blades at outlet is 7.2 ft . Expression for angular velocity of the runner. ω = 2 π n ˙ 60 ....(I) Here, the rotational speed is n ˙ . Expression for normal component of velocity at the inlet. V 2 n = Q 2 π r 2 b 2 ....(II) Here, the volume flow rate is Q , the radius at inlet is r 2 and the breadth at inlet is b 2 . Expression for tangential component of velocity at the inlet. V 2 t = ω r 2 − V 2 n tan β 2 ....(III) Here, the blade angle at inlet is β 2 . Expression for runner leading angle. α 2 = tan − 1 ( V 2 t V 2 n ) ....(IV) Expression for the normal component of velocity at the outlet. V 1 n = Q 2 π r 1 b 1 ....(V) Here, the volume flow rate is Q , the radius at outlet is r 1 and the breadth at outlet is b 1 . Expression for tangential component of velocity at the outlet. V 1 t = ω r 1 − V 1 n tan β 1 ....(VI) Here, the blade angle at outlet is β 1 . Expression for swirl angle. α 1 = tan − 1 ( V 1 t V 1 n ) ....(VII) Expression for power output. P = ρ ω Q ( r 2 V 2 t − r 1 V 1 t ) ....(VIII) Here, the density of the water is ρ . Expression for net head. H = P ρ g Q ....(IX) Expression for dimensionless specific speed. N st = ω ( P ) 1 / 2 ( ρ ) 1 / 2 ( g H ) 5 / 4 ....(X) Expression for specific speed in customary US unit. N st,US = 43.46 N st ....(XI) Calculation: Substitute 120 rpm for n ˙ in Equation (I). ω = 2 π × 120 rpm 60 = 4 π rad / s = 12.56 rad / s Substitute 4.7 × 10 6 gpm for Q 6.6 ft for r 2 and 2.6 ft for b 2 in Equation (II). V 2 n = ( 4.7 × 10 6 gpm ) 2 π × 6.6 ft × 2 .6 ft = 4.7 × 10 6 gpm 107.76 ft 2 ( 35.315 ft 3 / s 15850 gpm ) = 97.12 ft / s Substitute 12.56 rad / s for ω , 6.6 ft for r 2 , 97.12 ft / s for V 2 n and 82 ° for β 2 in Equation (III). V 2 t = ( 12.56 rad / s ) ( 6.6 ft ) − ( 97.12 ft / s ) tan 82 ° = 82.896 ft / s − 13.65 ft / s = 69.246 ft / s Substitute 97.12 ft / s for V 2 n and 69.246 ft / s for V 2 t in Equation (IV). α 2 = tan − 1 ( 69.246 ft / s 97.124 ft / s ) = tan − 1 ( 0.712 ) = 35.45 ° Substitute 4.7 × 10 6 gpm for Q 4.4 ft for r 1 and 7.2 ft for b 1 in Equation (V). V 1 n = ( 4.7 × 10 6 gpm ) 2 π × 4.4 ft × 7.2 ft = 4.7 × 10 6 gpm 198.95 ft 2 ( 35.315 ft 3 / s 15850 gpm ) = 52.6 ft / s Substitute 12.56 rad / s for ω , 4.4 ft for r 1 , 52.6 ft / s for V 1 n and 46 ° for β 1 in Equation (VI). V 1 t = ( 12.56 rad / s ) ( 4.4 ft ) − ( 52.6 ft / s ) tan 46 ° = 55.264 − 50

Fluid Mechanics: Fundamentals and Applications

4th Edition

ISBN: 9781259877827

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 107EP

To determine

The dimensionless specific speed.

The specific speed in US customary unit.

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Chapter 14, Problem 106P

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The dimensionless specific speed. The specific speed in US customary unit.

Solution Summary: The author explains the dimensionless specific speed in US customary unit, the blade angle at inlet, and the breadth of blades at outlet.