Explanation Given information: The temperature of the water is 60 ° C , density of the water is 983.3 kg / m 3 , dynamic viscosity of the water is 0.467 × 10 − 3 kg / m ⋅ s , diameter of the pipe is 1.2 cm , roughness of the plastic pipe surfaces is 0 , pump efficiency is 70 % , length of the pipe is 40 m and average velocity of the water is 2 m / s . Write the expression for the Reynolds number. Re = ρ v D μ ........... (I) Here, Reynolds number is Re , density of the water is ρ , velocity of the water is v , diameter of the pipe is D and dynamic viscosity of the water is μ . Write the expression for the volume flow rate of the water. Q ˙ = π 4 D 2 v ........... (II) Here, the volume flow rate of the water is Q ˙ . Write the expression for the friction factor for turbulent flow. 1 f = − 2 log ( ε D 3.7 + 2.51 Re f ) ........... (III) Here, friction factor for turbulent flow is f and roughness of the plastic pipe surfaces is ε . Write the expression for the head loss through the pipe. h f = f L v 2 2 g D ........... (IV) Here, length of the pipe is L , gravitational acceleration is g and the head loss through the pipe is h f . Write the expression for the pressure drop through the pipe. Δ P = ρ g h f ........... (V) Here, the pressure drop through the pipe is Δ P . Write the expression for the electric power consumption of the system for pumping of water. W ˙ = Q ˙ Δ P ........... (VI) Here, theoretical required power input for pump is W ˙ . Write the expression for the motor-pump efficiency. η pump = W ˙ W ˙ pump ........... (VII) Here, the pump efficiency is η pump and actual required power input for pump is W ˙ pump . Calculation: Substitute 2 m / s for v and 1.2 cm for D in Equation (II). Q ˙ = π 4 ( 1.2 cm ) 2 ( 2 m / s ) = π 4 ( ( 1.2 cm ) [ 1 m 100 cm ] ) 2 ( 2 m / s ) = 2.261 × 10 − 4 m 3 / s Substitute 983.3 kg / m 3 for ρ , 2 m / s for v , 0.467 × 10 − 3 kg / m ⋅ s for μ and 1.2 cm for D in Equation (I). Re = ( 983.3 kg / m 3 ) ( 2 m / s ) ( 1.2 cm ) ( 0.467 × 10 − 3 kg / m ⋅ s ) = ( 983.3 kg / m 3 ) ( 2 m / s ) ( 1.2 cm ) [ 1 m 100 cm ] ( 0.467 × 10 − 3 kg / m ⋅ s ) = 50533.618 Since value of the Reynolds number is greater than 4000 , therefore flow is turbulent. Substitute 50533.618 for Re , 0 for ε and 1.2 cm for D in Equation (III). 1 f = − 2 log ( ( 0 ) ( 1.2 cm ) 3.7 + 2.51 ( 50533.618 ) f ) − 1 2 f = log ( 2

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Author: CENGEL

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Chapter 8, Problem 99P

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Chapter 8 Solutions

EBK FLUID MECHANICS: FUNDAMENTALS AND A

Ch. 8 - How is the hydrodynamic entry length defined for...Ch. 8 - Why are liquids usually transported in circular...Ch. 8 - What is the physical significance of the Reynolds...Ch. 8 - Consider a person walking first in air and then in...Ch. 8 - Show that the Reynolds number for flow in a...Ch. 8 - Which fluid at room temperature requires a larger...Ch. 8 - What is the eneia1Iy accepted value of the...Ch. 8 - Consider the flow of air and wale in pipes of the...Ch. 8 - Consider laminar flow in a circular pipe. Is the...Ch. 8 - How does surface roughness affect the pressure...

Ch. 8 - What is hydraulic diameter? How is it defined?...Ch. 8 - Shown here is a cool picture of water being...Ch. 8 - What fluid property is responsible for the...Ch. 8 - In the fully developed region of flow in a...Ch. 8 - Someone claims that the volume flow rate in a...Ch. 8 - Someone claims that the average velocity in a...Ch. 8 - Someone claims that the shear stress at the center...Ch. 8 - Someone claims that in fully developed turbulent...Ch. 8 - How does the wall shear stress w , vary along the...Ch. 8 - How is the friction factor for flow in a pipe...Ch. 8 - Discuss whether fully developed pipe flow is one-,...Ch. 8 - Consider fully developed flow in a circular pipe...Ch. 8 - Consider fully developed laminar how in a...Ch. 8 - Explain why the friction factor is independent of...Ch. 8 - Consider laminar flow of air in a circular pipe...Ch. 8 - Consider fully developed laminar flow in a...Ch. 8 - How is head loss related to pressure loss? 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