Explanation Given information: The distance is 31 miles , the friction factor is 0.019 , the pressure in providence is 5 bargs, and the pressure in Narragansett is 1 barg . Refer from the table A.4 the properties of CH 4 from the book the value obtained of universal constant is 518 m 2 / s 2 , the diameter is 6.06 in , the actual size of pipe is 40 in , ambient pressure is 101 kPa . at 1 atm and 68 ° . Write the expression for the mass flow rate. ( m A ) 2 = P 1 2 − P 2 2 R T ( f L d + 2 ln ( P 1 P 2 ) ) ...(I) Here, the mass flow rate is m , the universal constant is R , the area is A , the length is L , the diameter is d , the temperature is T , the friction factor is f , the pressure in providence is P 1 , and the pressure in Narragansett is P 2 . Write the expression for the area. A = π d 2 4 ...(II) Write the expression for the exact pressure for Providence. P 1 = P ˙ 1 + 101 kPa ...(III) Write the expression for the exact pressure for Providence. P 2 = P ˙ 2 + 101 kPa ...(IV) Calculation: Substitute 5 bargs for P ˙ 1 in Equation (III). P 1 = 5 bargs + 101 kPa = 5 bargs ( 100 kPa 1 bargs ) + 101 kPa = 601 kPa Substitute 1 bargs for P ˙ 2 in Equation (IV). P 2 = 1 bargs + 101 kPa = 1 bargs ( 100 kPa 1 bargs ) + 101 kPa = 201 kPa Substitute 6.06 in for d , in Equation (II). A = π ( 6.06 in ) 2 4 = 28.82 in 2 ( 1 m 40 in ) 2 = 0.018 m 2 Substitute 0.018 m 2 for A , 601 kPa for P 1 , 201 kPa for P 2 , 518 m 2 / s 2 for R , 68 ° for T , 0

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ISBN: 9780073398273

Author: Frank M. White

Publisher: McGraw-Hill Education

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Chapter 9, Problem 9.100P

To determine

The mass flow if the pressure is 5 bargs in providence and 1 barg in Narragansett.

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Chapter 9, Problem 9.99P

Chapter 9, Problem 9.101P

Chapter 9 Solutions

Fluid Mechanics

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