Explanation Write the formula for inlet volume flow rate. Inlet volume flow rate = A 1 V 1 (I) Here, the cross sectional area is A and the velocity is V . Write the formula for inlet cross-sectional area of rectangular duct whose cross-section is rectangle. A 1 = l b (II) Here, the length of the rectangle is l and the breadth is b . Write the formula for inlet mass flow rate in terms of ideal gas equation. m ˙ 1 = A 1 V 1 p 1 ( R ¯ / M ) T 1 (III) Here, the inlet volumetric flow rate is A 1 V 1 , the pressure is p , universal gas constant is R ¯ , the molar mass of the air is M , the temperature is T and the suffix 1 indicates the inlet state. The molar mass of air ( M ) is 28.97 lb/lbmol . Conclusion: Substitute 6 in . for l and 4 in . for b in Equation (II). A 1 = ( 6 in . ) ( 4 in . ) = 24 in . 2 Substitute 24 in . 2 for A 1 and 3 ft/s for V 1 in Equation (I) Inlet volumetric flow rate = ( 24 in . 2 ) ( 3 ft/s ) = ( 24 in . 2 × 1 ft 2 144 in . 2 ) ( 3 ft/s ) = 0.5 ft 3 /s Thus, the inlet volumetric flow rate is 0.5 ft 3 /s . Substitute 0.5 ft 3 /s for A 1 V 1 , 16 lbf/in . 2 for p 1 , 1545 ft ⋅ lbf lbmol ⋅ °R for R ¯ , 28

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 4.12, Problem 23P

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The inlet volumetric flow rate, in ft3/s and mass flow rate, in kg/s. If possible find the volumetric flow rate and mass flow rate at exit also. If not possible explain why?

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Chapter 4.12, Problem 22P

Chapter 4.12, Problem 24P

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The inlet volumetric flow rate, in ft 3 /s and mass flow rate, in kg/s . If possible find the volumetric flow rate and mass flow rate at exit also. If not possible explain why?

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The inlet volumetric flow rate, in ft3/s and mass flow rate, in kg/s. If possible find the volumetric flow rate and mass flow rate at exit also. If not possible explain why?

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Chapter 4 Solutions