Assignment 2 Solution

# Assignment 2 Solution - triangle Using the half-triangle we...

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Chemical Engineering 100 Assignment #2 Due: October 1, 2004 Noon Solution: Question 1: Relationship between velocity (v), volumetric flow rate (Q) and cross sectional area (A): v = Q/A 2 2 2 2 2 2 2 2 ft 10 2.18 in) (12 ft) (1 in 3.14 in 3.1416 4 in) (2 4 d A - × = = = = = π /s ft 1.468 s 60 min min 1 ft 88.10 /min ft 88.10 lbm/ft 62.43 lbm/min 5500 Q 3 3 3 3 = = = = ft/s 67.3 ft 10 2.18 /s ft 1.468 v 2 2 - 3 = × = Question 2: Question 3:

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Question 4: Question 5:
Alternate solution to calculate A (as suggested by one 1 st year student, Fall 2003). This solution does not require solving any integral. Let S 1 be the surface A + surface of the triangle (S triangle ): 2 1 2 S 2 1 θ π = = Since A = S 1 – S triangle , we need to find expressions for θ and S

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Unformatted text preview: triangle . Using the half-triangle, we find: h-1 2 cos * 1 = That is θ = 2 cos-1 (1-h) The surface of the triangle is equal to twice the surface of the half right triangle. 2 h)-(1 L triange half of Surface = For right triangle: L 2 + (1-h) 2 = 1 2 that is 2 h)-(1-1 L = So, finally A = S 1 = S triangle = 2 1-h)-(1-1 h)-(1-h)-(1 cos 2 h)-(1 L 2-2 = Realizing that cos-1 (1-h) = π/2 – sin-1 (1-h), we obtain: h)-(1 sin-h)-(1-1 h)-(1-2 A 1-2 = A θ L h...
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## This note was uploaded on 08/05/2008 for the course CHE 100 taught by Professor Croiset during the Fall '07 term at Waterloo.

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Assignment 2 Solution - triangle Using the half-triangle we...

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