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CIVL2611 – Introductory Fluid Mechanics CIVL9611 – Introductory Fluid Mechanics Tutorial 7/8 Solutions Question 1: The total flow rate is /s m 75 . 15 5 . 1 5 . 1 2 1 2 5 . 1 2 3 3 = × × × + × × = = A Q v (a) The pipe flows full m 167 . 3 75 . 15 2 2 4 2 2 = × = = × = = π π π Q d d A Q v (b) The pipe flows half full m 48 . 4 75 . 15 4 4 2 4 2 2 = × = = × = = π π π Q d d A Q v (c) The pipe flows with a depth of half the radius Consider the shaded area as shown. The area is calculated as ( ) α α sin 2 2 = R A where π α α 3 2 120 2 1 2 2 cos o = = = = R R . Thus, ( ) 2 o 2 6142 . 0 120 sin 3 2 2 R R A = = π The radius of the pipe is determined as below: m 58 . 3 2 6142 . 0 75 . 15 6142 . 0 6142 . 0 2 = × = = = = v v v Q R R A Q So the diameter of the pipe is d = 2 R = 7.16 m α A 1
Question 2: Consider the control volume as shown. For steady flow of dry air, air dry , 2 3 m m = For steady flow of water, 4 water , 2 1 m m m + = Also, water , 2 air dry , 2 2 m m m + = From the above equations we obtained the rate of water evaporation as kg/s 1 19 20 3 2 water , 2 = = = m m m And the rate of cooled water flow as kg/s 5 . 30 1 5 . 31 water , 2 1 4 = = = m m m Question 3: Consider the control volume as shown. Surface is the inlet, Surfaces and are the outlets, and Surface consists of all solid surfaces. Since 30 0 A = 0.1 m 2 30 0 A = 0.1 m 2 Q = 0.1 m 3 /s A = 0.1 m 2 Control volume y, F Ay x, F Ax 2
m/s 1 1 . 0 1 . 0 1 1 1 = = = = A Q A Q v v From mass conservation (with constant density), 3 2 1 Q Q Q + = By symmetry (since the gravity is ignored), 3 2 Q Q = . Thus, /s m 05 . 0 2 1 3 1 3 2 = = = Q Q Q Accordingly, v 2 = v 3 = 0.5 m/s.