CHEMICAL FLUID MECHANIC - EXAMPLE 3.1 : Water flows through a pipe with circular cross sectional area at the rate of V / t = 80 L / s where V is the volume and t is time. Let Av = 80 L / s where A is cross sectional area and v is velocity of fluid. For point 1, the radius of the pipe is 16 cm. For point 2, the radius of the pipe is 8 cm. Find (a) the velocity at point 1; (b) the velocity at point 2; (c) the pressure at point 2 by using Bernoulli's equation where P + Rgy + 0.5 RV = constant. P is the pressure, R = density of fluid, V = square of fluid's velocity, g = gravitational constant of 9.81 N / kg and y = 2 m = difference of height at 2 points. The pressure of point 1 is 180 kPa.
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CHEMICAL FLUID MECHANIC - ANSWER 3.1 : Let V / t = 80 L / s = 0.08 cubic metres / s. V / t = Av : (a) pai = 3.142. A = (pai)(radius square). (3.142)(0.16)(0.16)v = 0.08, v = 0.995 m / s. (b) (3.142)(0.08)(0.08)v = 0.08, v = 3.98 m / s. (c) (180000) + (1000)(9.81)(0) + 0.5(1000)(0.995)(0.995) - (1000)(9.81)(2) - 0.5(1000)(3.98) = 158885 Pa at point 2. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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