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The Venturi tube shown in the figure below may be used as a fluid flowmeter. Suppose the device is used at a service station to
The Venturi tube shown in the figure below may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gasoline (p = 7.00 x 10- kg/m) through a hose having an outlet radius of 1.27 cm. The difference in pressure is measured to be P, - P, = 1.40 kPa and the radius of the inlet tube to the meter is 2.54 cm. P 1 (a) Find the speed of the gasoline as it leaves the hose. m/s (b) Find the fluid flow rate in cubic meters per second. m3/sPractice Another X The Venturi tube shown in the figure below may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gasoline (p = 7.00 x 102 kg/m) through a hose having an outlet radius of 1.16 cm. The difference in pressure is measured to be P, - P, = 1.10 kPa and the radius of the inlet tube to the meter is 2.32 cm. P1 (a) Find the speed of the gasoline as it leaves the hose. 3 X Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the fluid flow rate in cubic meters per second. 6 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m /sSolution or Explanation : Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values. (a) P2 VI V/2 A2 A : Since the tube is horizontal, y, = y, and the gravity terms in Bernoulli's equation cancel, leaving P 1 + = pv1= = P2 + = pv2 2 - - -- 2 2 . 2(P - P2) 2(1.10 x 10* Pa) P 7.00 x 102 kg/m3 land (,2 - v 2 = 3.14 m2/s2. [1] : From the continuity equation, A V, = A,v,, we find V 2 = A1)v1 = -1 = (2.32 cm 2 A , V1 1.16 cmor V, = 4V. [2] Substituting Equation [2] into [1] yields 15v 2 = 3.14 m /s , and v, = 0.458 m/s. Then, Equation [2] gives v, = 4(0.458 m/s) = 1.83 m/s. (b) The volume flow rate is 1 = Azz = (7r,")v, = (1.16 x 10-2 m)2(1.83 m/s) = 7.74 x 10-4 m3/s
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