Question

In the figure, water flows through a horizontal pipe and then out into the atmosphere at a speed v1 = 16.0 m/s. The diameters of the left and right sections of the pipe are 5.20 cm and 3.00 cm. (a) What volume of water flows into the atmosphere during a 10 min period? In the left section of the pipe, what are (b) the speed v2 and (c) the gauge pressure?

Question 1 In the figure, water flows through a horizontal pipe and then out into the atmosphere at a speed vi - 16.0 m/s. The diameters of the left and right sections of the pipe are 5.20 cm and 3.00 cm. (a) What volume of water flows into the atmosphere during a 10 min period? In the left section of the pipe, what are (b) the speed v2 and (c) the gauge pressure? V2 dc (a) Number (b) Number (c) Number Units Units Units
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Answer #1

The relationship between flow rate (Q )and velocity is

  O-Au.

Where, A is area of cross section of the pipe

Area of the left cross section :

  5.20 x 10-2 ) = 21.24 x 10-4 m?

Area of right cross section of pipe :

  A1 = π x (う 2 10-2 ) = 7.07 x 10-4 m 7.07 x 104m

From the continuity equation

Q2 = Q1

or,   A_{2}v_{2} = A_{1} v_{1}

Given,

v1- 16 m/s

So, 10-4x 7.07 x 16 21.24 x 10-4 5.3 m/s 02

For the right end of pipe

  Q1Alv17.07 x 10 x x16 113.1 m s

For left end of pipe ,

  Q2- A2t2 21.24 x 104 x x5.3 113.1 m3s

(a) The volume of water flows into the atmosphere during a 10 min period will be

V = Q_{1} t = 113.1 imes (10 imes 60) = 67858.4 , , m^{3} = 6.8 imes 10^{4} , , m^{3}

(b) The speed v_{2} , as calculated above, is 5.3 m/s .

(c) Gauge pressure is given by

P_{gauge} = P_{Absolute} - P_{atm}

Where, ho is density of fluid, for water it is 1000 kg/m3 .

To find the pressure in the left section, use Bernoulli’s equation:

P_{1} + ho g h_{1} + rac{1}{2} ho v_{1}^{2} = P_{2} + ho g h_{2} + rac{1}{2} ho v_{2}^{2}

Here, 11 = h2

So,

P_{1}+ rac{1}{2} ho v_{1}^{2} = P_{2} + rac{1}{2} ho v_{2}^{2}

or, P_{2}= P_{1} + rac{1}{2} ho ( v_{1}^{2} - v_{2}^{2})   

Here, P1 = Patm

So, P_{2}= P_{atm} + rac{1}{2} ho ( v_{1}^{2} - v_{2}^{2})

Hence, gauge pressure at the left end of pipe will be

P_{gauge}= rac{1}{2} ho ( v_{1}^{2} - v_{2}^{2}) = rac{1}{2} imes 1000 imes (16^{2} - (5.3)^{2}) = 113955 , , Pa

For any doubt please comment and please give an up vote. Thank you.

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