Gasoline with a density of 737 kg/m³ moves through a...

Gasoline with a density of 737 kg/m³ moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P? = 1.85 × 10? Pa, and the pipe diameter is 5.00 cm. At another point y = 0.30 m higher, the pressure is P? = 1.15 × 10? Pa and the pipe diameter is 2.50 cm. (a) Find the speed of flow (in m/s) in the lower section. m/s (b) Find the speed of flow (in m/s) in the upper section. m/s (c) Find the volume flow rate (in m³/s) through the pipe. m³/s

Transcript

00:01 In this question we are given that pressure and area of the two sections is given to us and elevation difference is also given.

00:09 We need to calculate the speed at both the sections and also the rate of volume flow.

00:14 So for this question we will apply bernoules's equation at those two sections.

00:23 This is section 2.

00:26 This is section 1.

00:30 P1 is given to us as 1 .85 into 10 to the power 4 pascal, 4 ,000.

00:38 Meters square we will say p2 is 1 .15 into 10 to the power 4 newton per meters square this is also given the elevation difference is 0 .3 meters this is 0 .3 meters area of this section is given by the radius is actually diameter is given to us that is 2 .5 centimeters and over here this is 5 centimeters times of a 2 as we know because the ratio of their diameter is twice so ratio of their area would be 4 time so if this is 4 time then velocity of section 2 would be 4 times of that of velocity of section at 1 why because area into velocity has to be constant because a 1 into b1 is equal a 2 into b2 is equal that is 2 q has to be constant so this must be equal from there we get v2 is equal to four times of v1 so now we will apply per knowledge equation at the two sections so p1 by gamma plus v1 square by 2g plus z1 is equal to p2 by gamma plus v2 square by 2g plus z now z2 is let us say 0 .3 this is 0 p2 is known to us p2 is 1 .15 1 1 .15 into 10 to the bar 4 this is also known to us this is 1 .85 into 10 to the bar 4 and gamma is 737 into 9 .8 1 737 into 9 .8 1 so this is 722 9 .97 7 2 .9 .97 7 2 .9 7 to 3 0 let us say this is 72230 72230 and also the v2 is replaced by 4 times v1 that means this would become to be 16 v1 squared by so after solving this we will get 15 times v1 square by 2g is equal to 1 .85 into 10 to the power 4 minus 1 .15 into 10 to the power 4 divided by 7 230 right minus 0 .3 so this is 0 .66 668.

03:31 So v1 will come out to be multiplied by 9 .81 divided by 2 divided by 15 and under root of it.

03:41 So this v1 is coming out to be 0 .935 meter per second and v2 would be four times of this.

03:51 So that would be 3 .7395 meter per second.

03:57 So that is a conference back in 2019 while subparts of a second...

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