QUESTION 2: Flow...

QUESTION 2: Flow Measurement/Bernoulli z1 z3 z4 z=0 z2 V2 Fig 2.1 Water funnel a) Develop an equation for the velocity exiting the funnel V2 in Figure 2.1, in terms of the dimensions z1 and z2, given the frictional head loss in the tube is 4.25v2"/(2g). b) Determine V2 and the volumetric rate Q (m3/s) given the tube diameter is 1.2 cm and the following dimensions (z1 = 53 cm; z2 = -22 cm; z3 = 100 cm, z4 = 40 cm).

Transcript

00:03 From the burn knowledge equation is p z1 divided by row plus g g g c j 1 plus v1 squared divided by 2 g c plus eta w p is equal to p z divided by ro plus g divided by g cz2 plus v2 squared divided by 2 g plus squared divided by 2gc plus hf where eta w p is the performance of the pump since there is no pump effect in this case this term can be neglective and hf is the fraction head loads so it given as 4 .25 v2 squared divided by 2 since the cross -sectional area at the point z 1 very compared to the cross -sectional area at the point z 2 so the velocity of the z -1 is very low and can be neglected hence we can write as v z1 is equal to 0 therefore the valuation it's become p z1 divided by row g plus z 1 is equal to p z 2 row g plus z 1 is equal to r g plus 2g plus 4 .25 v2 square divided by 2g now in si system whereas gc is equal to 1 since both the points are open to the atmospheric pressure and no additional force difference is there between therefore, we can write as p z1 row g minus p z2 row g is equal to 0, or v2 is equal to z2, or v2 is equal to z1 minus z2, 2g divided by 5 .25, and vzz2 per square is equal to v to square.

03:23 This is the required equation for the velocity at the point z2 in terms of dimensions z1 and z2.

03:34 So in the given data is tube diameter is 1 .2 cm which is 0 .01 meter and z1 is given as 53 cm.

03:51 This is also converting to meter that is 0 .53 meter j2 is equal to minus 22 centimeter 0 .22 meter z3 is equal to 100 centimeter is equal to 1 meter z4 is equal to 40 centimeter is equal to 0 .40 meter from the above equation we have v2 is equal to z1 minus z22g divided by 4 5 .25 and substituting the all values z1 is 0 .53 minus that is equal to plus minus minus minus plus 0 .22 to 2 gravity is 9 .8 divided by 5 .25, which is equal to 1 .67 .33 meter per second, or we can write as 167 .33 centimeters per second.

05:27 Now, cross -sectional area of the tube b is equal to alpha is equal to 5...

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