4 25 pts a centrifugal pump has r1 10 cm r2 18 cm b1 15 cm...

#4. (25 pts) A centrifugal pump has $r_1 = 10$ cm, $r_2 = 18$ cm, $b_1 = 15$ cm, $b_2 = 10$ cm, $\beta_1 = 25^\circ$, and $\beta_2 = 40^\circ$ and rotates at 1,160 r/min. If the fluid is water at $20^\circ$C and the flow enters the blades radially, estimate the theoretical (a) flow rate in L/min, (b) horsepower, and (c) head in m. For relevant equations, please see page 8. Hint: $\rho_{water} = 998$ kg/$m^3$

Transcript

00:01 A centrifugal pump is given with the value of r2 which is equal to 9 inch converting into fate.

00:09 This will be equal to 0 .75 feet.

00:13 Then the value of b2 is equal to the 2 inch.

00:18 Converting into fit this will be equal to 0 .1 ,667 feet.

00:26 Then the value of beta 2 this is equal to the 35 degree value of the angular speed this is given which is 160 rounds per minute converting into radian per second this will be 111 .0029 radian per second and the value of the height edge is given which is equal to 180 feet so in the equation a we have to find the theoretical flow rate so first we will apply the theoretical head formula this can be given by u2 square divided by 2 minus u2 cot beta 2 divided by 2 divided by 2 pi r2 b2 into multiplication of g this will be multiplied by q so we have to find the value of q then this will become u2 squared divided by 2 minus h divided by u2 cod beta 2 divided by 2 by r2 b2g so here the value of the g this will be given by 32 .2 feet per second square this is the gravitational escalation then here the tangential velocity formula can be given by the angular velocity multiply by r so we can found the value of u2 this will be w2 is equal to 111 .0 .029 multiplied by r which is 0 .75 feet so this will be equal to 83 .25 feet per second now further solving we will apply all this value in this equation q so therefore we can write this, the q will be equal to the 83 .25 divided by 32 .2.

03:01 This will be square of 85 .25 minus 180 divided by, this will be again we can write 83 .25 multiplied by cost of 35 degree divided by 2 multiplied by pi multiplied by 0 .75 multiplied by 0 .1667 multiplied by 32 .2.

03:37 By solving this, we will get the value of q.

03:41 This will be equal to 3 ,300.

03:46 64 .897 galleon per minute.

03:54 So this will be the answer of the first question and then for the second one where we have to find the theoretical horse power...

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