Suppose that the pump of Fig. P11.9 is run at 1100 r / min...

Key Concepts

Chart Interpretation

Chart interpretation involves using graphical data provided by manufacturers or experimental results to estimate pump performance parameters. These charts help in understanding how variations in operating conditions, such as pressure, flow rate, and speed, affect the actual delivery of a pump relative to its theoretical performance.

Overall Efficiency

Overall efficiency in pump systems quantifies the ratio of the actual output (such as actual flow rate or work done) to the theoretical output. It encompasses all forms of losses within the system, including hydraulic, mechanical, and volumetric losses, thereby providing a measure of how effectively the pump converts input energy into useful work.

Actual Delivery

Actual delivery is the real-world flow rate of a pump during operation, which is typically lower than the theoretical delivery due to factors such as leakage, frictional losses, and other inefficiencies. Engineers often use performance charts or empirical data to estimate the actual output of the pump when operating under load.

Measured Displacement

Measured displacement is the actual volume of fluid the pump moves during one complete revolution. It is determined through measurements or design specifications and is crucial for calculating the theoretical flow rate. This value is used as the basis for predicting the pump’s performance before accounting for real-world losses.

Theoretical Delivery

The theoretical delivery of a pump is the ideal volumetric flow rate that the pump would produce if there were no losses. It is calculated based on the pump’s displacement (the volume of fluid moved per revolution) and its operating speed. In essence, it represents the maximum possible performance under perfect conditions.

Transcript

00:02 In this problem, first calculating the angular speed, so angular speed omega is given by 2 pi n by 60.

00:10 This is the formula and n is the speed here.

00:13 So putting the data, we get omega is equal to 2 pi into 2875 by 60.

00:22 And on solving it, we get the angular speed is equal to 301 .0 .0.

00:30 07 radian per second.

00:34 So this is the angular speed.

00:36 Now going ahead and calculating the energy per unit mass.

00:42 So energy per unit mass is given by small h is equal to g multiplication capital h.

00:49 So just putting the data so h is equal to 9 .81 multiplication 40 and on solving we get h as 399 2 .4 meter square per second square.

01:06 Now going ahead and calculating specific speed of the pump and it is given by this formula, n s is equal to w, q to the power 1 by 2 by as to the power 3 by 4.

01:20 So going ahead and putting all the data in this formula, i can write ns is equal to 301 .07 multiplicity.

01:34 0 .016 hold to the power 1 by 2 by 392 hold to the power 3 by 4 and on solving it we get the value of ns is equal to 0 .4 433 radian per second this is the specific speed now going ahead and drawing the shape so drawing the diagram drawing the diagram so just look at it carefully i am drawing the diagram after that i will solve the problem further so this diagram is very important diagram in this question and i am now putting all the value here here it is 0 .3 and here in this side it is 0 .6 now now here here i can draw the diagram.

02:45 So just to look at it carefully, i am drawing a rough diagram so that you can understand how to draw the diagram.

03:00 This diagram is very important diagram.

03:05 So draw it more carefully while solving the problem...

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