In a test of the centrifugal pump shown in Fig. P11.12, the...

In a test of the centrifugal pump shown in Fig. P11.12, the following data are taken: $p_{1}=100 \mathrm{mmHg}$ (vacuum) and $p_{2}=500 \mathrm{mmHg}$ (gage). The pipe diameters are $D_{1}=12 \mathrm{cm}$ and $D_{2}=5 \mathrm{cm} .$ The flow rate is $180 \mathrm{gal} / \mathrm{min}$ of light oil $(\mathrm{SG}=0.91) .$ Estimate $(a)$ the head developed, in meters, and $(b)$ the input power required at 75 percent efficiency.

In a test of the centrifugal pump shown in Fig. P11.12, the following data are taken: $p_{1}=100 \mathrm{mmHg}$ (vacuum) and $p_{2}=500 \mathrm{mmHg}$ (gage). The pipe diameters are $D_{1}=12 \mathrm{cm}$ and $D_{2}=5 \mathrm{cm} .$ The flow rate is $180 \mathrm{gal} / \mathrm{min}$ of light oil $(\mathrm{SG}=0.91) .$ Estimate
$(a)$ the head developed, in meters, and $(b)$ the input power required at 75 percent efficiency.

Key Concepts

Pressure Head Conversion

This concept involves converting a pressure difference to an equivalent height of a fluid column using the relation h = p/(?g). It is fundamental in fluid mechanics as it allows engineers to relate pressure measurements to potential energy differences, which is especially useful in pump and piping system analyses.

Flow Rate Conversion

Flow rate conversion is the process of transforming a volumetric flow rate from one unit system (such as gallons per minute) to another (such as cubic meters per second). Accurate unit conversion is crucial for performing consistent calculations in fluid mechanics and pump design.

Pump Efficiency and Energy Conversion

This key concept explains how the hydraulic power delivered by a pump is related to the input power through an efficiency factor. It involves calculating the mechanical work done on the fluid (considering head and flow rate) and then adjusting for real-world losses by dividing by the pump efficiency, thereby determining the required input power.

Centrifugal Pump Operation

This concept covers the fundamentals of how centrifugal pumps work by imparting kinetic energy to the fluid, which is then converted to pressure energy (head). Understanding this mechanism is essential for analyzing pump performance, designing pump systems, and solving related energy and hydraulic problems.

Transcript

00:01 So given in the question, the flow, so for such kind of a pump arrangement, the flow rate of a liquid through the pump, let it be represented by capital q is given to be 120 gpm, which in si units can be written as 120 times 6 .309 into 10 to the power minus that is 7 .57 into 10 to the power minus 3 meter cube per second.

00:36 The diameter of the pipe, the specific gravity of the liquid sg is given to be 0 .9.

00:46 So the density of the liquid, let it be represented by rho l would be 0 .9 into 10 to the power 3 kg per meter cube.

00:56 The diameter of the pipe of the inlet pipe at a at 1 is d1 that is 110 mm which in si unit is 0 .110 meters.

01:15 The diameter of the pipe at 2 which is d2 is 55 mm that is 0 .055 meters.

01:24 The pressure at the point 1 p1 is 95 mm of hg and the pressure at the point 2 let it be represented by p2 is 80 kilo pascal which is 18 to 10 to the 3 newton per meter square it has been asked to find the actual head rise across the pump and and it has also been asked how to estimate the pump motor power requirement.

02:06 So, solving this problem, the area of cross section of the inlet pipe, of the inlet pipe that is at 1 would be a1 that is pi by 4 times d1 square that is 3 .142 by 4 times 0 .110 which comes out to be 9 .504 into 10 to the power minus 3 meters square.

02:40 Similarly, area of cross section of the outlet pipe at 2 would be a2.

02:49 Again, it is pi d2 square by 4, putting the value of pi and d2.

02:56 We get this value to be 2 .376 into 10 to the power minus 3 meters square.

03:05 Now the velocity of the liquid at the inlet pipe location 1 would be v1 that will be equal to q divided by a1, q is the flow rate of the liquid, it is 7 .57 into 10 to the power minus 3 meter q per second and a1 we just have calculated to be 9 .504 into 10 to the power minus 3 meter square.

03:28 Square.

03:29 It comes out to be 0 .796 meter per second.

03:32 If we use the equation of continuity, we get v1 a1 is equals to v2 a2, where v2 is the velocity of the liquid at 2.

03:43 So, v2 can be determined as v1 a1 by a2.

03:47 Now, we know v1, we know a1 and we know a2, putting the value of v1, a1 and a2, we get v2 to be equal to 3 .184 meter per second.

04:02 Therefore, v2 is 3 .184 meter per second...

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