Pump used to halve the time taken by fluid to flow out

In summary: Substituting this into the first equation, you get: $$\rho g h (2Q) +P= \frac{1}{2} \rho (\frac{2Q}{S_1})^2 (2Q)$$
  • #1
Soren4
128
2

Homework Statement


The tank in picture has section ##S_1## and it's linked with a tube of section ##S_2=\frac{1}{\sqrt{17}} S_1##. The tank is filled at an height ##h## and this level is kept constant by a sink ##R##, in such way that the exiting volumetric flow rate is ##Q=0.3 m^3/s## and constant. Determine the power ##P## of a pump located in the tube that can halve the time taken by the fluid to come out of the tank if the density of the fluid is ##\rho =1000kg/m^3##.
2345678909876543212345678.png

Homework Equations


Bernoulli equation

The Attempt at a Solution


What is unclear to me is how to interpret the condition of halving the time taken by the fluid to come out. I interpeted it as equivalent to double the coming out speed of the fluid, that is to have a volumetric flow rate of ##2Q## instead of ##Q##. Is that possibly correct?

In that case I can write $$\rho g h (2Q) +P= \frac{1}{2} \rho (\frac{2Q}{S_2})^2 (2Q)$$ And find ##P## from here. Is the reasoning correct for this problem?
 
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  • #2
Soren4 said:

Homework Statement


The tank in picture has section ##S_1## and it's linked with a tube of section ##S_2=\frac{1}{\sqrt{17}} S_1##. The tank is filled at an height ##h## and this level is kept constant by a sink ##R##, in such way that the exiting volumetric flow rate is ##Q=0.3 m^3/s## and constant. Determine the power ##P## of a pump located in the tube that can halve the time taken by the fluid to come out of the tank if the density of the fluid is ##\rho =1000kg/m^3##.
View attachment 102598

Homework Equations


Bernoulli equation

The Attempt at a Solution


What is unclear to me is how to interpret the condition of halving the time taken by the fluid to come out. I interpeted it as equivalent to double the coming out speed of the fluid, that is to have a volumetric flow rate of ##2Q## instead of ##Q##. Is that possibly correct?

In that case I can write $$\rho g h (2Q) +P= \frac{1}{2} \rho (\frac{2Q}{S_2})^2 (2Q)$$ And find ##P## from here. Is the reasoning correct for this problem?
I agree with your interpretation (of a poorly worded question). But note the information you have regarding S1. Think about how that might be relevant.
 
Last edited:
  • #3
Soren4 said:
What is unclear to me is how to interpret the condition of halving the time taken by the fluid to come out. I interpeted it as equivalent to double the coming out speed of the fluid, that is to have a volumetric flow rate of ##2Q## instead of ##Q##. Is that possibly correct?
I can think of no reasonable alternative interpretation. Sure!
In that case I can write $$\rho g h (2Q) +P= \frac{1}{2} \rho (\frac{2Q}{S_2})^2 (2Q)$$ And find ##P## from here. Is the reasoning correct for this problem?
You can write a second equation with P = 0 and eliminate h so that P = P(Q only).
 

1. How does a pump halve the time taken by fluid to flow out?

A pump uses mechanical energy to increase the velocity of the fluid, thereby reducing the time it takes for the fluid to flow out. This is achieved by creating a pressure difference between the inlet and outlet of the pump, which accelerates the fluid through the pump.

2. Why is it important to halve the time taken by fluid to flow out?

In many industries, time is a critical factor for production and efficiency. By halving the time taken for fluid to flow out, processes can be completed faster, leading to increased productivity and cost savings.

3. How is the flow rate affected by using a pump?

The flow rate, or volume of fluid flowing per unit of time, is directly affected by the pump. By increasing the velocity of the fluid, the flow rate also increases. Therefore, using a pump can significantly increase the flow rate of a fluid.

4. What types of pumps can be used to halve the time taken by fluid to flow out?

There are various types of pumps that can be used to halve the time taken by fluid to flow out, such as centrifugal pumps, positive displacement pumps, and axial flow pumps. The type of pump used will depend on the specific application and fluid properties.

5. Are there any limitations to using a pump to halve the time taken by fluid to flow out?

While pumps can significantly reduce the time taken by fluid to flow out, there are some limitations. The pump must be properly sized and designed for the specific application to achieve the desired results. Additionally, the fluid properties, such as viscosity and temperature, can also affect the performance of the pump.

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