Is pipe friction more important than velocity in determining pumping pressure?

In summary, the conversation discusses the relationship between pumping power and various factors such as pressure, volume rate, pipe length, and radius. It is mentioned that pumping power is directly proportional to volume rate, but also proportional to the length of pipe and inversely proportional to the 4th power of radius. The formula for volume rate is also mentioned and it is noted that increasing the radius by a factor of 2 will result in a 16-fold increase in volume rate. However, in Diagram 8-14, it is assumed that the volumetric flow rate is constant. The conversation also touches on the effect of pipe friction on pumping pressure.
  • #1
foo9008
678
4

Homework Statement


in the notes , we can see that the formula of pumping power per unit time is (pressure)(volume rate),
so pumping power is directly proportional to volume rate ...
but , the author told that the pumping power is proportional to the length of pipe , and inversely proportional 4th power of radius ...
As we can see , the volume rate has the formula of [delta(P) (R^4) / (8 μ L ) ] , so when R increases by factor of 2 , the volume rate should increases by factor of 16 , thus the pumping power is 16 times the pipe with radius R , am i right ? however , in diagram 8-14, it's different

Homework Equations

The Attempt at a Solution

 

Attachments

  • l1.jpg
    l1.jpg
    50.7 KB · Views: 325
  • l2.jpg
    l2.jpg
    51.9 KB · Views: 324
Physics news on Phys.org
  • #2
No, you are not right. No where does it say anything about changing the pump to a larger size.

Diagram 8-14 : the assumption is that the volumetric flow rate is constant.
 
  • Like
Likes foo9008
  • #3
256bits said:
No, you are not right. No where does it say anything about changing the pump to a larger size.

Diagram 8-14 : the assumption is that the volumetric flow rate is constant.
assuming the volumetric flow rate in both pipe is constant , thus W / time = P (volume rate ) ... now , only thr pressure is changing ...since in the large pipe , the velocity of water is slow , so the pressure is high ... thus , the pumping power should be higher than the thin pipe , right ?
 
  • #4
foo9008 said:
assuming the volumetric flow rate in both pipe is constant , thus W / time = P (volume rate ) ... now , only thr pressure is changing ...since in the large pipe , the velocity of water is slow , so the pressure is high ... thus , the pumping power should be higher than the thin pipe , right ?
You are forgetting that the pipe friction is greater in the smaller diameter pipe.
The pump has to produce a greater pressure pumping through a smaller pipe.
 
  • Like
Likes foo9008
  • #5
256bits said:
You are forgetting that the pipe friction is greater in the smaller diameter pipe.
The pump has to produce a greater pressure pumping through a smaller pipe.
the pipe friction outweigh the (v^2) / 2g ? , so the the pumping pressure in small pipe is higher?
 

What is pumping power in a pipe?

Pumping power in a pipe refers to the amount of energy required to move a fluid through a pipe from one point to another. It is a measure of the work done by the pump to overcome the resistance of the fluid and any friction in the pipe.

How is pumping power calculated?

Pumping power can be calculated using the following formula: Power = (Flow rate x Pressure) / Efficiency. The flow rate is measured in volume per unit time, pressure is measured in force per unit area, and efficiency is a dimensionless value that takes into account losses in the pumping system.

What factors affect pumping power?

The main factors that affect pumping power include the flow rate of the fluid, the pressure required to move the fluid, the efficiency of the pump, the length and diameter of the pipe, and the viscosity of the fluid. Other factors like elevation changes, bends in the pipe, and valves can also impact pumping power.

How is pumping power related to head and flow rate?

In a pump system, head refers to the pressure or height that the pump needs to overcome to move the fluid. The higher the head, the more pumping power is required. Similarly, as the flow rate increases, the pumping power also increases because more energy is needed to move a larger volume of fluid.

What are some ways to reduce pumping power?

To reduce pumping power, it is important to choose a pump with high efficiency, minimize pipe length and diameter, and use smooth, straight pipes without any obstructions. Proper maintenance, such as keeping the pump impeller clean, can also improve efficiency and reduce pumping power. Additionally, using a lower viscosity fluid or reducing the flow rate can also help reduce pumping power.

Similar threads

  • Introductory Physics Homework Help
4
Replies
116
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
3K
Replies
3
Views
2K
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
15K
  • Introductory Physics Homework Help
Replies
3
Views
882
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top