Is D2/D1 = 0 the Ideal Condition for Pressure Loss in a Pipe?

In summary, the author is discussing the concept of D2/D1 = 0 and its significance in fluid flow. This ratio applies in situations where fluid is flowing from a large volume into a small pipe. The limit of D2/D1 approaching 0 means that the diameter of the pipe is negligible compared to the open volume, and this can be seen in real-life examples such as a drain pipe in a swimming pool. The K factors mentioned are approximations and may not be accurate for critical applications.
  • #1
foo9008
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4

Homework Statement


what does the author mean by D2/ D1 = 0 ? when D2/ D1 = 0 , the pipe doesn't exist , right ?

Homework Equations

The Attempt at a Solution

 

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  • #2
foo9008 said:

Homework Statement


what does the author mean by D2/ D1 = 0 ? when D2/ D1 = 0 , the pipe doesn't exist , right ?

Homework Equations

The Attempt at a Solution

Right. The limiting case when D2/D1 = 0 applies when fluid is flowing from a wide-open volume, say a reservoir, into a pipe suddenly. The diameter of flow from the reservoir D1 is so huge in comparison to the diameter of the pipe D2 that the quantity D2/D1 → 0 in the limit.
 
  • #3
SteamKing said:
Right. The limiting case when D2/D1 = 0 applies when fluid is flowing from a wide-open volume, say a reservoir, into a pipe suddenly. The diameter of flow from the reservoir D1 is so huge in comparison to the diameter of the pipe D2 that the quantity D2/D1 → 0 in the limit.
so it's not exactly = 0 , it's approaching 0 , am i right ?
 
  • #4
foo9008 said:
so it's not exactly = 0 , it's approaching 0 , am i right ?

Right. Think about a drain pipe going straight down from the floor of a swimming pool. The area of the pipe is negligible in comparison to the area of the pool. The next entry in the table is D2/D1 = 0.1, so the big pipe is ten times the diameter of the small pipe (and you use K=0.45). Anything much bigger than that, use 0.5.

Keep in mind these K factors are approximate; they will give you "pretty close" results. For really critical applications, pressure losses are determined by testing. If you're designing something where K=0.45 gives acceptable results but 0.5 does not, you need to re-think your approach.
 

Related to Is D2/D1 = 0 the Ideal Condition for Pressure Loss in a Pipe?

1. What is loss due to contraction?

Loss due to contraction refers to the decrease in volume or size of a material or substance when it is exposed to certain conditions, such as changes in temperature or pressure. This decrease in volume can result in a loss of mass or density.

2. How does loss due to contraction occur?

Loss due to contraction occurs when the molecules of a material or substance become more closely packed together, causing a decrease in volume. This can happen as a result of cooling, where the molecules slow down and move closer together, or under pressure, where external forces push the molecules closer together.

3. What are some examples of loss due to contraction?

Some common examples of loss due to contraction include the shrinking of metals during welding, the decrease in volume of gases when they are compressed, and the decrease in volume of water when it freezes into ice.

4. How is loss due to contraction measured?

Loss due to contraction is typically measured by calculating the change in volume or density of a material or substance before and after it has been exposed to certain conditions. This can be done using specialized equipment, such as a displacement vessel, or through mathematical calculations.

5. Can loss due to contraction be reversed?

In most cases, loss due to contraction is reversible. This means that when the material or substance is returned to its original conditions, it will regain its original volume and density. However, there are some exceptions, such as in the case of irreversible chemical reactions, where the loss due to contraction cannot be reversed.

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