# Loss due to contraction

1. Mar 23, 2016

### foo9008

1. The problem statement, all variables and given/known data
what does the author mean by D2/ D1 = 0 ? when D2/ D1 = 0 , the pipe doesn't exist , right ?

2. Relevant equations

3. The attempt at a solution

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2. Mar 23, 2016

### SteamKing

Staff Emeritus
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. Mar 23, 2016

### foo9008

so it's not exactly = 0 , it's approaching 0 , am i right ?

4. Mar 23, 2016

### gmax137

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.