# Homework Help: Pumping power in pipe

1. Apr 8, 2016

### foo9008

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. The attempt at a solution

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2. Apr 9, 2016

### 256bits

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.

3. Apr 9, 2016

### foo9008

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. Apr 10, 2016

### 256bits

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.

5. Apr 10, 2016

### foo9008

the pipe friction outweigh the (v^2) / 2g ???? , so the the pumping pressure in small pipe is higher?