**Require same or more WORK to pump from a (6" into a 3") than of a (6" into a 6"pipe)?**
Instead of pumping from a **6"** cylinder up into a **6" **cylinder well i'm now pumping from a **6" **cylinder up into a **3"** diameter cylinder well.
I'm unsure if this would require the *same* or *more* **work** to pump than if the entire well was just a **(6"diameter well) ** due to the *added friction.* Nor do i even know where to begin on how to calculate the math of the added friction.
Below is a labeled drawing of a hand pump and well.
To avoid confusion both wells i speak of in the titled question do have the same amount of water in each as the only difference here is the changes of pipe dimension.
DRAWING:
The total amount of water in this well that's being lifted weighs **73.6 (lb).**
The work being done on the lever is:
(Pump rod/Piston/73.6lb) **(A)** ..** 2 ft **to Pump's **^(FULCRUM)^** **(B)**.. to **16ft Lever** with a **10 (lb)** load at the end **(C). ** On a seesaw these loads would be at least *equilibrium*. **WELL**: 20ft height 3" diameter cylinder labeled** (W).** **PUMP**: 1ft height 6" diameter cylinder pump labeled** (P).**
Total amount being pumped per stroke is *(1.5 gallons) * or **12.26 (lb)** *
Last edited by Rip; Aug 29th 2019 at 12:18 PM.
Reason: added photo
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