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Re: fluid analogs of battery, resistor, capacitor, inductor



William Beaty wrote:

A water pump can be a confusing electrical analogy because
real water pumps are usually CONSTANT FLOW devices,
whereas batteries are CONSTANT PRESSURE devices.

and he provided alternatives (see below).

Here is a way to visualize a constant pressure device. Imagine
a lake from which water is pushed to an elevated tank by a
wind operated pump. The pump is between the lake and the
tank and it stops automatically when the pressure from the
tank reaches a certain limit (too much p for this wind). Now
you have a fully charged battery. To have a closed circuit add
another pipe, it leads from the tank to the lake. You may
have a valve corresponding to an electric switch. Turn it "on"
and the current starts flowing, according to the diameter and
length of the pipe, for this setup.

The wind propeller of the pump corresponds to chemical
reactions inside the battery, it starts turning as soon the water
level in the tank starts dropping. The gravitational d.o.p. will
be nearly constant if the output current is not too large for
a given pump (and wind). Too much current will result in
a lower "voltage". You can even define the "internal r" as
the ratio of the drop in d.o.p. over the current.

If the unit of the gravitational d.o.p. is J/kg, and the unit of I
is kg/s, then the unit of r will be watt/kg^2. How many
W/kg^2 will a pipe have when its length and area are given?
I stop here. But this question can be answered, I suppose, if
the viscosity of water is given. A frictionless pipe and an ideal
fluid with be an analogy for a superconducting wire.

Ludwik Kowalski

A constant-pressure water pump is a strange device. It only runs itself
just enough to maintain a certain pressure difference across its inlet and
outlet. If you plug either hose of a constant-pressure water pump, the
pump will stop and sit quitely, with no overheating. If you allow a small
leak to occur between inlet and outlet, the pump will "see" this, and
start running just fast enough to keep the pressure up regardless of the
leak. If you connect the output solidly to the input, the pump will speed
up until it operates far beyond its design ratings, the bearings will heat
up, and the stored fuel will be rapidly exhausted.

However, you can create a simple and fairly obvious constant-pressure
water pump. Your "capacitor", if one of the springs is removed, becomes
much like a battery (you'd have to use specialized constant-force
springs.) Or perhaps remove ALL springs, turn the device upright, and
place a heavy weight on the piston. We get an added bonus: this "pump"
has limited energy storage, just like a battery. To "charge" this
"battery," pump the fluid out of the top and into the bottom, so the
piston is driven upwards. To use this "battery" as a power supply,
connect it to a resistor. The fluid then will flow slowly around the
circuit, through the resistor, and the weighted piston will slowly descend
until the "battery" is "dead," and needs to be "recharged". If at any
time you should pinch the tube closed at any point, the circuit will be
"open", and the piston will halt it's descent.