The construction of this stirling engine is quite simple. The acrylic container holds the working air. The air pushes the piston. The piston rotates the fly wheel. The fly wheel in turn pushes the displacer (the Styrofoam piece). The piston and the displacer are connected to the fly wheel in such a way that they are always out of phase by a quarter cycle. This automatically synchronizes the displacer to fulfill the important function of pushing the air to the hot end when it’s the air’s turn to push, and pushing the air to the cold end when it’s the fly wheel’s turn to push.
The diagrams above show the positions of the piston and the displacer at four stages. During abc, the displacer moves the air towards the hot end so that the air is heated up. This is when we extract work done by the air. During cda, the displacer moves the air towards the cold end so that the air is cooled down. This is when work is done by the piston. Since abc occurred at higher temperature and pressure, in one complete cycle, net work done is by the air.
– The diagram below shows what is most likely in the fountain boy: trapped air above a body of water.
Initially, surface tension helped to hold in the water. However, when hot water is poured onto the boy, the higher temperature leads to higher pressure of the trapped air, which pushes the water out of the boy.
All the bulbs are actually the same size. Some look bigger than others because of the magnifying effect of the curved glass walls. But they do have different weights (and thus densities) because of the slightly different weights of the attached metallic labels.
In theory, as the temperature of the liquid in the tube rises, its density drops, the upthrust it exerts on the bulbs drops, and the bulbs sink one by one.
In practice, some bulbs managed to cling on to the glass walls thanks to the extra friction, and sank only when the next higher bulb nudged them off the wall.
The terminal velocity were excruciatingly slow for bulbs that were not held back by friction. Bulbs whose onset of descent were delayed with the help of friction enjoyed a larger difference between gravitaional pull and upthrust during descent, and thus displayed higher terminal velocity.
The aluminum block is NOT colder than the piece of wood.
It FEELS cold to warm blooded mammals like us because heat is conducted from us into the aluminium block faster than if it were a piece of wood. Our sense of hotness/coldness is not based on actual temperature. Instead, it is based on the RATE (and direction) of heat transfer!
If there were a specie of humans whose body temperature were lower than room temperature, the aluminium block would feel hotter to them because heat will be conducted from the metal into them at a faster rate than if it were a piece of wood.
This toy consists of two glass bulbs connected by a glass tube. The yellow liquid is some kind of volatile liquid. So the two glass bulbs contain two bodies of trapped vapour separated by its own liquid form.
When the palms warm up the lower bulb, the increased temperature causes the yellow liquid to evaporate into the lower bulb. The increased number of vapour molecules in the lower bulb increases the pressure in the lower bulb, thus pushing the yellow liquid up the tube. When all the yellow liquid have been pushed out of the way, the vapour simply bubble their way up the glass tube.
So the yellow liquid was not boiling. The vapour forcing their way from the lower bulb to the upper bulb made it look like boiling.
Wrapping the palms around the upper bulb caused evaporation to happen at the upper bulb instead, thus reversing the direction of bubbles.
When dipped in hot water, evaporation in the lower bulb drove up the pressure in the lower bulb. On the other hand, when dipped in cold water, condensation in the lower bulb drove down the pressure in the lower bulb, causing the higher pressure in the upper bulb to force the liquid and subsequently vapour bubbles down the tube.
Further reading: http://en.wikipedia.org/wiki/Hand_boiler
This toy consists of two glass bulbs connected by a glass tube. The red liquid is some kind of volatile liquid. So the two glass bulbs contain two bodies of trapped vapour separated by its own liquid form.
Because the head is wet, evaporative cooling lowers the temperature in the upper bulb. (Earlier in the video, an ice cube was used to achieve the same cooling effect.) This causes the vapour in the bulb to condense, dramatically lowering the pressure in the upper bulb. The higher pressure at the lower bulb pushes the red liquid up the glass tube.
When the red liquid rises into the bird’s beak, the bird tips. The liquid column breaks and the pressures in the upper and lower bulbs equalize. Without the difference the pressure, the liquid falls back into the lower bulb, and the bird rights itself. The cycle repeats.
With a fan blowing, rate of evaporation was increased, resulting in higher evaporative cooling. The bird on the left thus “drinks” more frequently.
Further reading: http://en.wikipedia.org/wiki/Drinking_bird
As shown in this video (by the pitch of the whistle), as the balloon deflates, the balloon pressure decreases (expectedly), but increases (surprisingly) towards the end.
Is it possible for a smaller balloon to have higher pressure?
Consider the ideal gas equation pV=nRT. When we blow into a balloon, n increases. But so does V. The expansion can result in a drop in pressure instead.
Consider the forces acting on a patch of balloon membrane. The patch experiences an outward force due to balloon pressure and an inward force due to atmospheric pressure. The net outward pressure must be constrained by the tension of the balloon membrane (for static equilibrium).
(Pballoon.A = Patm.A + Tsinθ)
As the balloon expands, the tension increases. But since the curvature of the balloon becomes flatter, the resulting inward constraining force due to tension may actually decrease despite increasing tension. So it is possible for balloon pressure to drop as it expands.
There are are two ways to make a gas hot. This is summed up quite neatly by the First Law of Thermodynamics:
ΔU = Q + Won
This law says that ΔU, the increase in internal energy of a system, can be brought about by
(1) Q, supplying heat to the system, or/and
(2) Won, doing work on the system.
Macroscopically speaking, to do work on a gas is to compress it. Yup, work must be done against the gas pressure to force the gas into to a smaller volume.
Microscopically speaking, compressing a gas must involve a surface, let’s say a piston, charging into the gas. During this process, the piston is like a giant tennis racket smashing back all the trillions of tennis balls, i.e. the gas molecules, that happen to collide into it.
These gas molecules were sent flying back at higher speed than before. Since the average KE of the gas molecules increases, and the temperature of the gas increases.
The on/off mechanism of this “fountain” is effected through the pressure of the air space in the bottle, Pair.
When the bottle is not sealed
As the water drains through the holes, air from outside enters to “top up” the air space so Pair is always Patm. Since Pair + hρg > Patm, water squirts out continuously.
When the bottle is sealed
As the water drains, the air space expands and without any “top up” of the air molecules, Pair must drop. When Pair is low enough such that Pair + hρg = Patm, water stops squirting.
By how much must the air space expand? Well Patm, is about 760 mmHg, or about 1000 cmH20. If Boyle’s Law is roughly valid, then the air space only needs to expand by 1% for Pair to decrease by 1%. This to enough to make Pair 10 cmH2O below Patm, enough to “suck” and hold a water column of 10 cm in the bottle.
P.S. Actually, the required expansion is even less because surface tension also helps to keep the water in the bottle.