Last time we learned there are two formulas used to calculate gear ratio, R. Today we’ll see how to use them to calculate a gear reduction between gears in a gear train, a strategy which enables us to reduce the speed of the driven gear in relation to the driving gear. If you’ll recall from last time, our formulas to determine gear ratio are: R = N_{Driven} ÷ N_{Driving } (1) R = n_{Driving} ÷ n_{Driven } (2) Now let’s apply them to this example gear train to see how a gear reduction works. Here we have a driven gear with 23 teeth, while the driving gear has 18. For our example the electric motor connected to the driving gear causes it to turn at a speed, n_{Driving}, of 3600 revolutions per minute (RPM). Knowing these numerical values we are able to determine the driven gear speed, n_{Driven}. First we’ll use Formula (1) to calculate the gear ratio using the number of teeth each gear has relative to the other: R = N_{Driven} ÷ N_{Driving} R = 23 Teeth ÷ 18 Teeth R = 1.27 In gear design nomenclature, the gear train is said to have a 1.27 to 1 ratio, commonly denoted as 1.27:1. This means that for every tooth on the driving gear, there are 1.27 teeth on the driven gear. Interestingly, the R’s in both equations (1) and (2) are identical, and in our situation is equal to 1.27, although it is arrived at by different means. In Formula (1) R is derived from calculations involving the number of teeth present on each gear, while Formula (2)’s R is derived by knowing the rotational speeds of the gears. Since R is the common link between the two formulas, we can use this commonality to create a link between them and insert the R value determined in one formula into the other. Since we have already determined that the R value is 1.27 using Formula (1), we can replace the R in Formula (2) with this numerical value. As an equation this looks like: R = 1.27 = n_{Driving} ÷ n_{Driven} Now all we need is one more numerical value to solve Formula (2)’s equation. We know that the speed at which the driving gear is rotating, n_{Driving} , is 3600 RPM. We use basic algebra to calculate the driven gear speed, n_{Driven }: 1.27 = 3600 RPM ÷ n_{Driven} n_{Driven} = 3600 RPM ÷ 1.27 = 2834.65 RPM Based on our calculations, the driven gear is turning at a speed that is slower than the driving gear. To determine exactly how much slower we’ll calculate the difference between their speeds: n_{Driving} – n_{Driven} = 3600 RPM – 2834.65 RPM ≈ 765 RPM So in this gear reduction the driven gear turns approximately 765 RPM slower than the driving gear. Next time we’ll apply a gear reduction to a gear train and see how to arrive at a particular desired output speed. _______________________________________ |
Posts Tagged ‘licensed engineer’
Gear Reduction
Wednesday, March 5th, 2014Condensation Inside the Steam Turbine
Sunday, September 8th, 2013
Did you know that water droplets traveling at high velocity can take on the force of bullets? It can happen, particularly within steam turbines at a power plant during the process of condensation, where steam transforms back into water. The last couple of weeks in this blog series we’ve been talking about the steam and water cycle within electric utility power plants, how heat energy is added to water during the boiling process, and how turbines run on the sensible heat energy that lies within the superheated steam vapor supplied by boilers and superheaters. We learned that without a superheater there is a very real possibility that the steam’s temperature can fall to mere boiling point. When steam returns to boiling point temperature an undesirable situation is created. The steam begins to condense into water within the turbine. To understand how this happens, let’s return to our graph from last week. It illustrates the situation when there’s no superheater present in the power plant’s steam cycle. Figure 1
After consuming all the sensible heat energy in phase C in Figure 1, the only heat energy which remains available to the turbine is the latent heat energy in phase B. If you recall from past blog articles, latent heat energy is the energy added to the boiler water to initiate the building of steam. As the turbine consumes this final source of heat energy, the steam begins a process of condensation while it flows through the turbine. You can think of condensing as a process that is opposite to boiling. During condensation, steam changes back into water as latent heat energy is consumed by the turbine. When the condensing process is in progress, the temperature in phase B remains at boiling point, but instead of pure steam flowing through the turbine, the steam will now include water droplets, a dangerous mixture. As steam flows through the progressive chambers of turbine blades, more of its latent heat energy is consumed and increasingly more steam turns back into water as the number of water droplets increases. Figure 2 – Water Droplets Forming in the Turbine
The danger comes in when you consider that the steam/water droplet mixture is flying through the turbine at hundreds of miles per hour. At these high speeds water droplets take on the force of machine gun bullets. That’s because they act more like a solid than a liquid due to their incompressible state. In other words, under great pressure and at high speed water droplets don’t just harmlessly splash around. They hit hard and cause damage to rapidly spinning turbine blades. Without a working turbine, the generator will grind to a halt. So how do we supply the energy hungry turbine with the energy contained within high temperature superheated steam in sufficient quantities to keep things going? We’ll talk more about the superheater, its function and construction, next week.
________________________________________ |