
#1
Dec2008, 01:26 AM

P: 4,780

Every pilot knows the difference. Do you? The difference means life or death.
Ref [1] On a standard day: 15C, 101kPa, the atmosphere has known temperature, pressure and density variations. The pressure ratio varies with the altitude h, as given by equation [5.1] for a standard day. This means equation [5.1] already factors the variation of temperature with altitude (lapse rate) via the coefficients 5.265 and 6.876e6. The temperature changes with altitude on a standard day via the lapse rate given by equation [5.6][5.8]. Now, notice equation [5.9]. Here were see the PRESSURE ALTITUDE. This equation is found via the altimeter on the aircraft. It is calibrated based on the conditions on a standard day. If you pick an altitude, say 4,000 feet and plug it into equation [5.1], you will get a pressure ratio. If you take this pressure ratio and plug it into equation [5.9] you will get 4,000 again. Equation [5.1] is the inverse of equation [5.9]. In other words, given any pressure ratio, equation [5.9] will convert it into an equivalent pressure altitude. When you're sitting on the ground, the altimeter will read an (indicated) altitude equal to the elevation if it is a standard day (539' MSL here at KGAI airport). In other words, we are 539 feet above sea level. If a low pressure or high pressure cell has moved into the area due to weather changes, or if it's simply a hot or cold day, the pressure ratio will change. So despite the fact that you're sitting on the ground, the altimeter will read a new value that is NOT 539 feet MSL. In order to tare, or zero out, the altimeter, a small dial is turned on the altimeter setting knob of the kohlsman window. This is set by listening to the tower weather radio frequency, (called the AWOS), and turning the small knob on the bottom of the gauge. This turns a small card with indication marks of the pressure in inches of mercury (inHg), seen between the numbers 2 and 3. On a standard day, this reading is 29.92 or in pilot speak (two niner, ninetwo). At this point, it becomes very important to notice what the instrument is measuring. Pressure ratio. That's it. If the pressure ratio is 0.8637 at 4000ft on a standard day, the gauge will read 4000ft. Suppose its now a hot day, 86F, at 4000ft (or as pilots say, "at altitude"). Recall, the temperature at 4000 feet is normally 44.7F on a standard day. Since the temperature is now nonstandard we must use equation [5.11] to correct the density ratio, and plug that result into equation [5.10] to get the density altitude. What the density altitude is doing is taking the temperature and pressure observed at altitude, and turning it into an equivalent altitude, on a standard day. In our case, the pressure altitude is 4,000ft, but the density altitude is 6,400 feet. This means the airplane will have the performance of flying at 6,400 feet even though the pressure altitude is 4,000ft. This is a result of the effects of temperature on density. Let's say its a hot day today and we want to fly here (KGAI) to Pennsylvania (KLNS). We set the local pressure into the altimeter 29.92. At the end of the runway is a 50 foot obsticle (trees and power lines). We are full of fuel, and there are three passengers on board, so we are at maximum gross takeoff weight. Were today standard temperature 59F, we could expect climb performance of 680FPM (Feet per minute). Because of the high temperature 86F, we can now only climb at 450FPM (Data for a Cessna 172). We've flown out of this airport many times, but today is an unusually hot day. If we don't take into account the density altitude our takeoff distance may increase just enough, and our climb performance might decrease just enough, so that we cannot clear that obsticle. We will fly right into it, probably killing or seriously injuring whoever is on board in the process. Question: My altimeter measures a pressure altitude of 4,000, but its a hot day with a nonstandard temperature of 86F. This is equivalent to a density altitude of 6,400ft on a standard day. But wait, my instrument is calibrated for measurments on standard days! Does that mean when I fly in the airplane my gauge would read 4000ft, but in reality I'm all the way up at 6,400ft? Answer: No. Recall the equation of state: [tex]P=\rho R T[/tex] The altimeter is nothing more than a pressure gauge. It needs to read a pressure ratio of 0.8637 to think its at 4000ft. If its a standard day, it will read this value and the density altitude AND the pressure altitude will be equivalent. But is there another way I can create a pressure ratio of 0.8637? Yes. If I increase the temperature by x, I must decrease the density by 1/x to maintain the same pressure reading. So on a nonstandard day, I could get a pressure ratio of 0.8637, but my density ratio would now change. So all the gauge sees is a pressure ratio of 0.8637. It has no idea what the temperature is, and so it has no idea that the density rato has changed on us. So long as the right combination of temperature and density exists to form a pressure of 0.8637, the gauge will think its at 4000ft. But the performance of the aircraft, dynamic pressure etc, are all calculated according to the DENSITY. [tex]KE = \frac{1}{2} \rho V_\infty^2[/tex] 


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