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how to calculate mixed air temperature

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Question:I think I'm a little confused: How is it that warm air is less dense than cold air (i.e., temperature and density are inversely related), but both temperature and density decrease with increasing altitude (i.e., the two are directly related)? Can anyone help?

Answers:Its correct to say air density decreases with the increase of temprature but only in a constant pressure, but when we increase altitude, the main cause of a decrease in density is because of decrease in air pressure (PV=nRt) Or in other words, when P decreases, for a constant volume and gas, we have decrease in temperature You're not to mixed up, you just forgot to consider the air pressure factor! But that was a very good question!

Question:The mean air temperature in degrees Fahrenheit for a city is modeled by the function T(d) = 45 sin 2pi/365 (d - 108) + 30 where T is the temperature and d is the time in days with d=1 representing January 1. Between which of the days of the year is the mean temperature in the city at least 60 degrees? a) 132-266 b) 138-260 c) 144-254 d) 150-248 sorry... 45 sin[2pi/365] (d - 108) + 30 Can I easily calculate on a graphing calculator?

Answers:This is a sine curve with a vertical offset, a phase shift, a frequency change and a horizontal displacement. If you sketch the curve and view the Y axis as temperature, X axis as days of the year, the answer will leap off the paper.

Question:Hey guys, i was recently solving a fluid mechanics problem on finding out the internal temperature of the hot air inside the HOT AIR BALLOON. The details are simple and i kinda used the normal method solving it. The conditions mentioned were the balloon hovers in equilibrium at 4 KM height and the internal and external pressures are same. I arrived with an internal temperature of 50 deg C. I just wanted to cross check it. Can anyone tell me if this is a reasonable answer, like a logical one. I read in some forum that the maximum temperature is 120 deg C as the Nylon would melt by 230 deg C. If you can give me some real life readings or values, it could be helpful.. Thanks in advance.. :) Just wanna see if my answer is logically rite.. I calculated external pressure and temperature of the atmosphere using the gas laws with keeping the temperature and pressure at sea levels to be 1.0133 bar and 288 K respectively, with which i got the internal temperature to be 322.8K which in deg C is approx 50 deg C. In the problem defined to me it was given as the density at ground level is 1.2 kg/m^3 .. this also came in handy to calculate the density of the hot air inside the balloon.. Your answer was so very convincing.. Thank you so much for your time. Cheers Ramesh

Answers:This is not exactly a weather related question but ... what I can tell you is that the average adiabatic lapse rate (how cold it gets with altitude) is 0.65 C per 100 meters. The fall in pressure is not linear, it is one millibar per 8 meters, at sea level and at 4 km, it is about twice that; 16 meters for one millibar. A hot air balloon is not pressurized so we can assume that the air inside is similar to a parcel of warm air rising in the thermal convection of air heated by the ground exposed to the sunlight. If your balloon hoovers at 4 km, it means that the difference of the mass of the air inside and outside is equal to the weight of the balloon, with an eventual load. How do you come to an internal temperature of 50 C? Do you use the average temperature on earth at sea level of 15 C then reduce it by the standard lapse rate? In that case, the temperature at 4 km of altitude should be: 15 - (0.65 * 40) = -11 C. From your calculation, the difference of temperature in order to keep the balloon in equilibrium must be 61 C. How did you come to that? The thing is; as a balloon rises, it expands and that creates an adiabatic cooling of the air inside. One would then think that a hot air balloon needs a constant heating in order to keep it rising. But the fact is that the surrounding air is equally cooling down at the same rate ... unless there is e.g. an inversion, as it may happen in a high pressure where sinking air on the top is warmer than near the ground. But if we keep to the standard atmosphere (0.65 C per 100 meters, 15 C at sea level and 1,013 millibars) then we assume that the temperature sinks at that rate. I have no idea at what temperature nylon melts but I don't think air balloons is warmer than 100 C, do they? The need for now and then heating of the air inside is simply that, with time, the air cools down and some also escape since it is an open at the bottom balloon. Incidentally, I remember seeing an illustration of the past century showing a hot air balloon with passengers and flags and ... the flags were flagging! One thing is for sure; on a hot air balloon, there is no wind since it moves exactly at the speed of the wind! Likewise the Hollywood movie "Master and Commander" has an opening scene of a sail ship sailing downwind into a patch of fog to seek refuge from the enemy. ... how can a ship sail faster than the wind moving the fog patch? Ah, the magic of Hollywood! :-)

Question:because if the thermometer are put near the surface,. it will show ground temperature. where do i put the thermometer? and what the height to place it?

Answers:The measurement of true air temperature is a matter of extreme difficulty owing to the fact that a thermometer freely exposed to the air is affected by radiation of heat to and from sorrounding objects in addition to the direct radiation from the sun.It is therefore necessary to keep the thermometer in some form of enclosure or screen which will allow the air to circulate freely past the bulb of the thermometer and at the same time shield the thermometer from the radiation emitted by its sorroundings. The usual form of screen,Known as Stevenson's screen is a wooden box painted white with louvered sides.Now the air can freely circulate inside and external radiation also is prevented.The box is raised four feet above the ground and is erected in such a way as to open towards the north in the Northern hemisphere and towards south in the southern hemisphere.This is because, the apparent movement of the sun is limited by 23.5 degrees N and S latitudes and there is less chance of the themometer bulbs getting heated by the direct sunrays while opening the door.