- Exercise 5 I Was Changing 34 Around Too Much That I Just Made It A Separate Exercise Suppose 310 Grams Of Ethanol Et 1 (556.18 KiB) Viewed 40 times
- Exercise 5 I Was Changing 34 Around Too Much That I Just Made It A Separate Exercise Suppose 310 Grams Of Ethanol Et 2 (658.86 KiB) Viewed 40 times
- Exercise 5 I Was Changing 34 Around Too Much That I Just Made It A Separate Exercise Suppose 310 Grams Of Ethanol Et 3 (490.44 KiB) Viewed 40 times
Ex 11: Here's my version of "conductive losses." A) Use Equation 14-5 and the data in Table 14-4 to estimate the power loss through a window of thickness 3.00cm of glass when the area is 1.00 sq. meters and the inner surface temp.is 13.0C and the outer one is -5.0C. Ans: about half a kilowatt. B) Recalculate this when a set of drapes is between the warm room and the window, lowering the inner surface temp to 8.0C. Ans: 364 watts. C) How about the power loss from 6,00 inches (convert to m) of fiberglass if the inside is 20.00 and the outer is -5.0C? Much more reasonable power loss, right? Hope so. The next two exercises have to do with radiative power loss. As a reminder, the concept is that an object at temperature T gives off energy at a rate (power) given by P=e04T*. generally in the infrared part of the electromagnetic spectrum, unless its red hot or hotter. But that's only PART of the picture. The walls of a room give off energy at rate given by the same equation. When the object and the walls are at the same temp, then each is in equ net transfer of heat. But if, say, you are HOTTER than the walls, you can give off more energy than you get back, and you can cool down. But wait. You gotta give off whatever your metabolic rate is. So you WANT to give off more than you get back from the walls.... Ex 12: A) Confirm that a body of surface area 1.50square meters and surface temp of 307K and emissivity 0.700 has a radiative power output of 529 watts. If you were naked and in space, no walls give you anything back, and you LOSE ALL 529 watts. B) Assume a 2500 Cal-per-day diet (1Cal = 4186J) and show that the average power output (over one day) is about 121 watts. Of course, when you are AWAKE, you might give off 150 watts. In any case, note that it is MUCH less than 529, which is why astronauts in space are in a LOSE-LOSE situation via radiation loss. Early spacesuits were unheated and, even with reduced radiative efficiency, the loss was still too large, and astronauts got COLD. C) Now suppose this same person is in a room where the walls are at a temperature of 20.0C. Thus the walls give infrared radiation BACK. What the person gets is STILL 104T*, with the person's e, the person's area A, but the WALL'S temp T. Show that the power GAINED by the person from the walls is slightly less than 440watts. What this means is that your NET radiative loss is about 90 watts. Will you be cold? Not necessarily. Suppose when you are awake, you generate 150watts, so you need to DUMP energy at the rate of 150watts. The net loss from radiation (90 watts) is too little. You need to lose more or you are HOT. If conduction and convection is 60W, then you total losses match your generation. Aaah. Happy. But if air takes 80W, say, you are losing 170W while only generating 150. Now you are cold. D) This last part is to show why you can feel colder in winter than in summer even when the air temp in the room is the same. Suppose the air temp is the same as suggested in the end of Part C so that it takes away 60 watts through conduction and convection. But the walls are now at 10.0C instead of 20.0C, so that instead of having your generated power (150W) match your power loss (150W total), you are out of balance. By how much?
Ex 13: Cold Carl has a surface temp of 30.0C as he sits in a locker room with walls at 18.00. Assuming Carl generates body heat at 110 watts and loses 45 watts from blowing air (conduction + convection), will Carl feel cold in the locker room? Hint: you have to know Carl's emissivity is 0.71 and his surface area is 1.45 m². (This is another test question from last year.) Ex 14: Here's some fun with the radiation formula P=OAT* A) Find the power (in watts) output of the sun using this information: The radius is 6.96E8m (you need the area of a sphere) and the effective or average) surface temperature is about 5840K B) Since the area A is proportional to R squared, a star which has twice the radius as the sun and twice the surface temp. has (2²) (2)' = 64 times the power output of the sun. Astronomers say such a star has 64 solar luminosities, or about 64 sun's worth of power. A reasonable estimate for Alnilam, the star at the center of Orion's belt, is that it has a radius of about 1.4E10m and surface temperature of about 23,000K. Use ratio reasoning to find how many sun's worth of power this start puts out. (Hint: that's a LOT of sun's worth. Put Alnilam where the sun is and the Earth instantly starts vaporizing. You'll need SPF 1,000,000.) C) Try the same for Betelgeuse, the brightest star in Orion (as seen from Earth, anyway) if its surface temp is 3,300K and the radius is about 500 times the sun's radius. Again, use ratio reasoning. (Between 20,000 and 30,000 sun's worth. Hey, it's still less that Alnilam.) The sun is a star much more powerful than the average star, but there are a few monsters out there.