Скачать презентацию PHYSICS 231 Lecture 28 Conduction Convection Radiation

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PHYSICS 231 Lecture 28: Conduction, Convection & Radiation Remco Zegers walk-in: Monday 9: 15 -10: 15 Helproom PHY 231 1

Heat transfer to an object The amount of energy transfer Q to an object with mass m when its temperature is raised by T: Q=cm T Change in temperature Energy transfer (J or cal) Mass of object Specific heat (J/(kgo. C) or cal/(go. C) PHY 231 2

Phase Change GAS(high T) Gas liquid Q=m. Lv Q=csolidm T Solid (low T) Q=cgasm T liquid (medium T) Q=cliquidm T PHY 231 liquid solid Q=m. Lf 3

quiz (extra credit) A block of gold (room temperature 200 C) is found to just melt completely after supplying 4 x 103 J of heat. What was the mass of the gold block? Given: Lf=6. 44 x 104 J/kg Tmelt=1063 o. C cspecific=129 J/kg 0 C a) b) c) d) e) 0. 01 kg 0. 02 kg 0. 03 kg 0. 06 kg 10 kg Q=cm T+m. Lf =129*m*1043+m*6. 44 x 104 =2 x 105*m 4000=2 x 105 m m=0. 02 kg PHY 231 4

How can heat be transferred? PHY 231 5

Conduction Touching different materials: Some feel cold, others feel warm, but all are at the same temperature… PHY 231 6

Thermal conductivity metal T=200 C wood T=200 C The heat transfer in the metal is much faster than in the wood: (thermal conductivity) T=370 C PHY 231 7

Heat transfer via conduction Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P P=Q/ t (unit Watt) P=k. A(Th-Tc)/ x=k. A T/ x k: thermal conductivity Unit: J/(mso. C) metal k~300 J/(mso. C) gases k~0. 1 J/(mso. C) nonmetals~1 J/(mso. C) PHY 231 8

Example A glass window (A=4 m 2, x=0. 5 cm) separates a living room (T=200 C) from the outside (T=0 o. C). A) What is the rate of heat transfer through the window (kglass=0. 84 J/(mso. C))? B) By what fraction does it change if the surface becomes 2 x smaller and the temperature drops to -200 C? A) P=k. A T/ x=0. 84*4*20/0. 005=13440 Watt B) Porig=k. A T/ x Pnew=k(0. 5 A)(2 T)/ x=Porig The heat transfer is the same PHY 231 9

Another one. Heat sink Heat reservoir An insulated gold wire (I. e. no heat lost to the air) is at one end connected to a heat reservoir (T=1000 C) and at the other end connected to a heat sink (T=200 C). If its length is 1 m and P=200 W what is its cross section (A)? kgold=314 J/(ms 0 C). P=k. A T/ x=314*A*80/1=25120*A=200 A=8. 0 E-03 m 2 PHY 231 10

Water 0. 5 L 1000 C And another A=0. 03 m 2 thickness: 0. 5 cm. 1500 C A student working for his exam feels hungry and starts boiling water (0. 5 L) for some noodles. He leaves the kitchen when the water just boils. The stove’s temperature is 1500 C. The pan’s bottom has dimensions given above. Working hard on the exam, he only comes back after half an hour. Is there still water in the pan? (Lv=540 cal/g, kpan=1 cal/(ms 0 C) To boil away 0. 5 L (=500 g) of water: Q=Lv*500=270000 cal Heat added by the stove: P=k. A T/ x=1*0. 03*50/0. 005= =300 cal P=Q/ t t=Q/P=270000/300=900 s (15 minutes) He’ll be hungry for a bit longer… PHY 231 11

inside Isolation Tc Th L 1 L 2 L 3 A house is built with 10 cm thick wooden walls and roofs. The owner decides to install insulation. After installation the walls and roof are 4 cm wood+2 cm isolation+4 cm wood. If kwood=0. 10 J/(ms 0 C) and kisolation=0. 02 J/(ms 0 C), by what factor does he reduce his heating bill? Pbefore=A T/[0. 10/0. 10]=A T Pafter=A T/[0. 04/0. 10+0. 02/0. 02+0. 04/0. 10]=0. 55 A T Almost a factor of 2 (1. 81)! PHY 231 12

Convection T high low PHY 231 13

Radiation Nearly all objects emit energy through radiation: P= Ae. T 4 : Stefan’s law (J/s) =5. 6696 x 10 -8 W/m 2 K 4 A: surface area e: object dependent constant emissivity (0 -1) T: temperature (K) P: energy radiated per second. PHY 231 14

emissivity Ideal reflector e=0 no energy is absorbed Ideal absorber (black body) e=1 all energy is absorbed also ideal radiator! PHY 231 15

A BBQ The coals in a BBQ cover an area of 0. 25 m 2. If the emissivity of the burning coal is 0. 95 and their temperature 5000 C, how much energy is radiated every minute? P= Ae. T 4 J/s =5. 67 x 10 -8*0. 25*0. 95*(773)4=4808 J/s 1 minute: 2. 9 x 105 J (to cook 1 L of water 3. 3 x 105 J) PHY 231 16

radiation balance If an object would only emit radiation it would eventually have 0 K temperature. In reality, an object emits AND receives radiation. P= Ae(T 4 -T 04) where T: temperature of object T 0: temperature of surroundings. PHY 231 17

example The temperature of the human body is 370 C. If the room temperature is 200 C, how much heat is given off by the human body to the room in one minute? Assume that the emissivity of the human body is 0. 9 and the surface area is 2 m 2. P= Ae(T 4 -T 04) =5. 67 x 10 -8 * 2 * 0. 9 *(310. 54 -293. 54)= =185 J/s Q=P* T=185*60=1. 1 x 104 J PHY 231 18