Question

3.9 e wind chill, which is experienced on a cold indy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is 3mm thick and whose intenor surface is maintained at a temperature of 36°C. On a calm day the convection heat traneper coefficient at the exter surface is 25 W/m²K, but with 30km/h winds it reaches 65 W/m²K In both cases the ambient air temperature is -15°C. (a) what is the ratio of the rate of heat loss per unit area from the skin for the calm day to that for the windy day? - 1) what will be the skin outer spurface temperature for the calm day? For the windy day! What temperature would the air have to assume on the calm day to produce the same heat rate occurring with the air temperature at -15°C on the windy day?

Answer 1

wind To 15°C Human body - icaron as work k3mmy - - kuundy = 65 white Ty » 36°c As per darabook. K... = 0.370 W/mk (human skin) tissu it may nasy depending on dafabooks of different now, considering faffy tissue as a thin slab thicknes 3mm, the heap Rapecfions loss from the body : (9) a heat loss during calm day AT Ž Req I » Q 36 - (-15). I & I s naere s thermal Resigfore KA of tissue - Connectine LA Resistance for wind h.A 51 0.003 0.003. I Calmd 0.370xL 2581 » (Q) - 1060.11 W per m2 area CAM

SI nuclay 0.003 0.3701 + 1 6501 ( mindy & 2170 .ss wper m² area Wow, Rapio of heat loss per mich area for the calm day to that for » 1060.it from the the skin windy day & 0.488 2170.88 (6) To find T2 for Calm day & windy day now considering steady state heef framffer Condition, trascu For Calm day , To = -15°C T, - Iz - Tema lui n a hearty jy T2 - To SIKA he can choose amp two for finding the time to bw Tor more there alternatives. accuracy choose first two, 1,-12 T2 - To d/KA '1na , 36 - T2 0.003370 12-1-15) 1/25

(31-1) = (1+1) 95 370x12 - 37012 & 2572 + 375 4440-375 - (75 + 370) 12 4440-375 (2) ann laten day : 2746 4 how (1) during windy day of 25 w/on2K we have only in the place to put 65 W/m²K (36-72) 0.003/0.370 I/LA . In the Ta ti5 - 370 (36-7) - 65 (3+15) & 370412 - 370 72 6572 +975 4440-975 = 6583 + 370 12 2 3 4 6583 eur - 18.390 te

Air temperature on the calm day so that it will produce same effect as ou te mindy day by rise tupurature: - how me know that, heap loss during mindy day : 2170.88 W/m2 how same heat to be lost during Calm day then air temperapore will be I considering area T - To I again to be mif areas O and here > 2170.08 KA NA 890 in unknown, hah Calm we have to find 2170.58 36 - To - 0.003 tt 0.370 5 3-76 =2170.33 (0.00%*) 10. 370 25 B6 - To = 104.436 To = 36 -104.436 To – 68.436 °C An