Question

FIGURE P17-56E e develop imensional be used 7Clothing made of several thin la hat kindtrapped air in between, often called ski c layers of fahr clothing, is ly used in cold climates because it is light and a very effective thermal insulator. So it is n esistance monly used in cold climates because iti k asso such clothing has largely replaced thick a resist- ioned coats. hick and heavy old Consider a jacket made of five layers of o synthetic fabric (k = 0.13 W/mK) with 1.5-m space (k 0.026 W/m K) between the layers. mm-thick with cross inner surface temperature of the jacket to be 28C a ran surface area to be 1.25 m2, determine the rate of he ong through the jacket when the temperature of the ouud 0°C and the heat transfer coefficient at the outer surface 25 W/m2 K cm m. What would your response be if the jacket is made of single layer of 0.5-mm-thick synthetic fabric? What should h the thickness of a wool fabric (k ch 0.035 W/mK) if the person is to achieve the same level of thermal comfort wearing ah wool coat instead of a five-layer ski jacket? 17 59

Answer 1

Given, Ticknen of the jacket ,G4 0.1 mm 0.0015 m Tha thenal conductiviby f tha jacket is 013 w/ an nn surPoace temperature of tha JocketT Outdoor Temperature =1,,2-0"c On tha Outen susface the haat tan Pex coefftcient is Sur face e ap th jacket A 1.25 mi The thermal vesi stance ci cuit is as shoon

ne 2 Thae is ivvedsynt hutic Pabsic So,To calculate tho individual ℡istances ate as ollows OwS R,-R3-R5-R Fabric- k,A Substitula values m above 0.000 1 0.13 X1.25 6.153x10 abric 0,000 615.K/w k,A 0.000 0.026 X 1.25 ho A 25 X.25 31-2.5 0.032 K / W

ne3 we now that all ha sesatances ove in series So 'The tota. γ.esistance will be = 510.000615|十4)0.00307|十0.032 Rtotal = 0.0473 K/W the coll Thon the steady state rat op heat bransfer thso e come total Substibuting the voluos e 28 0 0.0 43 aer of 05 mm the jacket s made of a si ick synthetic Pabric, ha vote of heat transPex woould be Cal culated ak nC

ne 0.0005 fabstc 0. 13 xI25 .000 30% k/W 25 X 1.2S K/W 0.032 Total Resistance is; calculated as Pollows じ@.cause ot Y K Patric axe-in serie亼ノ 0, 00 303 +0032 =0.03507 k/w Heat txans fPer touh the wat become Rtotal B substibuting value get a (22-0) O. 03 5 0 198 403 W Tha jacket ls mada with single louer , the heat bransper is increased 2 times, this is ue to the a ct thot te jacket usith air space hawing high ses tance or los con ckckanceo Y loua Con to

e5 Conduct haat acyoss th jacket Ơn order to have the sorna level of the mal com rot et can be calculated a toith wool coat, tha thickness of the 0.04.73. wool k A 0,035 x 1.25 25 X1.2 O.03S X1.25 25 X 1 .2S +0032 0,043 3-0.032 o-043 t o. 000657 m Thickness of toool | t:0.6 5ว่า mm Coot