England's Stove Works 50TVL17 Stove User Manual


 
Page | 16
FLOORPROTECTION
RValueCalculation
Aneasymeansofdeterminingifaproposedalternatefloorprotectormeetsrequirementsis
tofollowthisprocedure:
1) ConvertspecificationtoRvalue:
i Rvalueisgivennoconversionisneeded
ii kfactorisgivenwitharequiredthickness(T)ininches:R=1/kxT
iii Cfactorisgiven:R=1/C
2) DeterminetheRvalueoftheproposedalternatefloorprotector:
i Usethecorrectformulagiveninstep1(above)toconvertvaluesnotexpressedas
“R.”
ii Formultiplelayers,addRvaluesofeachlayertodetermineoverallRvalue.
3) If the overallRvalueof thesystem isgreater thanthe Rvalue of thespecified floor
protector,thealternateisacceptable.
EXAMPLE:
The specified floor protector should be ¾” thick material with a kfactor of 0.84.The
proposed alternateis 4”brick witha Cfactorof 1.25over 1/8”mineral boardwith ak
factorof0.29.
Step(a): UseformulaabovetoconvertspecificationtoRvalue.
R=1/kxT=1/0.84x.75=0.893
Step(b): CalculateRofproposedsystem.
4”brickofC=1.25,thereforeRbrick=1/C=1/1.25=0.80
1/8”mineralboardofk=0.29,thereforeRmin.bd.=1/0.29x0.125=0.431
TotalR=R
brick
+R
mineralboard
=0.8+0.431=1.231
Step(c): CompareproposedsystemofRof1.231tospecifiedRof0.893.Sinceproposed
systemRisgreaterthanrequired,thesystemisacceptable.
Definitions:
Thermalconductance =C =_____Btu____ =____W____
(hr)(ft
2
)(degF) (m
2
)(degK)
Thermalconductivity =k =__(Btu)(inch)__ =___W____ =____Btu____
(hr)(ft
2
)(degF) (m)(degK)(hr)(ft)(degF)
Thermalresistance =R =(ft
2
)(hr)(degF) =(m
2
)(degK)
Btu W