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England's Stove Works 50TVL17 Stove User Manual

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FLOORPROTECTION

RValueCalculation

Aneasymeansofdeterminingifaproposedalternatefloorprotectormeetsrequirementsis

tofollowthisprocedure:

1) ConvertspecificationtoR‐value:

i R‐valueisgiven–noconversionisneeded

ii k‐factorisgivenwitharequiredthickness(T)ininches:R=1/kxT

iii C‐factorisgiven:R=1/C

2) DeterminetheR‐valueoftheproposedalternatefloorprotector:

i Usethecorrectformulagiveninstep1(above)toconvertvaluesnotexpressedas

“R.”

ii Formultiplelayers,addR‐valuesofeachlayertodetermineoverallR‐value.

3) If the overall R‐value of the system is greater than the R‐value of the specified floor

protector,thealternateisacceptable.

EXAMPLE:

The specified floor protector should be ¾” thick material with a k‐factor of 0.84.The

proposed alternate is 4” brick with a C‐factor of 1.25 over 1/8” mineral board with a k‐

factorof0.29.

Step(a): UseformulaabovetoconvertspecificationtoR‐value.

R=1/kxT=1/0.84x.75=0.893

Step(b): CalculateRofproposedsystem.

 4”brickofC=1.25,thereforeRbrick=1/C=1/1.25=0.80

 1/8”mineralboardofk=0.29,thereforeRmin.bd.=1/0.29x0.125=0.431

 TotalR=R

brick

+R

mineralboard

=0.8+0.431=1.231

Step(c): CompareproposedsystemofRof1.231tospecifiedRof0.893.Sinceproposed

systemRisgreaterthanrequired,thesystemisacceptable.

Definitions:

Thermalconductance =C =_____Btu____ =____W____

  (hr)(ft

2

)(degF) (m

2

)(degK)



Thermalconductivity =k =__(Btu)(inch)__ =___W____ =____Btu____

   (hr)(ft

2

)(degF) (m)(degK)(hr)(ft)(degF)



Thermalresistance =R =(ft

2

)(hr)(degF) =(m

2

)(degK)

Btu  W

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