Answer:
the value of conductivity in IP is
![8.406(Btu)/(ft.hr.F)](https://img.qammunity.org/2020/formulas/engineering/college/glks6l5uni5su89zgdwrc6o8o2eu8na3xl.png)
Step-by-step explanation:
Given that
Thermal conductivity K=14.54 W/m.K
This above given conductivity is in SI unit.
SI unit IP unit Conversion factor
m ft 0.3048
W Btu/hr 0.293
The unit of conductivity in IP is Btu./ft.hr.F.
Now convert into IP divided by 1.73 factor.
![0.57(Btu)/(ft.hr.F)=1 (W)/(m.K)](https://img.qammunity.org/2020/formulas/engineering/college/8uiq83dz4a2sqdgfwkkb88omto38sagvl5.png)
So
![0.57* 14.54(Btu)/(ft.hr.F)=14.54 (W)/(m.K)](https://img.qammunity.org/2020/formulas/engineering/college/dxil2rl5albmi2b0eq3utvn7czsev0cwtf.png)
![8.406(Btu)/(ft.hr.F)=14.54 (W)/(m.K)](https://img.qammunity.org/2020/formulas/engineering/college/v7q7gayyacub2u001wd1mqdrdscmye30j7.png)
So the value of conductivity in IP is
![8.406(Btu)/(ft.hr.F)](https://img.qammunity.org/2020/formulas/engineering/college/glks6l5uni5su89zgdwrc6o8o2eu8na3xl.png)