contestada

Given a thermal conductivity constant of a material of 0 4, a surface area of 100 square feet, a temperature on the far side of a material of 165°F and a temperature of 150°F on the near side, and a material thickness of 1.5 inches, what is the conductive heat transfer rate? 2400 F per ft. 4800 F per ft. 3600 F per ft. 6400 F per ft.

Respuesta :

Answer:4800

Explanation:

Given data

k=0.4

surface area(A)=100[tex]ft&^{2}[/tex]

Temprature on near side of material[tex](T_1)[/tex]=150°F

temprature on far side of material[tex](T_2)[/tex]=165°F

thickness(t)=1.5in=0.125[tex]ft^2[/tex]

Conductive heat transfer rate(Q) =[tex]\frac{k\times A\times\left ( T_2-T_1\right )}{y}[/tex]

Q=[tex]\frac{0.4\times 100\left ( 165-150\right )}{1.5}[/tex]

Q=4800°F per feet

Otras preguntas