A water-heater is covered up with insulation boards over a total surface area of 30 ft ^{2}. The inside board surface is at 175 F and the outside surface is at 70 F and the board material has a conductivity of 0.05 Btu/h ft F. How thick a board should it be to limit the heat transfer loss to 720 Btu/h ?
Question 3.298E: A water-heater is covered up with insulation boards over a t...
Steady state conduction through a single layer board.
\dot{Q}_{\text {cond }}= k \text { A } \frac{\Delta T }{\Delta x } \Rightarrow \Delta x = k A \Delta T / \dot{ Q }\Delta x=0.05 \frac{\text { Btu }}{\text { h-ft-R }} \times 30 ft ^{2}(175-70) R / 720 Btu / h
= 0.219 ft = 2.6 in
Related Answered Questions
C.V. A + B . Then { }_{1} Q _{2}=\emptys...
C.V. The 1 lbm water.
Continuty: m _{2}=...
C.V. The mass of R-134a. Properties in Table F.10....
C.V. Water. This is a control mass.
m _{2}=...
State 1: T _{1}, \quad v _{1}= V / m =\fra...
\dot{ Q }= k A \frac{\Delta T }{\Delta x }[...
One dimensional heat transfer by conduction, we do...
\dot{ W }= F V =300 lbf \times 40( mi / h...