Question

1. A bain Haick concepte wahle (k=0.50 BTR + E) 2o is exposed to Ais at 70.°F (h:- 2.0 BAR) ON one side And Aid Af 20,0F(bag 10.0 WR ) on the eppesite side. Eind the Location MEASURES from the 70E side where the tEMPERATURE is 32.°F.

Answer 1

Given data :

• Thickness of wall, t = 6 inch = 6/12 ft = 0.5 ft
• Thermal conductivity of wall, K = 0.50
• Temperature of air on left side of the wall, $T_i$ = 70 °F
• Convective coefficient on left side of the wall,
  hi
  = 2.0
• Temperature of air on right side of the wall,
  T.
  = 20 °F
• Convective coefficuent on right side of the wall , $h_o$ = 10.0

Left side Right side ho=10.0 BTU h;= 2.0 BTU he ft² of Ti = 70 °F heft 2 OF To = 20 ° Aiz > Ai q CHEAT FLUX) BTU hr ft² for 6in x Plone at which t= 32 °F Thermal Resistenie circuit LL hi Ww t K w I ho

Let at distance of X ft from the left side of the wall, temperature is 32 °F.

Now, let assume that the heat transfer from the wall is one directional and steady.

Steady heat flux through the wall is given by,

$\frac{Q}{A}= q= \frac{\Delta T}{\sum R_{th}} BTU/hr ft^2$   

Where, = Temperature difference between two planes or point

VI

$\sum R_{th}$ = Total thermal resistance between that two planes or points

So, Heat flux through wall  = Heat flux between air on left side and plane whose temperature is 32 °F

• Heat flux through air

$q_1= \frac{T_i - T_o }{\sum R_{th}}$

Now, $\sum R_{th}$ = Total Thermal resistance between air on both sides  

= Convective resistance of air left side + Conductive resistance of wall + Convective resistance of air on right side

So, $\sum R_{th} = \frac{1}{h_i}+ \frac{t}{K}+ \frac{1}{h_o}$

So, $\sum R_{th} = \frac{1}{2.0}+ \frac{0.5}{0.50}+ \frac{1}{10.0}= 1.6 hr ft^2^oF/BTU$

So, steady state heat flux from wall,

  $q_1= \frac{70 - 20 }{1.6}= 31.25 BTU/hr ft^2$

• Heat flux between air on left side and plane whose temperature is T = 32 °F

$q_2= \frac{T_i - T }{\sum R_{th}}$

Where, $\sum R_{th}$ = Total thermal resistance between Air on left side and plane whose temperature is 32 °F

So, $\sum R_{th}$ = Convective resistance of air on left side + Conductive resistance of wall till the plane  

So, $\sum R_{th} = \frac{1}{h_i}+ \frac{X}{K}$   

So, $\sum R_{th} = \frac{1}{2.0}+ \frac{X}{0.50}= 0.5 + 2X$   

So, heat flux, $q_2= \frac{70 - 32 }{0.5+2X}= \frac{38}{0.5+2X}$

Now, for steady state heat transfer,   $q_1=q_2$

So, $31.25= \frac{38 }{0.5+2X}$

So,  $0.5+2X= \frac{38}{31.25}= 1.216$

So,

So,  $X = 0.358 ft = 0.358\times 12 inch = 4.296 inch$

So, at X = 0.358 ft or 4.296 inch distance from left wall, temperature will be 32 °F.

Note : Please do ask in comment if you have any doubt. Thank you.