Conduction problem finding temperature?

Three building materials, plasterboard (k=0.30 J/ s.m. C degrees), brick (k=0.60 J/ s.m. C degrees), and wood (k=0.10 J/ s.m. C degrees) are sandwiched together. The temperatures at the inside surface is 27 degrees celcius and 0 degrees celcius on the outside. each material has the same thickness and cross sectional area. Find the temperature


(a) at the plasterboard brick interface and


(b) at the brick wood interface





i'm not sure how to work this problem...

Conduction problem finding temperature?
when the steady-state is reached then heat flowing across all intermittent junctions is same or Q-dot =same


Q-dot = - k A dT/dx = - dT/(x/kA)


- ve sign tells that fall in temp and dx are in opposite direction. This like current (motive is potential diff) anology driven here by temp difference (motive force - thermal causality)


I (current) = V/R %26gt;%26gt;%26gt; wherein


x/kA = thermal resistance


Q-dot (0r I same) series circuit


R = R1+R2+1/R3


R =(x /kpA)+(x /kbA)+(x /kwA)=(x/A)[1/kp+1/kb+1/kw] = 3x/K.A (right side composite slab of 3x thick, K conductivity)


as thickness (x) and A crosstion are same, the joint slab (3x) thickness and A crosssection, joint K of composite slab


3/K = 1/0.3 + 1/0.6 + 1/0.1= 15 %26gt;%26gt;


K = 0.2 composite


------------


let plaster-brick junction temp = Tpb


let brick-wood interface temp = Tbw


-----------------------


conduction %26gt;%26gt; Q-dot = - k A dT/dx


Q(27/0) = Q(27/Tpb) = Q(Tbw/0) steady state


-------------------


Q(27/0) = Q(27/Tpb)


(K A/3x) (27-0) = (kp A/x) (27-Tpb)


(K/3) (27-0) = (kp) (27-Tpb)


(0.2/3) (27) = (0.3) (27-Tpb)


(2) (27) = 9 (27-Tpb) =54


Tpb = 21 deg C


----------------------------------


Q(27/0) = Q(Tbw/0)


(0.2/3) (27) = (0.1) (Tbw - 0)


Tbw = 18 deg C


temps are


27 ---- 21 ----- 18 ----- 0


outside ---- Tpb ----- Tbw ----- inside


.........plaster .... brick ...... wood .....
Reply:The quantity of heat conducted per second through is given by





Q/t = k A (T1-T2) /d.





When steady state is reached, Q/t is the same for all the three slabs.





Let the temperatures from one end to the other end be, 27, T1, T2, and 0.


T1: the temperature at the plaster board brick interface


T2: the temperature at the brick wood interface.


Assuming, outside temp of plaster board is at 27 degree.








Q/t = k1 A (27-T1) /d = k2 A (T1-T2) /d = k3 A (T2-0) /d





Since A and d are the same, the above equation reduces to





0.3 (27-T1) = 0.6 (T1-T2) = 0.1T 2 or


3 (27-T1) = 6 (T1-T2) = T 2





Solving these equations





T1 = (7/6) T2





T1 = 21degree c and T2 =18 degree c.

thank you cards