3HEAT COMIN IN = HEAT FLOWIN OUT .............
LETS TAKE POINT B .............
k(√2T - T) = k(Tc - √2T)
2√2T - T = Tc
2√2 - 1 = Tc/T ........
NOT VERY SURE ..........
Three rods of identical cross sectional area are made up of same metal from the sides of an isosceles triangles ABC, right angled at B. The point A and B are maintained at temperatures T and √2T respectively.
In the steady state, the temperature of the point C is TC. Assuming that only heat conduction takes place
Tc/T=
-
UP 0 DOWN 0 0 10
10 Answers
3
1Tc/T=1
heat flow from A to B = heat flow from A to C + heat flow from C to B
simplifying u will get
(√2T - T )/a = (Tc - T)/√2a + (√2T - Tc )/a
u will get Tc = T ..
106
using electrical analogy... VA = 10 (=T degree) VB = 10√2 .. RAB = R = RBC, RAC = R√2
Let current be i... through RAB.. and i' through RBAC
VC/10 ... is required anser..
So, 10(√2-1) = iR ... (i)
10(√2-1) = i'R(√2+1) .... (ii)
==> (√2+1)i' = i .... (dividing (i) and (ii)) ..
Now, 10√2 - VC = i'R = [i/(√2-1)]*[10(√2-1)/i]
==> VC = 10(√2-1)
So, VC/10 = √2-1
OR TC/T = √2-1
1This Q cud be solved using nishant sir's post in conduction 3 post
It is also ∞ 1/x
Total heat transferred =k A (Tc - T) / √2 x + k A (Tc - √2 T ) / x
In steady state it is eaual to 0...
Solving it Tc/ T = 3 /√2 - 1 = 3(√2 -1)
