In: Other
1.
you can solve this Type of problems step by step
first write down given data
Here we are provided with
thermal conductivity (k),
Diameter of probe (D),
External ambient temp. ,
distance of probe from wall =X,
surface heat transfer coefficient (h),
radiant heat transfer is negligible
the problem which have combination of convection & conduction heat transfer total resistance to system is provided by these two
where U is overall heat transfer coefficient & 1/U gives overall resistance to the system
h is heat transfer coefficient
& 1/h is resistence to HT
similarly k is conductivity of material
x is distance across which temperature variation is observed
now finally the heat transfer problem can be solved by using
Q =U A
= log mean temperature = (T2-T1)/ln(T2/T1)
we can get area , overall HT coefficient from data also we have ambient temp now we can calculate probe temp
simply by putting all other values in equation & solving it .
always problems will be provided with all other data & by combination of formulaes we can solve our question .
2.
if you are asked for convection the you just simply need to use below formula
Q = hA
where A is heat transfer area .
similarly if we asked to solve for conduction the just use Q = k*A *dT/dX
3.
how to determine minimum length of probe to be consider thermally infinite
where is ambient temp. & Tx is temp at x distance from heated end
at which x
that length x can be considered as thermally infinite .
we can write it as
lim