Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 76 vol% carbide
1 answer:
Answer:
The answer is below
Explanation:
Given that:
Volume of carbide () = 76% = 0.76, Volume of Nickel (
) = 100% - 76% = 24% = 0.24, thermal conductivities of carbide (
) = 30 W/m-K and thermal conductivities of meta. (
) = 67 W/m-K
a) The maximum thermal conductivity is given by:
Max =
b) The minimum thermal conductivity is given by:
Min =
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Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for
while a is taken as 0.003m and Y is already known
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material