Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
The Biot number is given as;
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
The time for the heating process is given as;
Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
Answer:
T=151 K, U=-1.848*10^6J
Explanation:
The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:
pV=nRT
For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.
Using the given mass m, molar mass M, we can get the following equation:
pV=mRT/M
To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:
T=pVM/(Rm); so initial T=302.61K and final T=151.289K
Now we can calculate change of U:
U=3/2 mRT/M using T- difference in temperatures
U=-1.848*10^6 J
Note, that the energy was taken away from the system.
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
