Answer:
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Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are 295 K when 300 W of power are switched on. 1) Find the plate temperature after 10 minutes.
answer ; 311.36 k
Explanation:
Given data :
sum of mass * specific heat products for a base plate and components ( Mcp )
= 5000 J/K
effective heat transfer coefficient * surface area ( hA ) = 10 W/K
Initial temperature of plate and cooling air temperature( Tc ) = 295 k
power ( Q = W ) = 300 W
a) Determine plate temperature after 10 minutes
10 mins = 600 secs ( t )
heat supplied = change in temp + heat loss
Q * t = mCp ( ΔT ) + hA ( ΔT ) t
300*600 = 5000 * ( T -295 ) + 10 ( T -295 ) * 600
therefore ; T - 295 = 16.363
T = 311.36 K
Answer:critical stress= 20.23 MPa
Explanation:
Since there was an internal crack, we will divide the length of the internal crack by 2
Length of internal crack, a = 0.7mm,
Half length = 0.7mm/2= 0.35mm changing to meters becomes
0.35/ 1000= 0.35 x 10 ^-3m
The formulae for critical stress is calculated using
σC = (2Eγs /πa) ¹/₂
σC = critical stress=?
Given
E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²
γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m
a= Half Length of crack=0.35 x 10 ^-3m
σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂
=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂
=(4.0920 x10 ^14)¹/₂
σC=20.23 x10^6 N/m² = 20.23 MPa