Answer:
the heat loss from this insulated wire is less
Explanation:
Given data in question
diameter of cable (d) = 20 mm
( K ) = 1 W/m-k
heat transfer coefficient (h) = 50 W/m²-K
To find out
the heat loss from this insulated wire
solution
we will find out thickness of wire
heat loss is depend on wire thickness also
we have given dia 20 mm
so radius will be d/2 = 20/ 2 = 10 mm
Now we find the critical thickness i.e.
critical thickness = K / heat transfer coefficient
critical thickness = 1 / 50 = 0.02 m i.e. 20 mm
now we can see that critical thickness is greater than radius 10 mm
so our rate of heat loss will be decreasing
so we can say our correct option is (a) less
Explanation:
there are three stages of heat treatment
1.hit the metal slowly to ensure that the metal maintains a uniform temperature
2.soak or hold the metal at a specific temperature for a alloted period of time
3.cool the metal to room temperature
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
Answer:transmission
Explanation:
i’m not entirely familiar with this but i’m sure it’s transmission!