Solution :
Given :
External diameter of the hemispherical shell, D = 500 mm
Thickness, t = 20 mm
Internal diameter, d = D - 2t
= 500 - 2(20)
= 460 mm
So, internal radius, r = 230 mm
= 0.23 m
Density of molten metal, ρ =
=
The height of pouring cavity above parting surface is h = 300 mm
= 0.3 m
So, the metallostatic thrust on the upper mold at the end of casting is :
Area, A
= 7043.42 N
Answer:
Hello some parts of your question is missing below is the missing part
Convection coefficient = 11 w/m^2. °c
answer : 44.83 watts
Explanation:
Given data :
surface emissivity ( ε )= 0.95
head ( sphere) diameter( D ) = 0.25 m
Temperature of sphere( T ) = 35° C
Temperature of surrounding ( T∞ ) = 25°C
Temperature of surrounding surface ( Ts ) = 15°C
б = ( 5.67 * 10^-8 )
Determine the total rate of heat loss
First we calculate the surface area of the sphere
As =
= = 0.2 m^2
next we calculate heat loss due to radiation
Qrad = ε * б * As( ) ---- ( 1 )
where ;
ε = 0.95
б = ( 5.67 * 10^-8 )
As = 0.2 m^2
T = 35 + 273 = 308 k
Ts = 15 + 273 = 288 k
input values into equation 1
Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )
= 22.83 watts
Qrad ( heat loss due to radiation ) = 22.83 watts
calculate the heat loss due to convection
Qconv = h* As ( ΔT )
= 11*0.2 ( 35 -25 ) = 22 watts
Hence total rate of heat loss
= 22 + 22.83
= 44.83 watts