What kind of volcano usually forms over a hot spot?
2 answers:
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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
Answer:
The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.
Explanation:
Here the first think you have to consider is the definition of the Reynolds number () for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:
where is the density of the fluid,
is the viscosity, D is the pipe diameter and v is the velocity of the fluid.
Now, we know that Re=2100. So the velocity is:
For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is: