Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W
Volume=Hh(b1+b2)/2. b1 and b2 are the base of the trapezoid
1) 
2) 8.418
Explanation:
1)
The two components of the velocity field in x and y for the field in this problem are:
The x-component and y-component of the acceleration field can be found using the following equations:
The derivatives in this problem are:
Substituting, we find:
And
2)
In this part of the problem, we want to find the acceleration at the point
(x,y) = (-1,5)
So we have
x = -1
y = 5
First of all, we substitute these values of x and y into the expression for the components of the acceleration field:
And so we find:
And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:
