Calculations
To determine whether or not the block would make a decent heater, we
must calculate several properties of the system. These properties
include the amount of energy required by the system over
its operational period, the energy output in watts of the prism into
its surroundings, the mass and volume of paraffin needed to store the
energy required to maintain the internal temperature of the system
over the operational period, and the heat output of the thermal block
into the system in Watts.
First to determine the output from the prism to its surroundings, given
that the area of the prism is 70.57m
2 and the metric R value
is 3.52m
2K/W, and that the difference in temperature between
the prism and its surroundings is 50K.
Working
equation
P=A(T
prism - T
amb)/R
m
Solving
70.57m
2 (300K-250K)/3.52m
2K/W = 1002W
Now that the instantaneous heat output of the prism is found, the total
energy lost to surroundings during the operational period, Q, can be
found.
Working equation
Pt=Q
Solving
1002W(5400s) = 5.41*10^6 J
Now with the total energy known, we can solve for the mass of paraffin
wax necessary to replace the lost energy over the operational period.
With the specific heat of paraffin wax being 2.9 kJ/kg, and the
latent heat of fusion being 200kJ/ kg, the total mass , m, can be
solved for.
Working equation
C
wax (m) (T
high -
F
wax) +
C
wax (m) (
F
wax -T
high) + C
fwax (m) = Q
Solving
2.9kJ/kg*(m)*30 + 2.9kJ/kg*(m)*10 + 200kJ/kg*m = 5.41*10^6J
m=17.12kg
Now to solve for the volume, V, given the density of .93g/cm
3
Equation
m/D
wax = V
Solving
17.12kg/.93g/cm
3 = .0184 m
3
Assuming a block depth of .1m and a block width of .3m, the length of
the block will be .613m in length. Since the length by width gives us a
surface area of one side, the instantaneous heat output of the block
can be calculated.
Working equation
P= A(F
wax - T
prism)/R
wax
Solving
(.1839m
2)(310K-300K)/.4W/m
2 K = 4.6W