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Zulumike
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Hello everyone.
I've been doing a bit of independent study for this topic without much background and so my thermodynamic knowledge is fairly limited. I came across this problem and I'd like some assistance with it! If anyone can help out, I would be very grateful. A lot of the equations came from my own research, so I'm not sure if my starting point is altogether correct.
A 0.0985 m radius hollow sphere contains a spherical cavity with 0.0785 m radius. The sphere is made out of a material with a k-value of 0.003 W/m.K. Ambient temperature around the sphere is 25 C, and the initial temperature within the cavity is 5.5 C. Convection heat transfer coefficient of the air is 10.45 W/m2.K for both the internal and external air. Calculate the steady-state heat transfer rate at this given temperature.
Conductive heat transfer through spherical wall.
$$Q_{sphere} = -kA \frac{dT} {dr}$$
Convective heat transfer on a wall
$$q_{wall} = hA (T_{ambient} - T_{wall})$$
Looking at a few calculations online, I determined that the ##Q_{sphere}## can be estimated as
$$Q_{sphere} = 4 \pi k r_1 r_2 \frac {(T_{intwall} - T_{exwall})} {r_2-r_1}$$
Where ##r_1## is the interior radius and ##T_1## is the internal temperature.
Using this http://www.learnthermo.com/examples/ch04/p-4b-2.php link, I think I found a way to connect steady state calculations to combine convective and conductive heat transfer. Using a similar method to the one in the link, I assume that the internal convection heat transfer rate is equal to the heat conduction rate which is also equal to the external convection heat transfer rate.
Internal convection equations:
$$q_{internal} = 4 \pi h r_1^2 (T_1 - T_{intwall})$$
$$q_{external} = 4 \pi h r_2^2 (T_{exwall} - T_2)$$
Where ##T_2## is the external ambient air and ##T_1## is the internal ambient air.
Isolating these for the ##T_{intwall}## and ##T_{exwall}##:
$$T_{intwall} = T_1 - \frac {q} {4 \pi h r_1^2}$$
$$T_{exwall} = T_2 + \frac {q} {4 \pi h r_2^2}$$
Plugging these into the ##Q_{sphere}## equation:
$$q = 4 \pi k r_1 r_2 \frac {(T_1 - \frac {q} {4 \pi h r_1^2} - (T_2 + \frac {q} {4 \pi h r_2^2}))} {r_2-r_1}$$
Plug this into wolframalpha to isolate ##q##:
$$q = -\frac {(4 \pi h k r_1^2 r_2^2 (T_2 - T_1))} {(r_1^2 (k - h r_2) + h r_2^2 r_1 + k r_2^2)}$$
Plugging in values:
h=10.45
r1 = 0.0785
r2 = 0.0985
k = 0.003
T1 = 25
T2 = 5.5
Gives a final answer of
$$q = -0.276 W$$
Is this correct? Is my method at all correct? I'm really not entirely sure on the equations I used or whether I connected them correctly. Any help or pointing in the right direction would be much appreciated.
Thank you!
I've been doing a bit of independent study for this topic without much background and so my thermodynamic knowledge is fairly limited. I came across this problem and I'd like some assistance with it! If anyone can help out, I would be very grateful. A lot of the equations came from my own research, so I'm not sure if my starting point is altogether correct.
Homework Statement
A 0.0985 m radius hollow sphere contains a spherical cavity with 0.0785 m radius. The sphere is made out of a material with a k-value of 0.003 W/m.K. Ambient temperature around the sphere is 25 C, and the initial temperature within the cavity is 5.5 C. Convection heat transfer coefficient of the air is 10.45 W/m2.K for both the internal and external air. Calculate the steady-state heat transfer rate at this given temperature.
Homework Equations
Conductive heat transfer through spherical wall.
$$Q_{sphere} = -kA \frac{dT} {dr}$$
Convective heat transfer on a wall
$$q_{wall} = hA (T_{ambient} - T_{wall})$$
The Attempt at a Solution
Looking at a few calculations online, I determined that the ##Q_{sphere}## can be estimated as
$$Q_{sphere} = 4 \pi k r_1 r_2 \frac {(T_{intwall} - T_{exwall})} {r_2-r_1}$$
Where ##r_1## is the interior radius and ##T_1## is the internal temperature.
Using this http://www.learnthermo.com/examples/ch04/p-4b-2.php link, I think I found a way to connect steady state calculations to combine convective and conductive heat transfer. Using a similar method to the one in the link, I assume that the internal convection heat transfer rate is equal to the heat conduction rate which is also equal to the external convection heat transfer rate.
Internal convection equations:
$$q_{internal} = 4 \pi h r_1^2 (T_1 - T_{intwall})$$
$$q_{external} = 4 \pi h r_2^2 (T_{exwall} - T_2)$$
Where ##T_2## is the external ambient air and ##T_1## is the internal ambient air.
Isolating these for the ##T_{intwall}## and ##T_{exwall}##:
$$T_{intwall} = T_1 - \frac {q} {4 \pi h r_1^2}$$
$$T_{exwall} = T_2 + \frac {q} {4 \pi h r_2^2}$$
Plugging these into the ##Q_{sphere}## equation:
$$q = 4 \pi k r_1 r_2 \frac {(T_1 - \frac {q} {4 \pi h r_1^2} - (T_2 + \frac {q} {4 \pi h r_2^2}))} {r_2-r_1}$$
Plug this into wolframalpha to isolate ##q##:
$$q = -\frac {(4 \pi h k r_1^2 r_2^2 (T_2 - T_1))} {(r_1^2 (k - h r_2) + h r_2^2 r_1 + k r_2^2)}$$
Plugging in values:
h=10.45
r1 = 0.0785
r2 = 0.0985
k = 0.003
T1 = 25
T2 = 5.5
Gives a final answer of
$$q = -0.276 W$$
Is this correct? Is my method at all correct? I'm really not entirely sure on the equations I used or whether I connected them correctly. Any help or pointing in the right direction would be much appreciated.
Thank you!
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