T(t)=e−kt(∫ekt[KM(t)+H(t)+U(t)]dt+C)
M is the outside temperature, H is other things that affect temperature
in the tank(0 in this case), and U is the solar panel. K comes from the
time constant, and should be the inverse of the time constant I believe.
T is temperature, t is time.
T(t)=e−164t(∫e164t[164(80)+4t]dt
After integrating I keep getting
−16304+256t+Ce−164t
I calculate C to be 16414 setting t equal to 0 and using the initial conditions
8^2 /2+5(15-7)
=64/2+75-35
=32+40
=72
<span><span>3<span>(<span>5−9</span>)</span></span>+<span>4<span>(<span>4−9</span>)
</span></span></span><span>=<span><span><span>(3)</span><span>(<span>−4</span>)</span></span>+<span>4<span>(<span>4−9</span>)
</span></span></span></span><span>=<span><span>−12</span>+<span>4<span>(<span>4−9</span>)
</span></span></span></span><span>=<span><span>−12</span>+<span><span>(4)</span><span>(<span>−5</span>)
</span></span></span></span><span>=<span><span>−12</span>+<span>−20
</span></span></span><span>=<span>−32
</span></span><span><span>10<span>(<span>9−18</span>)</span></span>−<span>32
</span></span><span>=<span><span><span>(10)</span><span>(<span>−9</span>)</span></span>−<span>32
</span></span></span><span>=<span><span>−90</span>−<span>32
</span></span></span><span>=<span><span>−90</span>−9
</span></span><span>=<span>−<span>99
</span></span></span><span><span>−<span>12<span>(<span>5−7</span>)</span></span></span>−<span>10<span>(<span>2−5</span>)
</span></span></span><span>=<span><span><span>(<span>−12</span>)</span><span>(<span>−2</span>)</span></span>−<span>10<span>(<span>2−5</span>)
</span></span></span></span><span>=<span>24−<span>10<span>(<span>2−5</span>)
</span></span></span></span><span>=<span>24−<span><span>(10)</span><span>(<span>−3</span>)
</span></span></span></span><span>=<span>24−<span>(<span>−30</span>)
</span></span></span><span>=<span>54</span></span>
Wall height=wall above ground + wall below ground
or, 6 ft = 4 ft + wall below ground
or, wall below ground = 2 ft
It’s larger median one ages the graphs looks bigger
