A spherical shaped vessel of 2 m diameter is 90 mm thick. Find the rate of heat leakage, if the temperature difference between the inner and outer surfaces is 220°C Thermal conductivity of the material of the sphere is 0.09 W/m°C.
Solution.
r2=2m;r_2=2m;r2=2m;
r1=1.91m;r_1=1.91m;r1=1.91m;
Δt=220oC;\Delta t=220^oC;Δt=220oC;
k=0.09W/moC;k=0.09W/m^oC;k=0.09W/moC;
Q=ΔtR;Q=\dfrac{\Delta t}{R};Q=RΔt;
R=r2−r14πkr1r2;R=\dfrac{r_2-r_1}{4\pi k r_1r_2};R=4πkr1r2r2−r1;
R=2−1.914⋅3.14⋅0.09⋅2⋅1.91=0.094.32=0.021oC/W;R=\dfrac{2-1.91}{4\sdot3.14\sdot0.09\sdot2\sdot1.91}=\dfrac{0.09}{4.32}=0.021^oC/W;R=4⋅3.14⋅0.09⋅2⋅1.912−1.91=4.320.09=0.021oC/W;
Q=2200.021=10476W=10.476kW;Q=\dfrac{220}{0.021}=10476W=10.476kW;Q=0.021220=10476W=10.476kW;
Answer: Q=10.476kW.Q=10.476kW.Q=10.476kW.
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