hw6 - university of iowauser.engineering.uiowa.edu/~me_160/2019/hw solutions/hw6.pdfp3.150 in prob....
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HW6
steady
ρdw
dt= ρg + μ
d2w
dx2
ρg + μd2w
dx2= 0,
d2w
dx2= −
𝜌𝑔
𝜇
dw
dx= −
𝜌𝑔
𝜇𝑥 + 𝐶1, w(x) = −
𝜌𝑔
2𝜇𝑥2 + 𝐶1𝑥 + 𝐶2
Boundary condition 𝑤(−ℎ) = 0 → w(−h) = 0 = −𝜌𝑔
2𝜇ℎ2 − 𝐶1ℎ + 𝐶2
𝑤(ℎ) = 0 → w(h) = 0 = −𝜌𝑔
2𝜇ℎ2 + 𝐶1ℎ + 𝐶2
∴ 𝐶1 = 0 , 𝐶2 =𝜌𝑔
2𝜇ℎ2
▶ w(x) = −𝜌𝑔
2𝜇𝑥2 +
𝜌𝑔
2𝜇ℎ2
w(x) =𝜌𝑔
2𝜇(ℎ2 − 𝑥2) Ans.
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