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4.65 The velocity of the water entering the volume shown in Figure P4.65 is $V(t) = 10e^{-t/10}$ m/s. Assuming $h(0) = 0$, find $h(t)$ if the volume is: (a) A cone. The inlet is 40 mm in diameter. (b) A 10-m-long trough. The inlet is 40 mm high. Figure P4.65 $\frac{d V_{out}}{d V} - n_i - m_o$, $\frac{d h}{d t} = \rho (0.02)^2 \pi (10e^{-t/10})$ $A = \pi \left(0.02 + \frac{h}{\tan 30}\right)^2$ $\frac{d h}{d t} \pi \left(0.02 + \frac{h}{\tan 30}\right)^2 = 0.02^2 \pi (10e^{-t/10})$ $\frac{d h}{d t} \left(0.02 + \frac{h}{\tan 30}\right)^2 = 0.02^2 (10e^{-t/10}) dt$ $u = 0.02 + \frac{h}{\tan 30}$ $\int u^2 (\tan 30) du$ $\frac{\tan 30}{3} \left(0.02 + \frac{h}{\tan 30}\right)^3 = -0.04 e^{-t/10} + C$ $h(0) = 0$, $C = 0.01$ $\frac{\tan 30}{3} \left(0.02 + \frac{h}{\tan 30}\right)^3 = -0.04 e^{-t/10} + 0.01$ $\left(0.02 + \frac{h}{\tan 30}\right)^3 = \frac{3}{\tan 30} (-0.04 e^{-t/10} + 0.01)$ $\left(0.02 + \frac{h}{\tan 30}\right)^3 = -0.2078 e^{-t/10} + 0.05196$ $0.02 + \frac{h}{\tan 30} = \sqrt[3]{-0.2078 e^{-t/10} + 0.05196}$ $h(t) = \tan 30 \left(\sqrt[3]{-0.2078 e^{-t/10} + 0.05196} - 0.02\right)$ $h(t) = \sqrt[3]{-0.263 e^{-t/10} + 0.263} - 0.012 m$

          4.65 The velocity of the water entering the volume shown in Figure P4.65 is $V(t) = 10e^{-t/10}$ m/s. Assuming $h(0) = 0$, find $h(t)$ if the volume is:
(a) A cone. The inlet is 40 mm in diameter.
(b) A 10-m-long trough. The inlet is 40 mm high.
Figure P4.65
$\frac{d V_{out}}{d V} - n_i - m_o$, $\frac{d h}{d t} = \rho (0.02)^2 \pi (10e^{-t/10})$
$A = \pi \left(0.02 + \frac{h}{\tan 30}\right)^2$
$\frac{d h}{d t} \pi \left(0.02 + \frac{h}{\tan 30}\right)^2 = 0.02^2 \pi (10e^{-t/10})$
$\frac{d h}{d t} \left(0.02 + \frac{h}{\tan 30}\right)^2 = 0.02^2 (10e^{-t/10}) dt$
$u = 0.02 + \frac{h}{\tan 30}$
$\int u^2 (\tan 30) du$
$\frac{\tan 30}{3} \left(0.02 + \frac{h}{\tan 30}\right)^3 = -0.04 e^{-t/10} + C$
$h(0) = 0$, $C = 0.01$
$\frac{\tan 30}{3} \left(0.02 + \frac{h}{\tan 30}\right)^3 = -0.04 e^{-t/10} + 0.01$
$\left(0.02 + \frac{h}{\tan 30}\right)^3 = \frac{3}{\tan 30} (-0.04 e^{-t/10} + 0.01)$
$\left(0.02 + \frac{h}{\tan 30}\right)^3 = -0.2078 e^{-t/10} + 0.05196$
$0.02 + \frac{h}{\tan 30} = \sqrt[3]{-0.2078 e^{-t/10} + 0.05196}$
$h(t) = \tan 30 \left(\sqrt[3]{-0.2078 e^{-t/10} + 0.05196} - 0.02\right)$
$h(t) = \sqrt[3]{-0.263 e^{-t/10} + 0.263} - 0.012 m$

465 the velocity of the water entering the volume shown in figure p465 is vt 10e t10 ms assuming h0 0 find ht if the volume is a a cone the inlet is 40 mm in diameter b a 10 m long trough th 83298

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