/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 The \(\tan \mathrm{k},\) of diam... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 77

The \(\tan \mathrm{k},\) of diameter \(D,\) has a well-rounded nozzle with diameter \(d .\) At \(t=0,\) the water level is at height \(h_{0} .\) Develop an expression for dimensionless water height, \(h / h_{0},\) at any later time. For \(D / d=10,\) plot \(h / h_{0}\) as a function of time with \(h_{0}\) as a parameter for \(0.1 \leq h_{0} \leq 1 \mathrm{m} .\) For \(h_{0}=1 \mathrm{m},\) plot \(h / h_{0}\) as a function of time with \(D / d\) as a parameter for \(2 \leq D / d \leq 10\)

Short Answer

Expert verified
The equation that describes how the height of the water level decreases over time is \(\frac{dh}{dt} = -\frac{d^2}{D^2} \sqrt{2gh}\). It can be non-dimensionalized to \( \frac{1}{h/h_0} \frac{d(h/h_0)}{d\tau} = -\frac{d^2}{D^2} \sqrt{h/h0}\). Solving this equation will provide an expression that can be used to plot the height of the water as function of time for the given parameters.

Step by step solution

01

Develop an Expression

We start by developing an expression for the dimensionless water height. From Torricelli's law, the speed of water efflux is \(v=\sqrt{2gh}\). The flow rate \(Q\) is \(Q= Av\), where \(A\) is the cross-sectional area of the orifice and \(A= \pi \frac{d^2}{4}\). So, \(Q= \pi \frac{d^2}{4} \sqrt{2gh}\). Since the tank is cylindrical, decreasing volume of the water in the tank is given by: \(-\frac{dV}{dt} = Q\). So, \(-\pi \frac{D^2}{4} \frac{dh}{dt} = \pi \frac{d^2}{4} \sqrt{2gh} \). Solving for \(\frac{dh}{dt}\) gives \(\frac{dh}{dt} = -\frac{d^2}{D^2} \sqrt{2gh} \). This is the expression for how the water level decreases over time.
02

Non-dimensionalize the equation

To non-dimensionalize, we can divide the height \(h\) by the initial height \(h_0\) and introduce a non-dimensional time \(\tau= t \sqrt{2g/h_0}\). The equation becomes: \( \frac{1}{h/h_0} \frac{d(h/h_0)}{d\tau} = -\frac{d^2}{D^2} \sqrt{h/h_0} = -\frac{d^2}{D^2} \sqrt{h/h_0}\)
03

Solve the differential equation

The equation is a first-order non-linear differential equation. Solving it will give the dimensionless height \(h/h_0\) as a function of the dimensionless time \(\tau\). This should be done numerically using a scheme like Euler's method or a Runge-Kutta method.
04

Plot the results

After obtaining the function \(h / h_{0}(\tau)\), it can be plotted as a function of \(\tau\) for the given parameters. First, for \(D / d=10\) and \(0.1 \leq h_{0} \leq 1\). Then, for \(h_{0}=1\) and \(2 \leq D / d\leq 10\). This can be done by substituting these values into the obtained solution and using a software or tool that offers plotting facilities.

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