/*! 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 50 A conical flask contains water t... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 50

A conical flask contains water to height \(H=36.8 \mathrm{mm}\) where the flask diameter is \(D=29.4 \mathrm{mm}\). Water drains out through a smoothly rounded hole of diameter \(d=7.35 \mathrm{mm}\) at the apex of the cone. The flow speed at the exit is approximately \(V=\sqrt{2 g y},\) where \(y\) is the height of the liquid free surface above the hole. A stream of water flows into the top of the flask at constant volume flow rate, \(Q=3.75 \times 10^{-7} \mathrm{m}^{3} / \mathrm{hr}\) Find the volume flow rate from the bottom of the flask. Evaluate the direction and rate of change of water surface level in the flask at this instant.

Short Answer

Expert verified
The volume flow rate out of the flask (\(Q_{out}\)) and the rate of change of the water level can be calculated using the measurements and the equations provided. By comparing \(Q_{in}\) and \(Q_{out}\), you can decide whether the water level is rising, falling, or unchanging, and at what rate in \(m^3/s\).

Step by step solution

01

Determine the radius and area of the hole

First, convert the diameter \(d\) to meters: \(d = 7.35 \times 10^{-3} m\). The radius \(r\) is then \(\frac{d}{2} = 3.675 \times 10^{-3} m\). Using this radius, the area \(A\) of the hole can be calculated using the formula \(A = \pi r^{2}\).
02

Determine the flow speed \(V\) at the hole

To find the flow speed \(V\) at the hole, we use the given equation \(V = \sqrt{2gy}\), where \(g = 9.81 m/s^{2}\) is the gravity constant and \(y = 36.8 \times 10^{-3} m\) is the height of the water surface above the hole.
03

Determine the volume flow rate out of the flask (\(Q_{out}\))

To find the volume flow rate out of the flask \(Q_{out}\), you multiply the flow speed \(V\) by the area of the hole \(A\). The result is the volume flow rate in \(m^3/s\).
04

Determine the volume flow rate into the flask (\(Q_{in}\))

The flow rate of the water into the flask, \(Q_{in}\), is given as \(3.75 \times 10^{-7} \ m^3/hour\). This must be converted to \(m^3/s\) by dividing by \(3600\).
05

Determine the direction and rate of change of water surface level

By comparing \(Q_{in}\) and \(Q_{out}\), you can conclude the rate and direction of change of the water surface. If \(Q_{in} > Q_{out}\), the water surface level is rising with rate equal to the difference \(Q_{in} - Q_{out}\). If \(Q_{in} < Q_{out}\), the water level is falling with a rate equal to the difference \(Q_{out} - Q_{in}\). If \(Q_{in} = Q_{out}\), there is no net change in the water surface level in the flask.

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Most popular questions from this chapter

A pipe branches symmetrically into two legs of length \(L,\) and the whole system rotates with angular speed \(\omega\) around its axis of symmetry. Each branch is inclined at angle \(\alpha\) to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate \(Q\). The pipe diameter, \(D,\) is much smaller than \(L\) Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration \(\dot{\omega} ?\)

A centrifugal water pump with a 0.1-m-diameter inlet and a 0.1 -m-diameter discharge pipe has a flow rate of \(0.02 \mathrm{m}^{3} / \mathrm{s}\). The inlet pressure is \(0.2 \mathrm{m}\) Hg vacuum and the exit pressure is \(240 \mathrm{kPa}\). The inlet and outlet sections are located at the same elevation. The measured power input is \(6.75 \mathrm{kW}\). Determine the pump efficiency.

Incompressible liquid of negligible viscosity is pumped, at total volume flow rate \(Q,\) through two small holes into the narrow gap between closely spaced parallel disks as shown. The liquid flowing away from the holes has only radial motion. Assume uniform flow across any vertical section and discharge to atmospheric pressure at \(r=R\) Obtain an expression for the pressure variation and plot as a function of radius. Hint: Apply conservation of mass and the momentum equation to a differential control volume of thickness \(d r\) located at radius \(r\)

A cylindrical holding water tank has a \(3 \mathrm{m}\) ID and a height of \(3 \mathrm{m}\). There is one inlet of diameter \(10 \mathrm{cm},\) an exit of diameter \(8 \mathrm{cm},\) and a drain. The tank is initially empty when the inlet pump is turned on, producing an average inlet velocity of \(5 \mathrm{m} / \mathrm{s}\). When the level in the tank reaches \(0.7 \mathrm{m}\) the exit pump turns on, causing flow out of the exit; the exit average velocity is \(3 \mathrm{m} / \mathrm{s}\). When the water level reaches \(2 \mathrm{m}\) the drain is opened such that the level remains at \(2 \mathrm{m}\). Find (a) the time at which the exit pump is switched on, (b) the time at which the drain is opened, and (c) the flow rate into the drain \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\)

A cylindrical tank, \(0.3 \mathrm{m}\) in diameter, drains through a hole in its bottom. At the instant when the water depth is \(0.6 \mathrm{m}\) the flow rate from the tank is observed to be \(4 \mathrm{kg} / \mathrm{s}\). Determine the rate of change of water level at this instant.
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