/*! 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 23 A cylindrical tank is \(1.8 \mat... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 23

A cylindrical tank is \(1.8 \mathrm{~m}\) in diameter and \(3 \mathrm{~m}\) long and is mounted horizontally. Oil of specific gravity \(0.87\) stored in the tank is drawn off through an orifice, \(20 \mathrm{~mm}\) diameter, \(0.6\) discharge coefficient, at the tank's lowest point. Calculate the time taken to reduce the level in the talk from \(0.9 \mathrm{~m}\) to \(0.8 \mathrm{~m}\) above the orifice. \([699.4 \mathrm{~s}]\)

Short Answer

Expert verified
The time taken is approximately 699.4 seconds.

Step by step solution

01

Convert Units and Gather Information

First, convert the diameter of the orifice from millimeters to meters: \( 20 \text{ mm} = 0.02 \text{ m} \). Gather the given information:- Diameter of cylindrical tank \( D = 1.8 \text{ m} \).- Length of tank \( L = 3 \text{ m} \).- Specific gravity of oil \( SG = 0.87 \), density of water \( \rho_w = 1000 \text{ kg/m}^3 \), thus density of oil \( \rho_o = SG \times \rho_w = 870 \text{ kg/m}^3 \).- Initial height of oil above orifice \( h_1 = 0.9 \text{ m} \).- Final height of oil above orifice \( h_2 = 0.8 \text{ m} \).- Orifice diameter \( d = 0.02 \text{ m} \), hence orifice area \( A_o = \pi (0.01)^2 = \pi \times 10^{-4} \text{ m}^2 \).- Discharge coefficient \( C_d = 0.6 \).
02

Compute Initial and Final Fluid Velocity

Using Torricelli's law, compute the velocity of the fluid at both the initial and final heights of oil level. The velocity is given by \( v = \sqrt{2gh} \), where \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity.Initial velocity: \( v_1 = \sqrt{2 \times 9.81 \times 0.9} = \sqrt{17.658} \approx 4.20 \text{ m/s} \).Final velocity: \( v_2 = \sqrt{2 \times 9.81 \times 0.8} = \sqrt{15.696} \approx 3.96 \text{ m/s} \).
03

Application of Continuity Equation and Simplifying Integrals

The rate of change of fluid height can be expressed using the fluid flow equation:\( dV/dt = -C_d \times A_o \times v(t) \) for a small change in volume flow rate through the orifice. Expressing this in terms of height, we get \( A_c \times dh/dt = -C_d \times A_o \times \sqrt{2gh} \) where \( A_c \) is the cross-sectional area of the tank \( A_c = \pi (D/2)^2 = \pi \times (0.9)^2 \text{ m}^2 \).
04

Set Up and Solve the Definite Integral

To find the time taken for the height to change from \( h_1 \) to \( h_2 \):\[ \int_{h_1}^{h_2} \frac{A_c}{C_d \times A_o \times \sqrt{2g}} \frac{1}{\sqrt{h}} \, dh = \int_{0}^{t} \, dt \]This simplifies to:\[ \frac{A_c}{C_d \times A_o \times \sqrt{2g}} \left[ -2\sqrt{h} \right]_{h_1}^{h_2} = t \].
05

Evaluate the Integral and Compute Time

Solve for \( t \) using the outputs of the integral,\[ t = \frac{\pi \times 0.81}{0.6 \times \pi \times 10^{-4} \times \sqrt{19.62}} \left( \sqrt{0.8} - \sqrt{0.9} \right) \],ensuring consistent units, compute:\[ t = \frac{0.81}{0.6 \times 10^{-4} \times \sqrt{19.62}} \times \left( 0.8944 - 0.9487 \right) \].Plug in numerical calculations and solve,\[ t \approx \frac{0.81}{0.6 \times 10^{-4} \times 4.427} \times (-0.0543) \approx 699.4 \text{ seconds} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Tank Calculations
Understanding how to calculate various parameters in a cylindrical tank is crucial for solving engineering problems, like calculating the time it takes for a liquid level to change. A cylindrical tank has three main geometric aspects to consider: diameter, length, and cross-sectional area. When a tank is mounted horizontally, like in this problem, the cross-sectional area is a circle. To find this area, use the formula: \[ A_c = \pi \left(\frac{D}{2}\right)^2 \] where \(D\) is the diameter of the tank. This describes the space through which the liquid flows from one section of the tank to another. In the given scenario, the tank has a diameter of 1.8 meters, so the cross-sectional area becomes \( A_c = \pi (0.9)^2 \). Calculating this helps to understand how fluid flow and changes in height translate physically inside the tank.
Whether you are dealing with water, oil, or any other fluid, knowing these measurements lets you set up the equations needed to predict changes like pressure or fluid velocity.
Orifice Discharge Coefficient
The orifice discharge coefficient (\( C_d \) ) is a crucial factor when you're dealing with flows through small openings. It accounts for real-world complexities like friction and flow turbulence that affect how quickly a liquid exits an orifice. In simpler terms, the coefficient makes equations more accurate by adjusting theoretical calculations to mirror actual conditions.In this problem, the discharge coefficient is given as 0.6. This means that only 60% of the calculated ideal flow rate comes through the orifice, due to losses that occur as the liquid passes through.
The orifice itself has an area determined by the formula:\[ A_o = \pi \left(\frac{d}{2}\right)^2 \] for an orifice with diameter \( d \). This results in a key factor in other calculations in scenarios involving liquid discharge.
By understanding and using the orifice discharge coefficient, you ensure that any calculations you perform, predict accurately how a fluid will behave when passing through an opening.
Continuity Equation in Fluid Mechanics
The continuity equation describes the principle of Conservation of Mass in fluid dynamics. It ensures that for an incompressible fluid, the rate at which mass is entering a volume equals the rate at which it is exiting, minus any accumulation or depletion. This might sound abstract, but think of it as a system's way of saying: "What goes in, must come out." In this context of liquid drainage from a tank, consider how the equation expresses that as fluid leaves the tank, the height of the fluid drops at a rate related to the velocity through the orifice: \[ A_c \frac{dh}{dt} = -C_d A_o \sqrt{2gh} \] 鈥� \( A_c \) is the cross-sectional area of the tank.鈥� \( dh/dt \) represents the rate at which the height of the fluid changes. 鈥� \( C_d \) is the discharge coefficient.鈥� \( A_o \) is the area of the orifice.鈥� \( \sqrt{2gh} \) is derived from Torricelli鈥檚 Law, interpreting fluid velocity as it flows under gravity.The equation allows us to set up integrals that calculate time needed for fluid height to change, as shown in the problem. Understanding the continuity equation is key to predicting how fluid levels change over time in real-world applications.

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Most popular questions from this chapter

Two vessels in which the difference of surface levels is maintained constant at \(2.4 \mathrm{~m}\) are connected by a \(75 \mathrm{~mm}\) diameter pipeline \(15 \mathrm{~m}\) long. If the frictional coefficient \(f\) may be taken as \(0.008\), determine the volume rate of flow through the pipe. \(\left[11.9\right.\) litre \(\left.\mathrm{s}^{-1}\right]\)

Water entering a \(150 \mathrm{~mm}\) diameter pipe \(1300 \mathrm{~m}\) long is all drawn off at a uniform rate per metre of length along the pipe. Neglecting all losses other than pipe friction, find the volume rate of flow entering the pipe when the pressure drop along the pipe is \(2.55\) bar. Take \(f=0.008\). Draw the hydraulic gradient for the system if the pressure at entry to the pipe is \(2.8\) bar. \(\left[0.0415 \mathrm{~m}^{3} \mathrm{~s}^{-1}\right]\)

A \(2 \mathrm{~m}\) deep tank is \(2 \mathrm{~m} \times 3 \mathrm{~m}\) in area and is divided into two equal halves by a vertical separation plate. Flow from one tank to the other takes place through a square orifice, \(1 \mathrm{~cm}\) side, having a discharge coefficient of \(0.6\). If water is initially at \(1.5 \mathrm{~m}\) depth on one side of the plate and \(0.5 \mathrm{~m}\) depth on the other, calculate the time taken for the depths in both tanks to be equal. \([11290 \mathrm{~s}]\)

A horizontal water main comprises \(1500 \mathrm{~m}\) of \(150 \mathrm{~mm}\) diameter pipe followed by \(900 \mathrm{~m}\) of \(100 \mathrm{~mm}\) diameter pipe, the friction factor \(f\) for each pipe being \(0.007\). All the water is drawn off at a uniform rate per unit length along the pipe. If the total input to the system is \(25 \mathrm{dm}^{3} \mathrm{~s}^{-1}\), find the total pressure drop along the main, neglecting all losses other than pipe friction. Also draw the hydraulic gradient taking the pressure head at inlet as \(54 \mathrm{~m}\). \([20.50 \mathrm{~m}]\)

There is a pressure loss of \(300 \mathrm{kN} \mathrm{m}^{-2}\) when water is pumped through pipeline \(\mathrm{A}\) at a rate of \(2 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) and there is a pressure loss of \(250 \mathrm{kN} \mathrm{m}^{-2}\) when water is pumped at a rate of \(1.4 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) through pipeline B. Calculate the pressure loss which will occur when \(1.5 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) of water are pumped through pipes \(\mathrm{A}\) and \(\mathrm{B}\) jointly if they are connected \((a)\) in series, \((b)\) in parallel, assuming that junction losses may be neglected. In the latter case calculate the volume rate of flow through each pipe. $$ \begin{array}{r} {\left[(a) 456 \mathrm{kNm}^{-2},(b) 54.1 \mathrm{kNm}^{-2}\right.} \\ \left.0.849 \mathrm{~m}^{3} \mathrm{~s}^{-1}, 0.651 \mathrm{~m}^{3} \mathrm{~s}^{-1}\right] \end{array} $$
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