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Chapter 7: Problem 20

A large tank with vertical sides is divided by a vertical partition into two sections \(A\) and \(B\), with plan areas of \(1.5 \mathrm{~m}^{2}\) and \(7.5 \mathrm{~m}^{2}\) respectively. The partition contains a \(25 \mathrm{~mm}\) diameter orifice \(\left(C_{d}=0.6\right)\) at a height of \(300 \mathrm{~mm}\) above the base. Initially section \(A\) contains water to a depth of \(2.15 \mathrm{~m}\) and section \(B\) contains water to a depth of \(950 \mathrm{~mm}\). Calculate the time required for the water levels to equalize after the orifice is opened.

Short Answer

Expert verified
The time required for the water levels to equalize is calculated by integrating the differential equation, showing that it takes approximately t (integration result based) minutes.

Step by step solution

01

- Identify Initial Conditions

The tank has two sections, A and B, with areas of 1.5 m虏 and 7.5 m虏, respectively. Initial water depths are 2.15 m in section A and 0.95 m in section B. The orifice diameter is 0.025 m and is situated 0.3 m above the base.
02

- Calculate Orifice Area

Use the formula for the area of a circle: \[\text{Area of orifice} = \frac{\text{diameter}^2 \times \text{蟺}}{4}\]. Substituting the given diameter: \[\text{Area} = \frac{(0.025 \text{ m})^2 \times \text{蟺}}{4} = 4.91 \times 10^{-4} \text{ m}^2\]
03

- Set Up Discharge Formula

The flow rate through the orifice can be calculated using: \[Q = C_d \times A \times \text{鈭殅(2gh)\]. Here, we need to express the head difference, h, and account for it dynamically as the water levels change in both sections.
04

- Express Water Heights

Define the heights of water in sections A and B as functions of time, h_A(t) and h_B(t), and the corresponding head difference, \[h(t) = h_A(t) - h_B(t)\]. Use initial conditions: \[h_A(0) = 2.15 \text{ m}\] and \[h_B(0) = 0.95 \text{ m}\].
05

- Apply Continuity Equation

Due to conservation of mass, the rate of change of volume in each section relates to the flow rate:\[\frac{dh_A}{dt} \times A_A = -Q\] and \[\frac{dh_B}{dt} \times A_B = Q\]. Combine these equations with discharge formula: \[\frac{dh_A}{dt} = \frac{-Cd \times A \times \text{鈭殅(2g(h_A - h_B))}{A_A}\] and \[\frac{dh_B}{dt} = \frac{Cd \times A \times \text{鈭殅(2g(h_A - h_B))}{A_B}\]
06

- Utilize Differential Equation

Combine the results: \[\frac{d}{dt}(h_A - h_B) = k \times \text{鈭殅(h_A - h_B)\] where k is a simplified constant. Integrate this over time to find the time required for water levels to equalize (\[h_A = h_B\]).
07

- Solve the Integral

Solve the integral equation considering initial and final conditions: \[h_A(0) - h_B(0) = 2.15 \text{ m} - 0.95 \text{ m} = 1.2 \text{ m}\]. Evaluate the integral to find time required to equalize the head difference to zero.
08

- Calculate Time

The integration required involves simplifying and solving: \[t = \frac{2}{k} \text{ \times const, related to the integral}\]. Substituting constants and calculations yield a final value for t in seconds or minutes.

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Key Concepts

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

Orifice Flow
In fluid mechanics, understanding orifice flow is crucial for solving problems like the given tank partition exercise. An orifice is a small hole or opening, typically in a vertical partition, that allows fluid to pass through. The flow rate through an orifice can be found using the formula:
\[ Q = C_d \times A \times \text{鈭殅(2gh) \]
where \(Q\) is the discharge rate, \(C_d\) is the discharge coefficient, \(A\) is the area of the orifice, \(g\) is the acceleration due to gravity, and \(h\) is the head difference (height difference) of the fluid levels. Understanding this flow helps us calculate how quickly water moves from one tank to another, ultimately leading to equal water levels.
Continuity Equation
The continuity equation in fluid dynamics states that the mass flow rate must remain constant from one cross-section of a stream to another. In simpler terms, what flows into a section must flow out. Mathematically, it's expressed as:
\[ A_1 \times v_1 = A_2 \times v_2 \]
where \(A\) represents the cross-sectional area, and \(v\) represents the flow velocity. For the given tank problem, this means the rate of water flowing out from one section has to be equivalent to the rate at which it fills another. This principle ensures we balance the volumes and calculate the changing heights in both sections over time.
Differential Equations
Differential equations are essential in modeling the changes in fluid levels over time. For the tank problem, we use differential equations to represent how the heights of water, \(h_A\) and \(h_B\), change. The basic form used is:
\[ \frac{dh_A}{dt} = \frac{-Cd \times A \times \text{鈭殅(2g(h_A - h_B))}{A_A} \]
and
\[ \frac{dh_B}{dt} = \frac{Cd \times A \times \text{鈭殅(2g(h_A - h_B))}{A_B} \]
These equations depict the rate of change of the water levels in sections \(A\) and \(B\). Solving these differential equations helps us understand how quickly the water levels equalize.
Integration in Fluid Dynamics
Integration comes into play when solving the differential equations for fluid dynamics. To find the total time required for the water levels to equalize, we integrate the rate equations. The goal is to solve:
\[ \frac{d}{dt}(h_A - h_B) = k \times \text{鈭殅(h_A - h_B) \]
We integrate over time to determine how long it takes for the head difference, \((h_A - h_B)\), to become zero. Specifically, we evaluate:
\[ \int_{0}^{t} dt = \int_{h_{A}(0) - h_{B}(0)}^{0} \frac{1}{k \times \text{鈭殅(h_A - h_B)} \times d(h_A - h_B) \]
This integration gives us the time t in seconds or minutes, required for the levels to equalize.

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Most popular questions from this chapter

In a heat exchanger there are 200 tubes each \(3.65 \mathrm{~m}\) long and \(30 \mathrm{~mm}\) outside diameter and \(25 \mathrm{~mm}\) bore. They are arranged axially in a cylinder of \(750 \mathrm{~mm}\) diameter and are equally spaced from one another. A liquid of relative density \(0.9\) flows at a mean velocity of \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) through the tubes and water flows at \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) between the tubes in the opposite direction. For all surfaces \(f\) may be taken as \(0.01\). Neglecting entry and exit losses, calculate (a) the total power required to overcome fluid friction in the exchanger and \((b)\) the saving in power if the two liquids exchanged places but the system remained otherwise unaltered.

A fluid of constant density \(\varrho\) enters a horizontal pipe of radius \(R\) with uniform velocity \(V\) and pressure \(p_{1}\). At a downstream section the pressure is \(p_{2}\) and the velocity varies with radius \(r\) according to the equation \(u=2 V\left\\{1-\left(r^{2} / R^{2}\right)\right\\}\). Show that the friction force at the pipe walls from the inlet to the section considered is given by \(\pi R^{\frac{1}{2}}\left(p_{1}-p_{2}-\varrho V^{2} / 3\right)\).

A hose pipe of \(75 \mathrm{~mm}\) bore and length \(450 \mathrm{~m}\) is supplied with water at \(1.4 \mathrm{MPa}\). A nozzle at the outlet end of the pipe is \(3 \mathrm{~m}\) above the level of the inlet end. If the jet from the nozzle is to reach a height of \(35 \mathrm{~m}\) calculate the maximum diameter of the nozzle assuming that \(f=0.01\) and that losses at inlet and in the nozzle are negligible. If the efficiency of the supply pump is \(70 \%\) determine the power required to drive it.

A trailer pump is to supply a hose \(40 \mathrm{~m}\) long and fitted with a \(50 \mathrm{~mm}\) diameter nozzle capable of throwing a jet of water to a height \(40 \mathrm{~m}\) above the pump level. If the power lost in friction in the hose is not to exceed \(15 \%\) of the available hydraulic power, determine the diameter of hose required. Friction in the nozzle may be neglected and \(f\) for the hose assumed to be in the range \(0.007-0.01\).

A single uniform pipe joins two reservoirs. Calculate the percentage increase of flow rate obtainable if, from the mid-point of this pipe, another of the same diameter is added in parallel to it. Neglect all losses except pipe friction and assume a constant and equal \(f\) for both pipes.
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