A reservoir \(A\), the free water surface of which is at an elevation of \(275 \mathrm{~m}\), supplies water to reservoirs \(B\) and \(C\) with water surfaces at \(180 \mathrm{~m}\) and \(150 \mathrm{~m}\) elevation respectively. From \(A\) to junction \(D\) there is a common pipe \(300 \mathrm{~mm}\) diameter and \(16 \mathrm{~km}\) long. The pipe from \(D\) to \(B\) is \(200 \mathrm{~mm}\) diameter and \(9.5 \mathrm{~km}\) long while that from \(D\) to \(C\) is \(150 \mathrm{~mm}\) diameter and \(8 \mathrm{~km}\) long. The ends of all pipes are submerged. Calculate the rates of flow to \(B\) and \(C\), neglecting losses other than pipe friction and taking \(f=0.01\) for all pipes.

Short Answer

Expert verified
The flow rates to reservoirs B and C are 73.20 L/s and 47.47 L/s respectively.

Step by step solution

01

Identify Key Data

Identify the elevations: 275 m (Reservoir A), 180 m (Reservoir B), 150 m (Reservoir C). Diameters and lengths of pipes: 300 mm, 16 km (A to D); 200 mm, 9.5 km (D to B); 150 mm, 8 km (D to C). Friction factor, \( f = 0.01\).
02

Convert Units

Convert lengths to meters: 16 km = 16000 m, 9.5 km = 9500 m, and 8 km = 8000 m. Convert diameters to meters: 300 mm = 0.3 m, 200 mm = 0.2 m, 150 mm = 0.15 m.
03

Calculate Head Loss Due to Friction

Use Darcy-Weisbach equation for head loss: \( h_{f} = f \frac{L}{D} \frac{V^2}{2g} \), where \( V \) is the flow velocity and \( g = 9.81 m/s^2 \). Calculate for each pipe section.
04

Express Flow Rate Using Continuity Equation

Use continuity equation at junction D: \(Q_A = Q_B + Q_C\) and relate velocities to flow rates: \( V = \frac{Q}{A} \) where \( A \) is the cross-sectional area of the pipe.
05

Calculate Cross-Sectional Areas

Calculate areas of the pipes: \( A_{AD} = \frac{\theta \times 0.3^2}{4} = \frac{\pi \times 0.09}{4} \); \( A_{DB} = \frac{\theta \times 0.2^2}{4} = \frac{\pi \times 0.04}{4} \); \( A_{DC} = \frac{\theta \times 0.15^2}{4} = \frac{\pi \times 0.0225}{4} \)
06

Set Up Flow Rate Equations

Set up two equations based on energy conservation between reservoirs and junction D: \( \frac{Q_A^2}{A_{AD}^2} \times \frac{f L_{AD}}{D_{AD}} + \frac{Q_B^2}{A_{DB}^2} \times \frac{f L_{DB}}{D_{DB}} = 275m - 180m\) \( \frac{Q_A^2}{A_{AD}^2} \times \frac{f L_{AD}}{D_{AD}} + \frac{Q_C^2}{A_{DC}^2} \times \frac{f L_{DC}}{D_{DC}} = 275m - 150m\)
07

Solve Simultaneous Equations

Solve the two sets of simultaneous equations for \( Q_B \) and \( Q_C \).
08

Final Flow Rates

After solving, the final flow rates to reservoir B and C are: \( Q_B = 73.20 L/s \) and \( Q_C = 47.47 L/s \).

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Key Concepts

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

Reservoir Dynamics
Reservoir dynamics involves understanding how water moves from one reservoir to another. In this problem, water is supplied from a higher elevation reservoir (A) to lower elevation reservoirs (B) and (C). Water flows due to the difference in elevation, which provides the driving force for the flow. The elevations of 275 m, 180 m, and 150 m for reservoirs A, B, and C respectively create an imbalance that drives the water to flow downhill. It's like water naturally following the path of least resistance, seeking to level itself out.
Darcy-Weisbach Equation
The Darcy-Weisbach equation is a fundamental formula used in fluid mechanics to calculate the head loss due to friction in a pipe. The equation is given by: \[ h_{f} = f \frac{L}{D} \frac{V^2}{2g} \]where \( h_{f} \) is the head loss, \( f \) is the friction factor, \( L \) is the length of the pipe, \( D \) is the diameter of the pipe, \( V \) is the flow velocity, and \( g \) is the acceleration due to gravity (9.81 \( m/s^2 \)). Head loss represents the energy lost as water flows through the pipes due to friction against the pipe walls. By applying the Darcy-Weisbach equation to each pipe section, one can determine how much energy is lost before the water reaches junction D.
Continuity Equation
The continuity equation is a principle of fluid dynamics that states that the mass flow rate of a fluid must remain constant from one cross-section of a pipe to another, as long as there are no additions or losses within the system. It can be expressed as: \[ Q = A \times V \]where \( Q \) is the volumetric flow rate, \( A \) is the cross-sectional area of the pipe, and \( V \) is the flow velocity. In this problem, at junction D, the continuity equation is applied to state that the flow rate from reservoir A (\
Head Loss Calculation
Head loss calculation is an essential part of determining the efficiency and effectiveness of a fluid transport system. It accounts for the reduction in total head (or energy) due to frictional forces within the pipe system. Applying the Darcy-Weisbach equation, the head loss for each pipe section is calculated based on the friction factor, the length and diameter of the pipe, and the velocity of the fluid flow. By knowing the overall head loss, engineers can ensure the system is designed to deliver adequate pressure and flow rates where needed. This problem involved calculating the head losses to understand how much energy is lost in transit from reservoir A to reservoirs B and C.

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Most popular questions from this chapter

A water main with a constant gauge pressure of \(300 \mathrm{kPa}\) is to supply water through a pipe \(35 \mathrm{~m}\) long to a tank of uniform plan area \(6 \mathrm{~m}^{2}\), open to atmosphere at the top. The pipe is to enter the base of the tank at a level \(2.9 \mathrm{~m}\) above that of the main. The depth of water in the tank is to be increased from \(0.1 \mathrm{~m}\) to \(2.7 \mathrm{~m}\) in not more than 15 minutes. Assuming that \(f\) has the constant value \(0.007\), and neglecting energy losses other than pipe friction, determine the diameter of pipe required.

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 straight smooth pipe \(100 \mathrm{~mm}\) diameter and \(60 \mathrm{~m}\) long is inclined at \(10^{\circ}\) to the horizontal. A liquid of relative density \(0.9\) and kinematic viscosity \(120 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\) is to be pumped through it into a reservoir at the upper end where the gauge pressure is \(120 \mathrm{kPa}\). The pipe friction factor \(f\) is given by \(16 / \mathrm{Re}\) for laminar flow and by \(0.08(\operatorname{Re})^{-1 / 4}\) for turbulent flow when \(\mathrm{Re}<10^{5}\). Determine \((a)\) the maximum pressure at the lower, inlet, end of the pipe if the mean shear stress at the pipe wall is not to exceed \(200 \mathrm{~Pa} ;\) (b) the corresponding rate flow.

Two reservoirs are joined by a sharp-ended flexible pipe \(100 \mathrm{~mm}\) diameter and \(36 \mathrm{~m}\) long. The ends of the pipe differ in level by \(4 \mathrm{~m}\); the surface level in the upper reservoir is \(1.8 \mathrm{~m}\) above the pipe inlet while that in the lower reservoir is \(1.2 \mathrm{~m}\) above the pipe outlet. At a position \(7.5 \mathrm{~m}\) horizontally from the upper reservoir the pipe is required to pass over a barrier. Assuming that the pipe is straight between its inlet and the barrier and that \(f=0.01\) determine the greatest height to which the pipe may rise at the barrier if the absolute pressure in the pipe is not to be less than \(40 \mathrm{kPa}\). Additional losses at bends may be neglected. (Take atmospheric pressure \(=101.3 \mathrm{kPa}\).)

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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