A reservoir \(A\) feeds two lower reservoirs \(B\) and \(C\) through a single pipe \(10 \mathrm{~km}\) long, \(750 \mathrm{~mm}\) diameter, having a downward slope of \(2.2 \times 10^{-3}\). This pipe then divides into two branch pipes, one \(5.5 \mathrm{~km}\) long laid with a downward slope of \(2.75 \times\) \(10^{-3}\) (going to \(\left.B\right)\), the other \(3 \mathrm{~km}\) long having a downward slope of \(3.2 \times 10^{-3}\) (going to \(\left.C\right)\). Calculate the necessary diameters of the branch pipes so that the steady flow rate in each shall be \(0.24 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) when the level in each reservoir is \(3 \mathrm{~m}\) above the end of the corresponding pipe. Neglect all losses except pipe friction and take \(f=0.006\) throughout.

Short Answer

Expert verified
The calculated diameters of the branch pipes are approximately 0.488 m for the pipe to reservoir B and 0.322 m for the pipe to reservoir C.

Step by step solution

01

- Calculate the Head Loss in the Main Pipe

Use the Darcy-Weisbach equation to calculate the head loss (h_f) in the main pipe from reservoir A to the junction point. The formula is: \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] Given data: Length (L) of main pipe = 10,000 m, Diameter (D) = 0.75 m, friction factor (f) = 0.006. First, compute the velocity (V) using the flow rate (Q) and area (A). \[ Q = 0.48 \, m^3/s \] (since it splits equally) \[ A = \frac{\pi D^2}{4} = \frac{\pi \times (0.75)^2}{4} \approx 0.44 \, m^2 \] \[ V = \frac{Q}{A} = \frac{0.48}{0.44} \approx 1.091 \, m/s \] Now plug into the Darcy-Weisbach formula: \[ h_f = 0.006 \frac{10000}{0.75} \frac{(1.091)^2}{2 \times 9.81} \approx 11.0 \, m \]
02

- Calculate Head Loss for Branch Pipe B

Use the Darcy-Weisbach equation again to calculate the head loss (h_fB) in the branch pipe going to reservoir B. Given data: Length (L) = 5,500 m, slope (S) = 2.75 \times 10^{-3}. Since the flow rate in branch pipe B is 0.24 \, m^3/s and is split equally, find the diameter (D_B). Rearrange equation for diameter: \[ h_f = S \times L = 2.75 \times 10^{-3} \times 5500 = 15.125 \, m \] Using \[ h_f = f \frac{L}{D_B} \frac{V_B^2}{2g} \] and \[ V_B = \frac{0.24}{A_B} \] For V_B: \[ A_B = \frac{\pi D_B^2}{4} \] Solve for D_B: \[ 15.125 = 0.006 \frac{5500}{D_B} \frac{(0.24/A_B)^2}{2 \times 9.81} \approx 0.488 \, m \]
03

- Calculate Head Loss for Branch Pipe C

Use the Darcy-Weisbach equation for the branch pipe going to reservoir C. Given data: Length (L) = 3,000 m, slope (S) = 3.2 \times 10^{-3}. Similar steps are followed for the diameter (D_C). \[ h_f = S \times L = 3.2 \times 10^{-3} \times 3000 = 9.6 \, m \] Using \[ h_f = f \frac{L}{D_C} \frac{V_C^2}{2g} \] and \[ V_C = \frac{0.24}{A_C} \] Rearrange and solve for D_C: \[ 9.6 = 0.006 \frac{3000}{D_C} \frac{(0.24/A_C)^2}{2 \times 9.81} \approx 0.322 \, m \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Darcy-Weisbach Equation
The Darcy-Weisbach equation is fundamental in fluid mechanics for calculating head loss in a pipe. Head loss is the reduction in the total head (sum of potential energy, pressure energy, and kinetic energy per unit weight of fluid). The equation is given by: \ \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] ewline Here, \textbf{f} is the friction factor, \textbf{L} is the length of the pipe, \textbf{D} is the diameter, \textbf{V} is the flow velocity, and \textbf{g} is the acceleration due to gravity. To use this equation, the velocity \( V \) must be calculated using the flow rate \( Q \) and the cross-sectional area \( A \) of the pipe, where \ \[ A = \frac{\pi D^2}{4} \]. ewline Once you have all these variables, you can substitute them into the Darcy-Weisbach equation to find the head loss.
Head Loss Calculation
Head loss in a pipe system due to friction can significantly impact the design and efficiency of fluid transport systems. It is necessary to calculate the head loss to ensure proper flow rates and sufficient pressure. For instance, in the problem of calculating the necessary diameters for the pipes leading from reservoir A to reservoirs B and C, head loss had to be considered to meet specific flow requirements. The necessary steps involve: \ 1. Determining the flow velocity using the flow rate and the pipe's cross-sectional area. ewline2. Applying the Darcy-Weisbach equation: \ \[ h_f = f \frac{L}{D} \frac{V^2}{2g} \] ewline The goal is to find a balance where the head loss does not exceed the pressure head, ensuring steady flow to the reservoirs.
Pipe Friction
Pipe friction refers to the resistance to fluid flow within a pipe, caused by the interactions between the fluid and the pipe's inner surface. This friction results in energy loss, often expressed as head loss. The friction factor \( f \) in the Darcy-Weisbach equation accounts for this resistance and depends on factors like: ewline \begin{itemize} \item Pipe material and roughness - rougher pipes increase friction ewline \item Flow rate and velocity of fluid - higher flow rates increase turbulence and friction ewline \item Pipe diameter - smaller diameters lead to higher friction \end{itemize} Understanding pipe friction is crucial for designing efficient piping systems as it helps in selecting appropriate pipe sizes and materials to minimize energy losses while maintaining required flow rates.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A pipe \(600 \mathrm{~mm}\) diameter and \(1 \mathrm{~km}\) long with \(f=0.008\) connects two reservoirs having a difference in water surface level of \(30 \mathrm{~m}\). Calculate the rate of flow between the reservoirs and the shear stress at the wall of the pipe. If the upstream half of the pipe is tapped by several side pipes so that one-third of the quantity of water now entering the main pipe is withdrawn uniformly over this length, calculate the new rate of discharge to the lower reservoir. Neglect all losses other than those due to pipe friction.

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.

Two water reservoirs, the surface levels of which differ by \(1.5 \mathrm{~m}\), are connected by a pipe system consisting of a sloping pipe at each end, \(7.5 \mathrm{~m}\) long and \(75 \mathrm{~mm}\) diameter, joined by a horizontal pipe \(300 \mathrm{~mm}\) diameter and \(60 \mathrm{~m}\) long. Taking entry head losses as \(0.5 u^{2} / 2 g\) and \(f=0.005(1+25 / d)\) where \(d \mathrm{~mm}\) is the pipe diameter, calculate the steady rate of flow through the pipe.

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 \(0.007-0.01\).

Calculate the power required to pump sulphuric acid (dynamic viscosity \(0.04 \mathrm{~Pa} \cdot \mathrm{s}\), relative density \(1.83\) ) at \(45 \mathrm{~L} \cdot \mathrm{s}^{-1}\) from a supply tank through a glass-lined \(150 \mathrm{~mm}\) diameter pipe, \(18 \mathrm{~m}\) long, into a storage tank. The liquid level in the storage tank is \(6 \mathrm{~m}\) above that in the supply tank. For laminar flow \(f=16 / \mathrm{Re}\); for turbulent flow \(f=0.0014\left(1+100 \mathrm{Re}^{-1 / 3}\right)\) if \(\mathrm{Re}<10^{7}\). Take all losses into account.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free