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.

Short Answer

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Read items for pipes then calculate Step 1 through 4..

Step by step solution

01

Calculate Reynolds Number (Re)

To identify type of flow, calculate the Reynolds Number where Re = \[ \frac{\rho u D}{\mu} \]} \[ u \] where \[\rho\] is density of liquid, \[u\]is kinematic viscisoty of liquid and u is velocity .
02

Initialize

Consolidate items to evaluate \[u=120\ \text{mm}^{2}/s\] = u=120\ \text{mm}^2/s
03

Identify if flow is Laminar or Turbulent

Given Re, identify of any flow value whether < 103 or 105.
04

Determine angel on slope/inclination

Identify mean shear stress required E, Given mean shear stress if inlt <|max.
05

Rate of correcponding flow

Using Re calculation where f=0.16/Re for laminar flow
06

Verify

Insert value in f= re by repeating items above then.

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Key Concepts

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

Reynolds Number
The Reynolds Number (\text{Re}) is a key figure in fluid dynamics that helps predict flow patterns in different fluid systems. It is calculated using the formula \text{Re} =\[\frac{\rho u D}{\mu}\], where:
• \(\rho\) is the fluid density
• \(u\) is the flow velocity
• \(D\) is the pipe diameter
• \(\mu\) is the dynamic viscosity of the fluid

A high Reynolds number indicates turbulent flow, while a low Reynolds number indicates laminar flow. In this exercise, calculating the Reynolds number helps determine if the fluid flow is laminar or turbulent.
Laminar and Turbulent Flow
Fluid flow in pipes can be classified as either laminar or turbulent, based on the Reynolds number.

• **Laminar Flow**: This occurs when the Reynolds number is less than 2000. In laminar flow, the fluid particles move in parallel layers with minimal mixing.
• **Turbulent Flow**: This occurs when the Reynolds number is greater than 4000. Here, the flow becomes chaotic and features eddies and vortices.

In the transition region between 2000 and 4000, the flow can shift between laminar and turbulent. Understanding which type of flow is present is crucial for determining the friction factor and for solving problems like the one in this exercise.
Pipe Friction Factor
The pipe friction factor (\(f\)) quantifies the resistance that the fluid encounters as it moves through the pipe. For laminar flow, the friction factor can be calculated using the formula \(f = \frac{16}{\text{Re}}\). For turbulent flow, a common approximation is \(f = 0.08\times(\text{Re})^{-1/4}\), but only when \(Re < 10^5\).

In this specific exercise, knowing the friction factor allows us to compute the maximum pressure at the lower end of the pipe and the corresponding flow rate. Calculating the friction factor correctly is essential for accurately analyzing the fluid behavior in pipelines.
Shear Stress
Shear stress (\(\tau\)) in a fluid is the force per unit area exerted by the fluid against the walls of the pipe due to its motion. It is given by \(\tau = \mu \frac{du}{dy}\), where
• \(\mu\) is the dynamic viscosity of the fluid
• \(\frac{du}{dy}\) represents the velocity gradient perpendicular to the flow.

In this exercise, the problem states that the mean shear stress at the pipe wall should not exceed 200 Pa. This constraint is used to find the maximum allowable pressure at the lower, inlet end of the pipe.
Hydraulic Gradient
The hydraulic gradient is the slope of the energy grade line, which represents the total energy head per unit weight of fluid. It is a critical parameter in pipe flow analysis and illustrates the rate of energy loss as the fluid flows through the pipe.

The hydraulic gradient is usually represented as \(i = \frac{h_f}{L}\), where:
• \(h_f\) is the head loss
• \(L\) is the length of the pipe.

In this inclined pipe problem, the hydraulic gradient would influence the calculation of the pressure drop due to elevation change, friction, and shear stress. Understanding the hydraulic gradient helps in predicting how much work is needed to pump the fluid uphill.

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Most popular questions from this chapter

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.

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 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.

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 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.

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