/*! 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 142 Gasoline flows in a long. underg... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 142

Gasoline flows in a long. underground pipeline at a constant temperature of \(15^{\circ} \mathrm{C}\). Two pumping stations at the same elevation are located \(13 \mathrm{km}\) apart. The pressure drop between the stations is \(1.4 \mathrm{MPa}\). The pipeline is made from 0.6-m-diameter pipe. Although the pipe is made from commercial steel, age and corrosion have raised the pipe roughness to approximately that for galvanized iron. Compute the volume flow rate.

Short Answer

Expert verified
The volume flow rate of the gasoline is approximately \(1.42 m^3/s\).

Step by step solution

01

Understand the values given and what you need to find

We know that the pressure drop is \(1.4MPa = 1.4 \times 10^6Pa\), the diameter of the pipe \(d = 0.6m\), the distance between two pumping station \(L = 13,000m\). The relationship between these values we want to find the volume flow rate \(Q\). We will use the equation for volume rate of flow \(Q = \frac{\Delta P \pi (d^2)}{4fL} \, m^3 /sec\).
02

Estimate the friction factor

We're told that the pipe's roughness is approximately that of galvanized iron, which has a roughness coefficient \(k=0.00015\). We use the Moody chart to determine the friction factor \(f\). Given the Reynolds number \(Re = \frac{4Q}{\pi d \mu}\), where \(\mu\) is the dynamic viscosity and based on the given temperature \(15^{\circ} C\), we can look for \(\mu\) in the standard fluid tables, and it will lead us to approximately \(0.0025Pa*s\). Using the Moody chart with \(Re\) and relative roughness \(k/d\) we find that the friction factor \(f \approx 0.028\).
03

Calculate the volume flow rate

We have all the necessary elements to substitute into the formula for volume flow rate \(Q = \frac{(1.4 \times 10^6)(\pi)(0.6)^2}{4(0.028)(13000)} = 1.42 m^3/s.\)

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Most popular questions from this chapter

Consider blood flow in an artery. Blood is nonNewtonian; the shear stress versus shear rate is described by the Casson relationship: $$\left\\{\begin{array}{ll} \sqrt{\tau}=\sqrt{\tau_{c}}+\sqrt{\mu \frac{d u}{d r}} & \text { for } \tau \geq \tau_{c} \\ \tau=0 & \text { for } \tau<\tau_{c} \end{array}\right.$$ where \(\tau_{e}\) is the critical shear stress, and \(\mu\) is a constant having the same dimensions as dynamic viscosity. The Casson relationship shows a linear relationship between \(\sqrt{\tau}\) and \(\sqrt{d u / d r},\) with intercept \(\sqrt{\tau_{c}}\) and slope \(\sqrt{\mu} .\) The Casson relationship approaches Newtonian behavior at high values of deformation rate. Derive the velocity profile of steady fully developed blood flow in an artery of radius \(R\). Determine the flow rate in the blood vessel. Calculate the flow rate due to a pressure gradient \(d p / d x=-100 \mathrm{Pa} / \mathrm{m},\) in an artery of radius \(R=1 \mathrm{mm},\) using the following blood data: \(\mu=3.5 \mathrm{cP}\) \(\tau_{c}=0.05\) dynes \(/ \mathrm{cm}^{2}\)

Water flows from a horizontal tube into a large tank. The tube is located \(2.5 \mathrm{m}\) below the free surface of water in the tank. The head loss is \(2 \mathrm{J} / \mathrm{kg}\). Compute the average flow speed in the tube.

A fluid flows steadily between two parallel plates. The flow is fully developed and laminar. The distance between the plates is \(h\) (a) Derive an equation for the shear stress as a function of \(y\) Sketch this function. (b) For \(\mu=2.4 \times 10^{-5} \mathrm{lbf} \cdot \mathrm{s} / \mathrm{ft}^{2}, \quad \partial p / \partial x=-4.0 \mathrm{lbf} / \mathrm{ft}^{2} / \mathrm{n}\) and \(h=0.05\) in.., calculate the maximum shear stress, in \(\mathrm{lbf} / \mathrm{ft}^{2}\)

A water tank (open to the atmosphere) contains water to a depth of \(5 \mathrm{m} .\) A 25 -mm-diameter hole is punched in the bottom. Modeling the hole as square-edged, estimate the flow rate (L/s) exiting the tank. If you were to stick a short section of pipe into the hole, by how much would the flow rate change? If instead you were to machine the inside of the hole to give it a rounded edge \((r=5 \mathrm{mm})\), by how much would the flow rate change?

Consider fully developed laminar flow between infinite parallel plates separated by gap width \(d=0.2\) in. The upper plate moves to the right with speed \(U_{2}=5 \mathrm{ft} / \mathrm{s}\); the lower plate moves to the left with speed \(U_{1}=2 \mathrm{ft}\) /s. The pressure gradient in the direction of flow is zero. Develop an expression for the velocity distribution in the gap. Find the volume flow rate per unit depth (gpm/ft) passing a given cross section.
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