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During a heavy rainstorm, water from a parking lot completely fills an 18 -in.-diameter, smooth, concrete storm sewer. If the flowrate is \(10 \mathrm{ft}^{3} / \mathrm{s}\), determine the pressure drop in a \(100-\mathrm{ft}\) horizontal section of the pipe. Repeat the problem if there is a \(2-f t\) change in elevation of the pipe per 100 ft of its length.

Short Answer

Expert verified
For a horizontal pipe, the pressure drop is 3.315 psi. For the pipe with a 2-ft change in elevation per 100-ft length, the total pressure change is 4.795 psi (1.480 psi due to elevation change and 3.315 psi due to velocity change).

Step by step solution

01

Calculate the Velocity

Firstly, determine the cross-sectional area (A) of the pipe using the given diameter (D) by using the formula \(A = π(D/2)^2\), where D is in meters. Since the given diameter is in inches, convert it to feet. Then, calculate the velocity (v) of the water flow using the formula \(v = Q/A\), where Q is the flowrate.
02

Calculate the Pressure Drop for Horizontal Pipe

Now, calculate the pressure drop (∆P) in the 100-ft horizontal section of the pipe using the formula \(\Delta P = 0.5 * \rho * v^2\), where \(\rho\) is the density of water (which is approximately 1000 kg/m^3). The answer will be in Pascals, convert it to psi using the conversion factor 0.0001450377 psi/Pa.
03

Calculate the Pressure Drop for Pipe with Elevation Change

Next, determine the change in pressure due to the change in height. The pressure difference due to a change in fluid column height can be calculated using the formula \(\Delta P = \rho * g * h\), where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the change in elevation of the pipe.
04

Find the Total Pressure Change

Finally, to find the total pressure change for the pipe with elevation change, add the pressure changes calculated in step 2 and step 3.

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Key Concepts

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

Pressure Drop
Pressure drop, a crucial concept in fluid mechanics, measures how much pressure is lost as a fluid moves through a pipe. This loss can occur due to friction, elevation changes, or pipe bends. In our storm sewer problem, two scenarios were examined: a pipe with no elevation changes, and another with a two-foot elevation change over a hundred-foot length.

For a horizontal pipe, the primary cause of pressure loss is friction between the flowing water and the pipe walls. This friction-related pressure drop can be calculated using Bernoulli's principle. The calculation involves the formula: \[\Delta P = 0.5 * \rho * v^2\]where \(\Delta P\) represents the pressure drop, \(\rho\) is the fluid density, and \(v\) is the flow velocity.

When there's an elevation change, gravity's effect on the fluid introduces an additional pressure component. This can be determined using: \[\Delta P = \rho * g * h\]where \(g\) is the acceleration due to gravity, and \(h\) is the height difference. Understanding these components helps engineers know how to adjust pipe designs to minimize pressure drops.
Flowrate
Flowrate is a key parameter in fluid systems, reflecting the volume of fluid moving through a system per unit time. It is usually expressed in cubic feet per second (ft³/s) or liters per second (L/s). In the storm sewer context, flowrate dictates how much stormwater the sewer can handle before overflowing.

To find a flowrate, you can use the continuity equation, which relates speed, cross-sectional area, and volume flowrate. In our example, the flowrate is given as 10 ft³/s. To determine the velocity of water in the pipe, once we know the pipe's cross-sectional area, we rearrange the formula: \[v = \frac{Q}{A}\]where \(Q\) is the flowrate and \(A\) is the area. Such calculations help engineers ensure that a sewer system is adequate for heavy rain scenarios, preventing flooding and water damage.
Continuity Equation
The Continuity Equation is a foundational concept in fluid mechanics, enforcing mass conservation in a flowing fluid. It states that the mass flow rate must remain constant from one cross-section of a pipe to another. This principle aids in calculating variables like flowrate and velocity in systems like storm sewers.

The equation is expressed as: \[Q = v * A\]where \(Q\) represents the flowrate, \(v\) is the flow velocity, and \(A\) is the cross-sectional area. The equation highlights how, under constant flow conditions, a reduction in pipe area results in increased fluid velocity. Conversely, a larger area means slower flow if all other factors remain constant.

This equation is incredibly useful in situations where storm sewers need to be designed or analyzed, ensuring systems are capable of transporting stormwater efficiently without risk of overflow.
Storm Sewer Engineering
Storm sewer engineering focuses on designing effective systems to manage rainwater and eliminate flooding. This discipline uses principles from fluid mechanics to create comprehensive drainage systems that manage stormwater collection, conveyance, and discharge.

Key considerations for designing storm sewers include:
  • Capacity: Gauging the maximum flow rate the system can handle.
  • Pressure control: Minimizing frictional losses and unwanted pressure drops.
  • Elevation: Recognizing how changes in elevation affect flow and pressure.
  • Material selection: Choosing materials that minimize resistance and maximize durability.
In practical terms, engineers must calculate the expected flowrate, potential pressure changes, and ensure that pipes can effectively carry stormwater away during heavy rain events. Understanding fluid dynamics principles like pressure drops and the continuity equation aids in building robust sewer systems. Efficient storm sewer systems play a pivotal role in urban planning, reducing the risk of floods and water-related damage.

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Most popular questions from this chapter

A 10 -m-long \(, 5.042\) -cm I.D. copper pipe has two fully open gate valves, a swing check valve, and a sudden enlargement to a \(9.919-\mathrm{cm}\) I.D. copper pipe. The \(9.919 \mathrm{cm}\) copper pipe is \(5.0 \mathrm{m}\) Iong and then has a sudden contraction to another 5.042 -cm copper pipe. Find the head loss for a \(20^{\circ} \mathrm{C}\) water flow rate of \(0.05 \mathrm{m}^{3} / \mathrm{s}\)

A person with no experience in fluid mechanics wants to estimate the friction factor for 1 -in.-diameter galvanized iron pipe at a Reynolds number of 8,000 . The person stumbles across the simple equation of \(f=64 / \mathrm{Re}\) and uses this to calculate the friction factor. Explain the problem with this approach and estimate the error.

A 3 -in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in.) carries \(210^{\circ} \mathrm{F}\) SAE 40 crankcase oil at the rate of 6.0 gal/min. The oil specific gravity is \(0.89,\) and the absolute viscosity is \(6.6 \times 10^{-7} 1 \mathrm{b} \cdot \mathrm{sec} / \mathrm{ft}^{2} .\) Calculate the pipe size required to carry the same flow rate at approximately one-half the pressure drop of the 3 -in. pipe. Both pipes are horizontal.

A thief siphoned 15 gal of gasoline from a gas tank in the middle of the night. The gas tank is 12 in. wide, 24 in. long, and 18 in. high and was full when the thief started. The siphoning plastic tube has an inside diameter of 0.5 in. and a length of \(4.0 \mathrm{ft}\). Assume that at any instant of time, the steady-state mechanical energy equation is adequate to predict the gasoline flow rate through the tube. As 15 gal is 3465 in \(^{3}\), the gasoline level in the tank will drop 12.0 in. You may use the gasoline level after it has dropped 6.0 in. to estimate the average gasoline flow rate. Use this flow rate to estimate the time needed to siphon the 15 gal of gasoline. Compare your answer with the answer of 190 sec found in problem 3.107 using Bernoulli's equation. The siphon cischarges at the level of the bottom of the gasoline tank. You may find it useful to use the Blasius equation for smooth pipes found in problem 8.45

Assume a car's exhaust system can be approximated as \(14 \mathrm{ft}\) of 0.125 -ft-diameter cast-iron pipe with the equivalent of six \(90^{\circ}\) flanged elbows and a muffler. (See Video V8.14.) The muffler acts as a resistor with a loss coefficient of \(K_{t}=8.5 .\) Determine the pressure at the beginning of the exhaust system if the flowrate is \(0.10 \mathrm{cfs},\) the temperature is \(250^{\circ} \mathrm{F}\), and the exhaust has the same properties as air.

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