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Chapter 11: Problem 63

An airplane wing is designed so that the speed of the air across the top of the wing is \(251 \mathrm{m} / \mathrm{s}\) when the speed of the air below the wing is \(225 \mathrm{m} / \mathrm{s} .\) The density of the air is \(1.29 \mathrm{kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area \(24.0 \mathrm{m}^{2} ?\)

Short Answer

Expert verified
The lifting force on the wing is approximately 191,565.44 N.

Step by step solution

01

Understand Bernoulli's Principle

Bernoulli's principle relates the speed of a fluid and its pressure. For an airplane wing, the principle explains how different air speeds above and below the wing create a pressure difference, generating lift. The lift force can be calculated using the pressure difference and the area of the wing.
02

Calculate Pressure Difference Using Bernoulli's Equation

Bernoulli's equation in its general form is given by: \( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \), where \( P \) is the pressure, \( \rho \) is the density of the fluid, and \( v \) is the velocity of the fluid. Rearrange to find the pressure difference: \( \Delta P = \frac{1}{2} \rho (v_2^2 - v_1^2) \), where \( v_2 = 251 \text{ m/s} \) and \( v_1 = 225 \text{ m/s} \).
03

Substitute Values into Pressure Difference Formula

Substitute the values of air velocities and density into the pressure difference formula. \( \Delta P = \frac{1}{2} \times 1.29 \times (251^2 - 225^2) \).
04

Calculate the Pressure Difference

Calculate \( \Delta P \) using the equation from Step 3: \( \Delta P = \frac{1}{2} \times 1.29 \times (63001 - 50625) \approx \frac{1}{2} \times 1.29 \times 12376 = 7978.56 \text{ Pa} \).
05

Calculate the Lifting Force

The lift force is calculated as the product of the pressure difference and the wing area: \( F = \Delta P \times A \). Substitute the pressure difference and area: \( F = 7978.56 \times 24.0 \approx 191565.44 \text{ N} \).
06

Review the Calculations

Check each step for correctness and ensure that units are consistent. Confirm that calculated pressure difference and lift force make sense based on input values.

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Key Concepts

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

Lift Force
Lift force is a core concept in understanding how airplanes stay in the sky. It is the net force that acts perpendicular to the surface of an object, like an airplane wing, in a fluid. This force allows the aircraft to rise, stay aloft, and maneuver.

- Lift is generated by a difference in pressure between the top and bottom surfaces of the wing.
- The larger the surface area, the more air it can displace, potentially increasing lift.
- Lift force is crucial for achieving and maintaining flight. Without sufficient lift, an aircraft cannot overcome gravity.

In our exercise, we calculated lift force using Bernoulli's Principle, which shows how the differences in airflow speed over and under the wing result in a pressure difference. This forces the wing upward, counteracting the force of gravity.
Pressure Difference
Pressure difference is a key element in generating lift on an airplane wing. This pressure difference arises due to varying velocities of the air passing over and beneath the wing. According to Bernoulli's Principle, faster moving air results in lower pressure.

- Air flowing over the top of the wing moves faster than the air underneath.
- Faster airflow on top leads to lower pressure above the wing.
- The difference in speeds creates a pressure differential: lower pressure above, higher pressure below.

This pressure difference pushes the wing upwards, creating lift. The greater the difference in pressure, the more lift produced. In the exercise, the airspeed and density were used to calculate the pressure difference numerically, leading to our understanding of how lift force is generated.
Fluid Dynamics
Fluid dynamics is the study of fluids (liquids and gases) in motion. It involves understanding how forces interact with fluids to produce motion and other effects, like lift. Applying fluid dynamics principles helps us analyze how air moves around objects like wings.

- It includes studying the flow and behavior of air around the wing using principles such as Bernoulli's and Newton's laws.
- Understanding fluid dynamics allows engineers to design surfaces that optimize lift while minimizing drag.

In our problem, analyzing how air travels over and under the wing helped apply fluid dynamic principles. We used these principles to conclude how changes in velocity affect pressure and ultimately lift force. By mastering fluid dynamics, we can better anticipate and design for flight efficiency.

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Most popular questions from this chapter

One way to administer an inoculation is with a "gun" that shoots the vaccine through a narrow opening. No needle is necessary, for the vaccine emerges with sufficient speed to pass directly into the tissue beneath the skin. The speed is high, because the vaccine \(\left(\rho=1100 \mathrm{kg} / \mathrm{m}^{3}\right)\) is held in a reservoir where a high pressure pushes it out. The pressure on the surface of the vaccine in one gun is \(4.1 \times 10^{6}\) Pa above the atmospheric pressure outside the narrow opening. The dosage is small enough that the vaccine's surface in the reservoir is nearly stationary during an inoculation. The vertical height between the vaccine's surface in the reservoir and the opening can be ignored. Find the speed at which the vaccine emerges.

A water bed for sale has dimensions of \(1.83 \mathrm{m} \times 2.13 \mathrm{m} \times 0.229 \mathrm{m}\) The floor of the bedroom will tolerate an additional weight of no more than \(6660 \mathrm{N}\). Find the weight of the water in the bed and determine whether the bed should be purchased.

Water flows straight down from an open faucet. The crosssectional area of the faucet is \(1.8 \times 10^{-4} \mathrm{m}^{2},\) and the speed of the water is \(0.85 \mathrm{m} / \mathrm{s}\) as it leaves the faucet. Ignoring air resistance, find the crosssectional area of the water stream at a point \(0.10 \mathrm{m}\) below the faucet.

If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure, while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures can rupture an eardrum. Eardrums can rupture when the pressure difference is as little as \(35 \mathrm{kPa} .\) What is the depth at which this pressure difference could occur? The density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3}\).

Venturi meter is a device that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed \(v_{2}\) through a horizontal section of pipe whose cross-sectional area is \(A_{2}=0.0700 \mathrm{m}^{2} .\) The gas has a density of \(\rho=1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The Venturi meter has a cross-sectional area of \(A_{1}=0.0500 \mathrm{m}^{2}\) and has been substituted for a section of the larger pipe. The pressure difference between the two sections is \(P_{2}-P_{1}=120 \mathrm{Pa}\). Find (a) the speed \(v_{2}\) of the gas in the larger, original pipe and (b) the volume flow rate \(Q\) of the gas.
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