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Chapter 4: Problem 54

The experimental ground-effect machine has a total weight of 4200 lb. It hovers 1 or \(2 \mathrm{ft}\) off the ground by pumping air at atmospheric pressure through the circular intake duct at \(B\) and discharging it horizontally under the periphery of the skirt \(C\). For an intake velocity \(v\) of \(150 \mathrm{ft} / \mathrm{sec},\) calculate the aver age air pressure \(p\) under the 18 -ft-diameter machine at ground level. The specific weight of the air is \(0.076 \mathrm{lb} / \mathrm{ft}^{3}\).

Short Answer

Expert verified
The average air pressure under the machine is 16.5 lb/ft虏.

Step by step solution

01

Calculate the Area of the Air Intake

The circular intake duct has a diameter of 18 feet. The formula for the area of a circle is \(A = \pi (r^2)\). Using this, we have \[A = \pi \left(\frac{18}{2}\right)^2 = 254.47 \; \text{ft}^2.\]
02

Determine the Mass Flow Rate

The mass flow rate \( \dot{m} \) is given by the product of the specific weight \( \gamma \), the area \( A \), and the velocity \( v \). Thus, \[\dot{m} = \gamma \cdot A \cdot v = 0.076 \;\text{lb/ft}^3 \times 254.47 \; \text{ft}^2 \times 150 \; \text{ft/s} = 2898.44 \; \text{lb/s}.\]
03

Relate Mass Flow to Pressure

For the ground-effect machine to hover, the force from the pressure difference must equal the weight of the machine. The pressure \( p \) multiplied by the area gives the force:\[F = p \cdot A = 4200\;\text{lb}.\]
04

Solve for the Pressure

Since we need to solve for \( p \), rearrange the equation from Step 3 to get:\[p = \frac{F}{A} = \frac{4200}{254.47} = 16.5 \; \text{lb/ft}^2.\]
05

Verify Assumptions and Units

Ensure that the units throughout the calculation are consistent and that the assumptions about the flow and forces are valid for a hovering ground-effect machine in this simplified scenario.

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Key Concepts

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

Understanding Mass Flow Rate
Mass flow rate is a fundamental concept in fluid mechanics that describes the amount of mass passing through a given surface per unit of time. In our problem, the ground-effect machine requires a specific mass flow rate to function effectively. It keeps the machine hovering by providing enough air pressure beneath it.

To find the mass flow rate, we use the formula:
  • \[ \dot{m} = \gamma \cdot A \cdot v \]
Where:
  • \(\gamma\) is the specific weight of air, given as \(0.076\, \text{lb/ft}^3\).
  • \(A\) is the cross-sectional area of the air intake.
  • \(v\) is the velocity of the air intake, provided as \(150\, \text{ft/s}\).
For our calculations:
  • The area \(A\) was previously calculated as \(254.47\, \text{ft}^2\).
  • The resulting mass flow rate is \( 2898.44\, \text{lb/s} \).
This rate of air mass flow is crucial for maintaining the necessary pressure to keep the machine in flight.
Role of Pressure in Hovering
Pressure is the force exerted by the air per unit area upon contacting the surface beneath the hovering machine. This force needs to be sufficient to counterbalance the weight of the vehicle, allowing it to stay airborne in a controlled manner.

In our scenario, the total weight of the machine is \(4200\, \text{lb}\), so the pressure force must equal this weight to ensure it hovers.We can find the necessary pressure using the equation:
  • \[p = \frac{F}{A}\]
Here:
  •  \(F\) is the force exerted by the pressure, equivalent to \(4200\, \text{lb}\).
  •  \(A\) is the contact area of air, or \(254.47\, \text{ft}^2\).
From this, we determine the pressure \(p\) needed is \(16.5\, \text{lb/ft}^2\), establishing the sufficient force for the machine to remain airborne.
The Concept of Hovering
Hovering is when an object remains suspended in one place without any physical support from below. For the ground-effect machine, this involves balancing the downward gravitational force with the upward thrust generated by pressured air.

Hovering relies on the strategic control of air pressures, directing sufficient mass flow beneath the machine to create lift. It requires an even distribution of air pressure to stabilize the machine in flight.

Key factors for effective hovering include:
  • Maintaining an adequate mass flow rate to sustain necessary pressure.
  • Ensuring the exhaust air is uniformly distributed around the periphery.
  • Adjusting for changes in weight or external forces like wind gusts.
Hovering is a dynamic state that involves persistent adjustments due to varying conditions and design specifications of the machine.
Exploring the Ground-Effect Machine
A ground-effect machine is a vehicle that uses air cushions to hover slightly above the ground. It is specially designed to utilize the ground-effect principle, where lift increases effectively due to reduced drag close to the ground surface.

Such machines often have a circular design, allowing air to be forced through a central duct and expelled around the skirt, forming a cushion of air that lessens friction and supports weight.

Key points about ground-effect machines:
  • They exploit the unique aerodynamic properties of being near the ground, making them energy-efficient over short distances.
  • The skirt helps trap air, creating a pressure cushion beneath the vehicle.
  • The design alterations in skirt and duct size can significantly impact performance.
  • Ground-effect machines are used in various applications, from hovercraft for transport to industrial lifting devices.
These machines showcase innovative engineering by taking advantage of fluid dynamics to achieve hovering through efficient use of air pressure dynamics.

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Most popular questions from this chapter

The chain of mass \(\rho\) per unit length passes over the small freely turning pulley and is released from rest with only a small imbalance \(h\) to initiate motion. Determine the acceleration \(a\) and velocity \(v\) of the chain and the force \(R\) supported by the hook at \(A,\) all in terms of \(h\) as it varies from essentially zero to \(H\). Neglect the weight of the pulley and its supporting frame and the weight of the small amount of chain in contact with the pulley. (Hint: The force \(R\) does not equal two times the equal tensions \(T\) in the chain tangent to the pulley.)

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The chain of length \(L\) and mass \(\rho\) per unit length is released from rest on the smooth horizontal surface with a negligibly small overhang \(x\) to initiate motion. Determine (a) the acceleration a as a function of \(x,(b)\) the tension \(T\) in the chain at the smooth corner as a function of \(x,\) and \((c)\) the velocity \(v\) of the last link \(A\) as it reaches the corner.

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