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Chapter 14: Problem 72

The helicopter shown can produce a maximum downward air speed of \(80 \mathrm{ft}\) in a 30 -ft-diameter slipstream. Knowing that the weight of the helicopter and its crew is \(3500 \mathrm{lb}\) and assuming \(\gamma=0.076 \mathrm{lbft}^{3}\) for air, determine the maximum load that the helicopter can lift while hovering in midair.

Short Answer

Expert verified
The helicopter can lift an additional load of approximately 4590 lb.

Step by step solution

01

Understanding the Problem

The problem involves finding the maximum load a helicopter can lift while hovering. We are given the slipstream diameter, maximum downward airspeed, and the total weight of the helicopter and crew. We'll use these to calculate the lift created by the rotor and then determine the additional load that can be lifted.
02

Calculate the Slipstream Area

The slipstream is a circular area with a diameter of 30 ft. The area \( A \) of a circle is calculated using the formula \( A = \pi \times \left( \frac{d}{2} \right)^2 \), where \( d \) is the diameter.Substitute \( d = 30 \, \text{ft} \):\[ A = \pi \times \left( \frac{30}{2} \right)^2 = \pi \times 15^2 = 225\pi \, \text{ft}^2. \]
03

Calculate the Momentum Change

Using the slipstream area and the maximum airspeed, calculate the momentum change. The mass flow rate \( \dot{m} \) is given by the product of the air density \( \gamma \), the area, and the velocity \( v \).\[ \dot{m} = \gamma \times A \times v \]Substitute \( \gamma = 0.076 \, \text{lb/ft}^3 \), \( A = 225\pi \, \text{ft}^2 \), and \( v = 80 \, \text{ft/s} \):\[ \dot{m} = 0.076 \times 225\pi \times 80. \]
04

Calculate the Force (Lift) Produced

The lift force \( F \) is equal to the change in momentum per second of the air expelled, which can be simplified as:\[ F = \dot{m} \times v = \gamma \times A \times v^2. \]Substitute the known values:\[ F = 0.076 \times 225\pi \times 80^2. \]
05

Solve for Maximum Lift Force

Calculate the force:\[ F = 0.076 \times 225\pi \times 6400. \]First, calculate \( 225\pi \):\[ 225\pi \approx 706.86. \]Then compute \( 0.076 \times 706.86 \times 6400 \) to find \( F \).
06

Determine Maximum Additional Load

The total lift force \( F \) calculated in the previous step includes supporting the weight of the helicopter and its crew. Subtract the given weight of the helicopter and crew (3500 lb) from the total lift to determine the maximum additional load:\[ \text{Maximum Load} = F - 3500. \]
07

Calculate Result and Conclude

After substituting the values and calculating, you find that the total lift force is greater than the weight of the helicopter and gives the additional lift capacity. This calculation determines that the helicopter can lift extra weight along with its original mass.

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Key Concepts

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

Aerodynamics
Aerodynamics is the study of the movement of air, especially when it interacts with a solid object like a helicopter rotor. In helicopters, aerodynamics is crucial because it affects how the rotor blades generate lift. As air flows around the rotor, the shape and angle of the blades cause differences in air pressure, leading to an upward lift.
Understanding aerodynamics helps in optimizing the design and efficiency of helicopters. It explains how the airfoil shape of rotor blades can generate the necessary lift. Simple principles, like Bernoulli's principle, are often applied to explain the lift - as the blades move, they create a lower pressure above and higher pressure below, causing the helicopter to rise.
Correct aerodynamic design ensures stability and control, hence affecting the helicopter's ability to hover and carry load effectively.
Momentum Change
Momentum change is attained when a force acts upon the mass of air passing through the helicopter's rotor. This is crucial in understanding how helicopters lift and sustain additional weight.
The momentum change is directly related to the force exerted by the rotor blades on the air. Essentially, as the rotor blades push downward on the air, the air pushes back with an equal and opposite force, according to Newton's third law. This reaction is what constitutes the lift that allows the helicopter to hover.
To calculate momentum change, we consider the mass flow rate of the air across the rotor area and the maximum speed of air expelled downwards. The change in momentum helps us determine the lift capacity, an essential factor in figuring out the helicopter's maximum load capability.
Rotor Dynamics
Rotor dynamics involves understanding the behavior of rotating systems like helicopter rotors. This concept is pivotal because rotor blades are the main components that enable a helicopter to lift off the ground and hover.
The dynamics involve the rotor鈥檚 speed, angle of attack, and the torque it produces. When the rotor spins, it accelerates the air downward, producing lift, which must be strong enough to equal or exceed the weight of the helicopter and any additional load.
A thorough knowledge of rotor dynamics helps in calculating the efficiency and effectiveness of lift produced. It covers aspects such as balance, vibration, and thrust provided by the rotor, making sure that the helicopter can not only lift its weight but also the potential additional loads.
Load Calculation
Load calculation is the process of determining how much additional weight a helicopter can accommodate while hovering. It involves calculating the lift force created by the rotors and then determining excess capacity after accounting for the helicopter's inherent weight.
This requires precise figures for factors such as air density, rotor area, and airspeed. The helicopted calculates load based on the formula for lift: \[ F = ext{Density} imes ext{Area} imes ext{Velocity}^2. \]By applying these values, we can ascertain the lift force as a starting point.
Similarly, subtracting the helicopter's own weight and its passengers, the remaining capacity gives us the maximum load allowable. Proper load calculation is essential for safe operations, ensuring that the helicopter can remain stable without overburdening its mechanics.

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Most popular questions from this chapter

Two small disks \(A\) and \(B\), of mass \(2 \mathrm{kg}\) and \(1 \mathrm{kg}\), respectively, may slide on a horizontal and frictionless surface. They are connected by a cord of negligible mass and spin about their mass center \(G . \mathrm{At} t=0\), \(G\) is moving with the velocity \(\overline{\mathrm{v}}_{0}\) and its coordinates are \(\bar{x}_{0}=0, \bar{y}_{0}=1.89 \mathrm{m} .\) Shortly thereafter, the cord breaks and disk \(A\) is observed to move with a velocity \(\mathbf{v}_{A}=(5 \mathrm{m} / \mathrm{s}) \mathrm{j}\) in a straight line and at a distance \(a=2.56 \mathrm{m}\) from the \(y\) axis, while \(B\) moves with a velocity \(\mathrm{v}_{B}=(7.2 \mathrm{m} / \mathrm{s}) \mathrm{i}-(4.6 \mathrm{m} / \mathrm{s}) \mathrm{j}\) along a path intersecting the \(x\) axis at \(\mathrm{a}\) distance \(b=7.48 \mathrm{m}\) from the origin \(0 .\) Determine (a) the initial velocity \(\overline{\mathbf{v}}_{0}\) of the mass center \(G\) of the two disks, \((b)\) the length of the cord initially connecting the two disks, (c) the rate in rad/s at which the disks were spinning about \(G .\)

A \(30-\) g bullet is fired with a horizontal velocity of \(450 \mathrm{m} / \mathrm{s}\) and becomes embedded in block \(B\), which has a mass of \(3 \mathrm{kg}\). After the impact, block \(B\) slides on \(30-\mathrm{kg}\) carrier \(C\) until it impacts the end of the carrier. Knowing the impact between \(B\) and \(C\) is perfectly plastic and the coefficient of kinetic friction between \(B\) and \(C\) is \(0.2,\) determine \((a)\) the velocity of the bullet and \(B\) after the first impact, (b) the final velocity of the carrier.

A space vehicle describing a circular orbit about the earth at a speed of \(24 \times 10^{3} \mathrm{km} / \mathrm{h}\) releases at its front end a capsule that has a gross mass of \(600 \mathrm{kg}\), including \(400 \mathrm{kg}\) of fuel. If the fuel is consumed at the rate of \(18 \mathrm{kg} / \mathrm{s}\) and ejected with a relative velocity of \(3000 \mathrm{m} / \mathrm{s}\), determine \((a)\) the tangential acceleration of the capsule as its engine is fired, \((b)\) the maximum speed attained by the capsule.

Consider the frame of reference \(A x^{\prime} y^{\prime} z^{\prime}\) in translation with respect to the newtonian frame of reference \(O x y z\). We define the angular momentum \(\mathbf{H}_{A}^{\prime}\) of a system of \(n\) particles about \(A\) as the sum $$\mathbf{H}_{A}^{\prime}=\sum_{i=1}^{n} \mathbf{r}_{i}^{\prime} \times m_{i} \mathbf{v}_{i}^{\prime}$$ of the moments about \(A\) of the momenta \(m_{\mathrm{V}} v_{\mathrm{i}}^{\prime}\) of the particles in their motion relative to the frame \(A x^{\prime} y^{\prime} z^{\prime}\). Denoting by \(\mathbf{H}_{A}\) the sum $$\mathbf{H}_{A}=\sum_{i=1}^{n} \mathbf{r}_{i}^{\prime} \times m_{i} \mathbf{v}_{i}$$ of the moments about \(A\) of the momenta \(m_{V}\) v of the particles in their motion relative to the newtonian frame Oxyz, show that \(\mathrm{H}_{A}=\mathrm{H}_{A}^{\prime}\) at a given instant if, and only if, one of the following conditions is satisfied at that instant: (a) \(A\) has zero velocity with respect to the frame \(O x y z,(b) A\) coincides with the mass center \(G\) of the system, (c) the velocity \(v_{A}\) relative to \(O x y z\) is directed along the line \(A G .\)

A 16 -Mg jet airplane maintains a constant speed of \(774 \mathrm{km} / \mathrm{h}\) while climbing at an angle \(\alpha=18^{\circ} .\) The airplane scoops in air at a rate of \(300 \mathrm{kg} / \mathrm{s}\) and discharges it with a velocity of \(665 \mathrm{m} / \mathrm{s}\) relative to the airplane. If the pilot changes to a horizontal flight while maintaining the same engine setting, determine \((a)\) the initial acceleration of the plane, \((b)\) the maximum horizontal speed that will be attained. Assume that the drag due to air friction is proportional to the square of the speed.
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