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Chapter 4: Problem 8

A boat travelling at \(12 \mathrm{~m} \cdot \mathrm{s}^{-1}\) in fresh water has a \(600 \mathrm{~mm}\) diameter propeller which takes water at \(4.25 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) between its blades. Assuming that the effects of the propeller hub and the boat hull on flow conditions are negligible, calculate the thrust on the boat, the efficiency of the propulsion, and the power input to the propeller.

Short Answer

Expert verified
Thrust: 12,680 N. Efficiency: 84.7%. Power Input: 152,160 W.

Step by step solution

01

Determine the Cross-sectional Area of the Propeller

Convert the diameter of the propeller to meters: 600 mm = 0.6 m. The cross-sectional area, A, of the propeller can be calculated using the formula: \[ A = \pi \left( \frac{D}{2} \right)^2 \]Substitute D = 0.6 m: \[ A = \pi \left( \frac{0.6}{2} \right)^2 = \pi \left( 0.3 \right)^2 \approx 0.2827 \mathrm{~m}^2 \]
02

Calculate the Velocity of Water Through the Propeller

Using the flow rate, Q, and the cross-sectional area, A, the velocity, V, of water through the propeller can be calculated by: \[ V = \frac{Q}{A} \]Substitute Q = 4.25 m\textsuperscript{3}/s and A = 0.2827 m\textsuperscript{2}: \[ V = \frac{4.25}{0.2827} \approx 15.04 \mathrm{~m} \cdot \mathrm{s}^{-1} \]
03

Calculate the Thrust on the Boat

To find the thrust, T, use the conservation of momentum principle: \[ T = \dot{m} \times (V_{exit} - V_{boat}) \]Where - \( \dot{m} \) is the mass flow rate: \[ \dot{m} = Q \times \rho \]Assuming the density of water, \( \rho \) is 1000 kg/m\textsuperscript{3}: \[ \dot{m} = 4.25 \times 1000 = 4250 \text{kg/s} \]- Then, \[ T = 4250 \times (15.04 - 12) \approx 12,680 \mathrm{~N} \]
04

Calculate the Power Input to the Propeller

Power input, P, to the propeller can be calculated as: \[ P = T \times V_{boat} \]Substitute T = 12,680 N and \(V_{boat} = 12 \mathrm{~m} \cdot \mathrm{s}^{-1}\): \[ P = 12,680 \times 12 \approx 152,160 \mathrm{~W} \]
05

Calculate the Efficiency of the Propulsion

The efficiency, \( \eta \), is given by: \[ \eta = \frac{ \text{useful power output} }{ \text{total power input} } \]The useful power output is the work done by the boat, which is: \[ P_{out} = T \times V_{boat} \approx 152,160 \mathrm{~W} \]However, the thrust power (ideal power output) from the propeller is: \[ P_{thrust} = \dot{m} \cdot g \cdot \frac{V_{exit}^2 - V_{boat}^2}{2} \]Using \( V_{exit} = 15.04 \mathrm{~m} \cdot \mathrm{s}^{-1} \) and \( V_{boat} = 12 \mathrm{~m} \cdot \mathrm{s}^{-1} \): \[ P_{thrust} = 4250 \times 9.81 \times \frac{15.04^2 - 12^2}{2} \approx 128,960 \mathrm{~W} \]Then efficiency is: \[ \eta = \frac{128,960}{152,160} \approx 0.847 \ or \ 84.7\% \]

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Key Concepts

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

propeller thrust calculation
Thrust is the force exerted by the propeller to move the boat forward. To calculate the thrust, we leverage the conservation of momentum. The mass flow rate of water through the propeller, \(\backslashdot{m}\), is computed as the product of the flow rate and the density of water. After finding this key value, we use it with the velocities of water entering and exiting the propeller. The formula used is: \[ T = \backslashdot{m} \times (V_{exit} - V_{boat}) \] Here, \(T\) is the thrust, \( \backslashdot{m}\) the mass flow rate, \(V_{exit} \) the water velocity exiting the propeller, and \(V_{boat}\) the speed of the boat.
propeller efficiency
Efficiency helps us understand how effectively the propeller converts input power into useful thrust. We use the formula: \[ \eta = \frac{ \text{useful power output} }{ \text{total power input} } \]
Useful power is the power that propels the boat forward, calculated as: \[ P_{out} = T \times V_{boat} \]
The ideal power output (or thrust power) takes into account both the mass flow rate and the velocity: \[ P_{thrust} = \backslashdot{m} \times g \times \frac{V_{exit}^2 - V_{boat}^2}{2} \]
Finally, efficiency \( \eta \) is the ratio of these two powers, giving us a clear perspective on the performance of the propulsion system.
power input and output in propulsion
Power input to the propeller is the energy provided by the engine to turn the propeller and move the boat. It is calculated by the formula: \[ P = T \times V_{boat} \]
The power output, also known as useful power, is the actual work done by the boat moving through water. This is also represented by the same formula as it measures the boat's propulsion. To distinguish between the two, power input refers to the power supplied to the system while power output is the effective power converted into useful work.
conservation of momentum in fluids
Conservation of momentum is a fundamental principle in fluid mechanics. It states that the momentum in a system remains constant unless acted upon by external forces. In the context of propellers, when water flows through the propeller, its momentum changes due to the propeller’s blades exerting force on it. This change in momentum is directly related to the thrust generated. The formula to express this principle is: \[ T = \backslashdot{m} \times (V_{exit} - V_{boat}) \]
Where the change in momentum of the water (mass flow rate \( \backslashdot{m} \) times the velocity change \(V_{exit} - V_{boat} \)) equals the thrust force applied by the propeller.

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Most popular questions from this chapter

Air at constant density \(1.22 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) flows in a duct of internal diameter \(600 \mathrm{~mm}\) and is discharged to atmosphere. At the outlet end of the duct, and coaxial with it, is a cone with base diameter \(750 \mathrm{~mm}\) and vertex angle \(90^{\circ} .\) Flow in the duct is controlled by moving the vertex of the cone into the duct, the air then escaping along the sloping sides of the cone. The mean velocity in the duct upstream of the cone is \(15 \mathrm{~m} \cdot \mathrm{s}^{-1}\) and the air leaves the cone (at the \(750 \mathrm{~mm}\) diameter) with a mean velocity of \(60 \mathrm{~m} \cdot \mathrm{s}^{-1}\) paraltel to the sides. Assuming temperature changes and frictional effects to be negligible, calculate the net axial force exerted by the air on the cone.

A rocket sled of \(2.5 \mathrm{Mg}\) (tare) burns \(90 \mathrm{~kg}\) of fuel a second and the uniform exit velocity of the exhaust gases relative to the rocket is \(2.6 \mathrm{~km} \cdot \mathrm{s}^{-1}\). The total resistance to motion at the track on which the sled rides and in the air equals \(K V\), where \(K=1450 \mathrm{~N} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}\) and \(V\) represents the velocity of the sled. Assuming that the exhaust gases leave the rocket at atmospheric pressure, calculate the quantity of fuel required if the sled is to reach a maximum velocity of \(150 \mathrm{~m} \cdot \mathrm{s}^{-1}\).

A boat is driven at constant velocity \(c\) (relative to the undisturbed water) by a jet-propulsion unit which takes in water at the bow and pumps it astern, beneath the water surface, at velocity \(u\) relative to the boat. Show that the efficiency of the propulsion, if friction and other losses are neglected, is \(2 c /(c+u)\) Such a boat moves steadily up a wide river at \(8 \mathrm{~m} \cdot \mathrm{s}^{-1}\) (relative to the land). The river flows at \(1.3 \mathrm{~m} \cdot \mathrm{s}^{-1}\). The resistance to motion of the boat is \(1500 \mathrm{~N}\). If the velocity of the jet relative to the boat is \(17.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\), and the overall efficiency of the pump is \(65 \%\), determine the total area of the outlet nozzles, and the engine power required.

4.4 Two adjacent parallel and horizontal rectangular conduits A and B, of cross-sectional areas \(0.2 \mathrm{~m}^{2}\) and \(0.4 \mathrm{~m}^{2}\) respectively, discharge water axially into another conduit \(\mathrm{C}\) of crosssectional area \(0.6 \mathrm{~m}^{2}\) and of sufficient length for the individual streams to become thoroughly mixed. The rates of flow through \(A\) and \(B\) are \(0.6 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) and \(0.8 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) respectively and the pressures there are \(31 \mathrm{kPa}\) and \(30 \mathrm{kPa}\) respectively. Neglecting friction at the boundaries, determine the energy lost by each entry stream (divided by mass) and the total power dissipated.

To propel a light aircraft at an absolute velocity of \(240 \mathrm{~km} \cdot \mathrm{h}^{-1}\) against a head wind of \(48 \mathrm{~km} \cdot \mathrm{h}^{-1}\) a thrust of \(10.3 \mathrm{kN}\) is required. Assuming an efficiency of \(90 \%\) and a constant air density of \(1.2 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) determine the diameter of ideal propeller required and the power needed to drive it.
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