A boat travelling at 12 m ·s^-1 in fresh water has a 600 mm diameter propeller which takes wat

A boat travelling at 12 m ·s^-1 in fresh water has a 600...

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.

Key Concepts

Conservation of Momentum

This principle states that the change in momentum of a fluid, which occurs as the propeller accelerates water, results in a thrust force. It is used to relate the mass flow rate of the water and its change in velocity to the force generated, a fundamental concept in fluid mechanics applied to propulsion systems.

Mass Flow Rate

Mass flow rate is the quantity of mass passing through a given cross-sectional area per unit time. In the analysis of a propeller, it links the fluid density with the volumetric flow rate, providing a basis for calculating the change in momentum that results in thrust.

Propulsive Efficiency

Propulsive efficiency measures how effectively the energy input to a propulsion system is converted into useful work, such as the thrust that moves a boat. It is usually defined as the ratio of useful power output (often the product of thrust and boat speed) to the total power input supplied to the propeller.

Power Input and Kinetic Energy Transfer

This concept involves the calculation of the power required to accelerate the water through the propeller, which is determined by the rate at which kinetic energy is imparted to the fluid. Understanding the relationship between power, force, and velocity is key to assessing the performance of propulsion systems.

Transcript

00:01 In this problem, we have a boat traveling 12 meters per second, and i have them going to the left.

00:06 And the goal is probably really is nothing fancy.

00:08 It's just, see how everything matches up in terms of the formulas they give in the section on propellers.

00:17 So i'm going to basically show some practice with algebra, so we don't just blindly just go calculate every single number right off the bat.

00:28 And just see how that comes out if it shows any relationships between the quantities we are given the how fast the boat is moving now if the boat is moving to the left to the propeller state to the propeller who thinks of himself a stationary he's going to see the the water coming towards him 12 meters per second that's u1 that's why i call it u1 he's thinks of himself stationary and the water's coming toward him we're given the volume floor rate that goes through here through this region diameter of the propeller and let's start looking at what we need to get first thing they want the thrust so we're going to need a 2 for the thrust and we're going to need u2 for that now we can get a 2 very easily now u2 in that section was found to be u1 plus u4 over 2 but this is cute q over a.

01:37 Remember, this is defined flow rate through this region.

01:41 So it's q over a, just normal formulation that we've done.

01:44 And you could calculate this and it comes out to be 15 .03 meters per second.

01:50 It could, but let's leave everything alone.

01:55 Just look at symbols to a little practice with algebra and a straightforward question.

02:01 Rowe a, u2, u4x minus u1x.

02:09 So this is the, this obviously is the net rate of change of momentum and here.

02:18 And i'll make to the right, the positive direction, seems reasonable enough, doesn't really matter.

02:27 And so this becomes row q, u4 minus u1.

02:39 U4, and from this expression here is 2u2 minus u1.

02:47 So we could get a number for u4 also to use at this point.

02:54 That'd be 18 .06.

02:57 Nothing fancy about that.

03:00 So let's calculate, let's leave the thrust equation alone and put in symbols and see what we get to q over a minus u1 minus u1.

03:20 So i just putting in q over a for u2 and sticking it in here.

03:25 This is u4.

03:28 So we're going to get a 2 q over a and minus 2 u1.

03:32 So we can bring the two out, 2 row q, q over a.

03:37 Also, when you do this, u1, and i think, yeah, q over a minus u1, and we're all set.

03:53 Also, if you're thinking about roundoffice, errors and things of that nature if you can always go back to the base numbers you you save yourself that difficulty also save yourself the time of punching in number after number too so let's put in everything here to 1 ,000 kilograms cubic meter 4 .25 is cubed per second but also gives you practice when you do this if you have situations where you cannot punch in numbers and you've got to do sub -algebraic work or a pretty good amount of algebraic work and then finally you can get a number.

04:40 So more practice you have, more comfortable you are doing things like this with symbols, the better off it is for you.

04:47 That's down on mistakes.

04:49 It saves time also.

04:50 If you're in middle of a test, punching a lot of numbers can take a long time.

04:58 So there's putting in all the numbers...

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