What is the most effective nozzle size and water speed for maximum thrust?

[user079622]

Homework Statement

Stuck with thrust

Relevant Equations

m(dot)=r x v x A

Water jet comes from nozzle A=1m2, water speed is v=20m/s, density of water 1000km/m3, calculate thrust
Is it possible to find thrust from this data?

T=mass flow x v ?
= 20m/s x 1m2 x1000kg/m3 x 20m/s= 400 000N ?

Is more effective to use bigger nozzle A and slower water speed then use small nozzle with high water speed and why?

 

[haruspex]

T=mass flow x v ?
= 20m/s x 1m2 x1000kg/m3 x 20m/s= 400 000N ?

Yes.

Is more effective to use bigger nozzle A and slower water speed then use small nozzle with high water speed and why?

It depends how effectiveness is defined.
Please state the whole question exactly as given to you.

 

[user079622]

Yes.

How, if this is equation for thrust?https://www.grc.nasa.gov/www/k-12/airplane/rockth.html

It depends how effectiveness is defined.
Please state the whole question exactly as given to you.

To produce same thrust which engine use less power..

 

[erobz]

How, if this is equation for thrust?

The water jet expands to atmospheric pressure almost immediately after exiting the nozzle independent of nozzle geometry. The flow is virtually incompressible. This is different from the compressible gas flow exiting a chemical rocket.

 

[user079622]

The water jet expands to atmospheric pressure almost immediately after exiting the nozzle independent of nozzle geometry. The flow is virtually incompressible. This is different from the compressible gas flow exiting a chemical rocket.

Why rocket dont have atmospheric pressure as well immediately after exiting the nozzle
?

 

[user079622]

@haruspex

How F= mass flow x v is derived?

 

[erobz]

Why rocket dont have atmospheric pressure as well immediately after exiting the nozzle
?

Read the paper I linked in your other thread. Consider taking a course in Fluid Mechanics, and Thermodynamics. They are both required to understand the surrounding problem.

 

[haruspex]

How, if this is equation for thrust?
https://www.grc.nasa.gov/www/k-12/airplane/rockth.html

It is to do with compressibility. In the model used there, the exit pressure continues to accelerate the exhaust, making the effective exit velocity higher than that directly measured. Water being incompressible, there is no residual excess pressure.

How F= mass flow x v is derived?

In time dt you are accelerating a mass ##\dot m.dt## from 0 to v. What is the change in its momentum? What rate of change of momentum does that imply?

To produce same thrust which engine use less power..

Ok, so what is the formula for the power consumption?

 

[user079622]

It is to do with compressibility. In the model used there, the exit pressure continues to accelerate the exhaust, making the effective exit velocity higher than that directly measured. Water being incompressible, there is no residual excess pressure.

In time dt you are accelerating a mass ##\dot m.dt## from 0 to v. What is the change in its momentum? What rate of change of momentum does that imply?

Ok, so what is the formula for the power consumption?

Kinetic energy = 1/2 m v^2

instead m I will use mass flow (A density v)

P= 1/2 A density v^3

If I double A, I need double power, if double v I need 3times power, so it is more effective to increase nozzle A then velocity of fluid?

 

[haruspex]

Kinetic energy = 1/2 m v^2

instead m I will use mass flow (A density v)

P= 1/2 A density v^3

If I double A, I need double power, if double v I need 3times power, so it is more effective to increase nozzle A then velocity of fluid?

3 times?
To complete the comparison, you need to arrange that the thrust achieved is the same. I.e., for a given thrust, does increasing the area increase or reduce the power?

 

[user079622]

3 times?
To complete the comparison, you need to arrange that the thrust achieved is the same. I.e., for a given thrust, does increasing the area increase or reduce the power?

My mistake 2^3=8 , 8xtimes

For same thurst I get smaller power using bigger nozzle A.

 

[haruspex]

My mistake 2^3=8 , 8xtimes

For same thurst I get smaller power using bigger nozzle A.

Right.

 

[user079622]

Right.

I find this by plug in numbers in equtions, can I find this conclusions without using numbers?

 

[haruspex]

I find this by plug in numbers in equtions, can I find this conclusions without using numbers?

Sure. Write the equation for power as a function either of thrust and area or of thrust and velocity, instead of a function of area, velocity and density.

 

[user079622]

Sure. Write the equation for power as a function either of thrust and area or of thrust and velocity, instead of a function of area, velocity and density.

P=1/2 T v is this equation for power as a function of thrust and velocity?

A is inside T so how can I write that?

 

[Chestermiller]

The question asks for the "thrust." The thrust of what on what?

 

[haruspex]

P=1/2 T v is this equation for power as a function of thrust and velocity?

Ok, so for a given thrust, increasing v increases power. You can also write P as a function of T and A and show that increasing area reduces power.
That answers the question.

 

[user079622]

Ok, so for a given thrust, increasing v increases power. You can also write P as a function of T and A and show that increasing area reduces power.
That answers the question.

How do you get A in denominator?

 

[haruspex]

How do you get A in denominator?

In post #1 you used ##T=A\rho v^2##, and in post #15 an equation relating P, T, v. Combine them to eliminate v.

 

[user079622]

In post #1 you used ##T=A\rho v^2##, and in post #15 an equation relating P, T, v. Combine them to eliminate v.

How I can eliminate v if one equaiton has v^2 and second just v, an odd number of v

 

[erobz]

How I can eliminate v if one equaiton has v^2 and second just v, an odd number of v

You have for the power:

$$ P = \frac{1}{2}T v \tag{1} $$

And you have for the Thrust ##T##:

$$ T = \rho A v^2 \tag{2}$$

Solve ##(2)## for ##v##. Sub into ##(1)##.

 

[haruspex]

How I can eliminate v if one equaiton has v^2 and second just v, an odd number of v

So square both sides of the one with just v.

 

[user079622]

You have for the power:

$$ P = \frac{1}{2}T v \tag{1} $$

And you have for the Thrust ##T##:

$$ T = \rho A v^2 \tag{2}$$

Solve ##(2)## for ##v##. Sub into ##(1)##.

Then I get P = √ (1/4 T^3/rho A)

Why I get P=1/2 T v if P = F v ?

 

[haruspex]

Why I get P=1/2 T v if P = F v ?

If a force F acts on an object for time dt in the direction in which it is moving with velocity v then the power it exerts over that time is Fv. The work it does is therefore Fvdt.
The power pushing the jet out is not doing that. In time dt it accelerates a mass ##\dot m.dt## from 0 to v, doing work ##\frac 12 \dot m v^2 dt##, so ##P=\frac 12 \dot m v^2=\frac 12 T v##.
It's like the difference between final velocity and average velocity.

 

[user079622]

If a force F acts on an object for time dt in the direction in which it is moving with velocity v then the power it exerts over that time is Fv. The work it does is therefore Fvdt.
The power pushing the jet out is not doing that. In time dt it accelerates a mass ##\dot m.dt## from 0 to v, doing work ##\frac 12 \dot m v^2 dt##, so ##P=\frac 12 \dot m v^2=\frac 12 T v##.
It's like the difference between final velocity and average velocity.

So acceleration of water particles give us thrust, not velocity of particles..that make sense because f=m a

 

[haruspex]

So acceleration of water particles give us thrust, not velocity of particles..that make sense because f=m a

No, I didn't say that.
P=Fv is an instantaneous relationship. Over time, ##\int P.dt=\int F.v.dt##.
The power producing the water jet accelerates the water from stationary, so the average velocity for the force it exerts is only v/2.

 

[user079622]

P = √ (1/4 T^3/rho A)

Did I calculate OK?

 

[haruspex]

P = √ (1/4 T^3/rho A)

Did I calculate OK?

Yes (but you ought to put parentheses around the rho A to make it clear you are dividing by both).

 

[user079622]

Yes (but you ought to put parentheses around the rho A to make it clear you are dividing by both).

P = √ (1/4 T^3/(rho A))