Clarify power calculation

[Ripcrow]

TL;DR

Mass flow to power.

I have a nozzle through which a water vapour passes at 503 metres per second with a density of 0.051 kg per cubic metre giving a mass flow rate of 0.00129 kg per second. Using half mass times velocity squared I have an answer of 163. Is that answer 163 newtons and if it is how do I convert to Newton metres to find torque if that flow was directed onto a turbine

 

[Baluncore]

Power in watts is, the volume in metres cubed per second, multiplied by the pressure change in pascals.

Do you know the pressure drop across the nozzle?

 

[jack action]

Using half mass times velocity squared I have an answer of 163.

No, you use half mass flow rate times velocity squared to get that answer.

Is that answer 163 newtons

No, the "half mass times velocity squared" part is energy. But divided by time - from the mass flow rate - you get power, thus the unit is Watt. (Since everything is in SI units.)

how do I convert to Newton metres to find torque if that flow was directed onto a turbine

Power for a rotating machine is torque times the angular velocity. Thus the torque (N.m) from a turbine hit by that flow is found by dividing the power (W) of that flow by the angular velocity (rad/s) of that turbine. Assuming all the power from the flow is converted to mechanical work, obviously.

But if there is a density change in your process (compressible flow), things get a little more complicated than those simple calculations that assume constant static pressure.

 

[russ_watters]

I have a nozzle through which a water vapour passes at 503 metres per second with a density of 0.051 kg per cubic metre giving a mass flow rate of 0.00129 kg per second. Using half mass times velocity squared I have an answer of 163. Is that answer 163 newtons and if it is how do I convert to Newton metres to find torque if that flow was directed onto a turbine

Some gave specific answers but more broadly, you should use the units in the calculation so that the answer has units attached to it.

The short answer to your main question is that force is momentum change over time. But if you're trying to use that to calculate a turbine's performance it's not going to work. Turbine design is very non-trivial. The force will vary with turbine speed and geometry. If you're using an off-the shelf turbine life is easy: just read the power off the manufacturer's performance curve. If you're trying to design your own, you might just start with a target/guess of 30%-50% efficiency hope you hit that.

 

[Ripcrow]

Power in watts is, the volume in metres cubed per second, multiplied by the pressure change in pascals.

Do you know the pressure drop across the nozzle?

Pressure drop is from 7333 pascal to 867 pascal with a fluid density of 0.051 kg per cubic metre passing through a tube ( no restriction in diameter as in a nozzle ) with a diameter of 8 mm.

 

[Ripcrow]

No, you use half mass flow rate times velocity squared to get that answer.

No, the "half mass times velocity squared" part is energy. But divided by time - from the mass flow rate - you get power, thus the unit is Watt. (Since everything is in SI units.)

Power for a rotating machine is torque times the angular velocity. Thus the torque (N.m) from a turbine hit by that flow is found by dividing the power (W) of that flow by the angular velocity (rad/s) of that turbine. Assuming all the power from the flow is converted to mechanical work, obviously.

But if there is a density change in your process (compressible flow), things get a little more complicated than those simple calculations that assume constant static pressure.

Yes that half mass times velocity squared is half of the flow rate of 0.00129kg per second ( so that is 0.000645 kg per second ) multiplied by 503 metres per second squared. Looking for a max torque at stall so there is no angular momentum. As with all turbines max torque is at stall and power will be calculated using half available maximum torque and half maximum rpm so that the power output is calculated using a combination of rpm and torque. Power throughout the rev range can be calculated later just looking for maximum torque available( newtons ) from mass flow rate.

 

[Ripcrow]

No, you use half mass flow rate times velocity squared to get that answer.

No, the "half mass times velocity squared" part is energy. But divided by time - from the mass flow rate - you get power, thus the unit is Watt. (Since everything is in SI units.)

Power for a rotating machine is torque times the angular velocity. Thus the torque (N.m) from a turbine hit by that flow is found by dividing the power (W) of that flow by the angular velocity (rad/s) of that turbine. Assuming all the power from the flow is converted to mechanical work, obviously.

But if there is a density change in your process (compressible flow), things get a little more complicated than those simple calculations that assume constant static pressure.

Watts is equal to Newton and as the mass flow rate and velocity is calculated as per second that final answer of 163 watts ( newtons ) is divided by 1 so the answer remains 163. Am I missing something.

 

[jack action]

Watts is equal to Newton

No. Watt is a unit of power, Newton is a unit of force. 1 Watt is 1 Newton-meter per second.

As with all turbines max torque is at stall and power will be calculated using half available maximum torque and half maximum rpm so that the power output is calculated using a combination of rpm and torque.

With rotary machinery, Torque is a function of RPM and thus power is also a function of RPM. The product of half the maximum torque with half the maximum RPM is useless unless they coincide exactly, thus giving the power at half the maximum RPM.

The torque will depend on the geometry of the turbine (whether stalled or not). The direction of the flow at the outlet versus the inlet is important to evaluate the change in momentum. The reference is the Euler turbomachinery equation. You may want to check out the Euler equations for the complete compressible and incompressible flow characteristics.

For example, the torque of a Pelton wheel is derived here.

 

[Ripcrow]

No. Watt is a unit of power, Newton is a unit of force. 1 Watt is 1 Newton-meter per second.


With rotary machinery, Torque is a function of RPM and thus power is also a function of RPM. The product of half the maximum torque with half the maximum RPM is useless unless they coincide exactly, thus giving the power at half the maximum RPM.

The torque will depend on the geometry of the turbine (whether stalled or not). The direction of the flow at the outlet versus the inlet is important to evaluate the change in momentum. The reference is the Euler turbomachinery equation. You may want to check out the Euler equations for the complete compressible and incompressible flow characteristics.

For example, the torque of a Pelton wheel is derived here.

No. Watt is a unit of power, Newton is a unit of force. 1 Watt is 1 Newton-meter per second.


With rotary machinery, Torque is a function of RPM and thus power is also a function of RPM. The product of half the maximum torque with half the maximum RPM is useless unless they coincide exactly, thus giving the power at half the maximum RPM.

The torque will depend on the geometry of the turbine (whether stalled or not). The direction of the flow at the outlet versus the inlet is important to evaluate the change in momentum. The reference is the Euler turbomachinery equation. You may want to check out the Euler equations for the complete compressible and incompressible flow characteristics.

For example, the torque of a Pelton wheel is derived here.

Watts are watts per second also so they are the same.

The pelton turbine is an impulse turbine and uses different properties of flow than a gas turbine. All gas turbines generate maximum torque at stall as that is where the maximum gas deflection is occurring. At rotational speed the gas can’t act on the blades of the turbine to create maximum torque. If you imagine a mass flow of 1kg travelling at 2 metre per second hitting a blade then imagine that blade travelling at 1 metre per second then you can see the force acting on the blade is only going to act at half the velocity therefore only half the maximum force is felt by the blade. Not a turbine on earth that doesn’t make maximum torque at stall as that is where maximum flow deflection occurs.

 

[jack action]

Watts are watts per second also so they are the same.

??? This sentence does not make any sense.

$$1\ watt = 1\ \frac{newton\ \cdot\ meter}{second}$$(source)​

The pelton turbine is an impulse turbine and uses different properties of flow than a gas turbine.

You never specified what type of turbine you were referring to. Velocity triangles and the Euler turbomachinery equation still apply to gas turbines:

(source)​

At rotational speed the gas can’t act on the blades of the turbine to create maximum torque. If you imagine a mass flow of 1kg travelling at 2 metre per second hitting a blade then imagine that blade travelling at 1 metre per second then you can see the force acting on the blade is only going to act at half the velocity therefore only half the maximum force is felt by the blade. Not a turbine on earth that doesn’t make maximum torque at stall as that is where maximum flow deflection occurs.

If you want to have the torque at stall then you will need to know the force acting on the blade, which is the pressure times the area.

The maximum possible pressure will be half the density times the velocity squared. The area is the one from the outlet of the nozzle where that velocity is measured. That force times the radius of the blade it hits will give you the torque.

But the fluid never comes to a full stop when hitting the blade. The way the flow is deflected will affect the change in momentum and thus the actual force acting on the blade. So the geometry of the blade is still important. Even for a gas turbine. The Euler turbomachinery equation still applies; with the speed of the blade set to zero.

 

[Ripcrow]

??? This sentence does not make any sense.

$$1\ watt = 1\ \frac{newton\ \cdot\ meter}{second}$$(source)​

You never specified what type of turbine you were referring to. Velocity triangles and the Euler turbomachinery equation still apply to gas turbines:

View attachment 345327
(source)​

If you want to have the torque at stall then you will need to know the force acting on the blade, which is the pressure times the area.

The maximum possible pressure will be half the density times the velocity squared. The area is the one from the outlet of the nozzle where that velocity is measured. That force times the radius of the blade it hits will give you the torque.

But the fluid never comes to a full stop when hitting the blade. The way the flow is deflected will affect the change in momentum and thus the actual force acting on the blade. So the geometry of the blade is still important. Even for a gas turbine. The Euler turbomachinery equation still applies; with the speed of the blade set to zero.

Joules and watts are per second and are equal to 1 Newton metre. This is my original question as the units confused me. Maximum torque in a turbine is caused by deflection and therefore maximum torque is at stall where the blade rotation doesn’t lessen the deflection pressure. If the blade is rotating at 1 metre per second and the fluid is moving at 2 meters per second then deflection seen at the blade is equal to 1/2 mass flow times the velocity squared. If the mass flow is 2 grams we can calculate the energy available to be 4 watts / joules / newton. But because the blade is moving with the direction of flow at half the velocity of the flow the actual energy available to be imparted to the blade is 1 watt / joule / Newton. There is 4 times the torque at stall then there is at half the maximum rotational speed. If the turbine reaches 100% of its maximum speed ( the velocity of the fluid flow ) then there is no energy that gets imparted to the turbine as fluid flow and turbine speed is equal.

 

[Baluncore]

Joules and watts are per second and are equal to 1 Newton metre. This is my original question as the units confused me.

One joule, is the energy or work done, by a force of one newton, moving one metre.

Power is the rate of flow of energy. One watt, is one joule per second.

Power in watts, is equal to force in newtons, multiplied by velocity in metres per second.

 

[Ripcrow]

Yes the units are the same. Thanks for clarifying my question