Aircraft propulsion technology

Can anyone help me?

I'm doing a question on aircraft propulsion technology and i'm stuck.

The question is:

'If the mass of air through a propeller is 1000 Kg/s, the aircraft's velocity is 100 m/s and the slipstream velocity is 120 m/s, calculate the thrust?

Surely if F = ma then the equation should be F = 1000 * 100 = 10000 Kg/s

But I don't think this is the right answer as I have not taken the slipstream velocity in to consideration.

The only other equation I can find is

F = (m dot * V)e - (m dot * V)0 + (pe - p0) * Ae

But I don't have pe or p0 or Ae neither do I understand what they are.

Can someone please steer me in the right direction?

Thanks

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The plane's speed is misleading; the key is the speed change of the air. If there is no air speed change, there is no thrust, just air passing through the prop disc.

slipstream velocity is 120 m/s
So is this the exhaust velocity??
If so , then the relative exit velocity of the air will be = 100+120 = 220 m/s
From conservation of momentum equation :
F = (m dot * V)e - (m dot * V)i + (pe - pi) * Ae
F = (m dot * v)e - (m dot *v)i = 1000 * (220 - 100) = 1000 * 120 = 120000 N = 120 kN

I'm not so sure about that

But I don't have pe or p0 or Ae neither do I understand what they are.
Pe&P0 are the pressure of inlet and exhaust respectively

The plane is going through the air at 100 m/s; the prop adds 20 m/s to that.

I think thrust is mass time acceleration; not times speed - it is the difference between the freestream velocity and the speed of the air behind the prop. Acceleration takes time and is the result of a force. The formula from Makvegar is for a jet engine, including the area of the exhaust nozzle, which may not apply to a propellor - thrust for propellors includes however propellor diameter, which may be related to the exhaust area - food for thought...but both also include the density of the air, since that affects the mass.

And a change in speed is the result of acceleration.

Definitely - acceleration, though, is strictly a change in velocity which includes direction so, surprisingly, acceleration may be achieved without a change in speed. However in this thread we are principly concerned with a change in speed. If the air was still in the free steam in front of the plane, the calculation would be simpler. However the air may be moving in any direction relative to the prop during flight. So what you say is true - it is the airspeed that is accelerating, not the ground speed but we must remember that a tethered aircraft doing ground runs generates thrust, so it is not necessarily the plane going through the air.

have you seen newton's second law written in the form f=d(mv)/dt ?