Rate of Climb Calculations (w/ 4 engine jet powered aircraft)

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The forum discussion centers on calculating the rate of climb for a four-engine jet-powered aircraft at an altitude of 8000 meters. The aircraft has a total thrust of 600 kN, a mass of 330 tonnes, and a drag force of 500 kN, with a Mach number of 0.6. The key formula used is dh/dt = ((T-D)V)/W, where T is thrust, D is drag, V is airspeed, and W is weight. The user successfully resolved the calculation after initial confusion regarding measurement units and angles.

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I have this problem and it seems like it has a simple solution, however I can't seem to figure it out.

Using the following values and the previous equation calculate the rate of climb for a
typical airplane at an altitude of 8000m.

4 Engines each with 150kN of thrust.
Aircraft mass is 330 tonnes.
Drag is 500kN
Aircraft Mach number is 0.6.

I know I need to use something like dh/dt=((T-D)V)/W , but I seem to have a problem. I can figure out the airspeed, and get something like dh/dt=24,648/W , but I'm not sure where to go from there. I am not sure of the measurement units or if I need to convert anything.

Thank you
 
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Wellll... your question wasn't given too clearly. For us to help you, you should probably give more details. Regardless, I'm going to try to guess what you mean.

I'm assuming that the plane in facing at a slightly different angle from the one it is actually moving in.

So let \alpha be the angle from the horizontal to the plane nose, and let \theta be the angle from the horizontal to the direction of the planes velocity.

Let F_T be the thrust force (total--sum of forces from engines) and let F_D be the drag force. Then, you have to solve the following equations:
<br /> F_T \cos{\alpha} - F_D \cos{\theta} = 0<br />
<br /> F_T \sin{\alpha} - F_D \sin{\theta} - Mg = 0<br /> <br />
 
Thanks for the help, but I looked over the problem yesterday and figured it out. It was easier that I originally thought. However, I forgot to return to the forum to remove the question. Thank you for the response, though.
 

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