I have a question, regarding conversion of thrust to horsepower.

A few days ago i got to take a part 91 flight on a southwest mx flight. The flight crew let me work the throttles at 10,000 to try to remain at 250 knots. this 24,000lbs of thrust is by far the largest amount of power i've ever had control of. so my question is how many horsepower is that. The most i've had before this is a 230 hp Trinidad TB-20. This was quite a bit different  

The answer is a little muddy so maybe this link will help.

http://www.aerospaceweb.org/question/propulsion/q0195.shtml

Well, from Anderson's "Introduction to Flight," the power required for level, unaccelerated flight is:

PR = TR*V

where PR is the power required, TR is the thrust required, and V is the velocity.

Now, there's some information missing that you need to actually calculate the power in your situation, so I'll just make some assumptions.

Your thrust available at 10,000 feet is approx. the sea level static thrust multiplied by the density ratio between altitude and sea level.

TA = Tsealevel * (p_altitude/p_sealevel)

which in this case is 24,000*0.739 which yields 17,726 lb.

The part that makes it more complicated is that power depends on the thrust and the velocity. And a change in thrust will change the velocity as well. So since you most likely didn't have the levers fully forward, the engines weren't generating 17,726 lbs of thrust.

So just for fun let's say the engines were producing 10,000 lbs of thrust, which is just a number for the sake of argument - it's not even an educated guess. And 250 kt is equivalent to about 416.7 feet/second (I can't remember exactly how many feet are in a nautical mile, but it's close to 6,000).

So:

PR = 10,000 * 416.67 = 4,166,666 ft*lb/s

For horsepower, divide by 550

4,166,666 / 550 = 7,575.75 hp, which sounds reasonable to me.

I could have made some assumptions about the drag on a 737, and calculated the "actual" thrust required, but I'm too lazy. Let me know if you want any of that.

~Vik

wow that is very helpful. I'll plug in some numbers for you:

On southwest 737-700's, each engine is rated at 24,000lbs takeoff thrust. So total is 48,000lbs takeoff thrust. at 10,000 feet the temperature was around 6 celcius that day. In straight and level flight, the N1 was hovering around 77%.

i wouldn't worry about drag, but if you're extremely bored those numbers would be interesting as well. haha. thanks again.

It's your lucky (or unlucky) day, because I happen to be bored.

Glad I could be helpful. The thing is, the drag on the aircraft is what will tell you how much thrust is required, because for level, unaccelerated flight, the thrust will equal the drag. I don't know how to get the actual thrust from N1 or N2 or EPR or any of that. I'll try to make a decent estimate of the drag here though, just so you can see how it works (these are the basic engineering equations, so they will be good estimates, but not exact).

So there's what's called the parasite drag coefficient for the aircraft, which basically accounts for the friction drag due to the air, and this coefficient basically comes with the aircraft - that is, it depends on the exterior of the plane. So let's assume that the parasite drag coefficient (Cd) is 0.03. This is a dimensionless number, but we'll get the drag from it. I'll also assume the airplane was at a weight of 120,000 lbs. The total drag will be the sum of the parasite drag and induced drag (drag due to lift).

Now, lift and drag coefficients have the same form, which is:

Cd = Drag/(0.5*wingarea*density*velocity^2)
Cl = Lift/(0.5*wingarea*density*velocity^2)

where Lift = Weight

Cl = 120,000/(0.5*wingarea*density*velocity^2) = 0.59

Induced drag coefficient(Cdi) = Cl/(pi*efficiencyfactor*aspectratio)

pi = 3.1415...,
I will assume the efficiency factor is 0.85, though I don't know how good this is,
and the aspect ratio = wingspan^2/wingarea

So...Cdi = 0.59/(3.1415*0.85*10.22) = 0.022

To get the total drag coefficient, add the parasite and induced drag coefficients.

Cdtotal = 0.03+0.022 = 0.052

Total drag(D) = Cdtotal*0.5*wingarea*density*velocity^2
D = 0.052*0.5*1344*(1.7556*10^-3)*416.67^2 = 10,651 lb.

This will be your thrust required, as thrust required equals drag.

So, PR = 10,651*416.67 = 4,437,873 lb*ft/s

which equals 8068.9 hp. Wow...I was actually pretty close in my previous post! Didn't expect that.

If you made it through, congrats. I tried to be as clear as I could, but having not gone over this stuff in a long while, it took some time. Comments/corrections welcome.

~Vik

^^

Why SI is so much better.

777236ER,

I am in complete agreement. Don't understand why we haven't switched over here in the US. I actually understand things much better in SI units.

~Vik