How much horsepower does a space shuttle have after 10 seconds on liftoff?

Answer 1

This is a difficult question to ask, because it does not have a clearly defineable answer.

Power as physical quantity is defined as the amount of expended work divided by the time it took to do that work:

P = dW/dt

Work, however, is a more complex animal than you might have thought when setting the question. Work is a quantity that measures the expenditure of energy; any energy expended is measured as work:

W = -E

Thus, we need to consider how much energy the space shuttle is gaining (or how much work its engines are doing) at a given time.

The total mechanical energy of the space shuttle would in this example consist of potential energy and kinetic energy:

E(t) = Ep(t) + Ek(t) = mgh(t) + ½mv(t)2

where m is the mass of the vehicle, assumed to be close enough to constant during the ten seconds specified in the question;

g is the gravitational acceleration of 9.81 m/s2;

h(t) is the altitude of the vehicle as function of time, and

v(t) is the velocity of the vehicle as function of time.

To define the energy of the vehicle as function of time, we need to define the altitude and velocity as functions of time. To do this in meaningful way, we'll simplify the example with a few assumptions:

-the shuttle will travel upward only (constrained to one axis of motion)

-mass is constant or close enough to constant to not significantly matter, at 2,000,000 kg gross lift-off weight for the entire system stack

-the shuttle will have constant acceleration of 3g (it is limited to this acceleration for safety reasons).

Now to resolve the kinetic equations of the vehicle:

a = 3g

a = dv/dt

dv = a dt ∫(...)

∫dv = a ∫dt

v(t) = at + v0

v = dh/dt

dh/dt = at + v0

dh = at dt + v0 dt ∫(...)

∫dh = a∫t dt + v0∫dt

h(t) = ½ a t2 + v0t + h0

Thus, the energy of the vehicle at time t is

E(t) = mgh(t) + ½mv(t)2

E(t) = mg(½ a t2+ v0t + h0) + ½m(at + v0)2

To simplify things further, we can define initial velocity and altitude as zero:

h0 = 0, v0 = 0

E(t) = ½ mga t2 + ½m a2 t2 = (½ mga + ½ ma2) t2

Now, here's our function for the total mechanical energy of the vehicle. To determine the rate of energy consumption at ten seconds, we need to derivate the function by t:

E'(t) = (mga + ma2) t

Since E'(t) = dE/dt and P = dW/dt, we have our answer here by substituting our assumed values into this simple equation:

( 2,000,000 kg * 4 * 9.81 m/s2 + 2,000,000 kg * [3 * 9.81 m/s2]2 ) * 10 s

≈18.11 x 109 W = 18.11 GW

Based on this very simplified calculation, the space shuttle launch vehicle's total power output at T+10s is approximately 18 gigawatts, which translates to:

* 24,138,397.6 hp

* Full installed power output of Three Gorges Dam

* Eleven and a quarter of the projected power output of Olkiluoto 3 fission reactor (which will be the most powerful nuclear reactor in the world upon completion)

These comparisons should make it obvious why "power", especially axial power, is not especially valid or relevant quantity of capacity when you are talking about rockets, jet engines or other similar systems which rely on the reaction principle rather than mechanical traction for providing locomotion. Furthermore, the amount of chemical energy expended by the rocket motors of the space shuttle is significantly larger than the value achieved in this example; not all of the thermal energy generated in the chemical reactions is converted into potential energy and kinetic energy.

As a further exercise, you could try to define the power of a car with static 5 m/s2 acceleration. You will find that the "power" of the car varies as a function of time, rather than being a constant, clear-cut value.

Answer 2

A space shuttle’s main engines produce around 6.8 million pounds of thrust at liftoff, which is equivalent to about 37 million horsepower. After 10 seconds, the shuttle has not used up all of its fuel, so the horsepower output remains the same.