Problem#1
The first stage of a Saturn V space vehicle consumed fuel and oxidizer at the rate of 1.50 x 104 kg/s, with an exhaust speed of 2.60 x 103 m/s. (a) Calculate the thrust produced by these engines. (b) Find the acceleration of the vehicle just as it lifted off the launch pad on the Earth if the vehicle’s initial mass was 3.00 x 106 kg. Note: You must include the gravitational force to solve part (b).
Answer:
Given: ve = 2.60 x 103 m/s, dM/dt = 1.50 x 104 kg/s
(a) the thrust produced by these engines, we use
Fthrust = │vedM/dt│ = (2.60 x 103 m/s)(1.50 x 104 kg/s) = 3.90 x 107 N
(b) the acceleration of the vehicle just as it lifted off the launch pad on the Earth if the vehicle’s initial mass was M = 3.00 x 106 kg, given by
∑F = Fthrust – Mg = Ma
3.90 x 107 N – (3.00 x 106 kg)(9.80 m/s2) = (3.00 x 106 kg)a
a = 3.20 m/s2
Problem#2
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocket engine has an average thrust of 5.26 N, a fuel mass of 12.7 g, and an initial mass of 25.5 g. The duration of its burn is 1.90 s. (a) What is the average exhaust speed of the engine? (b) If this engine is placed in a rocket body of mass 53.5 g, what is the final velocity of the rocket if it is fired in outer space? Assume the fuel burns at a constant rate.
Answer:
Given: Fthrust = 5.26 N, dM/dt = 12.7 g/1.90 s = 6.68 x 10-3 kg/s.
(a) is the average exhaust speed of the engine is
Fthrust = │vedM/dt│
5.26 N = ve(6.68 x 10-3 kg/s)
ve = 787 m/s
(b) the final velocity of the rocket if it is fired in outer space, we use
vf – vi = veln(Mi/Mf)
with, Mi = 53.5 g + 25.5 g = 79.0g and Mf = 53.5 g + 25.5 g – 12.7g = 66.3g
vf – 0 = (787 m/s)ln(79.0g/66.3g)
vf = 138 m/s
Problem#3
A rocket for use in deep space is to be capable of boosting a total load (payload plus rocket frame and engine) of 3.00 metric tons to a speed of 10 000 m/s. (a) It has an engine and fuel designed to produce an exhaust speed of 2 000 m/s. How much fuel plus oxidizer is required? (b) If a different fuel and engine design could give an exhaust speed of 5 000 m/s, what amount of fuel and oxidizer
would be required for the same task?
Answer:
Given: Mf = 3.00 metric tons = 3.00 x 103 kg, vf = 10000m/s, ve = 2000 m/s. Then,
(a) we use
vf – vi = veln(Mi/Mf)
10000m/s – 0 = (2000 m/s)ln[Mi/(3.00 x 103 kg)]
5 = ln[Mi/(3.00 x 103 kg)]
e5 = Mi/(3.00 x 103 kg)
Mi = e5(3.00 x 103 kg) = 4.45 x 105kg
Then the mass of fuel and oxidizer is
∆M = Mi – Mf = 4.45 x 105kg – 3.00 x 103kg = 4.42 x 105kg
∆M = 442 metric tons
(b) amount of fuel and oxidizer would be required for the same task is
∆M = Mi – Mf
With Mi = ev/veMf
Mi = e2(3.00 x 103kg) = 2.22 x 104 kg
Then,
∆M = 2.22 x 104 kg – 3.00 x 103 kg = 1.92 x 104 kg = 19.2 metric tons
Because of the exponential, a relatively small increase in fuel and/or engine efficiency causes a large change in the amount of fuel and oxidizer required.
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