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Blababa [14]
3 years ago
14

explain each of the following kinds of rockets: Solid-Fuel Rocket, Liquid-Fuel Rocket, Ion Rocket and Plasma Rocket.

Engineering
1 answer:
Rudik [331]3 years ago
5 0

Answer:

ur answer friend

Explanation:

answer

<em>S</em><em>o</em><em>l</em><em>i</em><em>d</em><em>-</em><em>F</em><em>u</em><em>e</em><em>l</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>a solid-propellant rocket or solid rocket is a rocket with a rocket engine that uses solid propellants. The earliest rockets were solid-fuel rockets powered by gunpowder; they were used in warfare by the Chinese, Indians, Mongols and Persians, as early as 13th century.

<em>L</em><em>i</em><em>q</em><em>u</em><em>i</em><em>d</em><em>-</em><em>F</em><em>u</em><em>e</em><em>l</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>a liquid-propellant rocket or liquid rocket utilizes a rocket engine that use liqiud propellants. An inert gas stored in a tank at a high pressure is sometimes used instead of pumps in simpler small engines to force the propellants into the combustion chamber.

<em>I</em><em>o</em><em>n</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>an ion thruster or ion drive is a form of electric propulsion used for spacecraft propulsion. It creates thrust by accelerating ions using electricity. The Deep Space 1 spacecraft, powered by an ion thruster, changed velocity by 4.3 km/s ( 9600 mph ) while consuming less than 74 kg ( 163 lb ) of xenon.

<em>P</em><em>l</em><em>a</em><em>s</em><em>m</em><em>a</em><em> </em><em>R</em><em>o</em><em>c</em><em>k</em><em>e</em><em>t</em><em> </em><em>-</em><em> </em>in this type of rocket, a combination of electric and magnetic fields are used to break down the atoms and molecules of a propellant gas into a collection of particles that have either a positive charge (ions) or a negative charge (electrons). In other words, the propellant gas becomes a plasma.

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(TCO 4) A system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without
steposvetlana [31]

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

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3 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
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Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
Charging method .Constant current method​
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Answer:

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Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage.  The current will then taper down to a minimum value once that voltage level is reached.  The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating.  Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement.  This method is suitable for Ni-MH type of batteries.  The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods.  The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level.  The current then reduces as the battery becomes fully charged.  The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

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