What fuel do rockets use
2 answers:
Liquid Hydrogen is the fuel used by rockets.
Explanation:
- Liquid hydrogen which can be chemically denoted as "
" is often considered as the significant fuels for rocket.
- However rocket in its lower stages uses fuels such as Kerosene and oxygen where as in the higher stages such as second and third stages it uses liquid hydrogen.
- Liquid hydrogen is known to easily cool the nozzle and then also other parts of the rocket before mixing with the oxidizer such as the oxygen.
- Thus liquid hydrogen helps in preventing nozzle erosion and also reduces combustion chamber.
- Liquid hydrogen one the other hand is very expensive as 384,071 gallons of it will cost approximately $376,389.58 .
Thus liquid hydrogen is effectively used as a fuel for rocket.
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Answer:
Suction and exhaust processes do not affect the performance of Otto cycle.
Explanation:
Step1
Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.
Step2
Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.
Step3
The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:
Process 0-1 is suction process and process 1-0 is exhaust process.
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.