Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.
The Carnot cycle is a hypothetical cycle that takes no account of entropy generation. It is assumed that the heat source and heat sink have perfect heat transfer. The working fluid also remains in the same phase, as opposed to the Rankine cycle, in which the fluid changes phase. A practical thermodynamic cycle, such as the Rankine cycle, would achieve at most 50% of the Carnot cycle efficiency under similar heat source and heat sink temperatures.
<h3>What is Thermo-Electrochemical converter?</h3>
In a two-cell structure, a thermo-electrochemical converter converts potential energy difference during hydrogen oxidation and reduction to heat energy.
It employs the Ericsson cycle, which is less efficient than the Carnot cycle. In a closed system, it converts heat to electrical energy. There are no external input or output devices.
This means there will be no mechanical work to be done, as well as no exhaust. As a result, Carnot efficiency is not taken into account in this cycle. Carnot efficiency is accounted for by other options such as turbine and engine.
Learn more about Thermo-Electrochemical converter here:
brainly.com/question/13040188
#SPJ4
it is generated by two objects pushing against each other with equal/almost equal force, therefore contracting against each other
Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
Answer:
Approximately 
.
Assumption: air resistance on the rocket is negligible. Take 
.
Explanation:
By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.
.
Note that in this case, the uppercase letter 
in the units stands for "mega-", which is the same as 
times the unit that follows. For example, 
, while 
.
Convert the mass of the rocket and the thrust of its engines to SI standard units:
- The standard unit for mass is kilograms:
. - The standard for forces (including thrust) is Newtons:
.
At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:
- Thrust (which is supposed to go upwards), and
- Weight (downwards due to gravity.)
The thrust on the rocket is already known to be 
. Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately 
. Hence, the weight on the rocket would be approximately 
.
The magnitude of the net force on the rocket would be
.
Apply the formula to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)
.
