Answer:
The power developed by engine is 167.55 KW
Explanation:
Given that
![V_d=2.5\times 10^{-3} m^3](https://tex.z-dn.net/?f=V_d%3D2.5%5Ctimes%2010%5E%7B-3%7D%20m%5E3)
Mean effective pressure = 6.4 bar
Speed = 2000 rpm
We know that power is the work done per second.
So
![P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}](https://tex.z-dn.net/?f=P%3D6.4%5Ctimes%20100%5Ctimes%202.5%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20%5Cdfrac%7B2%5Cpi%20%5Ctimes2000%7D%7B120%7D)
We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.
P=167.55 KW
So the power developed by engine is 167.55 KW
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.
Answer:
I think reduce your following distance
Answer:
total width bandwidth = 8kHz
Explanation:
given data
transmitter operating = 3.9 MHz
frequencies up to = 4 kHz
solution
we get here upper side frequencies that is
upper side frequencies = 3.9 ×
+ 4 × 10³
upper side frequencies = 3.904 MHz
and
now we get lower side frequencies that is
lower side frequencies = 3.9 ×
- 4 × 10³
lower side frequencies = 3.896 MHz
and now we get total width bandwidth
total width bandwidth = upper side frequencies - lower side frequencies
total width bandwidth = 8kHz