Answer: The exit temperature of the gas in deg C is
.
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa, 

We know that for an ideal gas the mass flow rate will be calculated as follows.

or, m = 
=
= 10 kg/s
Now, according to the steady flow energy equation:




= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
= 
Therefore, we can conclude that the exit temperature of the gas in deg C is
.
They all share the way that they are fundamentally designed: if they are quite complex, they will share the same basic logic foundations, like the way that the programming languages work. They also all share the method of construction and common and fundamental electronic components, like resistors, capacitors and transistors. As we humans design them, they make logical sense to at least someone, and probably only discounting the internet, you can probably draw logic diagrams and whatever to represent how they work.
Because they are designed by Humans, in a way they all mimic how our brains and society work. Also, as yet there are no truly intelligent technological systems, and are only able to react to a situation how they have been programmed to do so.
In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:



Through Goodman's equations the combined effort by fatigue and compression is expressed as:

Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then

Re-arrenge for 

We know that the stress is representing as,

Then,
Where
=Max Moment
I= Intertia
The inertia for this object is

Then replacing and re-arrenge for 



Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm
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Answer:
Efficiency based on Otto cycle.
Effotto = 47.47%
Explanation:
Efficiency based on Otto cycle.
effotto = 1 – (V2 / V1)^γ-1
effotto = 1 – (1 / 5)^1.4 - 1
effotto = 47.47%