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lidiya [134]
3 years ago
7

A turbine produces shaft power from a gas that enters the turbine with a (static) temperature of 628 K, a velocity of 143 m/s an

d a stagnation pressure of 14 atm. The gas exits the turbine with a stagnation temperature of 275 K and a stagnation pressure of 0.95 atm. The gas passing through the turbine has a molecular weight of 16 and a specific heat ratio of 1.45. What is the stagnation temperature of the gas entering the turbine? Assuming the turbine is adiabatic, is it also reversible?

Engineering
1 answer:
drek231 [11]3 years ago
7 0

Answer:

Explanation:

The detailed step by step calculation and appropriate formula is as shown in the attached file.

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Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600 K. Calculate the
Fantom [35]

Solution:

Given:

T_{H} = 1200 K

T_{L} = 600 K

Q = 100 kJ

The Entropy change of the two reservoirs is given by the sum of entropy change of each reservoir system and is given by the formula:

\Delta s = \frac{-Q}{T_{H}}+\frac{Q}{T_{L}}

\Delta s = \frac{Q(T_{L}-T_{_{H}})}{T_{H}T_{L}}

\Delta s = \frac{-100(600-1200)}{1200\times 600}

[tex]\Delta s = 0.0833kJ/K

Since, the change in entropy is positive and according to the Increase in entropy principle, for any process the total change in entropy of a system is always greater than or equal to zero (with its enclosing adiabatic surrounding).

Therefore, the entropy principle is satisfied.

8 0
3 years ago
Can somebody help me with that
skelet666 [1.2K]

Answer:

I think it's 23 ohms.

Explanation:

Not entirely sure about it.

hope this helps

4 0
3 years ago
This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

6 0
3 years ago
What material property would still cause strain in a strain gauge that is positionedperpendicular to the direction of force if i
svetlana [45]

Answer:

oof

Explanation:

I don't know but please don't report me

I am trying to do a challenge

Thank you-

If you don't report me!

5 0
3 years ago
_______ is a material property that pertains to local resistance to plastic deformation, such as scratching or denting. It is of
Readme [11.4K]

Answer: hardness

Explanation:

Hardness is a measure of a material's ability to resist plastic deformation. In other words, it is a measure of how resistant material is to denting or scratching. Diamond, for example, is a very hard material. It is extremely difficult to dent or scratch a diamond. In contrast, it is very easy to scratch or dent most plastics.

7 0
3 years ago
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