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Verizon [17]
3 years ago
11

Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press

ure ratio is 10 and its isentropic efficiency is 85%. At the inlet to the turbine, the pressure is 950 kPa, and the temperature is 1400 K. The turbine has an isentropic efficiency of 88% and the exit pressure is 100 kPa. On the basis of an air-standard analysis,
a. develop a full accounting of the net exergy increase of the air passing through the gas turbine combustor, in kW.
b. devise and evaluate an exergetic efficiency for the gas turbine cycle.
Engineering
1 answer:
Zepler [3.9K]3 years ago
6 0

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

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The person that is correct based on the 2 statements from Tech A and Tech B is; Tech B

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However, for Tech A is incorrect and so the correct answer is that Tech B is right because his statement corresponds with the definition of mass flow sensor.

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2 years ago
A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at
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Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

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This is the answer for the question

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Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi
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Answer:

\vec F_{A} = -67500\,N\cdot (i + j)

Explanation:

The position of each point are the following:

A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

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\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j

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