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pentagon [3]
3 years ago
11

A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold

reservoir at TC = 400 K. For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
a.QH = 900 kJ, Wcycle= 450 kJ
b. QH = 900 kJ, Qc = 300 kJ
c. Weycle = 600 kJ, Qc= 400 kJ
d. η = 75%
Engineering
1 answer:
PIT_PIT [208]3 years ago
8 0

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle (\eta_{r}), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\% (1)

Where:

T_{C} - Temperature of the cold reservoir, in Kelvin.

T_{H} - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: (T_{C} = 400\,K, T_{H} = 1200\,K)

\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%

\eta_{r} = 66.667\,\%

In exchange, real efficiency for a power cycle (\eta), no unit, is defined by this expression:

\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\% (2)

Where:

Q_{C} - Heat released to cold reservoir, in kilojoules.

Q_{H} - Heat gained from hot reservoir, in kilojoules.

W_{C} - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for \eta < \eta_{r}, reversibily for \eta = \eta_{r} and it is impossible for \eta > \eta_{r}.

Now we proceed to solve for each case:

a) Q_{H} = 900\,kJ, W_{C} = 450\,kJ

\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 50\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

b) Q_{H} = 900\,kJ, Q_{C} = 300\,kJ

\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 66.667\,\%

Since \eta = \eta_{r}, the power cycle operates reversibly.

c) W_{C} = 600\,kJ, Q_{C} = 400\,kJ

\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%

\eta = 60\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

d) Since \eta >  \eta_{r}, the power cycle is impossible.

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2 years ago
Please answer fast. With full step by step solution.​
lina2011 [118]

Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

or

<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

Recall that for |<em>z</em>| < 1, we have

\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n

Replace <em>z</em> with 1/<em>z</em> to get

\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}

so that by substitution, we can write

\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}

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3 years ago
A vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr
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Answer:

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Explanation:

Given that Vacuum gauge pressure= 30 KPa

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We know that barometer always reads atmospheric pressure at given situation.So  atmospheric pressure is equal to  755 mm Hg.

We know that P= ρ g h

Density of Hg=13600 \frac{kg}{m^3}

So P=13600 x 9.81 x 0.755

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We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30   KPa

So

Absolute pressure=70.72 KPa

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Gtjffs
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