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VARVARA [1.3K]
3 years ago
13

Refrigerant-134a enters an adiabatic compressor at -30oC as a saturated vapor at a rate of 0.45 m3 /min and leaves at 900 kPa an

d 55oC. Determine (a) the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the rate of exergy destruction and the second-law efficiency of the compressor. Take T0
Engineering
1 answer:
saw5 [17]3 years ago
4 0

Answer:

a) 1.918 kw

b) 86.23%

c) 0.26 kw

Explanation:

Given data:

T1 = -30°C = 243 k  , T0 = 27°C

using steam tables

h1 = 232.19 KJ/kg

s1 = 0.9559 Kj/Kgk

T2 = 55°C   P2 = 900 kPa

Psat = 1492 kPa,  h2 = 289.95 Kj/Kg, s2 = 0.9819 Kj/kgk , m = 0.0332 kg/s

<u>a) Determine the power input to the compressor</u>

power input = 1.918 kw

<u>b) Determine isentropic efficiency of compressor</u>

Isentropic efficiency = 86.23%

<u>c) Determine rate of exergy destruction</u>

rate = 0.26 kw

Attached below is the detailed solution of the given problems

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This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
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<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>

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<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>

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

<em>z = xz = 0(p−k)10p </em>

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Answer:

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b. The temperature (in o F) is 200.8°F

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We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

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b.

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The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

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