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Vika [28.1K]
3 years ago
12

Thermodynamics 3-Hacmi 2 m3 olan kapalı bir kap 200 kPa basınçta 4 kg doymuş sıvı ve doymuş buhar karışımı su bulundurmaktadır.

Sıcaklığı 160⁰C olan çevreden kaptaki suyunu tamamı doymuş buhar oluncaya kadar ısı verilmektedir. Not: T(K) = 273 + ºC a)Sistemin entropi değişimini bulunuz. b)Kab'a yapılan ısı transferini bulunuz. c)Toplam entropi değişimini bulunuz
Engineering
1 answer:
Rufina [12.5K]3 years ago
7 0
I want say the best answer is C a sister-in-law Tran
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A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis
kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

V_1=0.3m^3,T_1=298 K

V_2=0.1m^3,T_1=1273 K

Mass of air=1 kg

For polytropic process  pv^n=C ,n is the polytropic constant.

  Tv^{n-1}=C

  T_1v^{n-1}_1=T_2v^{n-1}_2

298\times .3^{n-1}_1=1273\times .1^{n-1}_2

n=2.32

Work in polytropic process given as

       w=\dfrac{P_1V_1-P_2V_2}{n-1}

      w=mR\dfrac{T_1-T_2}{n-1}

Now by putting the values

w=1\times 0.287\dfrac{289-1273}{2.32-1}

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

  We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0
3 years ago
An irreversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP
Shkiper50 [21]

Answer:

Carnot heat pump

Explanation:

Carnot heat pump is an ideal heat pump in which all processes are reversible and that consume minimum amount of work to and produces maximum amount of heating effect compare to all real engine.And that is why COP of Carnot heat pump is more as compare to real heat pump.

All real heat pump are not perfectly reversible heat pump So this is also called irreversible heat pump .Due to irreversibility the COP of  irreversible heat pump is always  less than the COP of  Carnot heat pump.

6 0
4 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0
2 years ago
a stem and leaf display describes two-digit integers between 20 and 80. for one one of the classes displayed, the row appears as
allochka39001 [22]

Answer:

  52, 50, 54, 54, 56

Explanation:

The "stem" in this scenario is the tens digit of the number. Each "leaf" is the ones digit of a distinct number with the given tens digit.

  5 | 20446 represents the numbers 52, 50, 54, 54, 56

8 0
3 years ago
Which of the following would not be considered hot work? A chipping B soldering C
tankabanditka [31]
I believe the answer is D: brazing
Hope this helps you have a good night
5 0
3 years ago
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