Ask question

Ask question

All categories

3 years ago

10

Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper

ature reservoir at 1000 K (727 °C) and a large, cold reservoir at 308 K (35°C). If it withdraws 1.2 MJ/s from the high-temperature reservoir, what would be the rate of loss of entropy from that reservoir and what would be the rate of gain by the low-temperature reservoir? (4 marks) b. Express the work done by the engine in watts. (3 marks) What would be the total entropy gain of the system? d. Determine Carnot efficiency and recalculate the a, b, and c, accordingly.

1 answer:

Anvisha [2.4K]3 years ago

6 0

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

You might be interested in

In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26

scZoUnD [109]

Answer:

The strength coefficient is $625$ and the strain-hardening exponent is $0.435$

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find $(K)$ and the $(n)$ .

$\sigma =K\epsilon^n$

We will plug the values in the formula.

$250=K\times (0.12)^n\\350=K\times (0.26)^n$

We will solve these equation.

$K=\frac{250}{(0.12)^n}$ plug this value in $350=K\times (0.26)^n$

$350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n$

Taking a natural log both sides we get.

$ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435$

Now, we will find value of $K$

$K=\frac{250}{(0.12)^n}$

$K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625$

So, the strength coefficient is $625$ and the strain-hardening exponent is $0.435$ .

5 0

3 years ago

What is the governing ratio for thin walled cylinders?

Ann [662]

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

if $\frac{radius}{thickness} >10$ then you have a thin walled cylinder

or using the diameter:

if $\frac{diameter}{thickness} >20$ then you have a thin walled cylinder

3 0

3 years ago

Which of the following impulse responses correspond(s) to stable LTI systems? (a) h1(t) = e-(1-2j)ru(t) (b) h2(t) = e-r cos(2t)u

Illusion [34]

Answer:

LTI system is stable if Impulse response is finite.

so the correct answer is "b"

(b) h2(t) = e-r cos(2t)u(t)

6 0

4 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

<em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

.<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

<em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

<em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago

Roku internet service providet

podryga [215]

Answer:

What?? I do not understand?

6 0

3 years ago