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3 years ago

8

3. A power cycle operates between hot and cold reservoirs at 500 K and 310 K respectively. At steady state the power output deve

loped is: 0.1 MW. What is the minimum rate of energy rejection by heat transfer to the cold reservoir, in MW

1 answer:

grin007 [14]3 years ago

4 0

Answer:

0.163 MW

Explanation:

To get the minimum rate of energy rehection by heat transfer to cold reservoir, it implies that the power cycle operates in reversible cycle. The efficiency of reversible cycle whose value will be same as efficiency of power cycle will be given by

Efficiency of reversible cycle

n= (Th-Tc)/Th where T represent temperature, n efficiency, subscripts h and c hot and cold respectively

Substituting 500 and 310 for hot and cold temperatures respectively then efficiency

n=(500-310)/500=0.38

Efficiency of power cycle, n= Power output/Qh

Qh= 0.1/0.38= 0.263

Net power output, W= Qh- Qc

Qc=Qh-W= 0.263-0.1=0.163 MW

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Answer:

The distance between the station A and B will be:

$x_{A-B}=55.620\: km$

Explanation:

Let's find the distance that the train traveled during 60 seconds.

$x_{1}=x_{0}+v_{0}t+0.5at^{2}$

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

$x_{1}=\frac{1}{2}(0.6)(60)^{2}$

$x_{1}=1080\: m$

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

$v_{1}=v_{0}+at$

$v_{1}=(0.6)(60)=36\: m/s$

Then the second distance will be:

$x_{2}=v_{1}*1500$

$x_{2}=(36)(1500)=54000\: m$

The final distance is calculated whit the decelerate value:

$v_{f}^{2}=v_{1}^{2}-2ax_{3}$

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

$0=36^{2}-2(1.2)x_{3}$

$x_{3}=\frac{36^{2}}{2(1.2)}$

$x_{3}=540\: m$

Therefore, the distance between the station A and B will be:

$x_{A-B}=x_{1}+x_{2}+x_{3}$

$x_{A-B}=1080+54000+540=55.620\: km$

I hope it helps you!

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In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

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Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

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And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

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