Ask question

Ask question

All categories

2 years ago

7

2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compres

sor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rates of exergy destruction in the compressor and turbine, respectively, each in kW, for T0 5 300 K.

1 answer:

lutik1710 [3]2 years ago

5 0

You can see and download from the link

https://tlgur.com/d/GYYVL5lG

Please don't forget to put heart ♥️

You might be interested in

Suppose that a class CalendarDate has been defined for storing a calendar date with month, day and year components. (In our sect

laiz [17]

Answer:

note:

find the attachment

7 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

Fast plz-The mirror check may involve ______________.

barxatty [35]

Answer:

Realigning the mirror

Explanation:

mirrors should be aligned to minimize blind spots, not look at the tires.

6 0

2 years ago

Military glorification through mosaics, relief carvings and triumphant arches are often associated with?

emmasim [6.3K]

Military glorification through mosaics, relief carvings and triumphant arches are often associated with roman architectural monument and is the correct choice.

<h3>What is a Monument?</h3>

This can be defined as a structure which is usually large and is used to commemorate the history of an influential person. This helps to serve as an example and also a reminder as to why the performance of good deeds are important

The use of military glorification through mosaics, relief carvings and triumphant arches were also often used by the Romans in other to commemorate war victories and the succession of a new ruler which is known as the emperor.

Read more about Roman monuments here brainly.com/question/15786258

#SPJ1

8 0

2 years ago

den301095 [7]

Jae pain seems the most off

4 0

3 years ago