1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
dexar [7]
3 years ago
14

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred

until its temperature is 600 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine : (a) the final pressure, in bar. (b) the work, in kJ. (c) the amount of entropy produced, in kJ/K.
Physics
1 answer:
madam [21]3 years ago
5 0

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K   ,V₁ = 3 m³  ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas  P V = m R T

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

P_2=\dfrac{T_2}{T_1}\times P_1

P_2=\dfrac{600}{300}\times 2

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

m=\dfrac{PV}{RT}

m=\dfrac{200\times 3}{0.287\times 300}\ kg

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0  Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}

S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}

S₂ - S₁ = 3.42 KJ/K

 

You might be interested in
Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten
ladessa [460]

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

m_1v_1 + m_2v_2 = 0

now plug in all data into above equation

5(v) + 3(3)

5v = -9

v = -1.8 m/s

so correct answer is

A) - 1.8 m/s

3 0
3 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
What is precision?what is precision? ​
RUDIKE [14]

Answer:

According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."

Explanation:

Hope this helps! :)

6 0
2 years ago
Read 2 more answers
A ball is dropped from a window and takes two seconds to reach the ground .it starts from rest and reaches a final speed of 20m/
Akimi4 [234]
The ball can't reach the speed of 20 m/s in two seconds, unless you THROW it down from the window with a little bit of initial speed. If you just drop it, then the highest speed it can have after two seconds is 19.6 m/s .

If an object starts from rest and its speed after 2 seconds is 20 m/s, then its acceleration is 20/2 = 10 m/s^2 .

(Gravity on Earth is only 9.8 m/s^2.)
3 0
3 years ago
3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
andreev551 [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0
2 years ago
Other questions:
  • What injuries could occur due to incorrect movement and handling of equipment with any sport or activity? (Select ALL that apply
    8·1 answer
  • Which statement best describes the weather shown by the purple combination of semicircles and triangles on a line on a weather m
    7·2 answers
  • (01.02 MC) Nicole pushes her bike up a hill. Overhead, the sun exerts a gravitational force on Earth. Which statement is true ab
    9·2 answers
  • Some dams are designed to make electricity. How do dams make electricity?
    6·1 answer
  • Which of the following is an observation? A. The baby was crying very loudly; therefore, he must have been hungry. B. Jason star
    6·2 answers
  • A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra
    10·1 answer
  • How does this happen?
    12·1 answer
  • PLEASE HELP IT'S DUE IN LIKE 2 MINUTES
    8·1 answer
  • A 5 kg object moving to the right at 4 m/s collides inelastically with a 5 kg object
    12·1 answer
  • discuss four contributing factors that may lead to an increase of learners abusing substance in scools​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!