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Irina18 [472]
3 years ago
7

An ideal gas undergoes two processes: one frictionless and the other not. In both the cases, the gas is initially at 200 ℉ and 1

.5 atm, and when the process is over, the gas is at 700 °F and 22 atm. For which process does the gas experience a greater change in entropy?
Engineering
2 answers:
Lorico [155]3 years ago
8 0

Answer:

The change in entropy is the same in both process

Explanation:

The change is entropy is the same in both process because the entropy is about the randomness or disorderliness of the molecules of a substance due to chemical reactions or various arrangements given to them. it is a point function because it depends on the initial and final state of a substance.

In both processes the initial state of the ideal gas is the same for both process and the final state of the ideal gas for both processes are the same. hence the change in entropy is the same. also in ideal conditions there are no energy losses due to Friction.

Zarrin [17]3 years ago
6 0

Answer:

The process which has friction

Explanation:

The entropy is simply the change in the state of the things or the molecules in the system. It is simply the change in the energy of the system with a focus on the atoms in the system. This is also known as the internal energy of the system and is given the symbol, G. The friction contributes to the change in the energy of the system. This is because friction generates another form of energy - that is heat energy. This energy causes the internal temperature id the system to increase. Hence the greater change in the temperature.

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2. A thin vertical panel L = 3 m high and w = 1.5 m wide is thermally insulated on one side and exposed to a solar radiation flu
n200080 [17]

Answer: 383.22K

Explanation:

L = 3m, w = 1.5m

Area A = 3 x 1.5 = 4.5m2

Q' = 750W/m2 (heat from sun) ,

& = 0.87

Q = &Q' = 0. 87x750 = 652.5W/m2

E = QA = 652.5 x 4.5 = 2936.25W

T(sur) = 300K, T(panel) = ?

Using E = §€A(T^4(panel) - T^4(sur))

§ = Stefan constant = 5.7x10^-8

€ = emmisivity = 0.85

2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)

T(panel) = 383.22K

See image for further details.

5 0
3 years ago
At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates su
UkoKoshka [18]

Answer:

x=2.19in

Explanation:

This is the equation that relates the force and displacement of a spring

F=Kx

m=mass=12.5lbx1slug/32.14lb=0.39slug

F=mg=0.39*32.2=12.52Lbf

then we calculate the spring count in lbf / ft

K=F/x

K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft

Finally we calculate the displacement with the initial equation

X=F/k

x=12.52/68.4=0.18ft=2.19in

8 0
4 years ago
Khái niệm về môi trường nhiệt nóng và môi trường nhiện lạnh ?
I am Lyosha [343]

Explanation:

उह्ह्नमजज्ल्ह्ह्बनुतनकुहक्जो

7 0
3 years ago
The larger the Bi number, the more accurate the lumped system analysis. a)-True b)- False
mrs_skeptik [129]

Answer:

b). False

Explanation:

Lumped body analysis :

Lumped body analysis states that some bodies during heat transfer process remains uniform at all times. The temperature of these bodies is a function of temperature only. Therefor the heat transfer analysis based on such idea is called lumped body analysis.

                      Biot number is a dimensionless number which governs the heat transfer rate for a lumped body. Biot number is defined as the ratio of the convection transfer at the surface of the body to the conduction inside the body. the temperature difference will be uniform only when the Biot number is nearly equal to zero.  

                      The lumped body analysis assumes that there exists a uniform temperature distribution within the body. This means that the  conduction heat resistance should be zero. Thus the lumped body analysis is exact when biot number is zero.

In general it is assume that for a lumped body analysis, Biot number \leq 0.1

Therefore, the smaller the Biot number, the more exact is the lumped system analysis.

7 0
3 years ago
A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe ha
poizon [28]

Answer :  

The force needed to move the cylinder is 25.6 N

<h2>Further explanation  </h2>

Given that,  

Length of the cylinder, l = 0.5 m  

Outer diameter of the cylinder, d = 8 cm = 0.08 m  

Outer radius of the cylinder, r=0.04\ m  

Inside diameter of the pipe, d = 8.5 cm = 0.085 m  

Inside radius of the pipe, r=0.0425\ m  

Specific gravity of the oil, \rho=0.92  

Density of oil, d=\rho\times \rho_w

Kinematic viscosity of the oil, v=5.57\times 10^{-4}\ m^2/s  

Velocity of the cylinder, u = 1 m/s  

We need to find the force needed to move the cylinder. Let the force is F.  

Specific gravity is defined as the ratio of the density of the substance to the density of water.  

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

v=\dfrac{\mu}{d}

Where, d = density of oil

And d=\rho\times \rho_w (density of oil = specific gravity × density of water )

d=0.92\times 10^3\ kg/m^3

So,  

\mu=v\times d..............(1)

\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3

\mu=0.512\ Pa-s

The separation between the cylinder and pipe is given by :

dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m

d_p\ and\ d_c are diameter of pipe and cylinder respectively.  

The mathematical expression for the Newton's law of viscosity can be written as:  

\tau\propto\dfrac{du}{dy}  

\tau=\mu\times \dfrac{du}{dy}..........(2)  

Where  

\tau = Shear stress, \tau=\dfrac{F}{A}............(3)  

\mu = viscosity  

\dfrac{du}{dy} = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :  

\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}...............(4)  

A is the area of the cylinder, A=2\pi rl  

Equation (4) becomes :  

F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl..............(5)

A=\pi d\times l

A=\pi \times 0.08\ m\times 0.5\ m

A=0.125\ m^2

Now, equation (5) becomes :

F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl

F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2

F = 25.6 N

<h2>Learn more  </h2>

Kinematic viscosity : brainly.com/question/12947932

<h2>Keyword :  </h2>

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.  

7 0
3 years ago
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