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IrinaK [193]
3 years ago
11

An ideal gas, consisting of n moles, undergoes an irreversible process in which the temperature has the same value at the beginn

ing and end. If the volume changes from Vi to Vf , the change in entropy is given by:______
Engineering
1 answer:
Vlada [557]3 years ago
4 0

Answer:

n R ln(Vf/Vi)

Explanation:

Entropy is the loss of energy available to do work. Entropy is a state function (i.e. it depends only upon the current state of the system and is independent of how that state was prepared).

Since the temperature change of the ideal is constant, hence this is an isothermal expansion of a perfect gas. The change in entropy (ΔS) for an isothermal expansion of a perfect gas is given by:

\Delta S=nR*ln(\frac{V_f}{V_i})

Where n is the amount of gas molecules in mol and R is the gas constant in JK⁻¹mol⁻¹given by R = N_Ak, k is Boltzmann's constant in J K⁻¹ and Avogadro's constant N_A in mol⁻¹. Vf is the final volume and Vi the initial volume.

You might be interested in
Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati
Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0
3 years ago
Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new
Marina86 [1]

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))     where i=1 to N-1 and N=6

C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))

A area of the polygon can be found by the following formulaA=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i}) where i=1 to N-1

A=\frac{1}{2}[ (x_{1}  y_{2} -x_{2}  y_{1})+ (x_{2}  y_{3} -x_{3}  y_{2})+(x_{3}  y_{4} -x_{4}  y_{3})+(x_{4}  y_{5} -x_{5}  y_{4})+(x_{5}  y_{6} -x_{6}  y_{5})]

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(x_{2}+x_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(x_{3}+x_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(x_{4}+x_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(x_{5}+x_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{x} =15.36

For Cy

C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(y_{2}+y_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(y_{3}+y_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(y_{4}+y_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(y_{5}+y_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{y} =22.55

So coordinates for the fire station should be (15.36,22.55)

5 0
3 years ago
For a turbulent flow of a fluid in 0.6 m diameter pipe, the velocity 0.15 m from the wall is 2.7 m/s. Estimate the wall shear st
MAVERICK [17]

Answer:

If the turbulent velocity profile in a pipe of diameter 0.6 m may be approximated by u/U=(y/R)^(1/7), where u is in m/s and y is in m and 0.15 m from the pipe.

Explanation:

hope it helps

3 0
3 years ago
Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a
Veronika [31]

Answer:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero  is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

7 0
3 years ago
Write a naive implementation (i.e. non-vectorized) of matrix multiplication, and then write an efficient implementation that uti
erik [133]

Answer:

import numpy as np  

import time  

def matrixMul(m1,m2):      

   if m1.shape[1] == m2.shape[0]:  

       

       t1 = time.time()

       r1 = np.zeros((m1.shape[0],m2.shape[1]))

       for i in range(m1.shape[0]):

           for j in range(m2.shape[1]):

               r1[i,j] = (m1[i]*m2.transpose()[j]).sum()

       t2 = time.time()

       print("Native implementation: ",r1)

       print("Time: ",t2-t1)

       

       t1 = time.time()

       r2 = m1.dot(m2)

       t2 = time.time()

       print("\nEfficient implementation: ",r2)

       print("Time: ",t2-t1)

       

   else:

       print("Wrong dimensions!")

Explanation:

We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.

7 0
3 years ago
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