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3 years ago

What is a quasi-equilibrium process? What is its importance in engineering?

Engineering

1 answer:

nata0808 [166]3 years ago

8 0

Answer:

Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.

Explanation:

Step1

Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.

Step2

All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.

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A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank

Bad White [126]

Given:

mass of water, m = 2000 kg

temperature, T = $30^{\circ}C$ = 303 K

extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m $\bar{R}$ T (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

$V = \frac{m\bar{R}T}{P}$

$V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}$

$V = 2762.44 m^{3}$

Therefore, the volume of the tank is $V = 2762.44 m^{3}$

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = $30^{\circ}C$ = 303 K

At equilibrium, volume remain same

So,

P'V = m' $\bar{R}$ T'

$P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}$

Therefore, the final pressure is P' = 96.258 kPa

4 0

3 years ago

Do you understand entropy? Why the concept of entropy is difficult to engineering students?

Darya [45]

Answer:

Entropy:

Entropy is the measure of randomness of system.In other words the entropy is the measurement of tendency of system towards the disorder.

The concept of entropy arise from second law of thermodynamics.It is given as follows

$ds=\int \dfrac{dQ}{T}$

Entropy is a extensive property of system .

Entropy of universe = Entropy of system + Entropy of surrounding.

The entropy of the system can be zero,positive and negative.But entropy of the surrounding can not be negative,but it can be zero or positive.

Actually the concept of entropy is difficult to understand because we can not visualize because it is not like beam and like rods.Only we have to realize that there is entropy.

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2 years ago

A dual-fluid heat exchanger has 10 lbm/s water entering at 100 F, 20 psia and leaving at 50 F, 20 psia. The other fluid is glyco

kakasveta [241]

Answer:

Rate of internal heat transfer = 23.2 Btu/Ibm

mass flow rate = 21.55 Ibm/s

Explanation:

using given data to obtain values from table F7.1

Enthalpy of water at temperature of 100 F = 68.04Btu/Ibm

Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm

from table F.3

specific constant of glycerin $C_{p} = 0.58 Btu/Ibm-R$

<u>The rate of internal heat transfer ( change in enthalpy ) </u>

h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )

where ; T4 = 50 F

T3 = 10 F

Cp = 0.58 Btu/Ibm-R

substitute given values into equation 1

change in enthalpy ( h4 - h3 ) = 23.2 Btu/Ibm

<u>Determine mass flow rate of glycol</u>

attached below is the detailed solution

mass flow rate of glycol = 21.55 Ibm/s

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3 years ago

Which option distinguishes the members of a software deployment process team most likely involved in the following scenario?

Alchen [17]

Answer:

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5 0

2 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$