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NISA [10]
3 years ago
13

What is thermodynamics system? Briefly explain (i) control mass system (ii) control volume system. Comment briefly on control ma

ss system and control volume system at Steady state
Engineering
1 answer:
olga2289 [7]3 years ago
3 0

Explanation:

Thermodynamics system :

 Thermodynamics system is a region or space in which study of matters can be done.The system is separated from surroundings by a boundary this boundary maybe flexible or fixed it depends on situations.The out side the system is called surroundings.

Generally thermodynamics systems are of three types

1.Closed system(control mass system)

  Only energy transfer take place ,no mass transfer take place.

2.Open system(control volume system)

 Both mass as well as energy transfer take place.

3.Isolated system

   Neither mass or nor energy transfer take place.

At steady state ,property is did not changes with respect to time.

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What are the main factors for the high material removal rate in broaching operations?
Umnica [9.8K]

Answer:

Explanation:

Broaching:

  Broaching is a metal removal process or we can say that it is a machining process.Generally broach have three types teeth in which first one is called roughing teeth,second one semi finishing and third  one finishing teeth.By using these teeth metal remove so fast and that is why broaching is the high metal removal rate operation.

6 0
3 years ago
Consider a single crystal oriented such that the slip direction and normal to the slip plane are at angles 42.7° and 48.3°, resp
zlopas [31]

Answer:

Applied Stress > 58.29 MPa

Explanation:

  • Resolved shear stress should be greater than critically resolved shear stress in order to cause the single crystal to yield

Given angles are

∅ = 42.7 degree

Ф = 48.3 degree

Critically resolved shear stress = 28.5 MPa

If we consider

Critically resolved shear stress = resolved shear stress

Applied stress can be found by

Z_{R} = applied stress X cos\phi X cos\theta    (1)

Applied Stress = \frac{Z_{R} }{Cos\phi XCos\theta}

Applied Stress = \frac{28.5}{Cos(48.3)XCos(42.7)}

Applied Stress = 58.29 MPa

We got reference

  • By putting applied stress values of greater than 58.29 MPa in equation 1 we get

        Resolved Shear Stress = 60 x Cos(48.3) x Cos(42.7)

        Resolved Shear Stress = 29.33 MPa

Therefore, by the above calculation we conclude that applied stress should be greater than 58.29 MPa, In order to make resolved shear stress to be greater than critically resolved shear stress that is essential for single crystal to yield.

8 0
3 years ago
The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are
IrinaK [193]

Answer:

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum  = 9.93*10^{-8}

Number of vacant atoms = 5.854 * 10^{15} \ vacancies/cm^3

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = 6.022*10^{23

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

5.854*10^{15} = 9.93*10^{-8}  × Total number of sites(N)

Total number of sites (N) = \dfrac{5.854*10^{15}}{9.93*10^{-8}}

Total number of sites (N) = 5.895*10^{22}

From the expression of the total number of sites; we can determine the density of the electrum;

N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}

where ;

N_A = Avogadro's Number

\rho_{electrum} = density of the electrum

A_{electrum} = Atomic mass

5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}

5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}

8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}

\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}

\mathbf{  \rho _{electrum}=14.30 \ g/cm^3}

Thus; the density of the electrum is 14.30 g/cm³

7 0
3 years ago
Find the compressibility factor Z for oxygen at 3 MPa and 160 K.
saveliy_v [14]

Answer:

Z= 0.868

Explanation:

Given that

P= 3 MPa

T = 160 K

We know that

P v= Z R T

P= Pressure

v = specific volume

R= gas constant

T = Absolute temperature

Z=  Compressibility factor

Here specific volume of gas is not given so we assume that specific volume gas

v=0.012\ m^3/kg

We know that for oxygen gas constant

R = 0.259 KJ/kg.K

Now by putting the values

P v = Z R T

3000 x 0.012 = Z x 0.259 x 160

Z= 0.868

So  Compressibility factor is 0.868.

5 0
3 years ago
A steel loop ABCD of length 5ft and of 3/8" diameter is placed as shown around a 1" diameter aluminum rod AC. Cables BE and DF e
lorasvet [3.4K]

Answer:

a

Explanation:

ye men thats the answer<em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>#</u></em><em><u>%</u></em>

4 0
3 years ago
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