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Airida [17]
3 years ago
11

Explain the effect of heat transfer, reversibilities and irreversibilities on changes in entropy

Engineering
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer is given below

Explanation:

  • The second law of thermodynamics states that in the reversible process, the entropy of the universe is constant, while in the unchanging process, such as the transfer of heat from a hot object to a cold object, the reflection of the universe increases.
  • ΔS = ΔQ ÷ T  
  • The irreversible process increases the entropy of the universe. Since entropy is a state function, the change in entropy of the system is the same, the process is reversible or irreversible. The second law of thermodynamics can be used to determine whether a process is reversible.
  • for an irreversible process, ΔS is not same as reversible process, because more than 1 reversible process is required for 2  ΔS to be equal.
  • if the initial and final states are the same, ΔS for an irreversible process can be calculated as if it were a reversible process.
  • If the initial and final condition is same then ΔS can find out for irreversible process even if it were reversible process

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A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The
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3 years ago
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a
Ratling [72]

Answer:

Explanation:

From  the given question:

Using the distortion energy theory to determine the  factors of safety  FOS can be expressed  by the relation:

\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma  y^2-\sigma x \sigma y+3 \tau_{xy^2}}

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory  can be expressed as:

\tau_{max} = \dfrac{Syt}{2FOS}

where;

\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma  y}{2})^2+ \tau _{xy^2

a. Using distortion - energy theory formula:

\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}

\dfrac{350}{FOS}=160.35

{FOS}=\dfrac{350}{160.35}

FOS = 2.183

USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2

\dfrac{350}{2 FOS}  =88.51

\dfrac{350}{ FOS}  =2 \times 88.51

{ FOS}  =\dfrac{350}{2 \times 88.51}

FOS = 1.977

b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}

\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000

\dfrac{350}{FOS}=105.3565

FOS=\dfrac{350}{105.3565}

FOS =3.322

USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2

\dfrac{350}{2 FOS}  ={ (\dfrac{110-100}{2})^2

\dfrac{350}{2 FOS}  =25

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}

\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}

\dfrac{350}{FOS}= 88.88

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2

\dfrac{350}{2 FOS}  =\sqrt{ (35)^2+ (-20)^2

\dfrac{350}{2 FOS}  =\sqrt{ 1225+ 400

\dfrac{350}{2 FOS}  =40.31

FOS}  =\dfrac{350}{2*40.31}

FOS = 4.341

7 0
3 years ago
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