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GaryK [48]
3 years ago
6

Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i

sothermal compressibility and the volume expansivity coefficients.
Engineering
1 answer:
DedPeter [7]3 years ago
4 0

Answer:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

Explanation:

The internal energy is equal to:

dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV

The entropy is equal to:

dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }

Replacing:

dU=C_{v} dT+(T(\frac{\beta }{\kappa })  -P)dV

dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV

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3 years ago
Technician A says amperage cannot exist without both voltage and resistance. Technician B says if amperage is high, then you kno
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Answer:

Technician A

Explanation:

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3 years ago
A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm with single threads, and it is used to raise a loa
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Answer:

54mm.

Explanation:

So, we are given the following data or parameters or information that is going to assist in solving this type of question efficiently;

=> "A square-thread power screw has a major diameter of 32 mm"

=> "a pitch of 4 mm with single threads"

=> " and it is used to raise a load putting a force of 6.5 kN on the screw."

=> The coefficient of friction for both the collar and screw is .08."

=> "If the torque from the motored used to raise the load is limited to 26 N×M."

Step one: determine the lead angle. The lead angle can be calculated by using the formula below;

Lead angle = Tan^- (bg × T/ Jh × π ).

=> Jh = J - T/ 2. = 32 - 4/2. = 30mm.

Lead angle = Tan^- { 1 × 4/ π × 30} = 2.43°.

Step two: determine the Torque required to against thread friction.

Starting from; phi = tan^-1 ( 0.08) = 4.57°.

Torque required to against thread friction = W × Jh/2 × tan (lead angle + phi).

Torque required to against thread friction =( 6500 × 30/2) × tan ( 2.43° + 4.57°). = 11971.49Nmm.

Step three: determine the Torque required to against collar friction.

=> 2600 - 11971.49Nmm = 14028.51Nmm.

Step four = determine the mean collar friction.

Mean collar friction = 14028.51Nmm/0.08 × 6500 = 27mm

The mean collar diameter = 27 × 2 = 54mm.

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3 years ago
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Answer:

  See attached

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A schematic is attached.

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