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Ket [755]
3 years ago
13

An ideal gas initially occupying 0.020 m3 at 1.0 MPa is quasistatically expanded inside a piston-cylinder device at a constant p

ressure until its volume doubles. Next the expansion is continued at constant volume till the pressure reaches half of the initial pressure. Finally it is brought back to the initial state in a polytropic process with exponent n=1.6
a. Draw the processes on a P-v diagram and calculate the total work.
b. Calculate the total heat transfered, what is the difference between the initial and final temperature?
Engineering
1 answer:
egoroff_w [7]3 years ago
8 0
An ideal gas initially occupying 0.020 m3 at 1.0 MPa is quasistatically expanded inside a piston-cylinder device at a constant pressure until its volume doubles. Next the expansion is continued at constant volume till the pressure reaches half of the initial pressure. Finally it is brought back to the initial state in a polytropic process with exponent n=1.6

a. Draw the processes on a P-v diagram and calculate the total work.
b. Calculate the total heat transfered, what is the difference between the initial and final temperature?an answer is to present a question of how you are not able to join the world and how you can help please answer
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A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa
zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

  = 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

                                                 = 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

    = 29.5 Ksi

Ratio  = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa =  [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

     = 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

                                                      = 19.47 / 15.5 = 1.256

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These tadpoles are confined to a limited environment. What are they all competing for in that environment
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Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe
Ne4ueva [31]

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

\frac{p1}{p*} = 0.3636

\frac{T1}{T*} = 0.5289

\frac{T01}{T0*} = 0.7934

Isentropic Flow Chart:  M1 = 2.0 , \frac{T01}{T1} = 1.8

T1 = \frac{1}{0.7934} (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= \frac{T02}{T1}T1 = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = \frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K = 135.7*10^{3} J/Kg.

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