The work and heat transfer isothermal internally reversible process.
An isothermal procedure is a thermodynamic procedure in which the temperature of a gadget stays consistent. The transfer of heat into or out of the gadget occurs so slowly that thermal equilibrium is maintained. The melting of ice at 0 diploma is an example of isothermal technique. The reaction in a warmth pump is an example of isothermal manner. In Isothermal manner temperature is constant for the duration of the procedure and follows Boyles regulation.
Given,
Gas at initially P1= 2.8 bar
P2 = 14 bar
isothermal reversible process,
Compressed pressure at 60°C
(a) R = 134a
For Refrigerant 134a R134a
At P1 = 14 bar T1 = 600C
S1 = 0.9389 KJ/Kg K
V1 = 264.64 KJ/kg
At
P2 = 2.8 bar T2 = 600C
S2 = 1.1142 KJ/Kg K
V2 = 278.56 KJ/kg
T = 60 + 273 = 333K
Q= 2 Tds
2 Q=T ſ ds
= T(S2-S1)
= 333 ( 1.1142 - 0.9389 )
= 58.3749 KJ/Kg
Heat transfer Q = 58.3749 KJ/Kg
(b) air as an ideal gas.
Q= 2 Tds
2 Q=T ſ ds
= T(S2-S1)
=T-R In P2 P1
= 333 * -0.287 In 2.8 14
= 153.81 KJ/Kg
Heat transfer Q = 153.81 KJ/Kg
W = Q - ΔV
= 28.3749 - (278.56-264.64)
= 14.4549 KJ/Kg
Work W = 14.4549 KJ/Kg
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Answer:
Velocity after collision will be 5.56 m/sec
So option (a) will be correct answer
Explanation:
Mass of steel ball
Speed of steel ball before collision
Mass of large ball
Velocity of large ball
According to conservation of momentum
So velocity after collision will be 5.56 m/sec
So option (a) will be correct answer
Answer:
D
Explanation:
A) is not correct, because the gravitation potential energy will depend on the height the block is located at. It will be calculated with the formula:
U=mgh.
If we take the ground as a zero height reference, then on point 2 the potential energy will be:
While on point 3, the potential energy will be greater.
B) is not the right answer because the kinetic energy will vary with the height the block is located at in the fact that the energy is conserved (this is if we don't take friction into account or air resistance) so in this case:
We already know the potential energy at point 2. We can calculate the kinetic energy at point 3 like this:
So the kinetic energy at point 2 is given by the equation:
so:
As you may see the kinetic energy at point 2 is greater than the kinetic energy at point 3.
C) Is not correct because according to the first law of thermodinamics, energy is not lost, only transformed. So, since we are not taking into account friction or any other kind of loss, then we can say that the amount of mechanical energy at point 1 is exactly the same as the mechanical energy at point 3.
D) Because of what we talked about on part C, this will be the true situation, because the mechanical energy of the block will be the same no matter theh point you measure it at.