Thank you ,sir. I've got this question and it might seem somewhat crazy but it's really holding me back. You said that work done in a cyclic process is equal to the heat given but since the internal energy is changing in the intermediate steps in the cycle hence the heat supplied should be a little more to do the same work. Thanks in advance.
+Karan Bellic When the gas goes through a complete cycle, it ends in the exact same spot as when it started with the same T, P, and V and thus the internal energy change through a complete cycle is zero.
I don't know why, but my textbook says that there must be heat transfer out of the system in order to get a net work output. But in the video which is what I think is correct, states that Q = W for a cyclical process. So that means heat transfer into the system equals net work output. The statement in my book contradicts my point and your video. Why did that happen? Thank you. I actually think that maybe the situation is different. Mine is a PV diagram describing a heat engine. During the first adiabatic process, (Qh) heat transfer into the system; and during the second adiabatic process, which is the fourth and final process, (Qc) heat is transferred out of the system to return to the same state it started in.
I'm okay with this thing that internal energy at the end remains same. What I'm talking about is that within the INTERMEDIATE steps the internal energy changes and hence the heat supplied should be little more to do the work.
Israel, There are 4 main thermodynamic processes. Constant pressure (isobaric), constant volume (isochoric), constant temperature (isothermic), and adiabatic (nothing remains constant and Q = 0. None of them look like circles on a PV diagram. I suppose with some advanced math we could come up with a way to calculate the efficiency, but it would be a theoretical exercise.
***** That depends on how you define the first law of thermodynamics. Read the definition carefully. I like to define it like this: "The change in the internal energy of a gas is equal to the heat added TO the gas minus the work done BY the gas" This is the definition in many textbooks. With this definition, the work done is positive when the process on the PV diagram is clockwise. There are textbooks that use a different definition.
+Md. Mominul Islam Note that on the PV diagram we work with the state variables (P, V, T) and this does not include the efficiency of the engine only the portion of the energy added to the gas that is transferred to work. To calculate the efficiency, you will have to use a different equation.
at 4:16, why did W(4-1) you took pressure from P1 instead P4? but.. during W(2-3), you took p2?
P1 = P4 and P2 = P3
Thank you ,sir. I've got this question and it might seem somewhat crazy but it's really holding me back. You said that work done in a cyclic process is equal to the heat given but since the internal energy is changing in the intermediate steps in the cycle hence the heat supplied should be a little more to do the same work. Thanks in advance.
+Karan Bellic
When the gas goes through a complete cycle, it ends in the exact same spot as when it started with the same T, P, and V and thus the internal energy change through a complete cycle is zero.
I don't know why, but my textbook says that there must be heat transfer out of the system in order to get a net work output. But in the video which is what I think is correct, states that Q = W for a cyclical process. So that means heat transfer into the system equals net work output. The statement in my book contradicts my point and your video. Why did that happen? Thank you. I actually think that maybe the situation is different. Mine is a PV diagram describing a heat engine. During the first adiabatic process, (Qh) heat transfer into the system; and during the second adiabatic process, which is the fourth and final process, (Qc) heat is transferred out of the system to return to the same state it started in.
Thank you sir for your videos and for sharing your knowledge.
My pleasure
sir,
how to find the internal energies of the individual processes ?
I'm okay with this thing that internal energy at the end remains same. What I'm talking about is that within the INTERMEDIATE steps the internal energy changes and hence the heat supplied should be little more to do the work.
+Karan Bellic
When you add up the change of the internal energy of each step, they will add up to zero for a complete cycle.
Excellent
w=Q-U & also Q=n*cp*dt U=n*cv*dt so w=n*cp*dt-n*cv*dt (if cp-cv=R then how w can be 0j on the state1 to state2)
W is zero from state 1 to 2 because there is no change in volume. There can only be W if the volume changes.
why there cannot be a system that is describe by a circle instead of rectangle? how can we determine the efficiency of such cycle?
Israel,
There are 4 main thermodynamic processes. Constant pressure (isobaric), constant volume (isochoric), constant temperature (isothermic), and adiabatic (nothing remains constant and Q = 0. None of them look like circles on a PV diagram. I suppose with some advanced math we could come up with a way to calculate the efficiency, but it would be a theoretical exercise.
Thank you very much Michel
why is the work positive clockwise and negative anticlockwise?
By definition, the work is positive when done BY the gas.
Thus work can only be done by the gas when it expands.
Volume increases when gas expands.
*****
That depends on how you define the first law of thermodynamics.
Read the definition carefully.
I like to define it like this:
"The change in the internal energy of a gas is equal to the heat added TO the gas minus the work done BY the gas"
This is the definition in many textbooks.
With this definition, the work done is positive when the process on the PV diagram is clockwise.
There are textbooks that use a different definition.
sir, for this case what's the efficiency? 100% shouldn't possible.
+Md. Mominul Islam
Note that on the PV diagram we work with the state variables (P, V, T) and this does not include the efficiency of the engine only the portion of the energy added to the gas that is transferred to work. To calculate the efficiency, you will have to use a different equation.
Is V1 = V2 and V3 = V4?
Yes, that is correct.
@@MichelvanBiezen Thanks for the fast response!
Thank you thank you thank you like Jay Z says in Niggas in Paris Video Clip
2024?