Further discussion on question 13 from the 2013 case in the morning
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This post was last edited by zwp997 on 2015-6-4 at 12:50. It’s a further discussion regarding question 13 from the 2013 case; personally, I think the solutions available online are still incorrect. Since what is required in the question is the isentropic expansion work, all the solutions provided involve calculating the shaft work Ws of the actual compression process; these are two different concepts. △(PV) This is the expansion work. Seeking expert discussion. Original question: Ethylene gas with a flow rate of 88,700 Nm3/h, a pressure of 1135 kPa, and a temperature of 40°C expands in a turbine to a pressure of 1054 kPa. Assuming an efficiency of 0.75, the enthalpy change calculated based on an adiabatic reversible expansion process is 0.0123 KJ/kg. What is the value of the isentropic expansion work (W) for this process? (A) 5.05 (B) 283 (C) 53885 (D) 16584.68. There are several posts discussing this question: http://bbs.hcbbs.com/thread-1328998-1-13.html http://bbs.hcbbs.com/thread-1326627-1-36.html http://bbs.hcbbs.com/thread-1326263-1-20.html http://bbs.hcbbs.com/thread-1330037-1-27.html
There are a few concepts I’m not clear about here: is shaft work Ws the same as expansion work? Volume work? Shaft work is definitely not flow work. For others, the flow work is Δ(PV).
First, an adiabatic reversible process is called a isentropic process, and the work obtained in such a process is known as isentropic work. Wid = ΔH – TΔS; since ΔS = 0, Wid = ΔH = 0.0123 * (88700/22.4) * 28 KJ/h = 1363.76 KJ/h = 378.82 W. Therefore, W = 0.75 * Wid = 284 W. The correct answer is (B)
Please distinguish between expansion work, volume work, flow work, and shaft work.
Expansion work is a type of volume work; it represents the product of the external pressure P0 and the change in the fluid’s volume during its expansion. The flow work refers, in the process of fluid flow, to the difference in the work exerted by the fluid outside a certain section on its upstream and downstream interfaces, with that section being the object of study. Shaft work is a measure of the pressure energy within itself, and cannot be calculated directly. Could an expert please advise whether this way of phrasing is correct?
Seeking an expert’s explanation? ? ? ? Where are the experts?
I agree with you. An adiabatic reversible process is an isentropic process, and the work obtained is the isentropic work. The problem states that the enthalpy change for an adiabatic reversible process is 0.0123 kJ/kg; multiplying this value by the mass of the fluid in question gives the isentropic work (the theoretical maximum amount of work that can be produced). So why is it necessary to multiply by an efficiency factor as well? The final value, taking efficiency into account, represents the actual work done – is it still isentropic work?
Yes, because you calculated it based on the ideal conditions; in reality, the efficiency is only 0.75.
I wonder whether the concepts of \"adiabatic reversible process\" and \"isentropic process\" are completely equivalent, or whether one concept is included within the other but is not the only one In this question, does the isentropic process include not only \"adiabatic reversible processes\" but also other isentropic processes? At this point, the isentropic process represents the actual process work. Is that really the case? I’m completely puzzled.
An “adiabatic reversible process” is necessarily an “isentropic process”! The reverse is not necessarily true – an isentropic process does not have to occur through an adiabatic reversible process; it can also result from other processes!
What are the isentropic processes, master? Seeking advice.