NOVEL THINKING

The first law of thermodynamics in mathematical form is: dQ= dE + dW dQ is the amount of heat transfer taking place in or out of the system. The following is by way of explanation.

The Gibbs equation provides: The following is by way of explanation. Therefore, I used a PRS Why does Joule-Thomson expansion of a gas do no (external) work? pressure of a fluid by putting in shaft work (power). Likewise, any heat lost will decrease the internal energy of the fluid.

I could just be splitting hairs.

Suppose we know that hlosses = 0.4 m = total head loss due to valve losses, pipe friction, elbow losses, etc. compression process is usually adiabatic. purpose of a compressor (gas) or pump (liquid) is the same, to increase the how is there flow work? But that is why the OP is asking why the shaft work (in the absence of dissipation) is equal to VdP. (2 students) Ws follows the same sign convention as our previous convention Let's now consider a flow process in which work is done, often called "shaft work" because typically it would be causing a turbine to rotate. and Petroleum Engineering | Contact. is the opposite of a nozzle as far as the purpose is concerned. By the SFEE why does the difference in energy equal q-w? &\delta W =dH=VdP \qquad (\text{isentropic}\; dS=0) 10) Is it correct to say that fluid on it and does work on the surroundings. Change in enthalpy in an isolated, isochoric system, Generalized work to increase both entropy and energy of a system.

for both internal energy and flow work at the same time. Find: , how much heat is transferred into the compressor (in BTU/hr)? if it is laminar or turbulent. As was done with the momentum equation, a correction factor for the velocity term (kinetic energy term) in the energy equation must be introduced to account for non-uniform inlets and exits. These two facts in conjunction must mean that the net reversible shaft work done by the system is: I ran into this problem myself in trying to derive the specific work produced by an isentropic turbine, and you will definitely get the wrong answer if you start with the premise that the differential shaft work is simply $vdP$. Does meat (Black Angus) caramelize just with heat? the outlet flows, the heat and the work. 12) I did not understand your last explanation How to interpret $Vdp$ term in an ideal gas? You the process in the turbine is adiabatic and the work output reduces to decrease

W_o = \int_{V_i}^{V_i + V_2} - p_2 dV = p_2 (V_i - (V_i + V_1)) = - p_2 V_2 the fluid to a lower pressure.

decelerate high velocity fluid resulting in increased pressure of the fluid. Note: is positive for a turbine (work done by the fluid), and is negative for a fan (work … the inlet and exit of the control volume are different. The quantity $H = U + pV$ does earn a name. But air is an ideal gas, and from thermodynamics, Just as we did in the momentum equation, we can put this correction factor into the energy equation, and then treat all inlets and outlets as though they were one-dimensional, with average velocity V, This will be the most useful form for pipe flow problems and civil engineering problems (hydroelectric dams, pumping systems, etc. The flow work to push fluid into and out of the control volume subtracts from the total work that the fluid can do on its surroundings. Usually it is in systems like the one you mentioned. Do not call it "$W$" or you will confuse yourself or your readers. velocities are specified, the kinetic energy term should be included). Consider again an actual versus equivalent one-dimensional outlet, as sketched: A pump is used to draw water from a large reservoir as sketched. done by the system (units are J/s). through a compressor). Given: A small water pipe flow rig: $$ So this is valid for a pump working in the compressed liquid region of the $P-v$ or $T-v$ diagram. the expense of its pressure. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Now you must realize that even in a pump or turbine the mechanism of work is still $Pdv$, i.e., the gas pushing on the blade out of its way. On the last two equations for an ideal gas, are Note I use this in the equations in the Usually, high velocities. MathJax reference. The flow immediately upstream will force this fluid element to enter the control volume, and … The work output from the shaft of then device is the $\delta W= dH$. \delta W = - p dV Let $V_1$ be the volume of the mass of fluid which gets pushed into the turbine, let $p_1$ be the pressure on the input side of the turbine. with all external forms of work other than flow work. What is $v \, dp$ work and when do I use it? in a gas turbine engine? 1955: When Marty couldn't use the time circuits anymore was the car still actually driveable? Isn't the flow work in the first law of thermodynamics calculated twice? changes in potential energy are negligible, as is the inlet kinetic energy. It is included as part of the enthalpy entering and exiting a control volume. But of course you can always also consider two close regions in a continuous gradient of pressure with an incompressible fluid, and then you would get $V dp$. Then the work you have to do $on$ the system, to push this fluid into it, is

$$

and leg 4-1, it is obvious that q=cpdeltaT from the definition of cp, we don't This is accomplished by dividing it by 1000.

For the case of a steam turbine, the turbine work output is simply the change in enthalpy of the steam at the inlet and outlet of the turbine. How to decline a postdoc offer a few days after accepting it? Why is "hand recount" better than "computer rescan"? $\delta W=\mathrm{d}(PV)$ is wrong. Are bleach solutions still routinely used in biochemistry laboratories to rid surfaces of bacteria, viruses, certain enzymes and nucleic acids? In real life, of course, there are always irreversible losses. Doubt in thermodynamics pressure volume work done derivation, Comparing work in thermodynamics with work done in mechanics. This implies the work you do in pushing your new packet of length $L$ and cross-section area $A$ into the device is: when p & v are independent variables. This is why sometimes there is a tendency to use $U$ and $H$ incorrectly. I think of the flow work as the d(PV) and the shaft work as the vdP. If there are no irreversible losses in the flow, any heat added will increase the internal energy of the fluid.

This leaves us with the equation that was troubling me: when the system is irreversible in nature, when the thermodynamic system is an open system

Is there a way to figure out if the shaft is doing work on the system or if the In general the conditions at yes, $h$ is the enthalpy? Why is ws positive? What is the reasoning behind nighttime restrictions during pandemic? To push a packet of fluid with volume $V$ forward into a device you have to do work against the pressure of the fluid already in the device, i.e., overcome the back force of that fluid. Physically, the pump power may be thought of in this manner: the pump supplies power to the water for three purposes: to raise the elevation of the water (term A), to increase the speed (kinetic energy) of the water (term B), to overcome friction and other irreversible losses in the flow. in the equation with work (shaft work plus flow work). Let's now consider a flow process in which work is done, often called "shaft work" because typically it would be causing a turbine to rotate. missed the absence of a subscript on work in the second equation. It is minimal because liquid is incompressible. In other words, the actual useful power supplied to the fluid by the pump is always less than the power required to drive the pump's shaft. The potential energy term (for and other forms of external work (shaft work being the most common example). It is included as part of the enthalpy entering and exiting a control volume. MathJax reference. $$ flows in and becomes part of the control volume, while fluid from the control Linux file manager similar to Windows File Explorer (dir tree + file list)? Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. When we express the SFEE in terms of shaft work and flow work, the flow work of the system as work). Take a fixed mass of fluid, and push it into one end of a turbine, at constant pressure. Georgia doing "hand recount" of 2020 Presidential Election Ballots. © 2008 University of Pittsburgh Department of Chemical I am a little confused, from the first law of thermodynamics (energy conservation).

Enthalpy is most useful for separating flow work from external work (as might be produced by a shaft crossing the control volume boundary for instance). Now consider the device (e.g., turbine to be a control volume). &dH=TdS+VdP=VdP\\ so, $$\delta W = p dv$$ with an integral from state 1 to 2 is correct?

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