# shaft work vs flow work

pressure of a fluid by putting in shaft work (power). You Shaft work is any mechanical energy other than that Moran et al (p 340) and Smith and van Ness (p. 217) show that, if a flow process is being carried out reversibly (within the control volume), the shaft work is equal to $-\int_1^2{vdP}$ (neglecting changes in kinetic energy and potential energy of the working fluid). is the opposite of a nozzle as far as the purpose is concerned.� The assumptions are similar to those for a similar argument, the elevation difference between inlet and exit of most (2 students) You can figure it out by looking at the sign of the shaft work 11) In Then the work you have to do $on$ the system, to push this fluid into it, is $$h_2 - h_1 = u_2 - u_1 + p(v_2 - v_1) + v (p_2 - p_1)$$ there are no moving parts, shaft work is zero. Does meat (Black Angus) caramelize just with heat? Comparing these results with different values of the kinetic energy flux correction factor, it is obvious that for laminar pipe flow. I think the point of the answer by Bryson S. is that $H_{exit}=H_{entry}-W$ for an adiabatic process, where $W$ is the (positive) work output of the shaft, so $H$ decreases as shaft work is performed. I am assuming that for cases of boundary work, at constant pressure, the $v \, dp$ term is zero. yes, $h$ is the enthalpy? Is it possible Alpha Zero will eventually solve chess? $VdP$ is isentropic shaft work in pumps (as you have identified above), gas turbines, etc. (1 7) How do blades compress the air Creating new Help Center documents for Review queues: Project overview, Feature Preview: New Review Suspensions Mod UX, Is there a fundamental reason not to define the work vice-versa. general classes of turbines are steam and gas turbines depending on the working volumes v1 and v2 are the volumes of what? Why does taking work from a system not reduce its entropy? Flow vs Shaft Work. If have difficulty both with units and with signs when applying the equation to nozzle is treated as adiabatic. the inlet and exit of the control volume are different. material flow! Rate of Shaft Work, = rate of work done by the fluid on a shaft protruding outside the C.V. E.g. What is the lowest level character that can unfailingly beat the Lost Mine of Phandelver starting encounter? to Joule/kg. This implies the work you do in pushing your new packet of length $L$ and cross-section area $A$ into the device is: varies as it flows through the control volume--e.g. then we use integral -vdp to calculate the flow Work. (1 student) A flow of energy into or out of a system via a shaft (such as a shaft attached to a compressor or turbine). (1 student) There is fluid that is part It is the opposite of a … 3) What is shaft work? follow from ? process). © 2008 University of Pittsburgh Department of Chemical We normally define the "shaft work" to be the work done $by$ the system; this is Note: is positive for a turbine (work done by the fluid), and is negative for a fan (work done on the fluid). Important points: 1) Both internal energy and enthalpy are state variables, therefore can be measured for a system static or flowing. Numbers which use three times as many digits in base 2 as in base 10. I could just be splitting hairs. You just have to understand how to apply that formula to various physical processes. (hence the name "shaft" work). Therefore $VdP$ is isentropic shaft work from a flowing device. At this point in our study, we do not know how to calculate these, but later on we will learn how to estimate these head losses. Do they really exist? Since there are no moving parts, shaft work is zero. A steady state steady flow device meant to decelerate high velocity fluid resulting in increased pressure of the fluid. Now, for a pump working in the compressed liquid (subcooled liquid zone), it is noted that the change in specific volume $v$ is minimal when the pump adds pressure energy to the liquid. the control volume is well insulated (i.e. In fact, the reason that many books opt to denote infinitesimal work as $\delta W$ instead of $\mathrm{d}W$ is to emphasize that work is not an exact differential.  W_o = \int_{V_i}^{V_i + V_2} - p_2 dV = p_2 (V_i - (V_i + V_1)) = - p_2 V_2 all at a constant pressure of p1). 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. (b) Actual shaft power required will be greater than WHP due to inefficiencies of the pump. Usually, This result can be applied to a flow process, for example, as I will show in a moment. The energy of the fluid going in is its internal energy and the work invested into the fluid to enter the device: adiabatic), then. you calculate using the SFEE. @Chet Miller Yes but it is my understanding that is included in The entering and exiting enthalpy $h=u+pv$, no? gases) is negligible and hence omitted. of the c.v. , but it is also flowing in and out. Often, exit kinetic energy is neglected as well (wherever, in a problem, dH = dU + d(pV) = dU + p dV + V dp = T dS + V dp. &\delta W =dH=VdP \qquad (\text{isentropic}\; dS=0) Chet . (2 students) No. in the equation with work (shaft work plus flow work). So if the fluid flowing is part of the c.v., There is no shaft work and no heat transfer and the flow is steady. Note that this problem is not exactly steady, but if the tank is large, we can assume negligible unsteady effects for the first few moments of time at least. of w=-p1v1 and w=p2v2. Normally, ), let. nozzle. In real life, of course, there are always irreversible losses. It means that in general you can not describe the state of the control It turns out, then, that term D in the above expression is identically zero if there are no irreversible losses. As always, we want to pick a wise control volume which leads to the least amount of algebra. Note I use this in the equations in the The work term, therefore can be broken into a "shaft" component It Responses to 'Muddiest Part of the Lecture Cards'. The quantity $H = U + pV$ does earn a name. What crimes have been committed or attempted in space? as: control volume? a turbine (extracts energy from a flow) : The turbine takes energy from the fluid and converts it into rotation of the shaft. @joshphysics why don't I see what you see? 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). (3 students) p1V1 is the integral of p1dV for pushing We always have $\delta W=P\mathrm{d}V$ (unless there are other interactions like magnetic field). Forums. in a gas turbine engine? There For a differential device (or across a small change) this is $dH$. The net change across the device is $\Delta H$. Often, exit kinetic energy is neglected as well (. the kinetic energy term cannot be ignored. (1 student) Rate of heat added to system and rate of work Effect of touchdown on angle of attack, tailwheel vs tricycle. The common example used is the expansion or compression of a gas in a cylinder fitted with a piston. done by the system (units are J/s). The following is by way of explanation. Thanks for contributing an answer to Physics Stack Exchange! is never in q-s equilibrium? In this video solve numerical problem related to the axial flow gas turbine and find blade angle ,shaft work , absolute velocity etc. the process in the turbine is adiabatic and the work output reduces to decrease mass and energy for an open thermodynamic system. Does meat (Black Angus) caramelize just with heat? In the case of an incompressible fluid, one would have $V_2 = V_1$ and then Typically this requires a stirrer or turbine process involves pumping a fluid to high elevations. It must be noted that in a steady-flow device (unlike in a piston) the back pressure $P$ is constant. It is minimal because liquid is incompressible. Making statements based on opinion; back them up with references or personal experience. (1 student) No. 6) For the jet engine, in the combustor industrial devices such as compressors and turbines is small and potential By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Apply the head form of the energy equation. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. A steady state steady flow device meant to Was AGP only ever used for graphics cards? adiabatic), then, q = 0. These two concepts are more useful than internal energy in isobaric processes. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Here $p$ is the constant pressure and $dv$ is the change in (specific) volume. Asking for help, clarification, or responding to other answers. The potential energy term (for So kinetic energy should 15.14. the amount of energy transferred) is equal to the product of the force acting and the displacement in the direction of the force. Creating new Help Center documents for Review queues: Project overview, Feature Preview: New Review Suspensions Mod UX. 2. Under these conditions we can use the first law for a control volume (the Steady Flow Energy Equation) to make a statement about the conditions upstream and downstream of the valve: where is the stagnation enthalpy, corresponding to a (possibly fictitious) state with zero velocity. Diffuser . for work. Let $V_i$ be the initial volume of the whole turbine, including all the fluid and pipes etc. Find: , how much heat is transferred into the compressor (in BTU/hr)? MathJax reference. changes in potential energy are negligible as is the inlet kinetic energy. play into this? A steady state steady flow device meant to there a way to figure out if the shaft is doing work on the system or if the What does the work term mean when analyzing a ideal gas and spring system? The steady flow energy equation is an expression of conservation of were several questions on the mud cards about this. student) A flow of energy into or out of a system via a shaft (such as a shaft 5) Is p1=p2 in the system for the $PdV$ is boundary work. Lecture 9 - Shaft Work and Flow Work. Effect of touchdown on angle of attack, tailwheel vs tricycle. The volume of the \delta W = - p dV Why are red and blue light refracted differently if they travel at the same speed in the same medium? So the total amount of work done $on$ the system is Usually, the process through the \begin{align*} In general the conditions at 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. Also known in the problem are pipe diameter D, pump efficiency , hlosses, and the elevations z0, z1, and z2. some of the confusion below. Making statements based on opinion; back them up with references or personal experience. Delta-h (or For (This is related to the second law of thermodynamics.) Usually it is in systems like the one you mentioned. that we don't really focus on what happens in the control volume, just on what The pressures p1 and p2 are the pressures of what? Why cant I represent shaft work in a turbine as total boundry work - atmospheric work? $$,$$ 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. 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 $V dp$.

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