4. Energy balances Partly based on Chapter 4 of the De Nevers textbook.

Transcription

1 Lecture Notes CHE 31 Fluid Mechanics (Fall 010) 4 Energy balances Partly based on Chater 4 of the De Nevers textbook Energy fluid mechanics As for any quantity, we can set u an energy balance for a secific system (ie a volume with clearly defed boundaries) Before settg u energy balances it is wise to recall the first law of thermodynamics (conservation of energy) the various forms of energy the way energy is transferred The first law of thermodynamics says energy is a conserved quantity As a result energy balances have no birth and death terms Energy comes various forms The forms we will restrict ourselves to this fluid mechanics course are only three: ternal energy; related to temerature; measure of hotness; symbol U; secific ternal energy u is ternal energy er unit mass: U=m u Secific ternal energy usually can be written as heat caacity times temerature: u = c T 1 ketic energy; relates to velocity v: Ek otential energy; this fluid mechanics course this is related to gravity; = mv ; secific ketic energy: a vertical level relative to a ground level; secific otential energy: e ek E = gh 1 = mgh with h = v A basic examle of conservation of energy: release a mass m at a level z=h and let it fall Its velocity at level z=0 follows from equatg its (itial) otential energy to its fal ketic 1 energy: mgh = mv v = gh In a more formal manner we say that the sum of otential, ketic energy, and ternal energy situation 1 (mass havg a ternal energy U, hangg still at z=h) is the same as situation (mass still havg ternal energy U, havg a velocity 1 1 v at z=0): mgz + mv + U = mgz + mv + U See how much easier it is to solve for 1 the velocity terms of energy than it would be terms of momentum In terms of momentum it would go as follows: the mass is acceleratg with a rate g; the velocity as a 1 function of time is gt ; the osition as a function of time is z = h gt The moment the h mass reaches z=0 therefore is t 0 = The velocity at this moment is gt0 = gh Same g answer, more stes Energy is beg transferred by means of work and heat (we do not consider radiation this course) Heat largely relates to transfer of ternal energy If we brg a hot body to contact with a cold body energy starts to flow from the hot body to the cold body (the first law of thermodynamics does not rohibit a flow from cold to hot, the second law does) Work done

2 on a system, or done by a system is a force F actg over a ath s : W = F ds Please note: we take the ner roduct of force and ath (both are vectors); work (and energy) is a scalar Energy balance The general form of a balance equation for a quantity q was troduced Lecture Notes 3: dq = φq, φq, + Bq Dq (1) For q we now take the total energy: the sum of ternal, ketic and otential energy E = U + E + E t k Sce we consider the total energy a conserved quantity, the death and birth terms are absent the energy balance There are three ways the system exchanges energy with its surroundgs Energy enters and leaves the system because there is a mass flow enterg and leavg the system, that mass flow carries energy with it For stance: if a mass ilow φ m, has a secific ternal energy (ternal energy er unit mass) u this contributes to an energy ilux of φ u m, Heat enters or leaves the system Work is exerted on the system, or the system exerts work on its surroundgs With the above considerations the total energy balance becomes d m u + v + gz u + v + gz φm, u + v + gz + Q ɺ + W ɺ () Here Q ɺ and W ɺ are the net ( mus ) rates of heat and work (units Joule er second = Watt) Work In fluid mechanics, work is usually related to ressure forces actg on movg surfaces (like istons) Suose the system we are considerg the energy balance is under a ressure and exands its volume by an amount dv, the system exerts work on its surroundgs and therefore loses energy by an amount dv In an oen system, ie a system havg ilow and/or flow oengs, work enters and leaves the system with the ilow and flow This is called jection work It is quite an tricate subject but very relevant fluid mechanics; secifically if the ressure at the ilow is different from the ressure at the flow I will try to elucidate the concet with an examle In the De Nevers text book the concet is exlaed slightly differently Section 48 (ages ) Fd for yourselves which exlanation works best for you Suose we have a steady-state, constant density system with constant volume and with an ilow channel and an flow channel both havg the same cross sectional area A (see Figure 1) In steady state conservation of mass requires that φm, The ressure at the ilow and flow are and resectively Now consider the ilow and flow rocess as driven by istons a ste-by-ste rocess In the first ste, the iston at the ilow moves an amount dx durg a time to the right brgg an amount of mass Adx The

3 work done on the system as a result of the iston movg over dx is + Adx (force times dislacement) In the next ste, the same amount of mass enterg through the let leaves through the let, also by means of a movg iston Sce the let and let have the same cross sectional area, and the density is constant the iston has to move the same amount dx the same (ositive) x-direction It does so durg the same amount of time The work is now done on the surroundgs givg a negative sign to the work done on the system: Adx Figure 1 dx Now let us translate this back to the real system The mass flux is φm = A As a result, the rate of work done on the system at the let is dx the let is A = φm dx + A = φm, the rate of work done at We account for this the general energy balance (Eq ) by slittg the work rate term W ɺ to two arts: jection work, and non-jection (non flow, ) work = j + The jection work has been discussed above; its net ( ) contribution is (now with roof generalized to general systems) W φ φ ɺ j = m, m, (3) Substitutg this the general energy balance (Eq ) leads to: d m u + v + gz = φ u + v + gz + φ u + v + gz + + Qɺ + (4) m, m, A few remarks to connect the above to the De Nevers book and its notation Equation 4 the lecture notes is the same as Eq 411 (age 118) of the De Nevers book The differences are urely notation: Left-hand-side: De Nevers uses d; here we use d Right-hand-side: De Nevers uses dm and dm ; here we use φ,, φ, Right-hand-side: De Nevers uses v (secific volume); here we use 1 Right-hand-side: De Nevers uses dq and dw, here we use Qɺ and m m

4 Equation 418 (age 11 De Nevers book) looks more like our Eq 4, excet that Eq 418 the concet of enthaly h has been used: h u + This has been quite a theoretically oriented text In subsequent lecture notes and book chaters we will aly energy balances extensively to real (engeerg) roblems Here are already two examles Examle 1 We consider a hydro ower lant with water (constant density ) enterg the lant through a sgle let with a velocity of v 1 =3 m/s at a level z 1 =15 m at atmosheric ressure, and leavg the lant through a sgle let with velocity v =10 m/s at level z =0 m aga at atmosheric ressure What is the work done by the lant er kilogram water that asses through it? Neglect thermal effects For this we will use the energy balance as formulated Eq 4 As the volume we choose the ower lant between let and let We assume a steady (ie time deendent) situation Therefore the d term of Eq 4 is zero Neglectg thermal effects means that we do not consider ternal energy and heat Under these conditions Eq 4 reduces to v + gz + φm, v + gz + + A steady-state mass balance over the lant learns us that φm, As a result = v + gz + + v + gz + = v1 + v gz1 + gz = 1015 J/kg φm (ressure at let and let beg the same) is the work done on the system (ie the φm ower lant) er unit time, er unit mass flowg through the system the same amount of time The number beg negative means that the system does work on its surroundgs (which should be; a ower lant should deliver energy, not consume it) Please note that the unit J/kg is the same as m /s Examle Figure 3 Consider a um a water (constant density, = 1000 kg/m ) le, see Figure The ies connected to the um have diameter d=01 m At the ilow (left Figure 1) side the water has an average velocity of v=1 m/s The ower the um jects the flow is W ɺ =1 kw The whole thg is horizontal The whole thg is steady state Neglect heat effects (eg as a result of friction) First realize: Sce the density of the water does not change, and sce the ieles have constant diameter, the velocity at the flow side also is v=1 m/s

5 Set u a steady state energy balance over the um (ie with the um beg the control 1 1 volume): 0 v + gz + φm, v + gz + + The steady state is reresented by the zero the left-hand side of the equation; heat effects have been left by leavg the u (ternal energy) and Q ɺ terms From the mass balance it followed that φm, and that v = v Everythg is horizontal so that z = z The energy balance then can be written as =, and sce the density is constant φ m = = Pa=13 bar (1 bar is 10 5 Pa) φ m Please note: The (mechanical) energy serted by the um is not used to crease the ketic energy of the water (the velocity does not change due to the um), it is used to crease the ressure As we will see later the course, the ressure rovided by the um is needed to overcome friction the iele