# 3.8 The First Law of Thermodynamics and the Energy Equation

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1 CEE 3310 Control Volume Analysis, Sep 30, Review Conservation of angular momentum 1-D form ( r F )ext = [ˆ ] ( r v)d + ( r v) out ṁ out ( r v) in ṁ in t CV 3.8 The First Law of Thermodynamics and the Energy Equation The first law of thermodynamics tells us: Time rate of change of total system energy = Time rate of change by heat transfer + Time rate of change by work transfer or ( ) de = dt Q Ẇ sys where E is energy, Q is heat, and W is work. Recall that ( ) indicates time rate of change. Sign conventions: Q is the transfer by radiation, conduction, convection of heat. Transfer into the control volume is positive. W > 0 is work done by the system on the surroundings and W < 0 is work done on the system by the surroundings. Note that it is not uncommon to use the opposite sign convention for work, in which case the equation is written with the final terms as +Ẇ

2 66 (you may have seen it this way in a thermodynamics course). In CEE 3310 we will always define W>0 as work done by the system on the surroundings. We can decompose the rate of work into components: Ẇ = Ẇshaft + Ẇpressure + Ẇviscous stress where W shaft is the work done by a machine such as a pump, turbine, piston, generator, etc. Ẇ shaft = T shaft Ω, where T shaft is the torque and Ω is the angular velocity. Note that gravitational work will enter our energy budget through potential energy, which will show up linearly related to the height above a datum Stress Induced Work per Time (ower) Work occurs by applying a force over a distance. Therefore pressure (normal stress) and shear (viscous stress) can produce work. If we look at the rate of work, or work per unit time, we have power and we see that we can think of this as applying a force on a system with a given velocity. δẇ = δ F v Therefore for pressure we have δẇpres = n δa v = v n δa Therefore ˆ Ẇ pres = v n da CS This term is usually moved to the right-hand-side flux term of the energy equation as it is a flux, which is how we will treat it.

3 CEE 3310 Control Volume Analysis, Sep 30, For shear stress we have δẇvisc = τδa v = τ v δa Therefore ˆ Ẇ visc = τ v da CS But shear is internally self-canceling. On the control surface v =0at solid boundaries (no-slip boundary condition), if the control surface is normal to the flow then τ v and hence τ v =0. Hence it is often reasonable to assume that the shear stress induced work is small. What is a good environmental example of when this assumption breaks down? Let s consider the amount of work done! 3.9 The Energy Equation With our definitions of work and energy we are now ready to apply the Reynolds Transport Theorem to produce the conservation of energy equation. Let B = E and β = e = E/m, the energy per unit mass. We can decompose e into the following components e =û + v2 2 + gz + other where û is the internal energy per unit mass, v 2 /2 is the kinetic energy per unit mass, and gz is the potential energy per unit mass. We will ignore other sources of internal energy but from the definition they are easily included.

4 68 We can write the Reynolds Transport Theorem flux term (including the pressure work term) ˆ CS (û + v2 2 + gz + ) ( v n) da Therefore ( ) de = dt Q Ẇshaft = d ˆ sys dt CV ) ˆ (û + v2 2 + gz d + (û + v2 CS 2 + gz + ) ( v n) da The 1-D form is Q Ẇshaft = d dt CV ( û + v2 2 + gz ) d + outflows ( ) û + v2 + gz + ṁ ( ) û + v2 + gz + ṁ 2 2 inflows We sometimes will choose to write (û + /) =ĥ = enthalpy Forms of the Energy Equation We often find it convenient to cast the energy equation in alternate forms. Velocity Squared form: If the flow is at steady state then conservation of mass gives us that ṁ out = ṁ in = ṁ and we can normalized all of our quantities by ṁ which leaves our homogeneous equation with terms having units of velocity squared Q ṁ Ẇshaft ṁ =û out + v2 out 2 + gz out + out û in v2 in 2 gz in in Defining q = we can write the above as Q heat transfer = ṁ unit mass and w s = Ẇshaft ṁ shaft work = unit mass û out + v2 out 2 + gz out + out =û in + v2 in 2 + gz in + in + q w s

5 CEE 3310 Control Volume Analysis, Sep 30, Head form: Manometers have historically lead to a desire to think of energy in units of length, or head. We see that we can arrive at units of length by dividing our velocity squared form of the energy equation by gravity. Hence we have: û out g + v2 out 2g + z out + out γ = ûin g + v2 in 2g + z in + in γ + h q h s where h q and h s are q/g and w s /g, respectively the head forms of the heat transfer power and shaft power Incompressible 1-D Flow With No Shaft Work Rearranging our 1-D head form and setting w s =0we have: ( ) ( ) 2g + z = out 2g + z ûout û in q in g Now, defining û out û in q g = h loss = h f where we can think of h f as the friction losses and we see that h f > 0 (note that in adiabatic flow, say a perfectly insulated pipe, friction will heat the flow, therefore û out > û in and hence h f > 0). Thus we write ( ) ( ) 2g + z = out 2g + z h f in

6 Example Gas ipeline Consider the following pipe flow: If Q = 75 m 3 /s, the pipe radius is r =6cm, the inlet pressure is maintained at 24 atm by a pump, the outlet vents to the atmosphere, the pipe rises 150 m from inlet to outlet and the pipe length is 10 km, what is h f? What is the velocity head? h f =198 m Therefore the friction loss is greater than the z and the pump must drive against both! The velocity head is only 0.17 m! Note that the length did not come into our solution. h f includes the total losses along the pipe due to friction effects and hence includes the effect of length implicitly. We will see more about this a bit later in the course.

7 CEE 3310 Control Volume Analysis, Oct. 3, Review Work and ower The Energy Equation Q Ẇshaft = d ˆ dt CV ) ˆ (û + v2 2 + gz d + (û + v2 CS 2 + gz + ) ( v n) da where Q is the heat energy transfer rate, Ẇ shaft is the shaft power (work rate), û is the internal energy per unit mass, v 2 /2 is the kinetic energy per unit mass, and gz is the potential energy per unit mass Bernoulli Equation Bernoulli wrote down a verbal form of his famous equation in 1738 and Euler completed the analytic derivation in The differential form of the Bernoulli Equation is known as the Euler Equation. Consider our 1-D head form of the energy equation and let s apply it along a streamline of a flow. If the flow is steady then the integral over the control volume vanishes. Further, since by definition there is no flow normal to the streamline we only have flux terms at the starting and ending points along the streamline (really we are talking about a volume and hence a streamtube just a cylindrical volume element defined by a family of streamlines - a virtual pipe!). Clearly there is no shaft work along the streamline. If we assume the flow is frictionless (i.e., inviscid or ν =0) h f =0 and we have ( ) ( ) 2g + z = 2g + z out This is the Bernoulli equation. Clearly anywhere along a streamline, as long as no work is done between analysis points and the assumption of frictionless flow is good, we can write 2g + z = h 0 in

8 72 where the constant h 0 is referred to as the Bernoulli constant and varies across streamlines. The Bernoulli Equation can be derived by considering Newton s second law F = m a along a streamline (conservation of linear momentum). This leads to the steady form of the Bernoulli equation. If we add conservation of mass we can derive the unsteady form. Bernoulli Equation Assumptions Flow along single streamline different streamlines, different h 0. Steady flow (can be generalized to unsteady flow). Incompressible flow. Inviscid or frictionless flow, very restrictive! No w s between analysis points on streamline. No q between points on streamline Illustrations of Valid and Invalid Regions for the Application of the Bernoulli Equation

9 CEE 3310 Control Volume Analysis, Oct. 3, ressure form of Bernoulli Equation If we multiply our head form of the Bernoulli equation by the specific weight we arrive at the pressure form of the Bernoulli Equation: + v2 2 + γz = t where we call the first term the static pressure, the second the dynamic pressure, the third the hydrostatic pressure, and the right-hand-side the total pressure. Hence the Bernoulli Equation says that in inviscid flows the total pressure along a streamline is constant. If we remain at a constant elevation the above equation reduces to + v2 2 = s where we refer to s as the stagnation pressure. Thus by definition the stagnation pressure is the pressure along horizontal streamlines when the velocity is zero Stagnation oint and ressure Consider the flow around a circular cylinder: + v2 2 + γz = t We see that the stagnation pressure is simply the conversion of all kinetic energy to potential energy and hence there is a subsequent pressure rise. The elevation head simply accounts for any change in the potential energy due to vertical changes in elevation.

10 itot-static Tube The static and stagnation pressures can be measured simultaneously using a itot-static tube. Consider the following geometry: Now, we see the streamlines around the tube, either at the tip or away from the tip (but not around the curved front end), are horizontal. If the tube is not very long it is very reasonable to assume friction is negligible for this analysis. Now the velocity at the tip of the itot-static tube is zero hence the pressure at this point (and hence along the entire horizontal leg of the itot tube as this portion of the device is known) is the stagnation pressure. The holes perpendicular to the flow are similar to piezometers - they simply measure the static pressure in the fluid flow. If we write the equation for the itot tube we have 1 + v2 1 2 = 2 = γh Now, at the static tube we have the free-stream pressure, 2o but at this point, which is along the streamline from point 1 to point 2, the velocity is the same as it is for point 1 (assuming the itot-static tube is small and does not affect the flow) and there is no elevation change so the Bernoulli Equation gives us: 1 = 2o = γh Substituting this expression into the equation for the itot tube we arrive at the itot formula or in terms of heads V 1 = V = 2g(H h)

11 CEE 3310 Control Volume Analysis, Oct. 3, Example Flow accelerating out of a reservoir 2gh V 2 = 1 ( ) 2 A2 A and if A 1 A 2 : 1 ( A2 A 1 ) 2 1 V 2 2gh, again! Let s look at this a bit further by asking the question what speed will a parcel of fluid dropped a distance h be traveling at? Therefore t = v = ˆ t 0 2h g g dt = gt h = ˆ t and v = gt v = g 0 v dt = ˆ t 0 gt dt = 1 2 gt2 2h g = 2gh Thus we see that in an inviscid flow, which by definition has no frictional energy losses, we simply convert potential energy to kinetic energy and hence the same result, v = 2gh keeps showing up. This was first noted by Torricelli. aha!

12 Irrotational Flow and Bernoulli Consider h 4 γ γ 2g + z = h 0 0+ v2 2g + 0 = h 0 = v2 2g h 2 γ γ + v2 2g h 2 = h 0 h 3 γ γ + v2 2g h 3 = h 0 v 2 + (v/2)2 h 4 = 1 2g 4 2g h 0 h 5 γ γ +0 h 5 = 0 h 0 Notice that in the unsheared regions (uniform flow) h 0 = a constant across streamlines while where shear exists (e.g., shear is non-zero), h 0 varies across the streamlines. More strictly speaking we actually want to know if the flow is rotational. Our test is if we stick a small neutrally buoyant + shaped probe in the flow and see if it will rotate. In uniform flow it will not, in a linear shear, like the shear profile shown here, it will. Hence we say that h 0 is constant in irrotational (non rotational) flows. This allows us to connect Bernoulli points that are not on the same streamline in flows that are irrotational, further expanding the power of the Bernoulli equation but also the opportunities to misuse it!

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