3.8 The Steady-State Energy Equation When Eq. (3.3.6) is applied to steady how through a control volume similar to Fig. 3.15, the volume integral drops out and it becomes Since the flow is steady in this equation, it is convenient to divide through by the mass per second flowing through the system ρ 1 A 1 v 1 = ρ 2 A 2 v 2  (3.8.1) q H is the heat added per unit mass of fluid flowing, and w s is the shaft work per unit mass of fluid flowing. This is the energy equation for steady flow through a control volume.

Figure 3.15 Figure 3.15 Control volume with flow across control surface normal to surface

Example 3.7 The cooling-water plant for a large building is located on a small lake fed by a stream, as shown in Fig. 3.16a. The design low-stream flow is 150 L/s, and at this condition the only outflow from the lake is 150 L/s via a gated structure near the discharge channel for the cooling-water system. Temperature or the incoming stream is 27°C. The flow rate of the cooling system is 300 L/s, and the building's heat exchanger raises the cooling-water temperature by 5°C. What is the temperature of the cooling water recirculated through the lake, neglecting heat losses to the armosphere and lake bottom, if these conditions exist for a prolonged period? Solution The control volume is shown in Fig. 3.16b with the variables volumetric flow rate Q and temperature T. There is no change in pressure, density, velocity, or elevation from section 1 to 2. Equation (3.3.6) applied to the control volume is in which δQ H /δt is the time rate of heat addition by the heat exchanger.

The intrinsic energy per unit mass at constant pressure and density is a function of temperature only; it is u 2 – n 1 = c(T 2 – T 1 ), in which c is the specific heat or heat capacity of water. Hence, the energy equation applied to the control volume is Similarly, the heat added in the heat exchanger is given by in which ΔT = 10 is the temperature rise and Q e = 10 cft is the volumetric flow rate through the heat exchanger. Thus Since T 2 = T + ΔT, the lake temperature T is 32 °C.

Figure 3.16 Figure 3.16 Cooling-water system

The energy equation (3.8.1) in differential form, for flow through a stream tube (Fig. 3.17) with no shaft work, is ( ) For frictionless flow the sum of the first three terms equals zero from the Euler equation (3.5.8); the last three terms are one form of the first law of thermodynamics for a system, (3.8.4) Now, for reversible flow, entropy s per unit mass is defined by (3.8.5) in which T is the absolute temperature. Since Eq. (3.8.4) is for a frictionless fluid (reversible), dq H can be eliminated from Eqs. (3.8.4) and (3.8.5), (3.8.6) which is a very important thermodynamic relation; one form of the second law of thermodynamics. Although it was derived for a reversible process, since all terms are thermodynamic properties, it must also hold for irreversible-flow cases as well.

Figure 3.17 Figure 3.17 Steady-stream tube as control volume

3.9 Interrelations Between Euler's Equation and the Thermodynamic Relations The first law in differential form, from Eq. (3.8.3), with shaft work included, is (3.9.1) Substituting for du + p d(1/ρ) in Eq. (3.8.6) gives The Clausius inequality states that (3.9.3) Thus T ds – dq H ≥ 0. The equals sign applies to a reversible process

If the quantity called losses of irreversibilities is identified as d (losses) ≡ T ds – dq H (3.9.4) d (losses) is positive in irreversible flow, is zero in reversible flow, and can never be negative. Substituting Eq. (3.9.4) into Eq (3.9.2) yields (3.9.5) This is a most important form of the energy equation. In general, the losses must be determined by experimentation. It implies that some of the available energy is converted into intrinsic energy during all irreversible process. This equation, in the absence of the shall work, differs from Euler's equation by the loss term only. In integrated form, (3.9.6) If work is done on the fluid in the control volume, as with a pump, then w s is negative. Section 1 is upstream, and section 2 is downstream

3.10 Application Of The Energy Equation To Steady Fluid-flow Situations For an incompressible fluid Eq (3.9.6) may be simplified to (3.10.1) in which each term now is energy in metre-newtons per newton, including the loss term. The work term has been omitted but may be inserted if needed.

Kinetic-Energy Correction Factor In dealing with flow situations in open- or closed-channel flow, the so-called one-dimensional form of analysis is frequently used The whole flow is considered to be one large stream tube with average velocity V at each cross section. The kinetic energy per unit mass given by V 2 /2, however, is not the average of v 2 /2 taken over the cross section. It is necessary to compute a correction factor α for V 2 /2, so that αV 2 /2 is the he average kinetic energy per unit mass passing the section.

Fig shows the kinetic energy passing the cross section per unit time is in which ρv δA is the mass per unit time passing δA and v 2 /2ρ is the kinetic energy per unit mass. By equating this to the kinetic energy per unit time passing the section, in terms of αV 2 /2, By solving for α, the kinetic-energy correction factor, (3.10.2) The energy equation (3.10.1) becomes (3.10.3) For laminar flow in a pipe, α=2. For turbulent flow in a pipe, α varies from about 1.01 to 1.10 and is usually neglected except for precise work.

Figure 3.18 Figure 3.18 Velocity distribution and average velocity

All the terms in the energy equation (3.10.1) except the term losses are available energy. For real fluids flowing through a system, the available energy decreases in the downstream direction; It is available to do work, as in passing through a water turbine. A plot showing the available energy along a stream tube portrays the energy grade line. A plot of the two terms z + p/γ along a stream tube portrays the piezometric head, or hydraulic grade line. The energy grade line always slopes downward in real-fluid flow, except at a pump or other source of energy. Reductions in energy grade line are also referred to as head losses.

Example 3.8 The velocity distribution in turbulent flow in a pipe is given approximately by Prandtl's one-seventh-power law. with y the distance from the pipe wall and ro the pipe radius. Find the kinetic-energy correction factor. Solution The average velocity V is expressed by In which r = r 0 – y.

By substituting for r and v, or By substituting into Eq. (3.10.2)

3.11 Applications Of The Linear-momentum Equation Newton's second law, the equation of motion, was developed into the linear-momentum equation in Sec. 3.3, This vector can be applied for any component, say the x direction, reducing to (3.11.2)

In Fig with the control surface as shown and steady flow, the resultant force acting on the control volume is given by Eq. (3.11.2) as as mass per second entering and leaving is ρQ = ρ 1 Q 1 = ρ 2 Q 2. When the velocity varies over a plane cross section of the control surface, by introduction of a momentum correction factor β, the average velocity may be utilized, (3.11.3) in which β is dimensionless. Solving for β yields (3.11.4) In applying Eq. ( ) care should be taken to define the control volume and the forces acting on it clearly.

Figure 3.21 Figure 3.21 Control volume with uniform inflow and outflow normal to control surface

Example 3.9 The horizontal pipe of Fig is filled with water for the distance x. A jet of constant velocity V 1 impinges against the filled portion. Fluid frictional force on the pipe wall is given by τ 0 πDx, with τ 0 = ρfV 2 2 /8 [see Eq. (5.8.2)]. Determine the equations to analyze this flow condition when initial conditions are known; that is, t = 0, x = x 0, V 2 = V 20. Specifically, for V 1 = 20 m/s, D 1 = 60 mm, V 20 = 500 mm/s, D 2 = 250 mm, x 0 = 100 m, ρ = 1000 kg/m 3, and f = 0.02, find the rate of change of V 2 and x with time. Solution The continuity and momentum equations are used to analyze this unsteady-flow problem. Take as control volume the inside surface of the pipe, with the two end sections 1 m apart, as shown. The continuity equation

Becomes, using After simplifying, The momentum equation for the horizontal direction x, Becomes Which simplifies to

As t is the only independent variable, the partials may be replaced by totals. The continuity equation is By expanding the momentum equation, and substituting for dx/dt, it becomes These two equations, being nonlinear, can be solved simultaneously by numerical methods (such as Runge-Kutta methods described in Appendix B), when initial conditions are known. The rate of change of x and V 2 can be found directly from the equations for the specific problem

Figure 3.23 Figure 3.23 Jet impact on pipe flowing full over partial length

Example 3.10 In Fig 3.26 a fluid jet impinges on a body as shown; the momentum per second of each of the jets is given by M and is the vector located at the center of the jets. By vector addition find the resultant force needed to hold the body at rest. Solution The vector form of the linear-momentum equation (3.11.1) is to be applied to a control volume comprising the fluid bounded by the body and the three dotted cross sections. As the problem is steady, Eq (3.11.1) reduces to By taking M 1 and M 0 first, the vector M 1 – M 0 is the net momentum efflux for these two vectors, shown graphically on their lines of action. The resultant of these two vectors is then added to momentum efflux M 2, along its line of action, to obtain R. R is the momentum efflux over the control surface and is just equal to the force that must be exerted on the control surface. The same force must then be exerted on the body to resist the control-volume force on it.

Figure 3.26 Figure 3.26 Solution of linear-momentum problem by addition of vectors

The Momentum Theory for Propellers The action of a propeller is to change the momentum of the fluid within which it is submerged and thus to develop a thrust that is used for propulsion. Propellers cannot be designed according to the momentum theory, although some of the relations governing them are made evident by its application. Fig shows a propeller, with its slipstream and velocity distributions at two sections a fixed distance from it. The propeller may be either (1) stationary in a flow as indicated; (2) moving to the left with velocity V 1 through a stationary fluid, since the relative picture is the same. The fluid is assumed to be frictionless and incompressible.

Figure 3.28 Figure 3.28 Propeller in a fluid stream

The flow is undisturbed at section 1 upstream from the propeller and is accelerated as it approaches the propeller, owing to the reduced pressure on its upstream side. In passing through the propeller, the fluid gas its pressure increased, which further accelerates the flow and reduces the cross section at 4. The velocity V does not change across the propeller, from 2 to 3. The pressure at 1 and 4 is that of the undisturbed fluid, which is also the pressure along the slipstream boundary.

When the momentum equation (3.11.2) is applied to the control volume within sections 1 and 4 and the slipstream boundary of Fig. 3.28, the force F exerted by the propeller is the only external force acting in the axial direction, since the pressure is everywhere the same on the control surface. Therefore, (3.11.5) in which A is the area swept over by the propeller blades. The propeller thrust must be equal and opposite to the force on the fluid. After substituting and simplifying, (3.11.6) When Bernoulli's equation is written for the stream between sections 1 and 2 and between sections 3 and 4. since z 1 = z 2 = z 3 = z 4. In solving for p 3 - p 2, with p 1 = p 4, (3.11.7) Eliminating p 3 – p 2 in Eqs. ( ) gives (3.11.8)

The useful work per unit time done by a propeller moving through still fluid (power transferred) Is the product of propeller thrust and velocity. i.e., (3.11.9) The power input is that required to increase the velocity of fluid from V 1 to V 4. Since Q is the volumetric flow rate, ( ) Power input may also be expressed as the useful work (power output) plus the kinetic energy per unit time remaining in the slipstream (power loss) ( ) The theoretical mechanical efficiency e t is given by the ratio of Eqs (3.11.9) and ( ) or ( ) ( ) If ∆V = V 4 - V 1 is the increase in slipstream velocity, substituting into Eq. ( ) produces ( )

Example 3.11 An airplane traveling 400 km/h through still air, γ = 12 N/m 3, discharges 1000 m 3 /s through its two 2.25-m-diameter propellers. Determine (a) the theoretical efficiency, (b) the thrust, (c) the pressure difference across the propellers, and (d) the theoretical power required. Solution (a) From Eq. ( ) (b) From Eq. (3.11.8)

The thrust from the propellers is, from Eq. (3.11.5), (c) The pressure difference, from Eq. (3.11.6), is (d) The theoretical power is