Continuity Equation

8 Key excerpts on "Continuity Equation"

• 

  eBook - PDF

  Engineering Fluid Mechanics

  • Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Robertson(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  154 The Control Volume Approach and The Continuity Equation CHAPTER ROAD MAP This chapter describes how conservation of mass can be applied to a flowing fluid. The resulting equation is called the Continuity Equation. The Continuity Equation is applied to a spatial region called a control volume, which is also introduced. CHAPTERFIVE FIGURE 5.1 The photo shows an evacuated-tube solar collector that is being tested to measure its efficiency. This project was run by undergraduate engineering students. The team applied the control volume concept, the Continuity Equation, the flow rate equations, and knowledge from thermodynamics and heat transfer. (Photo by Donald Elger.) LEARNING OUTCOMES FLOW RATE (§5.1). ● Know the main ideas about mass and volume flow rate. ● Define mean velocity and know typical values. ● Solve problems that involve the flow rate equations. THE CONTROL VOLUME APPROACH (§5.2). ● Describe the six types of systems. ● Distinguish between intensive and extensive properties. ● Explain how to use the dot product to characterize net outflow. ● Know the main ideas of the Reynolds Transport Theorem. THE Continuity Equation (§5.3, §5.4). ● Know the main ideas about the Continuity Equation. ● Solve problems that involve the Continuity Equation. CAVITATION (§5.7). ● Know the main ideas about cavitation—for example, why cavitation happens, why cavitation matters, how to spot potential cavitation sites, and how to design to reduce the possibility of cavitation. 5.1 Characterizing the Rate of Flow Engineers characterize the rate of flow using the (a) mass flow rate, m · , and (b) the volume flow rate, Q. These concepts and the associated equations are introduced in this section. Volume Flow Rate (Discharge) Volume flow rate Q is the ratio of volume to time at an instant in time. In equation form, Q = ( volume of fluid passing through a cross-sectional area interval of time ) instant in time = lim Δt →0 ΔV Δt (5.1) EXAMPLE.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Physical Principles of Chemical Engineering

  International Series of Monographs in Chemical Engineering

  • Peter Grassmann, H. Sawistowski(Authors)
  • 2013(Publication Date)
  • Pergamon

    (Publisher)

  This makes it possible to represent the flow of a liquid by its thermal or electrical analogue (§ 8.16). The electrical voltage differences, which can be measured accurately by voltmeters, corres-pond to the pressure differences found on flow of liquids and the temperature differences on flow of heat. But the inertia which plays a critical part in turbu-lent flow cannot be represented directly by an electrical or thermal analogy (cf. § 6.4). The analogy between momentum, heat and mass transfer found in the differ-ential equations (3.13.1, 3.13.4 and 3.13.7) describing these three processes, is very widely used. We will deal with this in detail in Chapter 9. § 6.2. The Continuity Equation for Flow in Pipes In § 1.1 we applied the law of conservation of mass or of atoms to a balance region and thus obtained the balance equation (1.1.1). This states that the increase in the content of the balance region equals the difference between the quantities flowing in and flowing out. In this form it is applicable to all pro-perties for which a law of conservation is valid. But in theoretical fluid mecha-nics it is not the flows themselves but the velocities which are given directly. It is an advantage to calculate on the basis of these because they approach a finite limiting value when the balance region is reduced to a volume element. This limit transition has been found very fruitful in the whole field of hydro-and aerodynamics. But first we will apply the law of conservation of mass to a channel with solid walls. Through an element of surface dS oriented perpendicular to the velocity v the volume dV* = vdS enters per unit time. If the normal to the element of surface forms the angle oc with the direction of the velocity, then dV* = v cos oc dS = v n dS. (6.2.1) Here v n = v cos oc is the component of the vector v in the direction of the normal to the surface. The mass entering per unit time is given by dM* = ρ dV* = ρν cos oc dS = ρν η dS.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Engineering Fluid Mechanics

  • Donald F. Elger, Barbara A. LeBret, Clayton T. Crowe, John A. Roberson(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  If there are more than two ports and the accumulation term is zero, then Eq. (5.29) can be reduced to ∑ cs m ˙ i = ∑ cs m ˙ o (5.37) If the flow is assumed to have constant density, Eq. (5.37) can be written in terms of discharge: ∑ cs Q i = ∑ cs Q o (5.38) Summary. Depending on the assumptions of the problem, there are many ways to write the Continuity Equation. However, one can analyze any problem using the three equations summarized in Table 5.5. Thus, we recommend starting with one of these three equations because this is simpler than remembering many different equations. Example 5.8 shows how to apply continuity to flow in a pipe. *The mass accumulation term in a conduit can be nonzero for some unsteady flow problems, but this is rare. This topic is left to advanced textbooks. Control surface V 2 2 1 V 1 FIGURE 5.12 Flow through a conduit. The Continuity Equation (Application) 193 193 Example 5.9 shows how the Continuity Equation can be applied together with the Bernoulli equation. EXAMPLE 5.8 Applying the Continuity Equation to Flow in a Variable Area Pipe Problem Statement A 120 cm pipe is in series with a 60 cm pipe. The speed of the water in the 120 cm pipe is 2 m/s. What is the water speed in the 60 cm pipe? Define the Situation Water flows through a contraction in a pipe. State the Goal V 2 (m/s) mean velocity at section 2 Generate Ideas and Make a Plan Selection: Select the Continuity Equation because the problem variables are velocity and pipe diameter. Sketching: Select a fixed CV. Sketch this CV on the situation diagram. Label the inlet as section 1 and outlet as section 2. Analysis: Select the pipe flow form of continuity (i.e., Eq. 5.33) because the problem involves flow in a pipe:  A 1 V 1 =  A 2 V 2 (a) Assume density is constant (this is standard practice for steady flow of a liquid). The Continuity Equation reduces to A 1 V 1 = A 2 V 2 (b) Validate: To validate Eq. (b), notice that the primary dimensions of each term are L 3 /T.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Computational Hydraulics

  • C. A. Brebbia, A. J. Ferrante(Authors)
  • 2013(Publication Date)
  • Butterworth-Heinemann

    (Publisher)

  A stream tube with an infinitesimally small cross-section is called a stream filament Figure 2.4 Stream tube 2.3 Continuity Equation The idea of the stream tube allows one to formulate the Continuity Equation in a simple way. Considering the stream tube of Figure 2.4, and assuming the 18 Basic concepts of fluids in movement fluid to be incompressible, conservation of mass requires that the quantity of fluid which flows in at one end flows out at the other. This condition can be written as (2.2) where dF i and dr 2 are the surface differentials, and v t and v 2 are the velocities, at ends 1 and 2 respectively. Integration of this equation gives the total flux: (2.3) where V x and V 2 are the mean velocities, and r and T 2 are the areas, at ends 1 and 2, respectively. Then, for a stream tube, the Continuity Equation means that Q=Vr= constant (2.4) for any section T, having a mean velocity V. According to the previous expression the velocity is inversely proportional to the area of the section under consideration. Thus, streamlines tend to approach each other more closely in areas of larger velocity. 2.4 Bernouilli's equation Figure 2.5 shows a stream filament. The length of a differential element of the stream filament is ds, and the area of the cross-section is dT. The weight of that differential element will be y dsdT, where y is the specific weight of the fluid. The pressure acting on the left-hand side area is equal to p, and on the right-Figure 2.5 Forces acting on a stream filament Bernouillfs equation 19 hand side area is equal to p + (dp/ds)ds. Making use of Newton's law, which states that force is the product of mass and acceleration, the condition for equilibrium in the direction of s is expressed by (2.5) which reduces to (2.6) where p is the fluid density and t is time. From Figure 2.5 it follows that (2.7) where y is the vertical coordinate.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Computational Transport Phenomena for Engineering Analyses

  • Richard C. Farmer, Ralph W. Pike, Gary C. Cheng, Yen-Sen Chen(Authors)
  • 2009(Publication Date)
  • CRC Press

    (Publisher)

  Tis equation shows that the increase in mass flux of component i with distance, z , is from the formation of component i by chemical reaction, r i . For steady flow the overall Continuity Equation (Equation 2.176) is the same as that obtained previously, neglecting changes in gas flow rate due to reaction with the solid. Te overall Continuity Equation is the sum over the species equations. It can vary with position if there are chemical reac-tions between the gas and the char. To predict the pressure distribution, the procedure is the same as that discussed above. Te mass flux of gas, W , changes due to reaction with the solid phase, and it is included under the integral in Equation 2.180 as shown in d ⎛ d T ⎞ n 0 = ⎜ κ s ⎟ − ∑ H R xi i (2.191) d z ⎝ d z ⎠ i k = where ∑ k H R xi i i = 1 represents the heat absorbed or released by reactions occurring with the solid. Multiplying Equation 2.190 by the porosity, ε , Equation 2.191 by (1 − e ) and adding, the result is an equation that describes the energy transfer at any cross section in the z -direction in the solid and gas d T d ⎛ d T ⎞ n W C g ε p = ⎜ κ ⎟ − H R (2.192) d z d z ⎝ e d z ⎠ ∑ xi i i = 1 The Equations of Change ■ 101 where ∑ k xi i i = 1 H R represents the heat absorbed or released by chemical reactions. Te simplifications omitted potential energy changes, viscous dissipation, and internal heat generation. Heat transferred by diffusion in the direction of flow is very small compared to heat transfer by convection and is not included. For the solid matrix, there is no flow and the energy equation is where the overall Continuity Equation was used along with defining an effective thermal conductivity k = [ ek + (1 − e ) k s ]. Te term ∑ n e g i = 1 H R xi i represents the heat absorbed or released by chemical reactions in the gas phase, between the gas and solid, and in the solid phase.

  Sign up to read

  Learn more about book
• 

  eBook - ePub

  Turbulent Fluid Motion

  • R. Deissler(Author)
  • 2020(Publication Date)
  • CRC Press

    (Publisher)

  Fig. 3-1 . It thus appears that a continuum theory of turbulence that ignores differences among the molecular structures of various fluids is realistically based.

  3-2 EQUATION OF CONTINUITY (CONSERVATION OF MASS)

  The rate of mass flow per unit area in the x1 direction is

  m 1

  = ρ

  u 1

  ,

  (3-1)

  where ρ is the density and u1 is the velocity component in the x1 direction. In the xi direction this becomes

  m i

  = ρ

  u i

  .

  (3-2)

  If mi is measured at xi then at xi + Δxi , mi is replaced by

  m i

  +

  ∂

  m i

  ∂

  x

  ( i )

  Δ

  x

  ( i )

  + ⋯ ( no   sum   on   i ) ,

  where mi has been expanded in a Taylor series. For small Δxi only the first two terms need be retained. Then the change in mi in going from xi to xi + Δxi is (∂mi /∂x

  (i)

  ))Δx

  (i)

  ).

  1 There is an interesting analogy between the theory of liquids and the theory of strong turbulence. In both theories interactions among many modes or degrees of freedom must be considered.

  Figure 3-1

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Munson, Young and Okiishi's Fundamentals of Fluid Mechanics

  • Andrew L. Gerhart, John I. Hochstein, Philip M. Gerhart(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  The relationship between velocity and diameter is shown in Fig. E5.1. As expected, the velocity is inversely proportional to the square of the diameter. 5.1 Conservation of Mass—The Continuity Equation 151 EXAMPLE 5.2 Conservation of Mass—Steady, Compressible Flow Given Air flows steadily between two sections in a long, straight portion of 4-in.-inside-diameter pipe, as indicated in Fig. E5.2. The uniformly distributed temperature and pressure at each section are given. The average air velocity (nonuniform velocity distribution) at section (2) is 1000 ft/s. As mentioned, there is no written standard for air exchange rates, but changing the air every 3 minutes has been accepted by the industry. It is important for engineers to understand that individual industries and companies will have their own established safety precautions, depending on the nature of their work. As an engineer, one must heed these precautions. Safety is always a key component in the design and operation of engi- neering systems. Outlet diameter (in.) V out (ft/s) 6 8 10 12 4 2 0 25 0 5 10 15 20 (4 in., 22.9 ft/s) FIGURE E5.1 Control volume Flow Section (1) p 1 = 100 psia T 1 = 540 °R p 2 = 18.4 psia T 2 = 453 °R V 2 = 1000 ft/s D 1 = D 2 = 4 in. Section (2) Pipe FIGURE E5.2 Find Calculate the average air velocity at section (1). Solution The average fluid velocity at any section is the veloc- ity that yields the section mass flowrate when multiplied by the section average fluid density and section area (Eq. 5.7). We relate the flows at sections (1) and (2) with the control volume desig- nated with a dashed line in Fig. E5.2. Equation 5.5 is applied to the contents of this control vol- ume to obtain 0 (flow is steady) ∂ _ ∂t   cv ρd V+  cs ρ V · n ̂ dA = 0 The time rate of change of the mass of the contents of this con- trol volume is zero because the flow is steady. The control surface integral involves mass flowrates at sections (1) and (2), so that from Eq.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Munson, Young and Okiishi's Fundamentals of Fluid Mechanics

  • Philip M. Gerhart, Andrew L. Gerhart, John I. Hochstein(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  Section (2) Pipe ■ Figure E5.2 5.1 Conservation of Mass—The Continuity Equation 201 ∫ cs ρ V ∙ n ˆ dA = m ˙ 2 − m ˙ 1 = 0 or m ˙ 1 = m ˙ 2 (1) and from Eqs. 1, 5.6, and 5.7 we obtain ρ 1 A 1 V 1 = ρ 2 A 2 V 2 (2) or since A 1 = A 2 V 1 = ρ 2 ρ 1 V 2 (3) 202 Chapter 5 ■ ■ Finite Control Volume Analysis Air at the pressures and temperatures involved in this example problem behaves like an ideal gas. The ideal gas equation of state (Eq. 1.8) is ρ = p RT (4) Thus, combining Eqs. 3 and 4 we obtain V 1 = p 2 T 1 V 2 p 1 T 2 = (18.4 psia)(540 °R)(1000 ft/s) (100 psia)(453 °R) = 219 ft/s (Ans) COMMENT We learn from this example that the Continuity Equation (Eq. 5.5) is valid for compressible as well as incompress-ible flows. Also, nonuniform velocity distributions can be handled with the average velocity concept. Significant average velocity changes can occur in pipe flow if the fluid is compressible. EXAMPLE 5.3 Conservation of Mass—Two Fluids GIVEN The inner workings of a dehumidifier are shown in Fig. E5.3 a . Moist air (a mixture of dry air and water vapor) enters the dehumidifier at the rate of 600 lbm/hr. Liquid water drains SOLUTION The unknown mass flowrate at section (2) is linked with the known flowrates at sections (1) and (3) with the control volume designated with a dashed line in Fig. E5.3 b . The contents of the control volume are the air and water vapor mixture and the con-densate (liquid water) in the dehumidifier at any instant. out of the dehumidifier at a rate of 3.0 lbm/hr. A simplified sketch of the process is provided in Fig. E5.3 b . FIND Determine the mass flowrate of the dry air and the water vapor leaving the dehumidifier. Not included in the control volume are the fan and its motor, and the condenser coils and refrigerant. Even though the flow in the vicinity of the fan blade is unsteady, it is unsteady in a cycli-cal way.

  Sign up to read

  Learn more about book