A U-tube manometer is connected to the tank as shown in Fig. Determine the 8m weight of the cylinder. Note that the top of the cylinder is flush with the fluid surface.
A Figure P When such a gage is attached to the closed water tank of Fig. What is the ab- B Water 3 in. Assume standard atmospheric pressure of Determine the pressure at point 1 inside the closed tube.
Oil density 12 in. Figure P Determine the weight of the sandbag needed to keep the bottomless forms from lifting The liquid in the top part of the pip- ing system has a specific gravity of 0.
If the pressure gage read- 8 in. Hemispherical dome 10 in. Determine the minimum horizon- A tal force between the dam and the foundation required to keep 3m the dam from sliding at the water depth shown. Assume no fluid uplift pressure along the base.
Base your analysis on a Water unit length of the dam. The fluid in both pipes A and B is water, and the gage fluid in the manometer has a specific grav- ity of 2. What is the pressure in pipe B corresponding to the differential reading shown? The length of the stem Air protruding above the fluid surface of the liquid in which the Test tube hydrometer floats is a function of the specific gravity of the liquid.
If the mass of the hydrometer is 0. Water Plastic bottle Fluid Figure P For a liquid depth of 10 ft. The width of the wall is 6 ft. Show the magnitude, direction, and location of the force on a sketch. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause the test tube to sink to the bot- A tom of the bottle.
Explain this phenomenon. In this chapter we consider three equations that are mathematical representations of these laws—the momentum equation, the Bernoulli equation, and the mechanical energy equation. These equations deal with a flowing fluid, unlike the equations in the previous chapter that involved stationary fluids.
Thus, the objective of this chapter is to show the chapter objective use of these equations in thermal systems engineering.
These con- cepts include body and surface forces, viscosity, and the incompressible flow model. The only body force we body force consider is that associated with the action of gravity. We experience this body force as weight, w. For Example Similarly, the weight of a fluid may affect its motion. Surface forces can be written in terms of components normal and tangential to the surface.
As in- pA troduced in Sec. Although it Area A is possible to generate a shear stress in a stationary solid e. In fact, the definition of a fluid is that it is a material in which the application of fluid definition any shear stress no matter how small will cause motion.
The shear force concept is studied in the next section together with the property viscos- ity. We conclude the present discussion of surface forces by considering the role that gage pressure can play when evaluating the forces acting on a control volume. Figure Surface forces acting on the control volume include the force p1A1 acting in the direction of flow and the oppositely directed force p2A2.
By rearranging terms, Eq. Accordingly, in this section we provide background material required by these discussions. The character of the shear stress developed depends on the specific flow situation. Con- sider a flat plate of area A located a distance b above a fixed parallel plate and the gap between the plates filled with a viscous fluid.
As shown in Fig. Experimental observations show that the fluid condition sticks to both plates so that the fluid velocity is zero on the bottom plate and U on the top plate. Such fluids are called viscosity Newtonian fluids. For the flow shown in Fig. Thus, from Eq. Viscosity is a property. From Eq. In accordance with Eq. The actual value of the viscosity depends on the particular fluid, and for a particular fluid the viscosity is also highly dependent on temperature, as illustrated in Fig.
Values of viscosity for several common gases and liquids are listed in the tables of Appendix FM Values of kinematic viscos- ity for some common liquids and gases are given in Appendix FM In some applications, fluids are considered to be inviscid. That is, the fluid is considered inviscid to have zero viscosity. Shear stresses cannot play a role in such applications. Accordingly, to simplify evaluations involving the flow of liquids, the density is often taken as constant.
When this assumption is made, the flow is called an incompressible flow. The Momentum and Mechanical Energy Equations As we have seen in previous sections, the density of air, and other gases, can vary significantly. Still, flowing air can often be modeled as incompressible provided the ve- locity of the air is not too great and the temperature is nearly constant.
At higher velocities, density change becomes important, and then the compressible flow principles of Secs. In some discussions that follow, a flow may be modeled as both incompressible and steady. This usage is consistent with the steady state concept discussed in Sec. Consider flow through the control volume shown in Fig.
For simplicity, we assume that the control volume has one inlet, 1 , one outlet, 2 and that the flow is one-dimensional Sec. As discussed in Chaps. Similarly, the flow also transfers momentum into or out of the control volume.
In this expression, the momentum per unit of mass flowing across the boundary of the control volume is given by the velocity vector V. In accordance with the one-dimensional flow model, the vector is normal to the inlet or exit and oriented in the direction of flow.
In this text, components of vectors are resolved along rectangular coordinates. Thus, the x-, y-, and z- components of F are denoted Fx, Fy, and Fz, respectively. The components of the velocity vector V are denoted u, v, and w, respectively.
The applications have been selected to bring out important aspects of the momentum concept. Solution Known: The direction of a jet of water is changed by a vane.
Find: Determine the force needed to hold the vane stationary as a function of. The control volume shown on the accompanying figure is at steady state.
At the inlet and outlet of the control volume, sections 1 and 2 , the flow is one-dimensional and each cross-sectional area is 0. The pressure is atmospheric on the entire control surface. The flow occurs in the horizontal x-y plane. Analysis: a We select a control volume that includes the vane and a portion of the water see Fig. The x- and y-components of Eq.
Since the pressure is atmospheric on the entire control volume surface, the net pressure force on the control volume surface is zero. Thus, the only forces applied to the control volume contents are the x- and y-components of the anchoring force, FAx and FAy, respectively.
Although FAx and FAy are shown on the schematic as acting in the positive x- and y-directions, their magnitudes and directions will be determined as a part of the analysis.
We begin by evaluating the velocity components required in Eqs. Thus, Eqs. The plus sign in the expression for FAy indicates that this component is exerted in the positive y-direction. Al- though this product has the same magnitude at locations 1 and 2 , namely A change of direction requires a force, the components of which are FAx and FAy.
Thus, as shown in Fig. The pressure is uniform around the entire control surface and, therefore, provides no contribution to the force. In the next example the pressure is not uniform and is a factor in determining the anchoring force. The flow cross-sectional area is constant at a value of 0. The gage pres- sures at the entrance and exit of the bend are 30 psi and 24 psi, respectively. Calculate the horizontal x and y components of the anchoring force required to hold the bend in place.
Solution Known: Water flows under given conditions in a horizontal, pipe bend. Find: Determine the x- and y-components of the force needed to hold the bend in place. At 1 and 2 the flow is one-dimensional and each cross-sectional area is 0. The pressure is atmospheric on the outside of the pipe bend.
Analysis: Since we want to evaluate components of the anchoring force needed to hold the pipe bend in place, an appro- priate control volume see dashed line in Fig.
Note that the weight of the water is vertical in the negative z direction and does not contribute to the x and y components of the anchoring force.
Although FAx and FAy are shown as acting in the positive x- and y-directions, respectively, their magnitudes and directions will be determined as a part of the analysis. Since atmospheric pressure acts uniformly over the outside of the pipe bend, the effect of atmospheric pressure in the x direction cancels.
Accordingly, Eq. Accordingly, when Eq. Pressure is a compressive stress. Although this product has the same magnitude at locations 1 and 2 , the direction is different. Accordingly, as shown by the factor 2 in Eq.
The anchor- ing force must also oppose the net pressure force acting on the control volume. These forces and rates of momentum trans- fer are shown on Fig. The change in direction requires an anchoring force whether or not pressure plays an explicit role. In the next example, an anchoring force is evaluated for a case where the direction of the flow is unchanged, but the magnitude of the velocity increases in the flow direction.
Pressure and weight are also important. The nozzle weight is 1 N and the weight of the water in the nozzle at any instant is 0. The nozzle inlet and exit diameters are 16 mm and 5 mm, respectively. The nozzle axis is vertical and the gage pressures at sections 1 and 2 are kPa and 0, respectively. Solution Known: Water flows vertically at a known volumetric flow rate through a given nozzle.
Find: Determine the force needed to hold the nozzle in place. The control volume shown in the accompanying figure is at steady state.
At sections 1 and 2 the flow is one-dimensional. The water leaves the nozzle at atmospheric pressure zero gage pressure. Analysis: The anchoring force sought, FA, is the reaction force between the faucet and nozzle threads.
To evaluate this force we select a control volume that includes the entire nozzle and the water contained in the nozzle at an instant, as is indicated in Figs. All of the vertical forces acting on the contents of this control volume are identified in Fig. Since atmospheric pressure acts on the outside of the nozzle, the net pressure force in the z direction can be evaluated using gage pressures.
Thus, solving Eq. The water accelerates as it flows through the nozzle. As expected, the nozzle weight, wn, the water weight, ww, and the pressure force at section 1 , p1A1, all increase the anchoring force. Of these, the effect of the pressure at section 1 is far more important than the total weight. Equations 2 V2 7. As indicated in Fig. For steady flow a streamline can be thought of as the path along which a fluid particle moves when traveling from one location in the flow, point 1 , to another location, point 2.
Details of this development are provided in Sec. The first term, p, is the actual thermodynamic pressure of the fluid as it flows. Hence, p is normally termed the static pressure. Another way to measure the static static pressure pressure would be to drill a hole in a flat surface and fasten a piezometer tube as indicated by the location of point 3 in Fig. The third term in Eq. It is not actually a pressure, but does represent the change in pressure possible due to potential energy variations of the fluid as a result of elevation changes.
Its dynamic pressure interpretation can be seen in Fig. This type of tube is termed a Pitot tube. After the initial transient motion has died out, the liquid will fill the tube to a height of H as shown. The fluid in the tube, including that at its tip, 2 , will be stationary. It can be shown that there is a stagnation point on any stationary body that is placed into a flowing fluid. The Momentum and Mechanical Energy Equations 3 The sum of the static pressure, hydrostatic pressure, and dynamic pressure is termed the total pressure, pT.
The Bernoulli equation is a statement that the total pressure remains constant along a streamline. This is the principle on which the Pitot-static tube is based.
The center tube measures the stagna- V tion pressure at its open tip. The relationship between the stagnation, static, and dynamic pressures is then Figure The outer tube is made with several small holes at an appropriate distance from the tip so that they measure the static pressure. An alternate but equivalent form of the Bernoulli equation is obtained by dividing each term of Eq.
Each of the terms in this equation has the units of head length and represents a certain type of head. The elevation term, z, is related to the potential energy of the particle and is called the elevation, pressure, and elevation head. The Bernoulli equation states that the sum of the pressure head, the velocity head, and the elevation head is constant along a streamline.
Free Jets. Consider flow of a liquid from a large reservoir as is shown in Fig. A jet of liquid of diameter d flows from the nozzle with velocity V. Application of Eq. Confined Flows. In many cases the fluid is physically constrained within a device so that pressure cannot be prescribed on the boundary as was done for the free jet example above. For many such situations it is necessary to use the mass balance together with the Bernoulli equation as illustrated in the following example.
The difference in the static pressures at 1 and 2 is measured by the inverted U-tube manometer containing oil of specific gravity, SG, less than one. Determine the manometer reading, h, in terms of the volumetric flow rate and other pertinent quantities.
Solution Known: Water flows through a variable area pipe that has a manometer attached. Find: Determine the manometer reading in terms of the volumetric flow rate.
The flow is steady, inviscid, and incompressible. The pressure taps to which the manometer is fastened measure the difference in static pressure of the water between 1 and 2. That is, the flow rate, Q, in a pipe can be determined if the manometer reading is known. Note that the manometer reading is propor- tional to the square of the flow rate. In terms of heads, the Bernoulli equation states that the sum of the velocity head, the elevation head, and the pressure head is constant along a streamline.
To account for such effects the full mechanical energy equation must be applied. In Eq. The pump and turbine heads are zero if there is no pump and turbine within the control volume. Oth- erwise, these quantities must be positive. As can be seen from Eq. In Chap. In the present section three introductory examples are considered. The first of these involves the evaluation of head loss in a pipe. Determine the direction of flow and the head loss over the 6-m length of pipe.
Solution Known: Two pressure taps along a pipe indicate the pressure head in the constant diameter pipe. Find: Determine the direction of flow and the head loss for the flow. Schematic and Given Data: 0. The flow is steady and incompressible. The pipe diameter is constant. The two vertical liquid filled tubes measure the pressures p1 and p2. There are no pumps or turbines within the section of pipe of interest. Negative head losses cannot occur.
Thus, we assume the flow is uphill and use the mechanical energy equation to determine hL. The two following examples involve flows for which the turbine and pump heads play a significant role. Find: Determine the maximum power that the turbine can extract from the water. At 1 the velocity is essentially zero because the surface area is large; also the pressure is atmospheric.
At 2 the water exits at a specified velocity and as a free jet at atmospheric pressure. The elevation difference between the lake surfaces is 30 ft. Determine the head loss, in ft and in horsepower.
Solution Known: The pump power, the elevation difference, and the volumetric flow rate are known. Find: Determine the head loss. At each surface the pressure is atmospheric. Also, the water velocities on each surface are essentially zero because each surface area is large. Section 1 Pump Figure E The pump head is found from Eq. This mechanical energy is stored as potential energy.
These concepts include surface and body forces, viscosity, and the steady, incompressible flow model. For steady flow, the sum of all the forces acting on the contents of the control volume equals the difference between the out- flow and inflow rates of momentum across the control volume surface.
We have also considered two forms of the mechanical energy balance. The Bernoulli equa- tion is valid for steady, inviscid, incompressible flows and provides the relationship between pressure, elevation, and velocity for such flows.
The more general mechanical energy equa- tion can be used in situations where viscous effects are important and pumps or turbines add or remove mechanical energy to or from the flowing fluid.
The following checklist provides a study guide for this chapter. The subset of key terms listed here in the margin is particularly important. Also, except for the problems h U under the Compressible Flow heading, all problems are for y u steady, incompressible flow. The bottom Figure P If the distance between the two plates is 0. The effective area of the upper plate is in. Determine the mag- Plate A horizontal force of 9 N is required to hold the plate in place.
Determine the velocity 20 mm at the exit, V1, and the velocity within the pipe, V2. The tank rests on a smooth surface, and to pre- Q 0. What tensile force O does the cable have to support? Assume the flow to be Figure P Cable Fig. By how much is the thrust force along the cen- terline of the aircraft reduced for the case indicated com- Smooth pared to normal flight when the exhaust is parallel to the Figure P The pressure just up- the horizontal elbow and nozzle combination shown in stream of the valve is 90 psi and the pressure drop across the Fig.
Neglect gravity. The inside diameters of the valve inlet and exit pipes are 12 and 24 in. If the flow through the valve occurs in 6 in. The flow cross- sectional diameter is mm at the elbow inlet, section 1 , The elbow flow zontally in a circular arc as shown in Fig. If the pipe passage volume is 0. The elbow mass piping between sections 1 and 2 stationary. The steady flow is 12 kg. The loss in pressure due to fluid friction z direction anchoring forces required to hold the elbow in between sections 1 and 2 is 25 psi.
The water Figure P Determine the axial component of the Bernoulli Equation Section 2 Section 1 What would be the maxi- Determine a the pressure head in feet of water, b the velocity head.
If the nozzle is attached to a 3-in. If viscous effects are neglected, determine the height reached by a jet of water through a small hole in Determine the volumetric flow rate.
At one section, the static At the other section, the static pressure is Which way is the water flowing? If viscous effects are negligible, deter- mine the minimum value of h allowed without the siphon stopping. If the mercury manome- h ter reading, h, is mm, determine the volumetric flow rate Figure P Is the actual flow rate more or less than the frictionless value? Determine the maximum vol- mm umetric flow rate if the water is not to flow from the open ver- tical tube at A.
Calculate the stagnation pressure required the flow rate is as shown in Fig. Determine the 10 m is assumed. Determine the amount Pump 0.
The elevation difference between the reser- 50 ft voir surface and the turbine outlet is ft. What is the max- imum amount of power output possible? Why will the actual Section 1 amount be less?
Pump If all losses are negligible, determine a the elevation h, b the pressure difference across the turbine, and c if the turbine were Figure P What is the head loss associated with this flow?
If this p3 p4 same amount of head loss is associated with pumping the fluid h from the lower lake to the higher one at the same flow rate, estimate the amount of pumping power required.
T 12 in. Free jet Figure P A vacuum gage in the turbine discharge 3 m below the turbine inlet centerline reads mm Hg vacuum. If the turbine shaft output power is kW, calculate the frictional power loss through the turbine. The supply and discharge pipe Figure P The head loss is known to be 1. According to the pump The Momentum and Mechanical Energy Equations The block slides on a film of oil having a vis- cosity of 1.
If at steady state the velocity 3. Assume a outlet jet linear velocity distribution in the film. The lubricant that fills the 25 in. Assume the velocity distribution in the Determine the horizontal component Bearing Lubricant of the force that the blowing wind puts on the exhaust gases. Determine the average velocity Figure P Determine the ve- constant. V Three 0. It suspends a plate having a mass of 1. What is the vertical distance h?
For the conditions shown in the figure, what volumetric flow rate is needed to produce a lbf thrust? Assume the inlet and outlet jets of water are free jets at atmospheric pres- sure 0 gage. Ne- into the atmosphere as shown in Fig. When the pres- glect the weight of the air and all friction. Determine the volumetric flow rate Air 2 in. The flow is not frictionless. Open Free air jet 10 psi Bend Water 7 in. If viscous effects are negligible Pump P A Pitot tube connected to a water U-tube manometer is located in the free air jet.
Determine the horizontal Figure P Flo w 5 ft a Section 1 6 in. If the differential reading in ameter. The static pressure at section 2 , 10 ft below the tur- the mercury manometer is 3 ft, calculate the power that the bine inlet, is 10 in. Hg vacuum. If the turbine develops pump supplies to the oil if head losses are negligible. Hg vacuum 12 in. Approximately how far away was the lightning strike? The relation between the head added to the Assuming ideal gas given in Fig.
Mach number of 0. If the Re- pressure in the tank drops. Estimate the pressure in the tank, peat if the diverging portion acts as a supersonic diffuser. Determine the diameter of in the diverging section at a location where the pressure is 2 bar. Deter- 8 MPa, and 6 MPa, respectively. Determine pressure is 1 bar. For steady a the pressure, py, in bar.
All other upstream conditions remain the same. Up to the shock, the flow is isentropic. The exit plane area is 0. If the air behaves as an ideal gas with Determine b the stagnation pressure pox, in bar. That is, we consider steady, inviscid, incom- pressible flow as shown in Fig. The motion of each fluid particle is described in terms of its velocity vector, V, which is defined as the time rate of change of the position of the particle. If the flow is steady, each particle slides along its path, and its velocity vector is every- where tangent to the path.
The lines that are tangent to the velocity vectors throughout the flow field are streamlines. The particle has length ds along the streamline and cross-sectional area dA normal to the streamline. Note that, ds dA is the particle volume. Thus, the net force acting in the streamline direction on the particle shown in Fig. First, we note from Fig.
These ideas combined with Eq. The flows through the nozzles and diffusers of jet en- flow visualization gines are important examples. Other examples are the flows through wind tunnels, shock tubes, and steam ejectors.
These flows are known as compressible flows. In this section, we compressible flows introduce some of the principles involved in analyzing compressible flows. In this section we obtain an ex- pression that relates the velocity of sound, or sonic velocity, to other properties.
The veloc- ity of sound is an important property in the study of compressible flows. Modeling Pressure Waves. Let us begin by referring to Fig. The wave is generated by a small displacement of the piston.
It is eas- ier to analyze this situation from the point of view of an observer at rest relative to the wave, as shown in Fig. By adopting this viewpoint, a steady-state analysis can be applied to the control volume identified on the figure. If the disturbance is weak, the third term on the right of Eq. Since the thickness of the wave is small, shear forces at the wall are negligible.
The effect of gravity is also ignored. Hence, the only significant forces acting in the direction of flow are the forces due to pressure at the inlet and exit. For all sound waves, including the loudest jet engine and rock band sounds, the differences in pressure, density, and temperature across the wave are quite small.
Experiments also indicate that the relation be- tween pressure and density across a sound wave is nearly isentropic. Although we have assumed that sound propagates isen- tropically, the medium itself may be undergoing any process. Let us use Eq. In subsequent discussions, the ratio of the velocity V at a state in a flow- ing fluid to the value of the sonic velocity c at the same state plays an important role.
The term hypersonic is used for flows with Mach numbers subsonic much greater than one, and the term transonic refers to flows where the Mach number is close to unity. As we have discussed in stagnation state Sec. In a compressible flow we think of this occurring isentropically no friction and no heat transfer in a diffuser operating at steady state. The pressure po and temperature To at a stagnation state are called the stagnation pressure and stagnation temperature, respectively. Texts dealing with compressible flow should be consulted for discus- sion of other areas of application.
In the present section we determine the shapes required by nozzles and diffusers for sub- sonic and supersonic flow.
This is accomplished using mass, energy, entropy, and momen- tum principles, together with property relationships. In addition, we study how the flow through nozzles is affected as conditions at the nozzle exit are changed. The presentation concludes with an analysis of normal shocks, which can exist in supersonic flows.
This is accomplished using differential equations relating the principal variables that are obtained using mass and energy balances together with property relations, as consid- ered next. Governing Differential Equations. Let us begin by considering a control volume enclos- ing a nozzle or diffuser such as shown in Figs. In addition to Eqs. Assuming the flow occurs isen- tropically, the property relation Eq. Introducing Eq. Additional conclusions can be drawn by combining the above differential equations.
Com- bining Eqs. Eliminating dp between Eqs. The following four cases can be identified: Case 1: Subsonic nozzle. Case 2: Supersonic nozzle. Case 3: Supersonic diffuser. Case 4: Subsonic diffuser. The conclusions reached above concerning the nature of the flow in subsonic and super- sonic nozzles and diffusers are summarized in Fig. From Fig. These findings suggest that a Mach number of unity can occur only at the location in a nozzle or diffuser where the cross-sectional area is a minimum.
This location of minimum area is called the throat. The developments of this section have not required the specification of an equation of state; thus, the conclusions hold for all gases. Moreover, although the conclusions have been drawn under the restriction of isentropic flow through nozzles and diffusers, they are at least qualitatively valid for actual flows because the flow through well-designed nozzles and dif- fusers is nearly isentropic.
Isentropic nozzle efficiencies Sec. The back pressure is the pressure in the exhaust region outside the nozzle. The case of converging nozzles is taken up first and then converging—diverging nozzles are considered. Converging Nozzles. For the series of cases labeled a through e, let us consider how the mass flow rate m and nozzle exit pressure pE vary as the back pressure is decreased while keeping the inlet conditions fixed. This corresponds to case a of Fig.
If the back pressure pB is decreased, as in cases b and c, there will be flow through the noz- zle. As long as the flow is subsonic at the exit, information about changing conditions in the exhaust region can be transmitted upstream. Decreases in back pressure thus result in greater mass flow rates and new pressure variations within the nozzle. In each instance, the velocity is subsonic throughout the nozzle and the exit pressure equals the back pressure.
The exit Mach number increases as pB decreases, however, and eventually a Mach number of unity will be attained at the nozzle exit.
This case is represented by d on Fig. Since the velocity at the exit equals the velocity of sound, information about changing conditions in the exhaust region no longer can be transmitted upstream past the exit plane. Neither the pressure variation within the nozzle nor the mass flow rate is affected. Under these conditions, the nozzle is said to be choked.
When a nozzle is choked flow: choked, the mass flow rate is the maximum possible for the given stagnation conditions. The pressure variation outside the nozzle cannot be predicted using the one-dimensional flow model. Converging—Diverging Nozzles. The series of cases labeled a through j is considered next. When the back pressure is slightly less than po case b , there is some flow, and the flow is subsonic throughout the nozzle.
In accordance with the discussion of Fig. If the back pressure is start-up reduced further, corresponding to case c, the mass flow rate and velocity at the throat are greater than before.
Still, the flow remains subsonic throughout and qualitatively the same as case b. As the back pressure is reduced, the Mach number at the throat increases, and eventually a Mach number of unity is attained there case d. As before, the greatest velocity and lowest pressure occur at the throat, and the diverging portion remains a subsonic diffuser. However, because the throat velocity is sonic, the nozzle is now choked: The maximum mass flow rate has been attained for the choked flow: given stagnation conditions.
Further reductions in back pressure cannot result in an converging—diverging increase in the mass flow rate. Conditions within the diverging portion can be altered, however, as illustrated by cases e, f, and g. In case e, the fluid passing the throat continues to expand and becomes supersonic in the diverging portion just downstream of the throat; but at a certain location an normal shock abrupt change in properties occurs. This is called a normal shock. Across the shock, there is a rapid and irreversible increase in pressure, accompanied by a rapid de- crease from supersonic to subsonic flow.
Downstream of the shock, the diverging duct acts as a subsonic diffuser in which the fluid continues to decelerate and the pressure increases to match the back pressure imposed at the exit. If the back pres- sure is reduced further case f , the location of the shock moves downstream, but the flow remains qualitatively the same as in case e. With further reductions in back pressure, the shock location moves farther downstream of the throat until it stands at the exit case g.
Since the fluid leaving the nozzle passes through a shock, it is V In each of these cases, the flow through the nozzle is not af- fected. The adjustment to changing back pressure occurs outside the nozzle. In case h, the pressure decreases continuously as the fluid expands isentropically through the nozzle and then increases to the back pressure outside the nozzle. The compression that occurs outside the nozzle involves oblique shock waves.
In case i, the fluid expands isentropically to the back pressure and no shocks occur within or outside the nozzle. In case j, the fluid expands isentropically through the nozzle and then ex- pands outside the nozzle to the back pressure through oblique expansion waves.
The pressure variations outside the nozzle involving oblique waves cannot be predicted using the one-dimensional flow model. In a normal shock, this change of state occurs across a plane normal to the direction of flow. The object of the present V Modeling Normal Shocks. A control volume enclosing a normal shock is shown in Fig. Thus, there is no significant change in flow area across the shock, even though it may occur in a diverging passage, and the forces acting at the wall can be neglected relative Normal shock x y Vx, Mx, Tx, px Vy, My, Ty, py hx, sx hy, sy Figure Fanno and Rayleigh Lines.
The mass and energy equations, Eqs. Similarly, the mass and momentum equations, Fanno line Eqs. Fanno and Rayleigh lines are sketched on h—s coordinates Rayleigh line in Fig. It also can be shown that the upper and lower branches of each line correspond, respectively, to subsonic and supersonic velocities.
The downstream state y must satisfy the mass, energy, and momentum equations simul- taneously, so state y is fixed by the intersection of the Fanno and Rayleigh lines passing through state x. This conclusion is consistent with the discussion of cases e, f, and g in Fig. A significant increase in pressure across the shock accompanies the decrease in velocity.
The stagnation enthalpy does not change across the shock, but there is a marked decrease in stagnation pressure associated with the irre- versible process occurring in the normal shock region. Attention is now restricted to ideal gases with constant specific heats. This case is appropriate for many practical problems involving flow through nozzles and diffusers.
The assumption of constant specific heats also allows the derivation of relatively simple closed-form equations. For the case of an ideal gas with constant cp, Eq. Introducing the ideal gas equation of state, together with Eqs. Then with Eqs. This is consistent with the discussion of Fig. Equations Table Such a table facilitates the analysis of flow through nozzles and diffusers.
In Example The first step of the analysis is to check whether the flow is choked. Air enters the nozzle with negligible velocity at a pressure of 1. Solution Known: Air flows isentropically from specified stagnation conditions through a converging nozzle with a known exit area.
The control volume shown in the accompanying sketch operates at steady state. Flow through the nozzle is isentropic. Analysis: The first step is to check whether the flow is choked.
Thus, for back pressures of kPa or less, the Mach number is unity at the exit and the nozzle is choked. The mass flow rate is the maximum value that can be attained for the given stagnation properties. The exit Mach number can be found by solving Eq. The exit velocity is then 1. It is left as an exercise to develop a solution using this table. Also, observe that the first step of the analysis is to check whether the flow is choked. For this case, it follows from the energy equation, Eq.
Then, with Eq. Accordingly, tables can be Table In the next example, we consider the effect of back pressure on flow in a converging— diverging nozzle. Key elements of the analysis include determining whether the flow is choked and if a normal shock exists.
Solution Known: Air flows from specified stagnation conditions through a converging—diverging nozzle having a known throat and exit area. Find: The mass flow rate, exit pressure, and exit Mach number are to be determined for each of five cases. The T—s diagrams provided locate states within the nozzle. Flow through the nozzle is isentropic throughout, except for case e, where a shock stands in the diverging section. Analysis: a The accompanying T—s diagram shows the states visited by the gas in this case.
The diverging por- tion acts as a diffuser in the present part of the example; accordingly, the subsonic value is appropriate. The supersonic value is appropriate in part c. Thus, from Table This is the maximum mass flow rate for the specified geometry and stagnation conditions: the flow is choked. Using this, Table The pressure downstream of the shock is thus The mass flow is the same as found in part b.
Since the flow is choked, the mass flow rate is the same as that found in part b. Part c corresponds to case d of Fig. Part d corresponds to case g of Fig. An obvious goal of any experiment is to make the results as widely applicable as possible.
To achieve this end, the concept of similitude is often used so that measurements made on one system for example, in the laboratory can be used to describe the behavior of other similar systems outside the laboratory. The laboratory systems are usually thought of as models and are used to study the phenomenon of interest under carefully controlled conditions. From these model studies, empirical formulations can be developed, or specific predictions of one or more characteristics of some other similar system can be made.
The objective of this chapter is to determine how to use chapter objective similitude, dimensional analysis, and modeling to simplify the experimental investigation of fluid mechanics problems.
An important characteristic of this system, which would be of interest to an engineer designing a pipeline, is the pressure drop per unit length that develops along the pipe as a result of friction. Although this would appear to be a relatively simple flow prob- lem, it cannot generally be solved analytically even using large computers without the use of experimental data.
See Sec. Similitude, Dimensional Analysis, and Modeling To perform the experiments in a meaningful and systematic manner, it would be neces- sary to change one of the variables, such as the velocity, while holding all others constant, and measure the corresponding pressure drop.
This approach to determining the functional relationship between the pressure drop and the various factors that influence it, although log- ical in concept, is difficult. Some of the experiments would be hard to carry out—for ex- ample, it would be necessary to vary fluid density while holding viscosity constant. How would you do this? Fortunately, there is a much simpler approach to this problem that will eliminate the difficulties described above.
In the following sections we will show that rather than work- ing with the original list of variables, as described in Eq. The results of the experiment could then be represented by a single, universal curve see Fig. The qualitative aspect serves to identify the nature, or type, of the characteristics such as length, time, stress, and velocity , whereas the quan- titative aspect provides a numerical measure of the characteristics.
The quantitative de- scription requires both a number and a standard by which various quantities can be com- pared.
A standard for length might be a meter or foot, for time an hour or second, and units for mass a slug or kilogram. Such standards are called units, and systems of units in com- mon use are as described in Chapter 2. The qualitative description is conveniently given in terms of certain primary quantities, such as length, L, time, t, mass, M, and tempera- ture, T.
For a wide variety of problems involving fluid mechanics, only the three basic dimensions, L, t, and M, are required. Alternatively, L, t, and F could be used, where F is the basic dimension of force. Thus, secondary quantities ex- pressed in terms of M can be expressed in terms of F through the relationship above. It should be noted that in thermodynamic and heat transfer analyses, an additional basic dimension, temper- ature, T, is often involved.
Therefore, in such cases there are often four basic dimensions: L, t, M, and T rather than the three basic dimensions L, t, and M considered in this fluid mechanics chapter. Similitude, Dimensional Analysis, and Modeling That is, the dimensions of the left side of the equation must homogeneous be the same as those on the right side, and all additive separate terms must have the same dimensions.
Any reduction in the number of required variables represents a considerable simplification in the analysis of the problem. For Example… in the problem involving the pressure drop per unit length of smooth pipe discussed in Sec. As shown at the beginning of this section, it takes three basic dimensions either F, L, t or M, L, t to describe the physical variables of this problem. Control Volume Analysis Using Energy.
The Second Law of Thermodynamics. Using Entropy. Vapor Power and Refrigeration Systems. Gas Power Systems. Psychrometric Applications CD only. Save to Library Save. Create Alert Alert.
Share This Paper. Background Citations. Methods Citations. Results Citations. Citation Type. Has PDF. Publication Type. More Filters. We consider both Otto and Diesel heat engine cycles running upon the working substances modeled by the van der Waals fluid as a simple non-ideal gas model. We extensively perform the efficiency study … Expand.
View 3 excerpts, cites background. Thermo-elasto bulk-flow model for labyrinth seals in steam turbines. Abstract Over the last few decades, the increasing demand on efficiency and performance for steam turbines has resulted in OEMs operating machines near critical conditions of their structural and … Expand. Convective heat transfer is governed by a number of factors including various fluid properties, the presence of a thermal gradient, geometric configuration, flow condition, and gravity.
Heat transfer investigation in new cooling schemes of a stationary blade trailing edge. Abstract This paper describes an experimental study for a scaled model reproducing an innovative wedge shaped discharge trailing edge TE cooling scheme.
Three configurations were investigated: … Expand.
0コメント