Statics and dynamics demystified pdf download

Calculate the error in the pressure using the ideal-gas equation. Work will be calculated for several common situations. Heat transferred by conduction, convection, or radiation to systems or control volumes will either be given or information will be provided that it can be determined in our study of thermodynamics; it will not be calculated from temperature information, as is done in a heat transfer course.

The energy released during the combustion process is transferred to the crankshaft by means of the connecting rod, in the form of work. Thus, in thermodynamics, work can be thought of as energy being transferred across the boundary of a system, where, in this example, the system is the gas in the cylinder. Figure 3. The magnitude of the work is the product of that weight and the distance it could be lifted.

The schematic of Fig. The convention chosen for positive work is that if the system performs work on the surroundings it is positive. The units of work are quickly observed from the units of a force multiplied by a distance: in the SI system, newton-meters.

The rate of doing work, designated W, is called power. We will use the unit of horsepower because of its widespread use in rating engines. Note that if the system in Fig. We are primarily concerned with the work required to move a boundary against a pressure force. Consider the piston-cylinder arrangement shown in Fig.

There is a seal to contain the gas in the cylinder, the pressure is uniform throughout the cylinder, and there are no gravity, magnetic, or electrical effects. This assures us of a quasiequi- librium process, one in which the gas is assumed to pass through a series of equi- librium states. Now, allow an expansion of the gas to occur by moving the piston upward a small distance dl. The total force acting on the piston is the pressure times the area of the piston.

This pressure is expressed as absolute pressure since pressure is a result of molecular activity; any molecular activity will yield a pressure that will result in work being done when the boundary moves. The quantity Adl is simply dV, the differen- tial volume, allowing Eq. Typical pressure-volume diagrams are shown in Fig. The work W is the area under the P-V curve. The integration process highlights two important features in Eq.

First, as we proceed from state 1 to state 2, the area representing the work is very dependent on the path that we follow. That is, states 1 and 2 in Fig. Thus, work is a path function, as contrasted to a point function that is dependent only on the end points. The differential of a path function is called an inexact differential, whereas the differential of a point function is an exact differential.

PdV PdV Figure 3. We would never write W1 or W2, since work is not associated with a state but with a process. Work is not a prop- erty. The system passes through each equilibrium state shown in the P-V diagrams of Fig. An equilibrium state can usually be assumed even though the variables may appear to be changing quite rapidly. Combustion is a very rapid process that cannot be mod- eled as a quasiequilibrium process. For a system free of surface, magnetic, and electrical effects, the only work mode is that due to pressure acting on a moving boundary.

For such simple systems only two independent variables are necessary to establish an equilibrium state of the system composed of a homogeneous substance. Calculate the work done by the steam. Using Table C. A kg piston sits on top of the water. Solution The pressure in the cylinder is due to the weight of the piston and remains con- stant. The initial pressure and volume are kPa and 2 m3, respectively. Solution The work, using Eq. For nonequilibrium processes the work cannot be calculated using the integral of PdV.

Either it must be given for the particular pro- cess or it must be determined by some other means. Two examples will be given. Consider a system to be formed by the gas in Fig. In part a work is obviously crossing the boundary of the system by means of the rotating shaft, yet the volume does not change.

We could calculate the work input by multiplying the weight by the distance it dropped, neglecting friction in the pulley system. This would not, however, be equal to the integral of PdV, which is zero. The paddle wheel provides us with a nonequilibrium work mode. Suppose the membrane in Fig. The sudden expansion is a nonequilibrium process, and again we cannot use the integral of PdV to calculate the work for this nonequilibrium work mode.

The weight and the piston maintain a constant gage pres- sure of kPa. Determine the net work done by the gas on the surroundings. Neglect all friction. Solution The paddle wheel does work on the system, the gas, due to the kg mass dropping 3 m. The work necessary to stretch a linear spring Fig.

Note that the force is dependent on the variable x. The current i is related to the charge. Calculate the horsepower transmitted. Find the work done by the air on the frictionless piston. The spring is initially unstretched, as shown.

Transfer of energy that we cannot account for by any of the work modes is called heat transfer. A process in which there is zero heat transfer is called an adiabatic process. An insulated system in thermodynamics will always be assumed to have zero heat transfer.

Consider a system composed of a hot block and a cold block of equal mass. The hot block contains more energy than the cold block due to its higher temperature. Once in thermal equi- librium, there are no longer any temperature differences, therefore, there is no lon- ger any heat transfer. Energy in the hot block has simply been transferred to the cold block, so that if the system is insulated, the cold block gains the amount of energy lost by the hot block.

It should be noted that the energy contained in a system might be transferred to the surroundings either by work done by the system or by heat transferred from the system. In either case, heat and work are qualitatively equivalent and are expressed in units of energy. An equivalent reduction in energy is accomplished if J of heat is transferred from a system or if J of work is performed by a system.

In either case, energy is being added to the system. Although heat and work are similar in that they both represent energy crossing a boundary, they differ in that, unlike work, convention prescribes that heat trans- ferred to a system is considered positive.

Thus, positive heat transfer adds energy to a system, whereas positive work subtracts energy from a system. Because a system does not contain heat, heat is not a property. Thus, its differen- tial is inexact and is written as d Q, where Q is the heat transfer. For a particular process between state 1 and state 2 the heat transfer could be written as Q but it will generally be denoted simply as Q.

There are three heat transfer modes: conduction, convection, and radiation. Often, engineering applications involving heat transfer must consider all three modes in an analysis.

In each case, we can write rela- tionships between the rate of heat transfer and temperature. A course in heat transfer is dedicated to calculating the heat transfer in various situations. In thermodynamics, it is given in a problem or it is found from applying the energy equation.

How much heat must be trans- ferred to result in an equivalent effect? Quiz No. Which work mode is a nonequilibrium work mode? A Compressing a spring B Transmitting torque with a rotating shaft C Energizing an electrical resistor D Compressing gas in a cylinder 2.

A stop is located 20 mm above the piston at the position shown. If the mass of the frictionless piston is 64 kg, what work must the air do on the piston so that the pressure in the cylinder increases to kPa? Which of the following statements about work for a quasiequilibrium process is incorrect? A The differential of work is inexact B Work is the area under a P-T diagram C Work is a path function D Work is always zero for a constant volume process Questions 5—8 The frictionless piston shown has a mass of 16 kg.

The initial quality is 20 percent. The total mass of the water is nearest A 0. Air is compressed in a cylinder such that the volume changes from 0. The initial pressure is kPa. Estimate the work necessary to compress the air in a cylinder from a pressure of kPa to that of kPa.

The initial volume is cm3. An isothermal process is to be assumed. Estimate the work done by a gas during an unknown equilibrium process. Calculate the work needed to compress the spring from 0. A paddle wheel and an electric heater supply energy to a system. If the torque is 20 N.

What is the minimum horsepower engine that would be necessary if a maximum of A is anticipated from the V system? Which of the following does not transfer work to or from a system?

Find W1—2. A mass of 0. The pressure at state 2 is nearest A 8. The initial quality is nearest A The pressure at the beginning of the process is kPa. Air is expanded in a piston-cylinder arrangement at a constant pressure of kPa from a volume of 0.

Then the temperature is held constant during an expansion of 0. Determine the total work done by the air. Air undergoes a three-process cycle. A V electric resistance heater draws 10 A. It operates for 10 min in a rigid volume. Calculate the work done on the air in the volume. An electrical voltage of V is applied across a resistor providing a current of 12 A through the resistor. In previous courses, the study of conservation of energy may have empha- sized changes in kinetic and potential energy and their relationship to work.

A more general form of conservation of energy includes the effects of heat transfer and internal energy changes. Let a weight be attached to a pulley-paddle-wheel setup, such as that shown in Fig. Let the weight fall a certain distance thereby doing work on the system, contained in the tank shown, equal to the weight multiplied by the distance dropped.

Now, the system is returned to its initial state the completion of the cycle by transferring heat to the surroundings, as implied by the Q in Fig. This reduces the temperature of the system to its initial temperature. The paddle wheel then rotates until the spring is unstretched. Calculate the heat transfer necessary to return the system to its initial state. Figure 4.

Realizing that a cycle results when a system undergoes two or more processes and returns to the initial state, we could consider a cycle composed of the two processes represented by A and B in Fig. The quantity E is an extensive property of the system and represents the energy of the system at a particular state.

Equation 4. More often than not the subscripts will be dropped on Q and W when working problems. The property E represents all of the energy: kinetic energy KE, potential energy PE, and internal energy U, which includes chemical energy and the energy associ- ated with the molecules and atoms. Any other form of energy is also included in the total energy E. Its associated intensive property is designated e.

For simple systems in equilibrium, only two properties are necessary to establish the state of a pure substance, such as air or steam. Since internal energy is a property, it depends only on, say, pressure and temperature; or, for steam in the quality region, it depends on quality and tem- perature or pressure. Assuming a well insulated, sealed room, determine the internal energy increase after 1 hour of operation.

The negative sign results because the work is input to the system. Using steam table C. One plan for obtaining a solution is to guess a value for v2 and calculate u2 from the equation above. If this value checks with the u2 from the steam tables at the same temperature, then the guess is the correct one. Therefore, the guess must be revised. The actual v2 is a little less than 1.

From Eq. It is only the change in enthalpy or internal energy that is important; hence, we can arbitrarily choose the datum from which to measure h and u. Trial and error was unnecessary, and the solution was rather straightforward.

We illustrated that the quantity we invented, enthalpy, is not necessary, but it is quite handy. We will use it often in calculations. It is the change in enthalpy of the substance at the saturated conditions of the two phases. The heat of fusion and the heat of sublimation are relatively insensitive to pressure or temperature changes. The heat of vaporization of water is included as hfg in Tables C.

Since u is only a function of T, we see that h is also only a function of T for an ideal gas. In differential form Eq. Note that the difference between Cp and Cv for an ideal gas is always a constant, even though both are functions of temperature. E, which tabulate h T and u T , or integrate, using expressions for Cp T found in text books.

For water we will use the nominal value of 4. Compare with the steam tables. This linear relation would change, however, for each pressure chosen; hence, the steam tables are essential. Solution a Using the gas table in App. If T2 were closer to K, say K, the error would be much smaller. So, for large temperature differences, the tables should be used. Internal energy and enthalpy vary slightly with pressure for the isothermal process, and this variation must be accounted for in processes involving many substances.

For such a process the above three equations would not be valid. The study of such processes is, however, often postponed until after the second law of thermodynamics is presented.

This postponement is not necessary for an ideal gas, and because the adiabatic quasiequilibrium process is quite common, it is presented here. However, if an ideal gas can be assumed, Eq. For such a process involving an ideal gas, any of the equations in 4.

Assume a quasiequilibrium process. Solution For the constant-volume quasiequilibrium process the work is zero. The mass must be found. If the pressure is held constant, how much paddle- wheel work must be added to the air? Solution The process cannot be approximated by a quasiequilibrium process because of the paddle-wheel work. Thus, the heat transfer is not equal to the enthalpy change. Solution We will assume that the compression process is approximated by a quasiequilib- rium process, which is acceptable for most compression processes, and that the process is adiabatic due to the presence of the insulation we usually assume an adiabatic process anyhow since heat transfer is assumed to be negligible.

This restriction is acceptable for many problems of interest and may, in fact, be imposed on the power plant schematic shown in Fig. For a more complete analysis we must relate Win, Qin, Wout, and Qout to the pressure and temperature changes for the pump, boiler, turbine, and condenser, respectively. We must formulate equations that allow us to make the necessary calculations. Fluid mechanics treats the more general unsteady, nonuni- form situations in much greater detail. Solution From the superheat Table C.

Note that the energy consists of internal energy, kinetic energy, and potential energy. Substitute the expression for work W into Eq. A nozzle, or a diffuser, is a device in which the kinetic energy change cannot be neglected so Eq. Is it incompressible? A sketch of the pro- cess on a suitable diagram is often of use in the calculations. If the working sub- stance behaves as an ideal gas, then the appropriate equations may be used; if not, tabulated values must be used, such as those provided for steam.

For real gases that do not behave as ideal gases, properties can be found in App. Often heat transfer from a device or the internal energy change across a device, such as a pump, is not desired. For such situations, the heat transfer and internal energy change may be lumped together as losses.

Examples will illustrate. Two such devices are sketched in Fig. For this process [see Eq. Most valves are throttling devices, for which the energy equation takes the form of Eq. They are also used in many refrigeration units in which the sudden drop in pressure causes a change in phase of the working substance. This must equal the exiting enthalpy as demanded by Eq. Compressors and blowers also fall into this category but have the pri- mary purpose of increasing the pressure in a gas.

As a result there is a pressure drop from the inlet to the outlet of the turbine. In some situations there may be heat transferred from the device to the surroundings, but often the heat transfer is negligible. In addition, the kinetic and potential energy changes are neg- ligible.

For such devices operating in a steady-state mode, the energy equation takes the form [see Eq. For liquids, such as water, the energy equation 4. Neglect any heat transfer and kinetic energy change. Show that the kinetic energy change is negligible. Solution The energy equation in the form of Eq. Kinetic energy changes are usually omitted in the analysis of devices but not in a nozzle or a diffuser. Solution The energy equation 4. By neglecting the heat transfer and assuming no increase in internal energy, we establish the maximum pressure rise.

In most applications the inlet and exit areas are not given; but even if they are, as in this example, kinetic energy changes can be ignored in a pump or turbine. It does this by reducing the pressure. There is no work input into the devices and usually negligible heat transfer. Thus, we may have three unknowns at the exit, given the entering conditions. Only the more simple situations will be included here. This requires the density at the exit.

Referring to Eq. Energy is transferred from the hot gases after combustion in a power plant to the water in the pipes of the boiler, and from the hot water that leaves an automobile engine to the atmosphere by use of a radiator. The velocity does not nor- mally change, the pressure drop through the passage is usually neglected, and the potential energy change is assumed zero.

For a control volume including the combined unit, which is assumed to be insulated, the energy equation, as applied to the control volume of Fig. The Cp of sodium is 1. Also, calculate the rate of heat transfer. A Heat transfer equals the work done for a process. B Net heat transfer equals the net work for a cycle. C Net heat transfer minus net work equals internal energy change for a cycle. D Heat transfer minus work equals internal energy for a process.

How much heat must be added to a 0. The initial pressure is 1 MPa. A piston-cylinder arrangement provides a constant pressure of kPa on steam which has an initial quality of 0. One kilogram of steam in a cylinder accepts kJ of heat transfer while the pressure remains constant at kPa. Estimate the work required for the process of Prob. What is the equilibrium temperature?

Find the paddle wheel work if the rigid container is insulated. The air in the cylinder of an air compressor is compressed from kPa to 10 MPa. What exiting temperature is expected? What temperature change is expected? The inlet area is 10 cm2 and the outlet area is 50 cm2. The water enters through a cm-diameter pipe and exits through a cm-diameter pipe. Calculate the minimum horsepower required to operate the pump.

Determine the maximum power output. Determine the minimum power required to drive the insulated compressor. Calculate the rate of heat transfer if the power input is kW. A The heat transfer equals the internal energy change for an adiabatic process. B The heat transfer and the work have the same magnitude for a constant volume quasiequilibrium process in which the internal energy remains constant. C The total energy input must equal the total work output for an engine operating on a cycle.

D The internal energy change plus the work must equal zero for an adiabatic quasiequilibrium process. A system undergoes a cycle consisting of the three processes listed in the table. Compute the missing values a, b, c, d. All quantities are in kJ. Steam is contained in a 4-L volume at a pressure of 1.

How much heat must be added? How much heat is rejected? Calculate the heat transfer needed to increase the pressure to kPa. B represents the rate of change of energy between the inlet and outlet. C is often neglected in control volume applications. D includes the work rate due to the pressure forces. Estimate the exiting temperature. What is the expected temperature after the throttle?

To compress the air to kPa, 5 kW of energy is needed. Calculate the power output, neglecting heat transfer. It is the second law of thermo- dynamics that helps us establish the direction of a particular process. This, however, would be a violation of the second law and would thus be impossible. If the objective of the device is to perform work it is a heat engine; if its objective is to transfer heat to a body it is a heat pump; if its objective is to transfer heat from a body, it is a refrig- erator.

Generically, a heat pump and a refrigerator are collectively referred to as a refrigerator. A schematic diagram of a simple heat engine is shown in Fig. An engine or a refrigerator operates between two thermal energy reservoirs, entities that are capable of providing or accepting heat without changing tempera- tures. The atmosphere and lakes serve as heat sinks; furnaces, solar collectors, and burners serve as heat sources. Temperatures TH and TL identify the respective tem- peratures of a source and a sink.

The net work W produced by the engine of Fig. If the cycle of Fig. A heat pump would provide energy as heat QH to the warmer body e. The work would also be given by Eq. Figure 5. Each of the performance measures represents the desired output divided by the input energy that is purchased. The second law of thermodynamics will place limits on the above measures of performance.

The second law, however, estab- lishes limits that are surprisingly low, limits that cannot be exceeded regardless of the cleverness of proposed designs. The internal com- bustion engine is an example. The second law of thermodynamics can be stated in a variety of ways. Here we present two: the Clausius statement and the Kelvin-Planck statement. Neither is presented in mathematical terms.

We will, however, provide a property of the system, entropy, which can be used to determine whether the second law is being violated for any particular situation. Clausius Statement It is impossible to construct a device that operates in a cycle and whose sole effect is the transfer of heat from a cooler body to a hotter body.

This statement relates to a refrigerator or a heat pump. It states that it is impossi- ble to construct a refrigerator that transfers energy from a cooler body to a hotter body without the input of work; this violation is shown in Fig. Kelvin-Planck Statement It is impossible to construct a device that operates in a cycle and produces no other effect than the production of work and the transfer of heat from a single body.

In other words, it is impossible to construct a heat engine that extracts energy from a reservoir, does work, and does not transfer heat to a low-temperature reservoir. Note that the two statements of the second law are negative statements. They are expressions of experimental observations. No experimental evidence has ever been obtained that violates either statement of the second law. An example will demon- strate that the two statements are equivalent. Solution Consider the system shown.

The device in a transfers heat and violates the Clau- sius statement, since it has no work input. Let the heat engine transfer the same amount of heat QL. Con- versely, a violation of the Kelvin-Planck statement is equivalent to a violation of the Clausius statement.

The process obviously has to be a quasiequilibrium process; addi- tional requirements are: 1. No friction is involved in the process. Unrestrained expansion does not occur. The mixing of different substances and combustion also lead to irreversibilities, but the above three are the ones of concern in the study of the devices of interest.

The fact that friction makes a process irreversible is intuitively obvious. Consider the system of a block on the inclined plane of Fig. Weights are added until the block is raised to the position shown in Fig. Now, to return the system to its original state some weight must be removed so that the block will slide back down the plane, as shown in Fig.

Also, the block and plane are at a higher temperature due to the friction, and heat must be transferred to the surroundings to return the system to its original state. This will also change the surroundings. Because there has been a change in the surround- ings as a result of the process and the reversed process, we conclude that the process was irreversible. A reversible process requires that no friction be present.

To return the system to its original state, we must refrigerate the block that had its temperature raised. This will require a work input, demanded by the second law, resulting in a change in the surroundings since the surroundings must supply the work.

For an example of unrestrained expansion, consider the high-pressure gas con- tained in the cylinder of Fig. Pull the pin and let the piston suddenly move to the stops shown. Note that the only work done by the gas on the surroundings is to move the piston against atmospheric pressure. Now, to reverse this process it is necessary to exert a force on the piston. This will demand a considerable amount of work to be supplied by the surroundings.

In addition, the temperature will increase substantially, and this heat must be transferred to the sur- roundings to return the temperature to its original value.

Unrestrained expansion cannot occur in a reversible process. It is an ideal engine that uses reversible processes to form its cycle of operation; thus it is also called a reversible engine. The cycle associated with the Carnot engine is shown in Fig. Heat is transferred reversibly from the high-temperature reservoir at the constant temperature TH. The piston in the cylinder is withdrawn and the volume increases.

The cylinder is completely insulated so that no heat transfer occurs during this reversible process. The piston continues to be withdrawn, with the volume increasing. Heat is transferred reversibly to the low- temperature reservoir at the constant temperature TL. The piston compresses the working substance, with the volume decreasing. The completely insulated cylinder allows no heat transfer during this reversible process. The piston continues to compress the working substance until the original volume, temperature, and pressure are reached, thereby completing the cycle.

Let the heat transferred from the high-temperature reservoir to the engine be equal to the heat rejected by the refrigerator; then the work produced by the engine will be greater than the work required by the refrigerator i. Now, our system can be organized as shown in b. The engine drives the refrigerator using the rejected heat from the refrigerator. The net result is the conversion of energy from a single reservoir into work, a violation of the second law.

Let the heat rejected by the engine equal the heat required by the refrigerator. We can make this replacement for all reversible engines or refrigerators. Consequently, the relationship 5.

The Carnot engine, when operated in reverse, becomes a heat pump or a refrig- erator, depending on the desired heat transfer. The reversible cycles assumed are obviously unrealistic, but the fact that we have limits that we know cannot be exceeded is often very helpful.

Calculate the minimum percentage increase in work required, by assuming a Carnot refrigerator, for the same amount of energy removed. And this is a minimum percentage increase, since we have assumed an ideal refrigerator. Using Eqs. We let this differential be denoted by dS, where S depends only on the state of the system.

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