8 INTRODUCTION AND BASIC CONCEPTS for one hour.When dealing with electric power generation,the units kW and kWh are often confused.Note that kW or kJ/s is a unit of power, whereas kWh is a unit of energy.Therefore,statements like "the new wind turbine will generate 50 kW of electricity per year"are meaningless and incorrect.A correct statement should be something like"the new wind tur- bine with a rated power of 50 kW will generate 120,000 kWh of electricity per year." Dimensional Homogeneity We all know that apples and oranges do not add.But we somehow man- age to do it (by mistake,of course).In engineering,all equations must be dimensionally homogeneous.That is,every term in an equation must have the same unit.If,at some stage of an analysis,we find ourselves in a posi- tion to add two quantities that have different units,it is a clear indication that we have made an error at an earlier stage.So checking dimensions can serve as a valuable tool to spot errors. EXAMPLE 1-1 Electric Power Generation by a Wind Turbine ◆ A school is paying $0.12/kWh for electric power.To reduce its power bill, the school installs a wind turbine (Fig.1-12)with a rated power of 30 kW. If the turbine operates 2200 hours per year at the rated power,determine the amount of electric power generated by the wind turbine and the money saved by the school per year. SOLUTION A wind turbine is installed to generate electricity.The amount of electric energy generated and the money saved per year are to be determined. Analysis The wind turbine generates electric energy at a rate of 30 kW or 30 kJ/s.Then the total amount of electric energy generated per year becomes Total energy =(Energy per unit time)(Time interval) =(30kW)(2200h) =66,000kWh The money saved per year is the monetary value of this energy determined as Money saved =(Total energy)(Unit cost of energy) =(66.000kWh)($0.12kWh) =$7920 Discussion The annual electric energy production also could be determined in kJ by unit manipulations as FIGURE 1-12 3600s 1 kJ/s A wind turbine.as discussed in Total energy =(30 kW)(2200 h) =2.38×108kJ Example 1-1. Bear Dancer Studios/Mark Dierker RF which is equivalent to 66,000 kWh(1 kWh =3600 kJ). We all know from experience that units can give terrible headaches if they are not used carefully in solving a problem.However,with some attention and skill,units can be used to our advantage.They can be used to check formulas;sometimes they can even be used to derive formulas,as explained in the following example
8 INTRODUCTION AND BASIC CONCEPTS for one hour. When dealing with electric power generation, the units kW and kWh are often confused. Note that kW or kJ/s is a unit of power, whereas kWh is a unit of energy. Therefore, statements like “the new wind turbine will generate 50 kW of electricity per year” are meaningless and incorrect. A correct statement should be something like “the new wind turbine with a rated power of 50 kW will generate 120,000 kWh of electricity per year.” Dimensional Homogeneity We all know that apples and oranges do not add. But we somehow manage to do it (by mistake, of course). In engineering, all equations must be dimensionally homogeneous. That is, every term in an equation must have the same unit. If, at some stage of an analysis, we find ourselves in a position to add two quantities that have different units, it is a clear indication that we have made an error at an earlier stage. So checking dimensions can serve as a valuable tool to spot errors. EXAMPLE 1–1 Electric Power Generation by a Wind Turbine A school is paying $0.12/kWh for electric power. To reduce its power bill, the school installs a wind turbine (Fig. 1–12) with a rated power of 30 kW. If the turbine operates 2200 hours per year at the rated power, determine the amount of electric power generated by the wind turbine and the money saved by the school per year. SOLUTION A wind turbine is installed to generate electricity. The amount of electric energy generated and the money saved per year are to be determined. Analysis The wind turbine generates electric energy at a rate of 30 kW or 30 kJ/s. Then the total amount of electric energy generated per year becomes Total energy 5 (Energy per unit time)(Time interval) 5 (30 kW)(2200 h) 5 66,000 kWh The money saved per year is the monetary value of this energy determined as Money saved 5 (Total energy)(Unit cost of energy) 5 (66,000 kWh)($0.12/kWh) 5 $7920 Discussion The annual electric energy production also could be determined in kJ by unit manipulations as Total energy 5 (30 kW)(2200 h)a 3600 s 1 h b a1 kJ/s 1 kW b 5 2.38 3 108 kJ which is equivalent to 66,000 kWh (1 kWh = 3600 kJ). We all know from experience that units can give terrible headaches if they are not used carefully in solving a problem. However, with some attention and skill, units can be used to our advantage. They can be used to check formulas; sometimes they can even be used to derive formulas, as explained in the following example. FIGURE 1–12 A wind turbine, as discussed in Example 1–1. ©Bear Dancer Studios/Mark Dierker RF cen98179_ch01_001-050.indd 8 11/28/13 3:14 PM
9 CHAPTER 1 ■EXAMPLE1-2 Obtaining Formulas from Unit Considerations A tank is filled with oil whose density is p=850 kg/m3.If the volume of the tank is V=2 m3,determine the amount of mass m in the tank. U-2m3 SOLUTION The volume of an oil tank is given.The mass of oil is to be p=850kg/m3 determined. m=? Assumptions Oil is a nearly incompressible substance and thus its density is constant. Analysis A sketch of the system just described is given in Fig.1-13. FIGURE 1-13 Suppose we forgot the formula that relates mass to density and volume. Schematic for Example 1-2. However,we know that mass has the unit of kilograms.That is,whatever calculations we do,we should end up with the unit of kilograms.Putting the given information into perspective,we have p=850 kg/m3 and V=2m3 CAUTON It is obvious that we can eliminate m3 and end up with kg by multiplying these two quantities.Therefore,the formula we are looking for should be EVERY TERM IN AN m=pV EQUATION MUST HAVE Thus, THE SAME UNITS m=(850kgm3)(2m3)=1700kg Discussion Note that this approach may not work for more complicated for- mulas.Nondimensional constants also may be present in the formulas,and these cannot be derived from unit considerations alone. You should keep in mind that a formula that is not dimensionally homo- geneous is definitely wrong(Fig.1-14),but a dimensionally homogeneous formula is not necessarily right. Unity Conversion Ratios FIGURE 1-14 Just as all nonprimary dimensions can be formed by suitable combina- Always check the units in your tions of primary dimensions,all nonprimary units (secondary units)can be calculations. formed by combinations of primary units.Force units,for example,can be expressed as IN=1kg m amd11bf=32.1741bm3 ft 32.1741bm-fs2 1 kg'm/s2 They can also be expressed more conveniently as unity conversion ratios as 1w 1 kPa 1J/s1000Nm 1000N/m2 IN I kgm/?=1 and 1 lbf 32.1741bms=1 0.3048m 1 Ibm 1 ft 60s 0.45359kg Unity conversion ratios are identically equal to 1 and are unitless,and thus such ratios (or their inverses)can be inserted conveniently into any FIGURE 1-15 calculation to properly convert units (Fig.1-15).You are encouraged to Every unity conversion ratio(as well always use unity conversion ratios such as those given here when converting as its inverse)is exactly equal to one. units.Some textbooks insert the archaic gravitational constant g defined as Shown here are a few commonly used ge=32.174 lbm-ft/lbf-s2=1 kg-m/N-s2 =1 into equations in order to force unity conversion ratios
9 CHAPTER 1 EXAMPLE 1–2 Obtaining Formulas from Unit Considerations A tank is filled with oil whose density is r 5 850 kg/m3. If the volume of the tank is V 5 2 m3, determine the amount of mass m in the tank. SOLUTION The volume of an oil tank is given. The mass of oil is to be determined. Assumptions Oil is a nearly incompressible substance and thus its density is constant. Analysis A sketch of the system just described is given in Fig. 1–13. Suppose we forgot the formula that relates mass to density and volume. However, we know that mass has the unit of kilograms. That is, whatever calculations we do, we should end up with the unit of kilograms. Putting the given information into perspective, we have r 5 850 kg/m3 and V 5 2 m3 It is obvious that we can eliminate m3 and end up with kg by multiplying these two quantities. Therefore, the formula we are looking for should be m 5 rV Thus, m 5 (850 kg/m3 )(2 m3 ) 5 1700 kg Discussion Note that this approach may not work for more complicated formulas. Nondimensional constants also may be present in the formulas, and these cannot be derived from unit considerations alone. You should keep in mind that a formula that is not dimensionally homogeneous is definitely wrong (Fig. 1–14), but a dimensionally homogeneous formula is not necessarily right. Unity Conversion Ratios Just as all nonprimary dimensions can be formed by suitable combinations of primary dimensions, all nonprimary units (secondary units) can be formed by combinations of primary units. Force units, for example, can be expressed as 1 N 5 1 kg m s2 and 1 lbf 5 32.174 lbm ft s2 They can also be expressed more conveniently as unity conversion ratios as 1 N 1 kg·m/s2 5 1 and 1 lbf 32.174 lbm·ft/s2 5 1 Unity conversion ratios are identically equal to 1 and are unitless, and thus such ratios (or their inverses) can be inserted conveniently into any calculation to properly convert units (Fig. 1–15). You are encouraged to always use unity conversion ratios such as those given here when converting units. Some textbooks insert the archaic gravitational constant gc defined as gc 5 32.174 lbm·ft/lbf·s2 5 1 kg·m/N·s2 5 1 into equations in order to force FIGURE 1–13 Schematic for Example 1–2. Oil = 2 m3 m = ? ρ = 850 kg/m3 FIGURE 1–14 Always check the units in your calculations. FIGURE 1–15 Every unity conversion ratio (as well as its inverse) is exactly equal to one. Shown here are a few commonly used unity conversion ratios. 0.3048 m 1 ft 1 min 60 s 1 lbm 0.45359 kg 32.174 lbm?ft/s2 1 lbf 1 kg?m/s2 1 N 1 kPa 1000 N/m2 1 kJ 1000 N?m 1 W 1 J/s cen98179_ch01_001-050.indd 9 11/28/13 3:14 PM
10 INTRODUCTION AND BASIC CONCEPTS units to match.This practice leads to unnecessary confusion and is strongly discouraged by the present authors.We recommend that you instead use unity conversion ratios. Ibm EXAMPLE 1-3 The Weight of One Pound-Mass Using unity conversion ratios,show that 1.00 Ibm weighs 1.00 Ibf on earth (Fig.1-16). SOLUTION A mass of 1.00 Ibm is subjected to standard earth gravity.Its weight in Ibf is to be determined. FIGURE 1-16 Assumptions Standard sea-level conditions are assumed. A mass of I Ibm weighs 1 Ibf on earth. Properties The gravitational constant is g =32.174 ft/s2. Analysis We apply Newton's second law to calculate the weight (force)that corresponds to the known mass and acceleration.The weight of any object is equal to its mass times the local value of gravitational acceleration.Thus, Weight? 不ewton I thought gram Crinch-Cereal 1 Ibf was a unit of mass! W=mg=(1.001bm)(32.174ft/s2 =1.001bf 32.1741bm-ft/s2 Net weigl加t One pound (454 grams) Discussion The quantity in large parentheses in this equation is a unity con- version ratio.Mass is the same regardless of its location.However,on some other planet with a different value of gravitational acceleration,the weight of 1 Ibm would differ from that calculated here. When you buy a box of breakfast cereal,the printing may say "Net weight:One pound (454 grams)."(See Fig.1-17.)Technically,this means that the cereal inside the box weighs 1.00 lbf on earth and has a mass of 453.6 g(0.4536 kg).Using Newton's second law,the actual weight of the cereal on earth is 1kg W=mg=(453.6g)(9.81m/s2 IN =4.49N 、1kgm/s2 1000g 1-3-SYSTEMS AND CONTROL VOLUMES FIGURE 1-17 A quirk in the metric system of units. A system is defined as a quantity of matter or a region in space chosen for study.The mass or region outside the system is called the surroundings. The real or imaginary surface that separates the system from its surround- Surroundings ings is called the boundary (Fig.1-18).The boundary of a system can be fixed or movable.Note that the boundary is the contact surface shared by both the system and the surroundings.Mathematically speaking,the bound- System ary has zero thickness,and thus it can neither contain any mass nor occupy any volume in space. Systems may be considered to be closed or open,depending on whether a Boundary fixed mass or a fixed volume in space is chosen for study.A closed system (also known as a control mass or just system when the context makes it FIGURE 1-18 clear)consists of a fixed amount of mass,and no mass can cross its boun- System,surroundings,and boundary. dary.That is,no mass can enter or leave a closed system,as shown in
10 INTRODUCTION AND BASIC CONCEPTS units to match. This practice leads to unnecessary confusion and is strongly discouraged by the present authors. We recommend that you instead use unity conversion ratios. EXAMPLE 1–3 The Weight of One Pound-Mass Using unity conversion ratios, show that 1.00 lbm weighs 1.00 lbf on earth (Fig. 1–16). SOLUTION A mass of 1.00 lbm is subjected to standard earth gravity. Its weight in lbf is to be determined. Assumptions Standard sea-level conditions are assumed. Properties The gravitational constant is g 5 32.174 ft/s2. Analysis We apply Newton’s second law to calculate the weight (force) that corresponds to the known mass and acceleration. The weight of any object is equal to its mass times the local value of gravitational acceleration. Thus, W 5 mg 5 (1.00 lbm)(32.174 ft/s2 )a 1 lbf 32.174 lbm·ft/s2 b 5 1.00 lbf Discussion The quantity in large parentheses in this equation is a unity conversion ratio. Mass is the same regardless of its location. However, on some other planet with a different value of gravitational acceleration, the weight of 1 lbm would differ from that calculated here. When you buy a box of breakfast cereal, the printing may say “Net weight: One pound (454 grams).” (See Fig. 1–17.) Technically, this means that the cereal inside the box weighs 1.00 lbf on earth and has a mass of 453.6 g (0.4536 kg). Using Newton’s second law, the actual weight of the cereal on earth is W 5 mg 5 (453.6 g)(9.81 m/s2 )a 1 N 1 kg·m/s2 b a 1 kg 1000 gb 5 4.49 N 1–3 ■ SYSTEMS AND CONTROL VOLUMES A system is defined as a quantity of matter or a region in space chosen for study. The mass or region outside the system is called the surroundings. The real or imaginary surface that separates the system from its surroundings is called the boundary (Fig. 1–18). The boundary of a system can be fixed or movable. Note that the boundary is the contact surface shared by both the system and the surroundings. Mathematically speaking, the boundary has zero thickness, and thus it can neither contain any mass nor occupy any volume in space. Systems may be considered to be closed or open, depending on whether a fixed mass or a fixed volume in space is chosen for study. A closed system (also known as a control mass or just system when the context makes it clear) consists of a fixed amount of mass, and no mass can cross its boundary. That is, no mass can enter or leave a closed system, as shown in FIGURE 1–16 A mass of 1 lbm weighs 1 lbf on earth. lbm FIGURE 1–17 A quirk in the metric system of units. Net weight: One pound (454 grams) FIGURE 1–18 System, surroundings, and boundary. Surroundings Boundary System cen98179_ch01_001-050.indd 10 11/28/13 3:14 PM
11 CHAPTER 1 Fig.1-19.But energy,in the form of heat or work,can cross the boundary: and the volume of a closed system does not have to be fixed.If,as a special case,even energy is not allowed to cross the boundary,that system is called Mass No an isolated system. Closed Consider the piston-cylinder device shown in Fig.1-20.Let us say that system we would like to find out what happens to the enclosed gas when it is m constant heated.Since we are focusing our attention on the gas,it is our system.The inner surfaces of the piston and the cylinder form the boundary,and since Energy Yes no mass is crossing this boundary,it is a closed system.Notice that energy may cross the boundary,and part of the boundary (the inner surface of the piston,in this case)may move.Everything outside the gas,including the FIGURE 1-19 piston and the cylinder,is the surroundings. Mass cannot cross the boundaries of a An open system,or a control volume,as it is often called,is a prop- closed system,but energy can. erly selected region in space.It usually encloses a device that involves mass flow such as a compressor,turbine,or nozzle.Flow through these devices is best studied by selecting the region within the device as the control volume.Both mass and energy can cross the boundary of a con- Moving boundary trol volume. A large number of engineering problems involve mass flow in and out of Gas a system and,therefore,are modeled as control volumes.A water heater,a 2kg Gas car radiator,a turbine,and a compressor all involve mass flow and should 1.5m3 be analyzed as control volumes (open systems)instead of as control masses (closed systems).In general,any arbitrary region in space can be selected as a control volume.There are no concrete rules for the selec- tion of control volumes,but the proper choice certainly makes the analysis Fixed much easier.If we were to analyze the flow of air through a nozzle,for boundary example,a good choice for the control volume would be the region within FIGURE 1-20 the nozzle. The boundaries of a control volume are called a control surface,and A closed system with a moving boundary. they can be real or imaginary.In the case of a nozzle,the inner surface of the nozzle forms the real part of the boundary,and the entrance and exit areas form the imaginary part,since there are no physical surfaces there (Fig.1-21a). Imaginary boundary Real boundary Moving boundary CV CV (a nozzle) Fixed boundary FIGURE 1-21 (a)A control volume (CV)with real and (b)A control volume (CV)with fixed and A control volume can involve imaginary boundaries moving boundaries as well as real and fixed,moving,real,and imaginary imaginary boundaries boundaries
11 CHAPTER 1 Fig. 1–19. But energy, in the form of heat or work, can cross the boundary; and the volume of a closed system does not have to be fixed. If, as a special case, even energy is not allowed to cross the boundary, that system is called an isolated system. Consider the piston-cylinder device shown in Fig. 1–20. Let us say that we would like to find out what happens to the enclosed gas when it is heated. Since we are focusing our attention on the gas, it is our system. The inner surfaces of the piston and the cylinder form the boundary, and since no mass is crossing this boundary, it is a closed system. Notice that energy may cross the boundary, and part of the boundary (the inner surface of the piston, in this case) may move. Everything outside the gas, including the piston and the cylinder, is the surroundings. An open system, or a control volume, as it is often called, is a properly selected region in space. It usually encloses a device that involves mass flow such as a compressor, turbine, or nozzle. Flow through these devices is best studied by selecting the region within the device as the control volume. Both mass and energy can cross the boundary of a control volume. A large number of engineering problems involve mass flow in and out of a system and, therefore, are modeled as control volumes. A water heater, a car radiator, a turbine, and a compressor all involve mass flow and should be analyzed as control volumes (open systems) instead of as control masses (closed systems). In general, any arbitrary region in space can be selected as a control volume. There are no concrete rules for the selection of control volumes, but the proper choice certainly makes the analysis much easier. If we were to analyze the flow of air through a nozzle, for example, a good choice for the control volume would be the region within the nozzle. The boundaries of a control volume are called a control surface, and they can be real or imaginary. In the case of a nozzle, the inner surface of the nozzle forms the real part of the boundary, and the entrance and exit areas form the imaginary part, since there are no physical surfaces there (Fig. 1–21a). FIGURE 1–19 Mass cannot cross the boundaries of a closed system, but energy can. Closed system Yes m = constant Energy NoMass FIGURE 1–20 A closed system with a moving boundary. Gas 2 kg Gas 1.5 m3 2 kg 1 m3 Moving boundary Fixed boundary FIGURE 1–21 A control volume can involve fixed, moving, real, and imaginary boundaries. Real boundary (a) A control volume (CV) with real and imaginary boundaries Imaginary boundary CV (a nozzle) CV Moving boundary Fixed boundary (b) A control volume (CV) with fixed and moving boundaries as well as real and imaginary boundaries cen98179_ch01_001-050.indd 11 11/28/13 3:14 PM
12 INTRODUCTION AND BASIC CONCEPTS A control volume can be fixed in size and shape,as in the case of a nozzle,or it may involve a moving boundary,as shown in Fig.1-21b.Most control volumes,however,have fixed boundaries and thus do not involve any moving boundaries.A control volume can also involve heat and work interactions just as a closed system,in addition to mass interaction. As an example of an open system,consider the water heater shown in Fig.1-22.Let us say that we would like to determine how much heat we must transfer to the water in the tank in order to supply a steady stream of hot water.Since hot water will leave the tank and be replaced by cold water, it is not convenient to choose a fixed mass as our system for the analy- sis.Instead,we can concentrate our attention on the volume formed by the interior surfaces of the tank and consider the hot and cold water streams as mass leaving and entering the control volume.The interior surfaces of the tank form the control surface for this case,and mass is crossing the control surface at two locations. In an engineering analysis,the system under study must be defined care- fully.In most cases,the system investigated is quite simple and obvious, and defining the system may seem like a tedious and unnecessary task.In other cases,however,the system under study may be rather involved,and a proper choice of the system may greatly simplify the analysis. 1-4.PROPERTIES OF A SYSTEM FIGURE 1-22 An open system(a control volume) Any characteristic of a system is called a property.Some familiar proper- with one inlet and one exit. ties are pressure p temperature T,volume V,and mass m.The list can be McGraw-Hill Education,Christopher Kerrigan extended to include less familiar ones such as viscosity,thermal conductiv- ity,modulus of elasticity,thermal expansion coefficient,electric resistivity, and even velocity and elevation. Properties are considered to be either intensive or extensive.Intensive properties are those that are independent of the mass of a system,such as temperature,pressure,and density.Extensive properties are those whose values depend on the size-or extent-of the system.Total mass,total vol- ume,and total momentum are some examples of extensive properties.An easy way to determine whether a property is intensive or extensive is to T divide the system into two equal parts with an imaginary partition,as shown P in Fig.1-23.Each part will have the same value of intensive properties as the original system,but half the value of the extensive properties. Generally,uppercase letters are used to denote extensive properties (with mass m being a major exception),and lowercase letters are used for intensive properties (with pressure P and temperature T being the obvious 方m Extensive exceptions). properties Extensive properties per unit mass are called specific properties.Some examples of specific properties are specific volume (v=Vm)and specific P Intensive total energy (eE/m). properties Continuum FIGURE 1-23 Matter is made up of atoms that are widely spaced in the gas phase.Yet it Criterion to differentiate intensive and is very convenient to disregard the atomic nature of a substance and view it extensive properties. as a continuous,homogeneous matter with no holes,that is,a continuum
12 INTRODUCTION AND BASIC CONCEPTS A control volume can be fixed in size and shape, as in the case of a nozzle, or it may involve a moving boundary, as shown in Fig. 1–21b. Most control volumes, however, have fixed boundaries and thus do not involve any moving boundaries. A control volume can also involve heat and work interactions just as a closed system, in addition to mass interaction. As an example of an open system, consider the water heater shown in Fig. 1–22. Let us say that we would like to determine how much heat we must transfer to the water in the tank in order to supply a steady stream of hot water. Since hot water will leave the tank and be replaced by cold water, it is not convenient to choose a fixed mass as our system for the analysis. Instead, we can concentrate our attention on the volume formed by the interior surfaces of the tank and consider the hot and cold water streams as mass leaving and entering the control volume. The interior surfaces of the tank form the control surface for this case, and mass is crossing the control surface at two locations. In an engineering analysis, the system under study must be defined carefully. In most cases, the system investigated is quite simple and obvious, and defining the system may seem like a tedious and unnecessary task. In other cases, however, the system under study may be rather involved, and a proper choice of the system may greatly simplify the analysis. 1–4 ■ PROPERTIES OF A SYSTEM Any characteristic of a system is called a property. Some familiar properties are pressure P, temperature T, volume V, and mass m. The list can be extended to include less familiar ones such as viscosity, thermal conductivity, modulus of elasticity, thermal expansion coefficient, electric resistivity, and even velocity and elevation. Properties are considered to be either intensive or extensive. Intensive properties are those that are independent of the mass of a system, such as temperature, pressure, and density. Extensive properties are those whose values depend on the size—or extent—of the system. Total mass, total volume, and total momentum are some examples of extensive properties. An easy way to determine whether a property is intensive or extensive is to divide the system into two equal parts with an imaginary partition, as shown in Fig. 1–23. Each part will have the same value of intensive properties as the original system, but half the value of the extensive properties. Generally, uppercase letters are used to denote extensive properties (with mass m being a major exception), and lowercase letters are used for intensive properties (with pressure P and temperature T being the obvious exceptions). Extensive properties per unit mass are called specific properties. Some examples of specific properties are specific volume (v 5 V/m) and specific total energy (e 5 E/m). Continuum Matter is made up of atoms that are widely spaced in the gas phase. Yet it is very convenient to disregard the atomic nature of a substance and view it as a continuous, homogeneous matter with no holes, that is, a continuum. FIGURE 1–22 An open system (a control volume) with one inlet and one exit. © McGraw-Hill Education, Christopher Kerrigan FIGURE 1–23 Criterion to differentiate intensive and extensive properties. cen98179_ch01_001-050.indd 12 11/28/13 3:14 PM