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Solution - Koretsky/Solution - Koretsky/ch01.pdf Chapter 1 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu 2 1.2 An approximate solution can be found if we combine Equations 1.4 and 1.5: molecular keVm =22 1 r molecular kekT =2 3 m kTV 3≈∴ r Assume the temperature is 22 ºC. The mass of a single oxygen molecule is kg 1014.5 26−×=m . Substitute and solve: [ ]m/s 6.487=Vr The molecules are traveling really, fast (around the length of five football fields every second). Comment: We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is sketched in Figure 1.4. Looking up the quantitative expression for this expression, we have: dvvv kT m kT mdvvf 22 2/3 2 exp 2 4)( ⎭⎬ ⎫ ⎩⎨ ⎧−⎟⎠ ⎞⎜⎝ ⎛= ππ where f(v) is the fraction of molecules within dv of the speed v. We can find the average speed by integrating the expression above [ ]m/s 4498 )( )( 0 0 === ∫ ∫ ∞ ∞ m kT dvvf vdvvf V π r 3 1.3 Derive the following expressions by combining Equations 1.4 and 1.5: a a m kTV 32 =r b b m kTV 32 =r Therefore, a b b a m m V V = 2 2 r r Since mb is larger than ma, the molecules of species A move faster on average. 4 1.4 We have the following two points that relate the Reamur temperature scale to the Celsius scale: ( )Reamurº 0 C,º 0 and ( )Reamurº 80 C,º 100 Create an equation using the two points: ( ) ( ) Celsius º 8.0Reamurº TT = At 22 ºC, Reamurº 6.17 =T 5 1.5 (a) After a short time, the temperature gradient in the copper block is changing (unsteady state), so the system is not in equilibrium. (b) After a long time, the temperature gradient in the copper block will become constant (steady state), but because the temperature is not uniform everywhere, the system is not in equilibrium. (c) After a very long time, the temperature of the reservoirs will equilibrate; The system is then homogenous in temperature. The system is in thermal equilibrium. 6 1.6 We assume the temperature is constant at 0 ºC. The molecular weight of air is kg/mol 0.029g/mol 29 ==MW Find the pressure at the top of Mount Everest: ( ) ( ) [ ]( )( ) ( ) ⎥⎥ ⎥⎥ ⎦ ⎤ ⎢⎢ ⎢⎢ ⎣ ⎡ ⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅ −= K 73.152 Kmol J 314.8 m 8848m/s .819kg/mol029.0expatm 1P kPa 4.33atm 330.0 ==P Interpolate steam table data: Cº 4.71=satT for kPa 4.33=satP Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship given assumes that the temperature remains constant. In reality the temperature decreases with height as we go up the mountain. However, a solution in which T and P vary with height is not as straight-forward. 7 1.7 To solve these problems, the steam tables were used. The values given for each part constrain the water to a certain state. In most cases we can look at the saturated table, to determine the state. (a) Subcooled liquid Explanation: the saturation pressure at T = 170 [oC] is 0.79 [MPa] (see page 508); Since the pressure of this state, 10 [bar], is greater than the saturation pressure, water is a liquid. (b) Saturated vapor-liquid mixture Explanation: the specific volume of the saturated vapor at T = 70 [oC] is 5.04 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 508); Since the volume of this state, 3 [m3/kg], is in between these values we have a saturated vapor- liquid mixture. (c) Superheated vapor Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa], is 0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 511); Since the volume of this state, 0.05 [m3/kg], is greater than this value, it is a vapor. (d) Superheated vapor Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5 [MPa], is 6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state, 7.0592 [kJ/(kg K)], is greater than this value, it is a vapor. In fact, if we go to the superheated water vapor tables for P = 500 [kPa], we see the state is constrained to T = 200 [oC]. 8 1.8 From the steam tables in Appendix B.1: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 003155.0ˆ 3 criticalv ( )MPa 089.22 C,º 15.374 == PT At 10 bar, we find in the steam tables ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 001127.0ˆ 3 sat lv ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 19444.0ˆ 3 sat vv Because the total mass and volume of the closed, rigid system remain constant as the water condenses, we can develop the following expression: ( ) satvsatlcritical vxvxv ˆˆ1ˆ +−= where x is the quality of the water. Substituting values and solving for the quality, we obtain 0105.0=x or 1.05 % A very small percentage of mass in the final state is vapor. 9 1.9 The calculation methods will be shown for part (a), but not parts (b) and (c) (a) Use the following equation to estimate the specific volume: ( ) ( ) ( ) ( )[ ]Cº 250 MPa, 8.1ˆCº 250 MPa, 0.2ˆ5.0Cº 250 MPa, 8.1ˆCº 250 MPa, 9.1ˆ vvvv −+= Substituting data from the steam tables, ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 11821.0Cº 250 MPa, 9.1ˆ 3 v From the NIST website: ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 11791.0Cº 250 MPa, 9.1ˆ 3 NISTv Therefore, assuming the result from NIST is more accurate % 254.0%100 ˆ ˆˆ Difference % =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×−= NIST NIST v vv (b) Linear interpolation: ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 13284.0Cº 300 MPa, 9.1ˆ 3 v NIST website: ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 13249.0Cº 300 MPa, 9.1ˆ 3 NISTv Therefore, % 264.0Difference % = (c) Linear interpolation: ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 12406.0Cº 270 MPa, 9.1ˆ 3 v 10 Note: Double interpolation is required to determine this value. First, find the molar volumes at 270 ºC and 1.8 MPa and 2.0 MPa using interpolation. Then, interpolate between the results from the previous step to find the molar volume at 270 ºC and 1.9 MPa. NIST website: ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 12389.0Cº 270 MPa, 9.1ˆ 3 NISTv Therefore, % 137.0Difference % = With regards to parts (a), (b), and (c), the values found using interpolation and the NIST website agree very well. The discrepancies will not significantly affect the accuracy of any subsequent calculations. 11 1.10 For saturated temperature data at 25 ºC in the steam tables, ⎥⎦ ⎤⎢⎣ ⎡=⎥⎦ ⎤⎢⎣ ⎡= kg L 003.1 kg m 001003.0ˆ 3 sat lv Determine the mass: [ ] kg 997.0 kg L 003.1 L 1 ˆ = ⎥⎦ ⎤⎢⎣ ⎡== satlv Vm Because molar volumes of liquids do not depend strongly on pressure, the mass of water at 25 ºC and atmospheric pressure in a one liter should approximately be equal to the mass calculated above unless the pressure is very, very large. 12 1.11 First, find the overall specific volume of the water in the container: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡== kg m 2.0 kg 5 m 1ˆ 33 v Examining the data in the saturated steam tables, we find satv sat l vvv ˆˆˆ << at bar 2=satP Therefore, the system contains saturated water and the temperature is Cº 23.120== satTT Let lm represent the mass of the water in the container that is liquid, and vm represent the mass of the water in the container that is gas. These two masses are constrained as follows: kg 5=+ vl mm We also have the extensive volume of the system equal the extensive volume of each phase Vvmvm satvv sat ll =+ ˆˆ ( ) ( ) 333 m1/kgm8857.0kg/m 001061.0 =+ vl mm Solving these two equations simultaneously, we obtain kg 88.3=lm kg12.1=vm Thus, the quality is 224.0 kg 5 kg 12.1 === m mx v The internal energy relative to the reference state in the saturated steam table is satvv sat ll umumU ˆˆ += From the steam tables: 13 [ ]kJ/kg 47.504ˆ =satlu [ ]kJ/kg 5.2529ˆ =satvu Therefore, ( ) [ ]( ) ( ) [ ]( )kJ/kg 529.52kg .121kJ/kg 47.504kg 88.3 +=U kJ 4.4790=U 14 1.12 First, determine the total mass of water in the container. Since we know that 10 % of the mass is vapor, we can write the following expression ( ) ( )[ ]satvsatl vvmV ˆ1.0ˆ9.0 += From the saturated steam tables for MPa 1=satP ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 001127.0ˆ 3 sat lv ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 1944.0ˆ 3 sat vv Therefore, ( ) ( ) ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡+⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡=+= kg m 1944.01.0 kg m 001127.09.0 m 1 ˆ1.0ˆ9.0 33 3 sat v sat l vv Vm kg 9.48=m and ( ) ( )( )kg/m 1944.0kg 9.481.0ˆ1.0 3== satvv vmV 3m 950.0=vV 15 1.13 In a spreadsheet, make two columns: one for the specific volume of water and another for the pressure. First, copy the specific volumes of liquid water from the steam tables and the corresponding saturation pressures. After you have finished tabulating pressure/volume data for liquid water, list the specific volumes and saturation pressures of water vapor. Every data point is not required, but be sure to include extra points near the critical values. The data when plotted on a logarithmic scale should look like the following plot. The Vapor-Liquid Dome for H2O 0.0010 0.0100 0.1000 1.0000 10.0000 100.0000 0.0001 0.001 0.01 0.1 1 10 100 1000 Specific Volume, v (m3/kg) Pr es su re , P (M Pa ) Vapor-Liquid VaporLiquid Critical Point 16 1.14 The ideal gas law can be rewritten as P RTv = For each part in the problem, the appropriate values are substituted into this equation where ⎥⎦ ⎤⎢⎣ ⎡ ⋅ ⋅= Kmol barL 08314.0R is used. The values are then converted to the units used in the steam table. (a) ⎥⎦ ⎤⎢⎣ ⎡=⎥⎦ ⎤⎢⎣ ⎡ ⎥⎦ ⎤⎢⎣ ⎡ ⎥⎦ ⎤⎢⎣ ⎡ == ⎥⎦ ⎤⎢⎣ ⎡= − kg m 70.1 L m 10 mol kg 018.0 mol L 7.30 ˆ mol L 7.30 33 3 MW vv v For part (a), the calculation of the percent error will be demonstrated. The percent error will be based on percent error from steam table data, which should be more accurate than using the ideal gas law. The following equation is used: %100 ˆ ˆˆ Error% ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×−= ST STIG v vv where IGv is the value calculated using the ideal gas law and STv is the value from the steam table. %62.1% kg m 6729.1 kg m 6729.1 kg m 70.1 Error% 3 33 = ⎟⎟ ⎟⎟ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎜⎜ ⎜⎜ ⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡− ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ = This result suggests that you would introduce an error of around 1.6% if you characterize boiling water at atmospheric pressure as an ideal gas. 17 (b) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= kg m 57.3ˆ mol L 3.64 3 v v =Error% 0.126% For a given pressure, when the temperature is raised the gas behaves more like an ideal gas. (c) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= kg m 0357.0ˆ mol L 643.0 3 v v %Error = 8.87% (d) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= kg m 0588.0ˆ mol L 06.1 3 v v %Error = 0.823% The largest error occurs at high P and low T. 18 1.15 The room I am sitting in right now is approximately 16 ft long by 12 feet wide by 10 feet tall – your answer should vary. The volume of the room is ( )( )( ) 33 m 4.54ft 1920ft 10ft 16ft 12 ===V The room is at atmospheric pressure and a temperature of 22 ºC. Calculate the number of moles using the ideal gas law: ( )( ) ( )K 15.295 Kmol J 314.8 m 4.54Pa 1001325.1 35 ⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅ ×== RT PVn mol 2246=n Use the molecular weight of air to obtain the mass: ( ) ( )( ) kg 2.65kg/mol 029.0mol 27.2246 === MWnm That’s pretty heavy. 19 1.16 First, find the total volume that one mole of gas occupies. Use the ideal gas law: ( ) ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡=× ⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅== mol m 0244.0 Pa 101 K 15.293 Kmol J 314.8 3 5P RTv Now calculate the volume occupied by the molecules: ⎟⎠ ⎞⎜⎝ ⎛=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛×⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= 3 3 4 molcule one of Volume moleper molecules ofNumber rNv A occupied π ( ) ⎥⎦⎤⎢⎣⎡ ×⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×= − 310 23 m 105.1 3 4 mol 1 molecules 10022.6 πoccupiedv ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡×= − mol m 1051.8 3 6occupiedv Determine the percentage of the total volume occupied by the molecules: % 0349.0%100Percentage = ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ×= v voccupied A very small amount of space, indeed. 20 1.17 Water condenses on your walls when the water is in liquid-vapor equilibrium. To find the maximum allowable density at 70 ºF, we need to find the smallest density of water vapor at 40 ºF which satisfies equilibrium conditions. In other words, we must find the saturation density of water vapor at 40 ºF. From the steam tables, ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 74.153ˆ 3 sat vv (sat. water vapor at 40 ºF = 4.44 ºC) We must correct the specific volume for the day-time temperature of 70 ºF (21.1 ºC ) with the ideal gas law. 3.294 K 294.3 6.277 K 6.277, ˆˆ T v T vsatv = ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡== kg m 163 kg m 74.153 K 6.277 K 3.294ˆˆ 33 K 6.277, 6.277 3.294 K 294.3 sat vvT Tv Therefore, ⎥⎦ ⎤⎢⎣ ⎡×== − 33 m kg 1013.6 ˆ 1ˆ sat vv ρ If the density at 70 ºF is greater than or equal to [ ]33 kg/m 1013.6 −× , the density at night when the temperature is 40 ºF will be greater than the saturation density, so water will condense onto the wall. However, if the density is less than [ ]33 kg/m 1013.6 −× , then saturation conditions will not be obtained, and water will not condense onto the walls. 21 1.18 We consider the air inside the soccer ball as the system. We can answer this question by looking at the ideal gas law: RTPv = If we assume it is a closed system, it will have the same number of moles of air in the winter as the summer. However, it is colder in the winter (T is lower), so the ideal gas law tells us that Pv will be lower and the balls will be under inflated. Alternatively, we can argue that the higher pressure inside the ball causes air to leak out over time. Thus we have an open system and the number of moles decrease with time – leading to the under inflation. 22 1.19 First, write an equation for the volume of the system in its initial state: Vvmvm satvv sat ll =+ ˆˆ Substitute ( )mxml −= 1 and xmmv = : ( )[ ] Vvxvxm satvsatl =+− ˆˆ1 At the critical point Vvm critical =ˆ Since the mass and volume don’t change ( ) satvsatlcritical vxvxv ˆˆ1ˆ +−= From the steam tables in Appendix B.1: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 003155.0ˆ 3 criticalv ( )MPa 089.22 C,º 15.374 == PT At 0.1 MPa, we find in the steam tables ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 001043.0ˆ 3 sat lv ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 6940.1ˆ 3 sat vv Therefore, solving for the quality, we get 00125.0=x or 0.125 % 99.875% of the water is liquid. 23 1.20 As defined in the problem statement, the relative humidity can be calculated as follows saturationat water of mass waterof massHumidity Relative = We can obtain the saturation pressure at each temperature from the steam tables. At 10 [oC], the saturation pressure of water is 1.22 [kPa]. This value is proportional to the mass of water in the vapor at saturation. For 90% relative humidity, the partial pressure of water in the vapor is: [kPa] 10.122.19.0 =×=Waterp This value represents the vapor pressure of water in the air. At 30 [oC], the saturation pressure of water is 4.25 [kPa]. For 590% relative humidity, the partial pressure of water in the vapor is: [kPa] 12.263.55.0 =×=Waterp Since the total pressure in each case is the same (atmospheric), the partial pressure is proportional to the mass of water in the vapor. We conclude there is about twice the amount of water in the air in the latter case. 24 1.21 (a) When you have extensive variables, you do not need to know how many moles of each substance are present. The volumes can simply be summed. ba VVV +=1 (b) The molar volume, v1, is equal to the total volume divided by the total number of moles. ba ba tot nn VV n Vv + +== 11 We can rewrite the above equation to include molar volumes for species a and b. b ba b a ba a ba bbaa v nn nv nn n nn vnvnv +++=+ +=1 bbaa vxvxv +=1 (c) The relationship developed in Part (a) holds true for all extensive variables. ba KKK +=1 (d) bbaa kxkxk +=1 Solution - Koretsky/Solution - Koretsky/ch02.pdf Chapter 2 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University koretsm@engr.orst.edu 2 2.1 There are many possible solutions to this problem. Assumptions must be made to solve the problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the towel when you dry yourself. In other words, let [ ]kg 5.0 2 =OHm Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy balance and neglecting potential and kinetic energy effects reveals hq ˆˆ ∆= Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find the minimum energy required for drying the towel, assume that the temperature of the towel remains constant at K 298.15Cº 25 ==T . In the drying process, the absorbed water is vaporized into steam. Therefore, the expression for heat is l OH v OH hhq 22 ˆˆˆ −= where is v OHh 2 ˆ is the specific enthalpy of water vapor at bar 01.1=P and K 15.298=T and l OHh 2 ˆ is the specific enthalpy of liquid water at bar 01.1=P and K 15.298=T . A hypothetical path must be used to calculate the change in enthalpy. Refer to the diagram below P = 1 [atm] liquid vapor liquid vapor ∆ ˆ h 1 P ∆ ˆ h 2 ∆ ˆ h 3 ∆ ˆ h Psat 1 atm 3.17 kPa By adding up each step of the hypothetical path, the expression for heat is ( ) ( )[ ] ( ) ( )[ ] ( ) ( )[ ]Cº 25ˆbar 01.1 C,º 25ˆ Cº 25 ˆCº 25ˆbar 1.01 C,º 25 ˆCº 25ˆ ˆ , ,,, 321 22 2222 satv OH v OH satl OH satv OH l OH satl OH hh hhhh hhhq −+ −+−= ∆+∆+∆= 3 However, the calculation of heat can be simplified by treating the water vapor as an ideal gas, which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the hypothetical path containing the pressure change of the liquid can be neglected. This leaves ( ) ( )kPa 3.17 C,º 25ˆ kPa .173 C,º 25ˆˆ 22 l OH v OH hhq −= From the steam tables: ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 2.2547ˆ , 2 satv OHh (sat. H2O vapor at 25 ºC) ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 87.104ˆ , 2 satl OHh (sat. H2O liquid at 25 ºC) which upon substitution gives ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 3.2442qˆ Therefore, [ ]( ) [ ]kJ 2.1221 kg kJ 442.32kg 5.0 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡=Q To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20 minutes (1200 s) to dry the towel. From the definition of electrical work, [ ]( ) [ ]( ) [ ]( ) [ ]kJ 7488s 1200V 208A 30 === IVtW Therefore, the efficiency is [ ][ ] %3.16%100kJ 7488 kJ 221.21%100 =⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ×=⎟⎠ ⎞⎜⎝ ⎛ ×= W Qη There are a number of ways to improve the drying process. A few are listed below. • Dry the towel outside in the sun. • Use a smaller volume dryer so that less air needs to be heated. • Dry more than one towel at a time since one towel can’t absorb all of the available heat. With more towels, more of the heat will be utilized. 4 2.3 In answering this question, we must distinguish between potential energy and internal energy. The potential energy of a system is the energy the macroscopic system, as a whole, contains relative to position. The internal energy represents the energy of the individual atoms and molecules in the system, which can have contributions from both molecular kinetic energy and molecular potential energy. Consider the compression of a spring from an initial uncompressed state as shown below. M State 1 State 2 ∆x uncompressed compressed Since it requires energy to compress the spring, we know that some kind of energy must be stored within the spring. Since this change in energy can be attributed to a change of the macroscopic position of the system and is not related to changes on the molecular scale, we determine the form of energy to be potential energy. In this case, the spring’s tendency to restore its original shape is the driving force that is analogous to the gravity for gravitational potential energy. This argument can be enhanced by the form of the expression that the increased energy takes. If we consider the spring as the system, the energy it acquires in a reversible, compression from its initial uncompressed state may be obtained from an energy balance. Assuming the process is adiabatic, we obtain: WWQE =+=∆ We have left the energy in terms of the total energy, E. The work can be obtained by integrating the force over the distance of the compression: 2 2 1 kxkxdxdxFW ==⋅−= ∫∫ Hence: 2 2 1 kxE =∆ We see that the increase in energy depends on macroscopic position through the term x. It should be noted that there is a school of thought that assigns this increased energy to internal energy. This approach is all right as long as it is consistently done throughout the energy balances on systems containing springs. 5 2.4 For the first situation, let the rubber band represent the system. In the second situation, the gas is the system. If heat transfer, potential and kinetic energy effects are assumed negligible, the energy balance becomes WU =∆ Since work must be done on the rubber band to stretch it, the value of the work is positive. From the energy balance, the change in internal energy is positive, which means that the temperature of the system rises. When a gas expands in a piston-cylinder assembly, the system must do work to expand against the piston and atmosphere. Therefore, the value of work is negative, so the change in internal energy is negative. Hence, the temperature decreases. In analogy to the spring in Problem 2.3, it can be argued that some of the work imparted into the rubber band goes to increase its potential energy; however, a part of it goes into stretching the polymer molecules which make up the rubber band, and the qualitative argument given above still is valid. 6 2.5 To explain this phenomenon, you must realize that the water droplet is heated from the bottom. At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The water vapor forms an insulation layer between the skillet and the water droplet. At low temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures. 7 2.6 Apartment System Fr id ge + -W Q HOT Surr. If the entire apartment is treated as the system, then only the energy flowing across the apartment boundaries (apartment walls) is of concern. In other words, the energy flowing into or out of the refrigerator is not explicitly accounted for in the energy balance because it is within the system. By neglecting kinetic and potential energy effects, the energy balance becomes WQU +=∆ The Q term represents the heat from outside passing through the apartment’s walls. The W term represents the electrical energy that must be supplied to operate the refrigerator. To determine whether opening the refrigerator door is a good idea, the energy balance with the door open should be compared to the energy balance with the door closed. In both situations, Q is approximately the same. However, the values of W will be different. With the door open, more electrical energy must be supplied to the refrigerator to compensate for heat loss to the apartment interior. Therefore, shutajar WW > where the subscript “ajar” refers the situation where the door is open and the subscript “shut” refers to the situation where the door is closed. Since, shutajar QQ = shutshutshutajarajarajar WQUWQU +=∆>+=∆ shutajar TT ∆>∆∴ The refrigerator door should remain closed. 8 2.7 The two cases are depicted below. System H ea te r Q Surr. W System H ea te r Q Surr. Heater off Heater on Let’s consider the property changes in your house between the following states. State 1, when you leave in the morning, and state, the state of your home after you have returned home and heated it to the same temperature as when you left. Since P and T are identical for states 1 and 2, the state of the system is the same and ∆U must be zero, so 0=+=∆ WQU or WQ =− where -Q is the total heat that escaped between state 1 and state 2 and W is the total work that must be delivered to the heater. The case where more heat escapes will require more work and result in higher energy bills. When the heater is on during the day, the temperature in the system is greater than when it is left off. Since heat transfer is driven by difference in temperature, the heat transfer rate is greater, and W will be greater. Hence, it is cheaper to leave the heater off when you are gone. 9 2.8 The amount of work done at constant pressure can be calculated by applying Equation 2.57 QH =∆ Hence, hmQH ˆ∆==∆ where the specific internal energy is used in anticipation of obtaining data from the steam tables. The mass can be found from the known volume, as follows: [ ]( ) [ ]kg 0.1 kg m0.0010 L m001.0L1 ˆ 3 3 = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ == v Vm As in Example 2.2, we use values from the saturated steam tables at the same temperature for subcooled water at 1 atm. The specific enthalpy is found from values in Appendix B.1: [ ]( ) [ ]( ) ⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡−⎥⎦⎤⎢⎣⎡=−=∆ kgkJ 15.314kgkJ 87.104kgkJ 02.419C25at ˆC100at ˆˆ o1,o2, ll uuu Solve for heat: [ ]( ) [ ]kJ 15.314 kg kJ 05.314kg 0.1ˆ =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡=∆= umQ and heat rate: [ ][ ]( ) [ ] [ ]kW 52.0 min s 60min. 01 kJ 15.314 = ⎟⎠ ⎞⎜⎝ ⎛== t QQ& This value is the equivalent of five strong light bulbs. 10 2.9 (a) From Steam Tables: ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 8.2967ˆ1u (100 kPa, 400 ºC) ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 8.2659ˆ2u (50 kPa, 200 ºC) ⎥⎦ ⎤⎢⎣ ⎡−=−=∆ kg kJ 0.308ˆˆˆ 12 uuu (b) From Equations 2.53 and 2.63 ( )∫∫ −==−=∆ 2 1 2 1 12 T T P T T v dTRcdTcuuu From Appendix A.2 )( 322 ETDTCTBTARcP ++++= − [ ]∫ −++++=∆ −2 1 1322 T T dTETDTCTBTARu Integrating ⎥⎦ ⎤⎢⎣ ⎡ −+−−−+−+−−=∆ )( 4 )11()( 3 )( 2 ))(1( 41 4 2 12 3 1 3 2 2 1 2 212 TT E TT DTTCTTBTTARu The following values were found in Table A.2.1 0 1021.1 0 1045.1 470.3 4 3 = ×= = ×= = − E D C B A Substituting these values and using 11 K 473.15K) 15.273200( K 673.15K) 15.273400( Kmol J 314.8 2 1 =+= =+= ⎥⎦ ⎤⎢⎣ ⎡ ⋅= T T R provides ⎥⎦ ⎤⎢⎣ ⎡−=⎟⎠ ⎞⎜⎝ ⎛⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡−=∆ ⎥⎦ ⎤⎢⎣ ⎡−=∆ kg kJ 1.308 J 1000 kJ 1 kg 1 g 1000 O]H g[ 0148.18 O]H [mol 1 mol J 5551ˆ mol J 5551 2 2u u The values in parts (a) and (b) agree very well. The answer from part (a) will serve as the basis for calculating the percent difference since steam table data should be more accurate. ( ) %03.0%100 0.308 1.308308% =×− −−−=Difference 12 2.10 (a) Referring to the energy balance for closed systems where kinetic and potential energy are neglected, Equation 2.30 states WQU +=∆ (b) Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal 0=∆U According to Equation 2.77 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= 1 2 11 1 2 lnln P PRTn P PnRTW From the ideal gas law: 1111 VPRTn = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= 1 2 11 ln P PVPW Substitution of the values from the problem statement yields ( )( ) [ ]J 940 bar 8 bar 5lnm 105.2Pa 108 335 −= ⎟⎠ ⎞⎜⎝ ⎛××= − W W The energy balance is [ ]J 940 J 0 =∴ += Q WQ (c) Since the process is adiabatic 0=Q The energy balance reduces to WU =∆ 13 The system must do work on the surroundings to expand. Therefore, the work will be negative and 12 0 0 2 1 TT TcnU U v T T <∴ <∆=∆ <∆ ∫ T2 will be less than 30 ºC 14 2.11 (a) (i). 2 P [b ar ] v [m3/mol] 1 2 3 0.01 0.030.02 1 Path A Path B (ii). Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal 0=∆u Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore, 0=∆h Performing an energy balance and neglecting potential and kinetic energy produces 0=+=∆ wqu For an isothermal, adiabatic process, Equation 2.77 states ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= 1 2ln P PnRTW or ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛== 1 2ln P PRT n Ww Substituting the values from the problem statement gives 15 ( ) ⎟⎠ ⎞⎜⎝ ⎛+⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅= bar 3 bar 1lnK )15.27388( Kmol J 8.314w ⎥⎦ ⎤⎢⎣ ⎡−= mol J 3299w Using the energy balance above ⎥⎦ ⎤⎢⎣ ⎡=−= mol J 3299wq (b) (i). See path on diagram in part (a) (ii). Since the overall process is isothermal and u and h are state functions 0=∆u 0=∆h The definition of work is ∫−= dvPw E During the constant volume part of the process, no work is done. The work must be solved for the constant pressure step. Since it is constant pressure, the above equation simplifies to )( 12 vvPdvPw EE −−=−= ∫ The ideal gas law can be used to solve for 2v and 1v ( ) ( ) ⎥⎦ ⎤⎢⎣ ⎡=× +⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅ == ⎥⎦ ⎤⎢⎣ ⎡=× +⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅ == mol m 010.0 Pa 103 K )15.27388( K molJ 314.8 mol m 030.0 Pa 101 K )15.27388( K molJ 314.8 3 5 1 1 1 3 5 2 2 2 P RTv P RTv Substituting in these values and realizing that 1PPE = since the process is isobaric produces 16 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡− ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡×−= mol m 010.0 mol m 030.0Pa) 103( 33 5w ⎥⎦ ⎤⎢⎣ ⎡−= mol J 6000w Performing an energy balance and neglecting potential and kinetic energy results in 0=+=∆ wqu ⎥⎦ ⎤⎢⎣ ⎡=−=∴ mol J 6000wq 17 2.12 First, perform an energy balance. No work is done, and the kinetic and potential energies can be neglected. The energy balance reduces to QU =∆ We can use Equation 2.53 to get ∫= 2 1 T T vdTcnQ which can be rewritten as ∫= 2 1 T T PdTcnQ since the aluminum is a solid. Using the atomic mass of aluminum we find mol 3.185 mol kg 0.02698 kg 5 = ⎥⎦ ⎤⎢⎣ ⎡=n Upon substitution of known values and heat capacity data from Table A.2.3, we get ( ) ( )∫ −×+⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅= K 15.323 K 15.294 3 1049.1486.2 Kmol J 314.8mol 3.185 dTTQ [ ]kJ 61.131=Q 18 2.13 First, start with the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes WQU +=∆ The value of the work will be used to obtain the final temperature. The definition of work (Equation 2.7) is ∫−= 2 1 V V EdVPW Since the piston expands at constant pressure, the above relationship becomes ( )12 VVPW E −−= From the steam tables ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 02641.0ˆ 3 1v (10 MPa, 400 ºC) [ ]33111 m 07923.0kgm 02641.0kg) 3(ˆ =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡== vmV Now 2V and 2v are found as follows [ ]36312 m 4536.0Pa 100.2 J 748740m 07923.0 =×−−=−= EPWVV [ ][ ] ⎥⎥⎦⎤⎢⎢⎣⎡=== kgm 1512.0kg 3 m 4536.0ˆ 33 2 2 2 m Vv Since 2vˆ and 2P are known, state 2 is constrained. From the steam tables: [ ]Cº 4002 =T ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ kg m 0.1512 bar, 20 3 Now U∆ will be evaluated, which is necessary for calculating Q . From the steam tables: 19 ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 2.2945ˆ2u ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ kg m 0.1512 bar, 20 3 ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 4.2832ˆ1u ( )Cº 400 bar, 001 ( ) [ ]( ) [ ]kJ 4.338 kg kJ 4.2832 kg kJ 2.2945kg 3ˆˆ 121 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡−⎥⎦ ⎤⎢⎣ ⎡=−=∆ uumU Substituting the values of U∆ and W into the energy equation allows calculation of Q WUQ −∆= [ ]( ) [ ]J 1009.1J 748740 [J] 338400 6×=−−=Q 20 2.14 In a reversible process, the system is never out of equilibrium by more than an infinitesimal amount. In this process the gas is initially at 2 bar, and it expands against a constant pressure of 1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible. To solve for the final temperature of the system, the energy balance will be written. The piston- cylinder assembly is well-insulated, so the process can be assumed adiabatic. Furthermore, potential and kinetic energy effects can be neglected. The energy balance simplifies to WU =∆ Conservation of mass requires 21 nn = Let 21 nnn == The above energy balance can be rewritten as ∫∫ −= 2 1 2 1 V V E T T v dVPdTcn Since vc and EP are constant: ( ) ( )1212 VVPTTnc Ev −−=− 2V and 1T can be rewritten using the ideal gas law 2 2 2 P nRTV = nR VPT 111 = Substituting these expressions into the energy balance, realizing that 2PPE = , and simplifying the equation gives nR VPP T 2 7 2 5 112 2 ⎟⎠ ⎞⎜⎝ ⎛ + = Using the following values 21 [ ] [ ] [ ] [ ] ⎥⎦ ⎤⎢⎣ ⎡ ⋅ ⋅= = = = = Kmol barL 08314.0 mol 0.1 L 10 bar 1 bar 2 1 2 1 R n V P P results in [ ]K 2062 =T To find the value for work, the energy balance can be used ( )12 TTncUW v −=∆= Before the work can be calculated, 1T must be calculated [ ]( ) [ ]( ) [ ]( ) [ ]K 241 Kmol barL 08314.0mol 1 L 10bar 211 1 = ⎟⎟⎠ ⎞⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ ⋅ ⋅== nR VPT Using the values shown above [ ]J 727−=W 22 2.15 The maximum work can be obtained through a reversible expansion of the gas in the piston. Refer to Section 2.3 for a discussion of reversible processes. The problem states that the piston assembly is well-insulated, so the heat transfer contribution to the energy balance can be neglected, in addition to potential and kinetic energy effects. The energy balance reduces to WU =∆ In this problem, the process is a reversible, adiabatic expansion. For this type of process, Equation 2.90 states [ ]11221 1 VPVP k W −−= From the problem statement (refer to problem 2.13), [ ] [ ] [ ]bar 1 L 10 bar 2 2 1 1 = = = P V P To calculate W, 2V must be found. For adiabatic, reversible processes, the following relationship (Equation 2.89) holds: constPV k = where k is defined in the text. Therefore, kkV P PV 1 1 2 1 2 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= Noting that 5 7== v P c ck and substituting the proper values provides [ ]L 4.162 =V Now all of the needed values are available for calculating the work. [ ] [ ]J 900barL 9 −=⋅−=W From the above energy balance, [ ]J 900−=∆U 23 The change in internal energy can also be written according to Equation 2.53: ∫=∆ 2 1 T T vdTcnU Since vc is constant, the integrated form of the above expression is ( )122 5 TTRnU −⎟⎠ ⎞⎜⎝ ⎛=∆ Using the ideal gas law and knowledge of 1P and 1V , [ ]K 6.2401 =T and [ ]K 3.1972 =T The temperature is lower because more work is performed during the reversible expansion. Review the energy balance. As more work is performed, the cooler the gas will become. 24 2.16 Since the vessel is insulated, the rate of heat transfer can be assumed to be negligible. Furthermore, no work is done on the system and potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes 0ˆ =∆u or 12 ˆˆ uu = From the steam tables ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 2.2619ˆ1u (200 bar, 400 ºC) ⎥⎦ ⎤⎢⎣ ⎡=∴ kg kJ 2.2619ˆ2u The values of 2uˆ and 2P constrain the system. The temperature can be found from the steam tables using linear interpolation: Cº 5.3272 =T [ ] ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 2.2619ˆ ,bar 100 2u Also at this state, ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 02012.0ˆ 3 2v Therefore, [ ]( ) [ ]332 m 020.0kgm 02012.0kg 0.1 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡== mvVvessel 25 2.17 Let the entire tank represent the system. Since no heat or work crosses the system boundaries, and potential and kinetic energies effects are neglected, the energy balance is 0=∆u Since the tank contains an ideal gas K 300 0 12 12 == =− TT TT The final pressure can be found using a combination of the ideal gas law and conservation of mass. 22 2 11 1 VP T VP T = We also know 12 2VV = Therefore, bar 5 2 1 2 == PP 26 2.18 (a) First, as always, simplify the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance is WQU +=∆ Since, this system contains water, we can the use the steam tables. Enough thermodynamic properties are known to constrain the initial state, but only one thermodynamic property is known for the final state: the pressure. Therefore, the pressure-volume relationship will be used to find the specific volume of the final state. Since the specific volume is equal to the molar volume multiplied by the molecular weight and the molecular weight is constant, the given expression can be written constvP =5.1ˆ This equation can be used to solve for 2vˆ . 5.1 1 5.1 1 2 1 2 ˆˆ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= v P Pv Using [ ] [ ] [ ] [ ]bar 100 kg m 1.0 kg 10 m 1.0 ˆ bar 20 2 33 1 1 = ⎥⎦ ⎤⎢⎣ ⎡== = P v P gives ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 0.0342 ˆ 3 2v Now that the final state is constrained, the steam tables can be used to find the specific internal energy and temperature. [ ] ⎥⎦ ⎤⎢⎣ ⎡= = kg kJ 6.3094ˆ K 7.524 2 2 u T To solve for the work, refer to the definition (Equation 2.7). 27 ∫−= 2 1 V V EdVPW or ∫−= 2 1 ˆ ˆ ˆˆ v v E vdPw Since the process is reversible, the external pressure must never differ from the internal pressure by more than an infinitesimal amount. Therefore, an expression for the pressure must be developed. From the relationship in the problem statement, constvPvP == 5.1115.1 ˆˆ Therefore, the expression for work becomes ∫∫ −=−= 2 1 2 1 ˆ ˆ 5.1 5.1 11 ˆ ˆ 5.1 5.1 11 ˆ ˆ 1ˆˆ ˆ ˆˆ v v v v vd v vPvd v vPw Integration and substitution of proper values provides ⎥⎦ ⎤⎢⎣ ⎡= ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ ⋅= kg kJ 284 kg mbar 840.2ˆ 3 w [ ]( ) [ ]kJ 2840 kg kJ 284kg 10 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡=∴W A graphical solution is given below: 2 P [b ar ] v [m3/kg] 20 60 100 0.02 0.100.06 1 Work 28 To solve for Q , U∆ must first be found, then the energy balance can be used. ( ) [ ]( ) [ ]kJ 4918 kg kJ 8.2602 kg kJ 6.3094kg 10ˆˆ 12 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡−⎥⎦ ⎤⎢⎣ ⎡=−=∆ uumU Now Q can be found, [ ] [ ] [ ]kJ 2078kJ 2840kJ 4918 =−=−∆= WUQ (b) Since the final state is the same as in Part (a), U∆ remains the same because it is a state function. The energy balance is also the same, but the calculation of work changes. The pressure from the weight of the large block and the piston must equal the final pressure of the system since mechanical equilibrium is reached. The calculation of work becomes: ∫−= 2 1 ˆ ˆ ˆ v v E vdmPW All of the values are known since they are the same as in Part (a), but the following relationship should be noted 2PPE = Substituting the appropriate values results in [ ]kJ 6580=W Again we can represent this process graphically: 2 P [b ar ] v [m3/kg] 20 60 100 0.02 0.100.06 1 Work 29 Now Q can be solved. [ ] [ ] [ ]kJ 1662kJ 6580kJ 4918 −=−=−∆= WUQ (c) This part asks us to design a process based on what we learned in Parts (a) and (b). Indeed, as is characteristic of design problems there are many possible alternative solutions. We first refer to the energy balance. The value of heat transfer will be zero when WU =∆ For the same initial and final states as in Parts (a) and (b), [ ]kJ 4918=∆= UW There are many processed we can construct that give this value of work. We show two alternatives which we could use: Design 1: If the answers to Part (a) and Part (b) are referred to, one can see that two steps can be used: a reversible compression followed by an irreversible compression. Let the subscript “i" represent the intermediate state where the process switches from a reversible process to an irreversible process. The equation for the work then becomes [ ]J 4918000ˆ ˆ 1ˆ)ˆˆ( ˆ ˆ 5.1 5.1 1122 1 =⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ −−−= ∫i v v i vd v vPvvPmW Substituting in known values (be sure to use consistent units) allows calculation of ivˆ : ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 0781.0ˆ 3 iv The pressure can be calculated for this state using the expression from part (a) and substituting the necessary values. [ ]bar 0.29 5.1 1 1 = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= i i i P v vPP 30 Now that both iP and ivˆ are known, the process can be plotted on a P-v graph, as follows: 2 P [b ar ] v [m3/kg] 20 60 100 0.02 0.100.06 1 i Work Design 2: In an alternative design, we can use two irreversible processes. First we drop an intermediate weight on the piston to compress it to an intermediate state. This step is followed by a step similar to Part (b) where we drop the remaining mass to lead to 100 bar external pressure. In this case, we again must find the intermediate state. Writing the equation for work: [ ] [ ]J 4918000)ˆˆ()ˆˆ( 221 =−−−−= iii vvPvvPmW However, we again have the relationship: 5.1 1 1 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= i i v vPP Substitution gives one equation with one unknown vi: [ ]J 4918000)ˆˆ()ˆˆ( 221 5.1 1 1 =⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ −−−⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛−= ii i vvPvv v vPmW There are two possible values vi to the above equation. Solution A: ⎥⎦ ⎤⎢⎣ ⎡= kg m 043.0ˆ 3 iv 31 which gives [ ]bar 8.70=iP This solution is graphically shown below: 2 P [b ar ] v [m3/kg] 20 60 100 0.02 0.100.06 1 i Work Solution B ⎥⎦ ⎤⎢⎣ ⎡= kg m 0762.0ˆ 3 iv which gives [ ]bar 0.30=iP This solution is graphically shown below: 2 P [b ar ] v [m3/kg] 20 60 100 0.02 0.100.06 1 i Work 32 2.19 (a) Force balance to find k: Piston Fspring=kx Fatm=PatmA Fgas=PgasA Fmass=mg Pgas = Patm + mgA − kxA since ∆V=Ax Pgas = Patm + mgA + k∆VA2 since ∆V is negative. Now solve: ( )( ) ( )( )22 3 2 2 55 m 1.0 m 02.0 m 1.0 m/s 81.9kg 2040Pa 101Pa 102 k−+×=× ⎥⎦ ⎤⎢⎣ ⎡×=∴ m N1001.5 4k Work can be found graphically (see P-V plot) or analytically as follows: Substituting the expression in the force balance above: WA = Patm + mgA + k∆VA2 ⎛ ⎝ ⎞ ⎠ ∫ dV ( )∫ ∫ ∆ ∆ ∆∆+⎟⎠ ⎞⎜⎝ ⎛ += f i f i V V V V atmA Vd A VkdV A mgPW 2 ( )( ) [ ]( ) ( ) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ −×+−×= 2 0m02.0 m 1.0 N/m 1001.5m 05.003.0Pa1000.3 223 22 4 25 AW 33 [ ]J 105 3×−=AW V [m3] 1 2 3 0.01 0.03 0.05 Patm mg A k∆V A2 -W 10squares 0.5kJsquare = 5kJ Work P [b ar ] 2 1 = (b) You need to find how far the spring extends in the intermediate (int) position. Assume PVn=const (other assumptions are o.k, such as an isothermal process, and will change the answer slightly). Since you know P and V for each state in Part (a), you can calculate n. PiPf = Vi V f ⎛ ⎝ ⎞ ⎠ n or 10 5 Pa 2×105 Pa = 0.03 m3 0.05 m3 ⎛ ⎝ ⎞ ⎠ n ⇒ n = 1.35 Now using the force balance Pgas,int = Patm + mgA + k∆VA2 with the above equation yields: Pint = Pi ViV int( )n = Patm + mgA + k Vint −Vi( )A2 This last equality represents 1 equation. and 1 unknown (we know k), which gives 3int m 0385.0=V Work can be found graphically (see P-V plot) or analytically using: 34 WB = Pgas Vi V int∫ dV + Pgas Vint V f ∫ dV expanding as in Part (a) WB = 2 ×105 Nm2 ⎛ ⎝ ⎞ ⎠ Vi V int∫ dV + 5 ×106 N m5 ∆V ∆Vi ∆Vint∫ d ∆V( )+ 3 ×105 Nm2⎛ ⎝ ⎞ ⎠ Vint Vf ∫ dV + 5 ×106 N m 5 ∆V ∆Vint ∆Vif ∫ d ∆V( ) Therefore, [ ]kJ85.3−=BW just like we got graphically. V [m3] 1 2 3 0.01 0.03 0.05 Patm mg A P [b ar ] 2 1 Work mg A (c) The least amount of work is required by adding differential amounts of mass to the piston. This is a reversible compression. For our assumption that PVn = const, we have the following expression: ∫∫ == f i f i V V V V gasrevC dV V constdVPW 35.1, Calculate the constant from the initial state 35 ( )( ) 335.13 5 1075.1m05.0Pa101. ×=×=const Therefore, WC,rev = 1.75 ×103 .05 .03 ∫ dV V1.35 = − 1.75×1030.35 V −0.35 .05 .03 [ ]J2800−= V [m3] 1 2 3 0.01 0.03 0.05 Patm P [b ar ] 2 1 Workrev 36 2.20 Before this problem is solved, a few words must be said about the notation used. The system was initially broken up into two parts: the constant volume container and the constant pressure piston-cylinder assembly. The subscript “1” refers to the constant volume container, “2” refers the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f” denotes the final state. To begin the solution, the mass of water present in each part of the system will be calculated. The mass will be conserved during the expansion process. Since the water in the rigid tank is saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water is constrained to a specific state. From the steam tables, [ ] kPa 8.1553 kg kJ 3.2595ˆ kg kJ 64.850ˆ kg kJ 12736.0ˆ kg kJ 001156.0ˆ ,1 ,1 ,1 ,1 = ⎥⎦ ⎤⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= sat v i l i v i l i P u u v v (Sat. water at 200 ºC) Knowledge of the quality of the water and the overall volume of the rigid container can be used to calculate the mass present in the container. ( ) ( )vil i vmvmV ,11,111 ˆ95.0ˆ05.0 += Using the values from the steam table and [ ]31 m 5.0=V provides [ ]kg 13.41 =m Using the water quality specification, [ ] [ ]kg 207.005.0 kg 92.395.0 11 11 == == mm mm l v For the piston-cylinder assembly, both P and T are known. From the steam tables 37 ⎥⎦ ⎤⎢⎣ ⎡= ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg kJ 9.2638ˆ kg m 35202.0ˆ ,2 3 ,2 i i u v (600 kPa, 200 ºC) Enough information is available to calculate the mass of water in the piston assembly. [ ]kg 284.0 ˆ2 2 2 == v Vm Now that the initial state has been characterized, the final state of the system must be determined. It helps to consider what physically happens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final temperature and pressure of the entire system will match the surroundings’. In other words, ⎥⎦ ⎤⎢⎣ ⎡== kg kJ 9.2638ˆˆ ,2 if uu (600 kPa, 200 ºC) Thus, the change in internal energy is given by l i l i v i v iif l i v i umumumummmU ,1,1,1,1,22,1,12 ˆˆˆˆ)( −−−++=∆ Substituting the appropriate values reveals [ ]kJ 0.541=∆U To calculate the work, we realize the gas is expanding against a constant pressure of 600 kPa (weight of the piston was assumed negligible). From Equation 2.7, )( ifE V V E VVPdVPW f i −−=−= ∫ where [ ] [ ][ ] [ ] [ ]333 3 ,2,1,12 m 6.0m 5.0m 0.1 m 55.1ˆ)( Pa 600000 =+= =++= = i i l i v if E V vmmmV P 38 Note: iv ,2ˆ was used to calculate fV because the temperature and pressure are the same for the final state of the entire system and the initial state of the piston-cylinder assembly. The value of W can now be evaluated. [ ]kJ 570−=W The energy balance is used to obtain Q. [ ] [ ]( ) [ ]kJ 1111kJ 570kJ 0.541 =−−=−∆= WUQ 39 2.21 A sketch of the process follows: 10 cm50 cm A B p1,B = 20 bar T1,B = 250 oC p1,A = 10 bar T1,A = 700 oC H2O H2O system boundary Process A B p2 T2 H2O system boundary p2 T2 H2O The initial states are constrained. Using the steam tables, we get the following: State 1,A State 1,B p 10 [bar] 20 [bar] T 700 [oC] 250 [oC] v 0.44779 [m3/kg] 0.11144 [m3/kg] u 3475.35 [kJ/kg] 2679.58 [kJ/kg] V 0.01 m3 0.05 m3 m = V v 0.11 [kg] 0.090 [kg] All the properties in the final state are equal. We need two properties to constrain the system: We can find the specific volume since we know the total volume and the mass: [ ][ ] ⎥⎥⎦⎤⎢⎢⎣⎡==++= kgm 30.0kg 0.20 m 06.0 33 ,1,1 ,1,1 2 BA BA mm VV v We can also find the internal energy of state 2. Since the tank is well insulated, Q=0. Since it is rigid, W=0. An energy balance gives: ∆U = Q −W = 0 Thus, U2 =U1 = m1,Au1, A + m1,Bu1,B or ⎥⎦ ⎤⎢⎣ ⎡=+ +== kg kJ 3121 ,1,1 ,1,1,1,1 2 2 2 BA BBAA mm umum m Uu We have constrained the system with u2 and v2, and can find the other properties from the steam Tables. Very close to T2 = 500 [oC] and P2 = 1200 [kPa] Thus, 40 [ ]( ) [ ]m 267.0kg 0.09 kg m 30.0 3 ,2,2,2 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡== AAA mvV and ( ) [ ]m 167.01.0267.0 =−=∆x 41 2.22 We start by defining the system as a bubble of vapor rising through the can. We assume the initial temperature of the soda is 5 oC. Soda is usually consumed cold; did you use a reasonable estimate for T1? A schematic of the process gives: State 1: T1 = 278 K P1 = 3 bar State 2: T2= ? P2= 1.01 bar where the initial state is labeled state 1, and the final state is labeled state 2. To find the final temperature, we perform an energy balance on the system, where the mass of the system (CO2 in the bubble) remains constant. Assuming the process is adiabatic and potential and kinetic energy effects are negligible, the energy balance is wu =∆ Expressions for work and internal energy can be substituted to provide ( ) ( )1212 vvPdvPTTc EEv −−=−=− ∫ where cv = cP – R. Since CO2 is assumed an ideal gas, the expression can be rewritten as ( ) ⎥⎦ ⎤⎢⎣ ⎡ −−=⎥⎦ ⎤⎢⎣ ⎡ −−=− 1 12 2 1 1 2 2 12 P TPTR P T P TRPTTc Ev where the equation was simplified since the final pressure, P2, is equal to the external pressure, PE. Simplifying, we get: ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ +=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + 1 2 12 11 Pc RPT c RT vv or K 2371 1 2 12 =⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ += P v v c c Pc RPTT 42 2.23 The required amount of work is calculated as follows: VPW ∆−= The initial volume is zero, and the final volume is calculated as follows: ( ) 3333 m0436.0ft 54.1ft 5.0 3 4 3 4 ==== ππrV Assuming that the pressure is 1 atm, we calculate that ( )( ) J4417m 0m 0436.0Pa1001325.1 33 5 =−×=W This doesn’t account for all of the work because work is required to stretch the rubber that the balloon is made of. 43 2.24 (a) Since the water is at its critical point, the system is constrained to a specific temperature, pressure, and molar volume. From Appendix B.1 ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 003155.0ˆ 3 cv Therefore, [ ] [ ]kg 17.3 kg m 003155.0 m 01.0 ˆ 3 3 = ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡== cv Vm (b) The quality of the water is defined as the percentage of the water that is vapor. The total volume of the vessel can be found using specific volumes as follows ( )[ ] ( ) vlvvll vxmvmxvmvmV ˆˆ1ˆˆ +−=+= where x is the quality of the water. To solve for the quality, realize that starting with saturated water at a pressure of 1 bar constrains the water. From the steam tables, ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 001043.0ˆ kg m 6940.1ˆ 3 3 l v v v (sat. H2O at P = 1 bar) Now the quality can be found 00125.0=x Thus, the quality of the water is 0.125%. (c) To determine the required heat input, perform an energy balance. Potential and kinetic energy effects can be neglected, and no work is done. Therefore, QU =∆ 44 where ( )[ ] ( )[ ]vl uxmumxumU 112 ˆˆ1ˆ +−−=∆ From the steam tables ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 58.2029ˆ2u (H2O at its critical point) ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= kg m 1.2506ˆ kg m 33.417ˆ 3 1 3 1 v l u u (sat. H2O at P=1 bar) Evaluation of the expression reveals [ ] [ ]J 1010.5kJ 6.5102 6×==∆U 45 2.25 (a) Consider the air in ChE Hall to be the system. The system is constant volume, and potential and kinetic energy effects can be neglected. Furthermore, disregard the work. The energy balance is q dt du &= since the temperature of the system changes over time. Using the given expression for heat transfer and the definition of dU , the expression becomes ( )surrv TThdt dTc −−= We used a negative sign since heat transfer occurs from the system to the surroundings. If vc is assumed constant, integration provides ( ) ChtTTc surrv +−=−ln where C is the integration constant. Therefore, ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −+= tc h surr veCTT 1 where C1 is a constant. Examining this equation reveals that the temperature is an exponential function of time. Since the temperature is decreasing, we know that the plot of temperature vs. time shows exponential decay. time Te m pe ra tu re T0 Tsurr 46 (b) Let time equal zero at 6 PM, when the steam is shut off. At 6 PM, the temperature of the hall is 22 ºC. Therefore, ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −+= tc h surr veCTT 1 )0(1Cº 2 Cº 22 eC+= Cº 201 =∴C After 10 PM, ( hr 4=t ), the temperature is 12 ºC. ( ) ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ −+= hr) 4(Cº 20Cº 2 Cº 21 vc h e 1-hr 173.0=∴ vc h At 6 AM, hr 12=t . Substitution of this value into the expression for temperature results in Cº 5.4=T 47 2.26 The gas leaving the tank does flow work as it exits the valve. This work decreases the internal energy of the gas – lowering the temperature. During this process, water from the atmosphere will become supersaturated and condense. When the temperature drops below the freezing point of water, the water forms a solid. Attractive interactions between the compressed gas molecules can also contribute to this phenomena, i.e., it takes energy to pull the molecules apart as they escape; we will learn more of these interactions in Chapter 4. 48 2.27 Mass balance inoutin mmmdt dm &&& =−= Separating variables and integrating: ∫∫ = t in m m dtmdm 0 2 1 & or ∫=− t indtmmm 0 12 & Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is ∑ ∑ ++−=⎟⎠⎞⎜⎝⎛ out sin ininoutoutsys WQhmhmdt dU &&&& The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to inin sys hm dt dU & =⎟⎠ ⎞⎜⎝ ⎛ The following math is performed ( ) in t inin t inin U U hmmumumUU dtmhdthmdU ˆˆˆ 12112212 00 2 1 −=−=− == ∫∫∫ && where the results of the mass balance were used. Both 2m and 1m can be calculated by dividing the tank volume by the specific volume 49 1 1 2 2 ˆ ˆ v Vm v Vm = = Substitution of these relationships and simplification results in ( ) ( ) 0 ˆ ˆ ˆ ˆ 1 1 2 2 =−−− v hu v hu inin From the steam tables: ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= kg m 19444.0ˆ kg kJ 6.2583ˆ 3 1 1 v u (sat. H2O vapor at 1 MPa) ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 2.3177iˆnh (6 MPa, 400 ºC) There are still two unknowns for this one equation, but the specific volume and internal energy are coupled to each other. To solve this problem, guess a temperature and then find the corresponding volume and internal energy values in the steam tables at 6 MPa. The correct temperature is the one where the above relationship holds. Cº 600=T : Expression = 4427.6 Cº 500=T : Expression = 1375.9 Cº 450=T : Expression = -558.6 Interpolation between 500 ºC and 450 ºC reveals that the final temperature is Cº 4.4642 =T 50 2.28 We can pick room temperature to be 295 K Tin = T1 = 295 K[ ] Mass balance inoutin nnndt dn &&& =−= Separating variables and integrating: ∫∫ = t in n n dtndn 0 2 1 & or ∫=− t indtnnn 0 12 & Energy balance Neglecting ke and pe, he unsteady energy balance, written in molar units is written as: WQhnhn dt dU outoutinin sys &&&& −+−=⎟⎠ ⎞⎜⎝ ⎛ The terms associated with flow out, heat and work are zero. inin sys hn dt dU &=⎟⎠ ⎞⎜⎝ ⎛ Integrating both sides with respect to time from the initial state where the pressure is 10 bar to the final state when the tank is at a pressure of 50 bar gives: dtnhdthndU t inin t inin U U ∫∫∫ == 00 2 1 && since the enthalpy of the inlet stream remains constant throughout the process. Integrating and using the mass balance above: ( ) inhnnunun 121122 −=− 51 Now we do some math: ( ) inhnnunun 121122 −=− ( ) ( )inin hunhun −=− 1122 By the definition of h 11 RTuRTuvPuh inininininin +=+=+= so ( ) ( ) 1111112122 TRnuunRTnuun −−=−− ( ) 1112122 RTnRTnTTcn v −=−− Since RRcc Pv 2 3=−= n2 3 2 T2 − T1( )− n2T1 = −n1T1 or 111222 253 TnTnTn −=− dividing by n1: 11 1 2 2 1 2 253 TT n nT n n −=− Using the ideal gas law: 21 12 1 2 TP TP n n = so 11 21 12 2 21 12 253 TT TP TPT TP TP −=⎥⎦ ⎤⎢⎣ ⎡−⎥⎦ ⎤⎢⎣ ⎡ or 52 [K] 434 23 5 1 2 1 12 2 = ⎥⎦ ⎤⎢⎣ ⎡ +⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎥⎦ ⎤⎢⎣ ⎡ = P P P TP T (b) Closed system ∆u = q − w = q ( ) ( ) ⎥⎦ ⎤⎢⎣ ⎡=−=−=∆ kg kJ 9.28 2 5ˆ 1212 TTMW RTT MW cu v ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 9.28q (c) P2T2 = P3T3 [ ]bar 34 3 22 3 == T TPP 53 2.29 Mass balance inoutin nnndt dn &&& =−= Separating variables and integrating: ∫∫ = = t in n n dtndn 00 2 1 & or ∫= t indtnn 0 2 & Energy balance Neglecting ke and pe, the unsteady energy balance, in molar units, is written as: WQhnhn dt dU outoutinin sys &&&& ++−=⎟⎠ ⎞⎜⎝ ⎛ The terms associated with flow out and heat are zero. Whn dt dU inin sys && +=⎟⎠ ⎞⎜⎝ ⎛ Integrating both sides with respect to time from the empty initial state to the final state gives: WnhWdtnhdtWdthndU in t inin tt inin U U +=+=+= ∫∫∫∫ 2 000 2 1 &&& since the enthalpy of the inlet stream remains constant throughout the process. The work is given by: 22122 )( vPnvvPnW extext −=−−= 54 [ ]2222 vPhnun extin −= Rearranging, 222 vPvPuvPhu extinininextin −+=−= 22 vPvPuu extininin −=− ( ) 2 2 2 P TPRRTTTc extininv −=− so ( ) [ ]K 333 2 2 = ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ + += in ext v v T R P Pc RcT 55 2.30 valve maintains pressure in system constant T1 = 200 oC x1 = 0.4 V = 0.01 m3 v l Mass balance outoutin mmmdt dm &&& −=−= Separating variables and integrating: ∫∫ −= t out m m dtmdm 0 2 1 & or ∫−=− t outdtmmm 0 12 & Energy balance Qhm dt dU outout sys && +−=⎟⎠ ⎞⎜⎝ ⎛ ˆ Integrating [ ]∫ ∫∫∫ +−=+−= t tt outoutoutoutum um dtQdtmhdtQhmdU 0 00 ˆ ˆ ˆˆ 22 11 &&&& Substituting in the mass balance and solving for Q ( ) outhmmumumQ ˆˆˆ 121122 −−−= 56 We can look up property data for state 1 and state 2 from the steam tables: ⎥⎦ ⎤⎢⎣ ⎡=×+×=+−= kg m 0.0511274.04.0001.6.0ˆˆ)1(ˆ 3 1 gf vxvxv ⎥⎦ ⎤⎢⎣ ⎡= kg m 1274.0ˆ 3 2v So the mass in each state is: [ ] [ ]kg 196.0 kg m 0.051 m 01.0 ˆ 3 3 1 1 1 = ⎥⎦ ⎤⎢⎣ ⎡== v Vm [ ] [ ]kg 0785.0 kg m 0.1274 m 01.0 ˆ 3 3 2 2 2 = ⎥⎦ ⎤⎢⎣ ⎡== v Vm [ ]kg 1175.012 −=− mm And for energy and enthalpy ⎥⎦ ⎤⎢⎣ ⎡=×+×=+−= kg kJ 54915.25974.064.8506.0ˆˆ)1(ˆ1 gf uxuxu ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 3.2595ˆ2u ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 2.2793oˆuth Solving for heat, we get ( ) [ ]kJ 228ˆˆˆ 121122 =−−−= outhmmumumQ 57 2.31 Consider the tank as the system. Mass balance inoutin mmmdt dm &&& =−= Separating variables and integrating: ∫∫ = t in m m dtmdm 0 2 1 & or ∫=− t indtmmm 0 12 & Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is ∑ ∑ ++−=⎟⎠⎞⎜⎝⎛ out sin ininoutoutsys WQhmhmdt dU &&&& The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to inin sys hm dt dU & =⎟⎠ ⎞⎜⎝ ⎛ The following math is performed in t inin t inin U U hmumU dtmhdthmdU ˆˆ 2222 000 2 1 == == ∫∫∫ = && where the results of the mass balance were used. Thus, inhu ˆˆ2 = From the steam tables 58 ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 5.3632ˆ2u (9 MPa, 800 ºC) so ⎥⎦ ⎤⎢⎣ ⎡= kg kJ 5.3632iˆnh We can use the value of inh and the fact that the steam in the pipe is at 9 MPa to find the temperature. Cº 600=inT 59 2.32 (a) First, the energy balance must be developed. Since the problem asks how much energy is stored in the battery after 10 hours of operation, the process is not steady-state. Let the battery be the system. Potential and kinetic energy effects can be neglected. Furthermore, heating of the battery as it is charged can be ignored. The energy balance is s sys WQ dt dU && +=⎟⎠ ⎞⎜⎝ ⎛ No shaft work is performed, but electrical is supplied to the battery, which must be accounted for in sW& . The value of Q& is given explicitly in the problem statement. Both of these values remain constant over time, so integration provides ( )tWQU s&& +=∆ From the problem statement [ ] [ ] [ ]s 36000 kW 1 kW 5 = −= = t Q Ws & & Substituting these values allows the calculation of the amount of energy stored: [ ] [ ]MJ 144kJ 000,144 ==∆U (b) To calculate the velocity of the falling water, an energy balance must be developed with the water passing through the electricity generator (probably a turbine) as the system, where the water enters with a velocity 1V r and leaves with a negligible velocity, which will be approximated as 0. Assume that potential energy changes can be neglected. Furthermore, assume that the temperature of the water does not change in the process, so the change in internal energy is zero. Also, view the process as adiabatic. The energy balance reduces to riverK WE && =∆ where riverW& is the power of the flowing water. The actual power being provided by the stream can be calculated using the efficiency information. Let η represent the efficiency. 60 [ ] [ ]kW 10 5.0 kW 5 ===∴ = η η s river river s WW W W && & & The value of riverW& should be negative since the water is supplying work that is stored electrical energy. Therefore, the energy balance becomes [ ]W 10000−=∆ KE& This expression can be rewritten as ( ) [ ]W 10000 2 1 2 1 2 2 −=−VVm rr & From the problem statement and the assumptions made, ⎥⎦ ⎤⎢⎣ ⎡= ⎥⎦ ⎤⎢⎣ ⎡= s m 0 s kg 200 2V m r & Therefore, ⎥⎦ ⎤⎢⎣ ⎡= s m 101V r There are a number of reasons for the low conversion efficiency. A possible potential energy loss inherent in the design of the energy conversion apparatus decreases the efficiency. Heat is lost to the surroundings during conversion. Some of the energy is also lost due to friction (drag) effects. 61 2.33 Considering the turbine to be the system, rearrangement of the steady-state, open system energy balance provides ∑ ∑ +=++−++ out s in inpkinoutpkout WQeehneehn &&&& )()( Performing a mass balance reveals 21 nnnn outin &&&& === Assuming the rate of heat transfer and potential energy effects are negligible and realizing that there is one inlet and one outlet allows the simplification of the above equation to ( ) ( )[ ]1,2,122 KKs eehhnW −+−= && ( )12 hh − can be rewritten using Equations 2.58 and Appendix A.2 ( ) ( )∫∫ ++++==− −2 1 322 12 T T p dTETDTCTBTARdTchh Since the quantity ( )1,2, kk ee − is multiplied by n& , it is rewritten as follows for dimensional homogeneity ( ) ( )21221,2, )(21 VVMWee airkk rr −=− To solve for n& , the ideal gas law is used 2 22 2 21222 RT VPn RTnVP & & && = = To solve for the volumetric flow rate, the fluid velocity must be multiplied by the cross-sectional area ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= 4 2 2 2 2 VDV r & π The energy balance is now 62 ( ) ( ) ⎥⎥ ⎥ ⎦ ⎤ ⎢⎢ ⎢ ⎣ ⎡ −+⎟⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎜ ⎝ ⎛ ++++⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= ∫ − 2122322222 2 2 )( 2 1 4 2 1 VVMWdTETDTCTBTARVD RT PW air T T s rrr& π Substituting values from Table A.2.1 and the problem statement results in [MW] -4.84[W] 1084.4 6 =×−=W 63 2.34 First, a sketch of the process is useful: q 30 bar 100 oC 20 bar 150 oC To find the heat in we will apply the 1st law. Assuming steady state, the open system energy balance with one stream in and one stream out can be written: ( ) Qhhn && +−= 210 which upon rearranging is: 12 hhn Q −=& & Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values for the other properties (like enthalpy!) are already constrained. From Table A.2.1, we have an expression for the ideal gas heat capacity: 263 10392.410394.14424.1 TT R c p −− ×−×+= with T in (K). Since this expression is limited to ideal gases any change in temperature must be under ideal conditions. From the definition of heat capacity: ( )∫ −− ×−×+=−= 2 1 263 12 10392.410394.14424.1 T T dTTThh n Q & & By integrating and substituting the temperatures, we obtain: ⎥⎦ ⎤⎢⎣ ⎡= mol J5590 n Q & & 64 2.35 A schematic of the process follows: T1 = 350 oC v1 = 600 cm3/mol P2= 1 atm Ý W s T1 = 308 oC To solve for nWs && / we need a first law balance. With negligible eK and eP, the 1st law for a steady state process becomes: ( ) sWQhhn &&& ++−= 210 If heat transfer is negligible, h n Ws ∆=& & We can calculate the change in enthalpy from ideal gas heat capacity data provided in the Appendix.
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