ENGR Thermodynamics

Transcription

1 ENGR Thermodynamis Baratui Problem : 1 - Mass, Fore, Density and Aeleration - 4 pts 6-Apr-11 A losed system onsists of 0.5 lbmole of liquid water and oupies a volume of ft 3. Determine the weiht of the system, in lb f, and the averae density, in lb m /ft 3, at a loation where the aeleration of ravity is = 30.5 ft/s 2. The key ideas here are the meanin of lbmole, the use of in Newton's 2nd Law of Motion and the definition of density. Given : n 0.5 lbmole 30.5 ft/s 2 V ft lb m -ft/lb f -s 2 Find : a.) W??? lb f b.)??? lb m /ft 3 MW lb m /lbmole Diaram : None. Assumptions: 1 - Gravitational aeleration is uniform. 2 - Density of the water is uniform. Part a.) The key here is the ralationship between mass and fore, in this ase weiht, iven by Newton's 2nd Law of Motion. F ma Eqn 1 W m Eqn 2 We an solve Eqn 2 for the weiht of the water, W. W m Eqn 3 The next step in determinin the weiht of the water is to onvert from the iven number of moles to mass usin: m n MW Eqn 4 Pluin values into Eqn 1 yields : m lb m Now, we an plu values into Eqn 3 to evaluate the weiht of the water. W lb f Part b.) This is a straihtforward appliation of the definition of density. m V Eqn 5 Sine we determined m in part (a) and the volume of the water was iven, all we need to do is plu values into Eqn lb m /ft 3 Verify : None of the assumptions made in the solution of this problem an be verified based on the iven information. Answers : a.) b.) W 8.54 lb f lb m /ft 3 Dr. Baratui - ChemE 260 hw1-sp11.xlsm, WB-1 4/11/2011

2 ENGR Thermodynamis Baratui Problem : 1.7E - Mass, Weiht and Aeleration - 3 pts 6-Apr-11 A man weihs 210 lb f at a loation where = ft/s 2. Determine his weiht on the moon where = 5.47 ft/s 2. The keys to this problem are to understand that mass does not depend on loation, but weiht does and how to apply Newton's 2nd Law of Motion. Given : W Earth 210 lb f moon 5.47 ft/s 2 earth ft/s ft-lb m /lb f -s 2 Find : W moon??? N Diaram : None. Assumptions: None. The key here is the ralationship between mass and fore, in this ase weiht, iven by Newton's 2nd Law of Motion. F ma Eqn 1 W ma Eqn 2 We an now apply Eqn 2 to the objet when it is on the moon and aain to the objet when it is on the surfae of the Earth. W m moon moon Eqn 3 W m Eqn 4 Earth Earth The key is that and the mass of the objet are onstants. So, if we take the ratio of Eqn 3 to Eqn 4, these terms will anel! W m m moon moon W Earth Earth Eqn 5 Wmoon moon This leaves us with : Eqn 6 W Earth Now, we an solve Eqn 6 for the unknown variable W moon. Earth moon Wmoon WEarth Earth Eqn 7 Pluin values into Eqn 7 yields : W moon lb f Verify : None. Answers : W moon 35.8 lb f Dr. Baratui - ENGR 224 hw1-sp11.xlsm, 1.7E 4/11/2011

3 ENGR Thermodynamis Baratui Problem : 1.11E - Speifi Heat: Unit Conversions - 4 pts 6-Apr-11 The onstant-pressure speifi heat of air at 25 o C is kj/k- o C. Express this value in kj/k-k, J/- o C, kal/k- o C, and Btu/lb m - o F. This problem is an exerise in unit onversions. The key to the problem is to reonize that the temperature unit in the heat apaity orresponds to a hane in temperatureof a manitude of one with the iven units. Given : C P kj/k- o C Find : a.) C P??? kj/k-k.) C P??? kal/k- o C b.) C P??? J/- o C d.) C P??? Btu/lb m - o F Assumptions: None. Part a.) The only unit we need to onvert in this part is the temperature unit. We need to et from o C to Kelvins. This is an easy one sine a hane in T of 1 o Celsius is the same as a hane in T of 1 o Kelvin. So, here is the onversion fator. 1 o C = 1 K Eqn 1 kj 1C CP = k C 1 K Eqn 2 C P kj/k-k Part b.) In this part, we need to onvert units of both enery and mass. We need to et from kj to J and from k to. Here are the onversion fators. 1 kj = 1000 J Eqn 3 1 k = 1000 Eqn 4 kj 1 k 1000 J C P = k C kj Eqn 5 C P J/- o C Dr. Baratui - ChemE 260 hw1-sp11.xlsm, 1.11E 4/11/2011

4 Part.) In this part, we only need to onvert units enery. We need to et from kj to kal. Here is the onversion fator kj = 1 kal Eqn 6 kj 1kal C P = k C kj Eqn 7 C P kal/k- o C Part d.) In this part, we need to onverat all three units from SI to AE. Here are the onversion fators. 1 kj = Btu Eqn 8 1 k = lb m Eqn 9 1 o C = 1.8 o F Eqn 10 kj 1 k Btu 1 C CP = k C lb m 1 kj 1.8 F Eqn 11 C P Btu/lb m - o F Verify : None. Answers : a.) C P??? kj/k-k.) C P??? kal/k- o C b.) C P??? J/- o C d.) C P??? Btu/lb m - o F Dr. Baratui - ChemE 260 hw1-sp11.xlsm, 1.11E 4/11/2011

5 ENGR Thermodynamis Baratui Problem : 1.37E - Temperature Conversions - 2 pts 6-Apr-11 What is the temperature of heated air at 150 o C in o F and in o R? This problem provides pratie onvertin temperature units. Given : T( o C) = 150 Ratio = 1.8 o F / o C T freezin 32 o F R zero F = o R Find : T( o F) =??? T( o R) =??? Assumptions: None. Part a.) The key here is the relationship between the Celsius and Fahrenheit temperature sales. Pluin values into Eqn 1 yields : o o o F o T F 150 C F o C o o T F T C T = 302 o F Eqn 1 Part b.) There are seveal ways to solve this part of the problem. One is to use the relationship between the Fahrenheit and Rankine temperature sales. o o o R o T R T F R o F Eqn 2 Pluin values from part (a) into Eqn 2 yields : T = o R Verify : None. Answers : a.) T = 302 o F b.) T = 762 o R Dr. Baratui - ENGR 224 hw1-sp11.xlsm, 1.37E 4/11/2011

6 ENGR Thermodynamis Baratui Problem : 1.41E - Temperature Chane - 3 pts 6-Apr-11 The temperature of water hanes by 10 o F durin a proess. Express this temperature hane in Celsius, Kelvin and Rankine units. It is very important to understand that a hane in temperature is not the same as a value of the temperature. It seems obvious, but it is ommon point of onfusion. Given : T 10 o F Find : a.) T??? o C.) T??? o R b.) T??? K Assumptions: None. Part a.) A hane in temperature of 1 o C is equal to a hane in temperature of 1.8 o F. Ratio = 1.8 o F / o C 10F Therefore : D T = Eqn 1 1.8F / C T 5.56 o C Part b.) A hane in temperature of 1 o C is exatly the same as a hane in temperature of 1 K. Ratio KC = 1 K / o C Therefore : D T = 5.56C 1 K/ C Eqn 2 Therefore it is easy to write the answer to part (b). T 5.56 K Part.) One way to do this part of the problem is to use the result from part (b) and the onverion fator for Kelvins and derees Rankine. Ratio RK = 1.8 o R / K o o R T R TK1.8 K Eqn 3 Pluin values into Eqn 3 yields : T 10 o R Of ourse, it would be even easier to reonize that a hane in temperature of 1 o F is exatly the same as a hane in temperature of 1 o R. o o o F T F T R 1 o R Eqn 4 Pluin values into Eqn 3 yields : T 10 o R Verify : None. Answers : a.) T 5.6 o C ) T 10 o R b.) T 5.6 K Dr. Baratui - ENGR hw1-sp11.xlsm, 1.41E 4/11/2011

7 ENGR Thermodynamis Baratui Problem : Differential, Multi-Fluid Manometer - 6 pts 6-Apr-11 Fresh water and seawater flowin in parallel, horizontal pipelines are onneted to eah other by a double u- tube manometer, as shown in the fiure. Determine the pressure differene between the two pipelines in kpa. Assume the density of seawater to be 1035 k/m 3. Can the air olumn be inored in this analysis? h 4 = = h 3 1 H k/m 3 = h 1 h 2 = This problem requires the areful appliation of the Manometer Equation. By workin from one pipe to the other we an determine the differene in the pressure between the two pipes. The key is that none of the fluid in the manometer tubes is movin. Given : h m FW 1000 k/m 3 h m SW 1035 k/m 3 h m H k/m 3 h m Find : P SW -P FW??? kpa Assumptions: 1 - The aeleration of ravity is uniform. 2 - The density of the merury is uniform. 3 - The density of the water is uniform The density of the air is neliible. Therefore, P 1 = P 2. None of the fluids inside the tubin are movin. Let's use the Manometer Equation to work our way around from the seawater pipe to the fresh water pipe. P P h SW 1 SW 4 Eqn 1 Beause the density of air is three orders of manitude smaller than the density of liquid water or merury : P P 2 1 P P h 3 2 H 2 Eqn 2 Eqn 3 Beause the elevation at points 3 and 4 is the same, the manometer fluid is not movin and points 3 and 4 are onneted by a tube filled with merury only: P 4 = P 3. P P h 4 FW FW 1 Eqn 4 Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

8 Combinin Eqns 1 & 2 yields : P P h SW 2 SW 4 Eqn 5 Solvin Eqn 3 for P 2 yields : P P h P h 2 3 H 2 4 H 2 Eqn 6 Combinin Eqns 5 & 6 yields : P P h h SW 4 H 2 SW 4 Eqn 7 Combinin Eqns 4 & 7 yields : P P h h h SW FW FW 1 H 2 SW 4 Eqn 8 Solvin Eqn 7 for P 2 - P FW yields : P P h h h SW FW FW 1 H 2 SW 4 Eqn 9 Now, we an plu values into Eqn 9 to omplete the solution of this problem. P SW -P FW N/m 2 P SW -P FW kpa Verify : None of the assumptions made in the solution of this problem an be verified based on the iven information. Answers : P SW -P FW kpa Could you have uessed just by lookin at the fluid levels and fluid densities where the pressure was hiher, in the fresh or salt water pipe? I ould have. Think about it. Look at the diaram and the equations. Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

9 ENGR Thermodynamis Baratui Problem : WB-2 - Absolute and Gaue Pressures - 5 pts 6-Apr-11 Tank A lies inside of Tank B, as shown in the fiure. Pressure aue A is loated inside Tank B and reads 1.4 bar. Both tanks ontain air. The manometer onneted to Tank B ontains merury ( = /m 3 ). The manometer readin is h = 20 m, atmospheri pressure is 101 kpa and = 9.81 m/s 2. Determine the absolute pressures inside Tank A and Tank B in kpa The fous of this problem is the meanin of aue pressure. The key is that a pressure uae indiates the differene between the pressure inside the pipe or vessel to whih it is attahed and the pressure outside of the pressure aue. In this ase, the pressure aue indiates P A - P B, where P A and P B are absolute pressures. Given : P A,aue 1.4 bar aue P atm 101 kpa H /m m/s 2 h 20 m 1 k-m/n-s 2 Find : P A??? kpa P B??? kpa Assumptions: 1 - The aeleration of ravity is uniform. 2 - The density of the merury is uniform. 3 - The density of the air is neliible. Therefore, the pressure inside eah tank is uniform. 4 - The merury in the manometer is not movin. Let's use the Manometer Equation to work our way around from point 1 on the diaram all the way to the inside of tank A. Beause the manometer tube is open at the end, P 1 = P atm. P kpa The eneral form of the Manotmeter Equation is : P P z z down up fluid up down Eqn 1 Applyin Eqn 1 to our manometer ives us : P2 P1 H h Eqn 2 Beause the elevation at points 2 and 3 is the same, the manometer fluid is not movin and points 2 and 3 are onneted by a tube filled with merury only: P 3 = P 2. Dr. Baratui - ENGR 224 hw1-sp11.xlsm, WB-2 4/11/2011

10 Now, beause the density of the air in Tank B is neliible, P B = P 3 = P 2! Therefore : PB Patm H h Eqn 3 Pluin values into Eqn 3 yields : P 101kPa m m m 1km/ Ns 100m 10 N/m k 10 m 9.81m / s 1m kpa B Eqn 4 P B kpa The key to determinin P A from P B is the meanin of aue pressure. Usually, we use : P P P Eqn 5 abs aue atm But in this problem the pressure on the outside of the aue is not atmospheri, but P B instead. Also, the P abs that we are lookin for here is P A. As a result, the form of Eqn 5 that we need to use here is : P P P A A,aue B Eqn 6 Pluin values into Eqn 4 yields : 100 kpa PA 1.4bar kPa 1bar P A kpa Verify : None of the assumptions made in the solution of this problem an be verified based on the iven information. Answers : P A kpa P B kpa Dr. Baratui - ENGR 224 hw1-sp11.xlsm, WB-2 4/11/2011

11 ENGR Thermodynamis Baratui Problem : Steam Table Fundamentals - 4 pts 6-Apr-11 Complete the followin table for water. T ( o C) P (kpa) H (kj/k) x (k vap/k) Phase Desription This problem is desined to test how well you understand how to use tables of thermodynami properties. It also tests your understandin and ability to use quality, x. Finally, it makes you think about the relationship between the data tables and the physial system (phases) that they represent. Given : Two piees of data for eah part, (a) throuh (e). Find : Complete the table. Assumptions: The system is in an equilibrium state. Part a.) Sine we are iven the quality, x = 0.7, we know that vapor and liquid are both present in the system at equilibrium. Beause both vapor and liquid are present at equailibrium, the temperature must be equal to the saturation temperature at P = 200 kpa. We an find this temperature in the Saturation Pressure Table of the steam tables. Here is the relevant data : Pressure Temp. H (kj/k) (kpa) ( o C) Sat. Liq Sat. Vap T = T sat o C Beause both saturated liquid and saturated vapor are present in the system at equilibrium, we must use the quality to evaluate the overall averae speifi enthalpy, as follows. The key equation that relates the speifi enthalpy of the system to the speifi enthalpy of the saturated liquid and of the saturated vapor is: Hˆ x Hˆ 1x Hˆ sat sat sat mix vap liq Eqn 1 Part b.) Pluin values into Eqn 1 yields : H kj/k The first step here is to use the iven T to obtain data from the Saturation Temperature table. This will allow us to ompare the iven value of H to the values of H sat liq and H sat vap in order to determine the phase or phases present in the system at equilibrium. Here is the relevant data : Sine : Hˆ Hˆ Hˆ sat liq sat vap Temp. Pressure H (kj/k) ( o C) (kpa) Sat. Liq Sat. Vap the system ontains a saturated mixture and the pressure must be equal to the saturation pressure. P= P sat kpa Last, we must use the iven value of the enthalpy to determine the quality of the water in the system. The key equation is: Hˆ Hˆ x Hˆ Hˆ sat vap sat liq sat liq Eqn 2 Pluin values into Eqn 2 yields : x k vap/k Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

12 Part.) Beause the quality is zero, we immediately know the system ontains a saturated liquid. This tells us that T = T sat, H = H sat liq and P sat = P = 950 kpa (iven). So, we an look up 950 kpa in the Saturation Pressure table. Here is the relevant data : Pressure Temp. H (kj/k) (kpa) ( o C) Sat. Liq Sat. Vap T = T sat o C H = H sat liq kj/k Part d.) Here we are iven both the T and P. Let's use the iven P, and the Saturation Pressure table, to determine the T sat assoiated with P. Then, we an determine the phases present by omparin the iven T to T sat (P). T sat (500 kpa) = o C Sine T = 80 o C is less than T sat, we an onlude that the sytem ontains a subooled or ompressed liquid. The quality is undefined! So, we need to look at the Subooled Liquid tables. Here thins et a bit diffiult beause the first pressure table in the Subooled Liquid tables is assoiated with 5 MPa. Our iven P of 500 kpa is less than this value, so it seems we do not have 2 values to interpolate between to determine H. This is not true. It just means that we must interpolate between the saturated liquid state at T = 80 o C and the subooled liquid state at 80 o C and 5 MPa. Here is the relevant data from the Steam Tables, arraned in a onvenient way. Temp. Pressure H ( o C) (kpa) (kj/k) (saturated liquid) ??? (subooled liquid) (subooled liquid) Now, we an determine H from this table by interpolation. H kj/k Part e.) The first step here is to use the iven P to obtain data from the Saturation Pressure table. This will allow us to ompare the iven value of H to the values of H sat liq and H sat vap in order to determine the phase or phases present in the system at equilibrium. Here is the relevant data : Pressure Temp. H (kj/k) (kpa) ( o C) Sat. Liq Sat. Vap H kj/k Sine : Hˆ Hˆ sat vap the system ontains a superheated vapor, quality is undefined and we must use the Superheated Vapor Table to determine T. We are fortunate that one of the Superheated Steam tables is assoiated with a P of 800 kpa. Pressure Temp. H (kpa) ( o C) (kj/k) T 350 o C Verify : None of the assumptions made in the solution of this problem an be verified based on the iven information. Answers : T ( o C) P (kpa) H (kj/k) x (k vap/k) Phase Desription Sat'd Mixture (VLE) Sat'd Mixture (VLE) Sat'd Liquid N/A Subooled Liquid N/A Superheated Vapor Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

13 ENGR Thermodynamis Baratui Problem : 3.29E - R-134a Table Fundamentals - 4 pts 6-Apr-11 Complete the followin table for R-134a. T P ( o F) (psia) H (Btu/lb m ) x (lb m vap/lb m ) Phase Desription This problem is desined to test how well you understand how to use tables of thermodynami properties. It also tests your understandin and ability to use quality, x. Finally, it makes you think about the relationship between the data tables and the physial system (phases) that they represent. Given : Two piees of data for eah part, (a) throuh (e). Find : Complete the table. Assumptions: The system is in an equilibrium state. Part a.) P 80 psia H 78 Btu/lb m The first step here is to use the iven P to obtain data from the Saturation Pressure table. This will allow us to ompare the iven value of H to the values of H sat liq and H sat vap in order to determine the phase or phases present in the system at equilibrium. Here is the relevant data : Pressure Temp. H (Btu/lb m ) (psia) ( o F) Sat. Liq Sat. Vap Sine : Hˆ Hˆ Hˆ sat liq sat vap the system ontains a saturated mixture and the temperature must be equal to the saturation temperature. T = T sat o F Last, we must use the iven value of the enthalpy to determine the quality of the water in the system. The key equation is: Hˆ Hˆ x Hˆ Hˆ sat vap sat liq sat liq Eqn 1 Pluin values into Eqn 1 yields : x k vap/k Part b.) T 15 o F x 0.6 lb m vap/lb m Beause the quality lies between 0 and 1, we immediately know the system ontains a saturated mixture. This tells us that T = T sat = 15 o F (iven) and P = P sat. So, we an look up 15 o F in the Saturation Temperature table. Here is the relevant data : Temp. Pressure H (Btu/lb m ) ( o F) (psia) Sat. Liq Sat. Vap P = P sat psia Beause both saturated liquid and saturated vapor are present in the system at equilibrium, we must use the quality to evaluate the overall averae speifi internal enery, as follows. The key equation that relates the speifi internal enery of the system to the speifi internal enery of the saturated liquid and of the saturated vapor is: Hˆ x Hˆ 1x Hˆ sat sat sat mix vap liq Pluin values into Eqn 2 yields : H kj/k Eqn 2 Dr. Baratui - ENGR 224 hw1-sp11.xlsm, 3.29E 4/11/2011

14 Part.) T 10 o F P 70 psia Here we are iven both the T and P. Let's use the iven P, and the Saturation Pressure table, to determine the T sat assoiated with P. Then, we an determine the phases present by omparin the iven T to T sat (P). Temp. Pressure H ( o F) (psia) (Btu/lb m ) (saturated liquid) 10 70??? (subooled liquid) (subooled liquid) Now, we an determine H from this table by interpolation. T sat (70 psia) = o F Sine T = 10 o F is less than T sat, we an onlude that the sytem ontains a subooled or ompressed liquid. The quality is undefined! So, we need to look at the Subooled Liquid tables. Here thins et a bit diffiult beause (1) CB did not ive you a subooled liquid table for R-134a and (2) the first pressure table in the LT Subooled Liquid tables is assoiated with 100 psia. Our iven P of 70 psia is less than this value, so it seems we do not have 2 values to interpolate between to determine H. This is not true. It just means that we must interpolate between the saturated liquid state at T = 10 o F and the subooled liquid state at 10 o F and 100 psia. Here is the relevant data from the R-134a Tables, arraned in a onvenient way. H Btu/lb m Part d.) P 180 psia H Btu/lb m The first step here is to use the iven P to obtain data from the Saturation Pressure table. This will allow us to ompare the iven value of H to the values of H sat liq and H sat vap in order to determine the phase or phases present in the system at equilibrium. Here is the relevant data : Pressure Temp. H (Btu/lb m ) (psia) ( o F) Sat. Liq Sat. Vap Hˆ Hˆ the system ontains a superheated vapor, quality is undefined and we must use the sat Sine : vap Superheated Vapor Table to determine T. We are amazinly fortunate that one of the Superheated R-134a tables is assoiated with a P of 180 psia and H = Btu/lb m at T = 160 o F! Part e.) T 110 T 160 o F x 1.0 lb m vap/lb m Beause the quality is 1.0, we immediately know the system ontains a saturated vapor. This tells us that T = T sat = 100 o F (iven), H = H sat vap and P = P sat. So, we an look up 110 o F in the Saturation Temperature table. o F Here is the relevant data : Temp. Pressure H (Btu/lb m ) ( o F) (psia) Sat. Liq Sat. Vap Verify : Answers : P = P sat psia H = H sat vap Btu/lb m None of the assumptions made in the solution of this problem an be verified based on the iven information. T ( o F) P (psia) H (Btu/lbm) x (lb m vap/lb m ) N/A N/A Phase Desription Sat'd Mixture (VLE) Sat'd Mixture (VLE) Subooled Liquid Superheated Vapor Saturated Vapor Dr. Baratui - ENGR 224 hw1-sp11.xlsm, 3.29E 4/11/2011

15 ChemE Thermodynamis Baratui Problem : 3.37E - Expansion of R-134a in a Sprin-Loaded P&C Devie - 5 pts 6-Apr-11 A sprin-loaded piston-and-ylinder devie is initially filled with 0.2 lb m of a vapor-liquid mixture of R-134a at -30 o F and a quality of 80%. The sprin onstant in the sprin fore relation F = k x is 37 lb f /in, and the piston diameter is 12 in. The R-134a underoes a proess that inreases its volume by 40%. Calulate the final temperature and enthalpy of the R-134a. The key to this problem is that the mass of R-134a inside the ylinder remains onstant. The initial state of the R-134a is fixed beause we know both T 1 and x 1. This will allow us to determine the speifi volume and then the total volume in the initial state. We an then determine the total volume and speifi volume in the final state. One we know the total volume in the final state, we an use the sprin onstant to determine P 2. Finally, knowin P 2 and V 2, we will be able to determine all the other intensive properties for state 2, inludin T 2 and H 2 that we need to answer the question. Given : m 0.2 lb m k 37 lb f /in T 1 30 o F D 12 in x k vap/k V 2 / V Find : T 2??? o F Assumptions: - Both the initial and final states are equilibrium states. Lets bein by determinin the total volume of the R-134a in the initial state. We know the mass, so if we an determine the speifi volume, we an determine the total volume usin: V = m V ˆ 1 1 Eqn 1 We an determine the initial speifi volume beause we know both T 1 and x 1. Here is the relevant data from the Saturation Temperature table for R-134a. Vˆ x Vˆ 1x Vˆ sat sat sat mix vap liq Temp. Pressure V (ft 3 /lb m ) ( o F) (psia) Sat. Liq Sat. Vap Eqn V ft 3 /lb m V ft 3 Next, we an determine V 2 beause we know it is 40% larer than V 1. V ft 3 /lb m V ft 3 Now, we need to determine P 2. We an aomplish this usin the linear sprin equations iven in the problem statement. As the R-134a expands, the pressure inreases beause the sprin exerts additional fore opposed to the expansion. The key to the relationship between the fore exerted by the sprin and the pressure inside the ylinder is the ross-setional area of the piston, A ross. F k (h -h ) P P P sprin = 1+ = 1+ Aross Aross Eqn 3 p V 2 Where : A Eqn 4 Eqn 5 ross = D h = 4 Aross We don't how muh fore the sprin exerts on the piston in the initial state, but that is not a problem. What we need to know is how muh additional fore the sprin exerts on the pistons as the R-134a expands. Pluin values into Eqns 4, 5 & 3 yields : A ross ft 2 1 ft 12 in in 2 1 ft 2 = 144 in 2 h in h in Be very areful with ft and in units when usin Eqn 3! P psia Dr. Baratui - ENGR 224 hw1-sp11.xlsm, 3.37E 4/11/2011

16 Now, we know the values of two intensive variable for the final state: V 2 and P 2, so we an determine the values T 2 and H 2 by interpolation on the R-134a tables. Here is the relevant data from the Saturation Temperature table for R-134a. Pressure Temp. V (ft 3 /lb m ) (psia) ( o F) Sat. Liq Sat. Vap Sine V 2 is reater than V sat vap even at the lower P of 10 psia, we an onlude that the R-134a is a superheated vapor in the final state. So, we need to look at the Superheated table for R-134a. Beause there is no table for psia in the Superheated R-134a table and our value of V 2, ft 3 /lb m, does not appear in both the 10 psia and 15 psia tables, double interpolation is required to solve this problem. Here is the double interpolation table for T : This double interpolation is a little different beause V 2 is known and T 2 is unknown. I hose to interpolate on pressure first and then to interpolate on volume. I don't think you an do this problem in usin a different interpolation order. Pressure (psia) T( o F ) T o F Here is the double interpolation table for H : I hose to interpolate on pressure first and then to interpolate on temperature. If you do the interpolations in the opposite order, you will et a slihtly different answer. Either method is satisfatory. Pressure (psia) T( o F ) H Btu/lb m Verify : None. Answers : T o F H Btu/lb m Notes: The reason I haned T 1 from -30 o F to +30 o F was to redue the error that is an unavoidable problem in all interpolations, but espeially in double interpolations. The orret answers for this problem usin T 1 = 30 o F are: T o F H Btu/lb m As you an see, the errors due to interpolation are sinifiant, but not disasterous. If we had used T1 = -30 o F, the errors would have been disasterous! Dr. Baratui - ENGR 224 hw1-sp11.xlsm, 3.37E 4/11/2011

17 ENGR Thermodynamis Baratui Problem : Isobari Expansion of Water - 6 pts 6-Apr-11 A piston-and-ylinder devie ontains m 3 of liquid water and 0.9 m 3 of water vapor in equilibrium at 600 kpa. Heat is added to the water at onstant pressure until the temperature reahes 200 o C. a.) What is the initial temperature of the water in the ylinder in o C? b.) What is the total mass of the water in the ylinder in k?.) Calulate the final volume in m 3. d.) Show the proess on a ompletely labeled PV Diaram. Be sure to inlude the two-phase envelope. The keys to this problem are that the proess is isobari and that the ylinder is a losed system. The mass in the system is the same in the initial and final states. We an determine the mass of water in the system in the initial state usin the iven T, P and V. Beause the proess is isobari, state 2 is a saturated vapor at 300 kpa and we an lookup all of its properties in the saturation pressure table. Given : V 1,sat liq m 3 P 1 = P kpa V 1,sat vap 0.9 m 3 T o C Find : a.) T 1??? b.) m??? k.) V 2??? m 3 d.) Construt a fully labeled PV diaram of this proess. Assumptions : 1 - The system is losed. No mass enters or leaves the ylinder. 2 - The intial and final states are equiibrium states. o C Diaram : 1 2 P 1, V 1, T 1 Isobari Heatin P2 = P1 V 2 > V 1 T 2 > T 1 Part a.) In the initial state, liquid and vapor exist in equilibrium at 600 kpa. Therefore the initial temperature must be equal to the saturation temperature at 600 kpa. We an look this value up in the saturated steam pressure table. T sat (600kPa) = o C T o C Part b.) We know the volume of the saturated liquid and the volume of the saturated vapor in the ylinder, so we an determine the mass of eah from the speifi volume usin : m sat vap V ˆV sat vap sat vap Vsat liq Eqn 1 msat liq Eqn 2 ˆV sat liq We an look up the speifi volumes of the saturated liquid and saturated vapor in the saturated steam pressure table and then use Eqns 1 & 2 to determine the mass o f eah phase in the initial state. Pressure Temp. V (m 3 /k) m 1,sat liq k (kpa) ( o C) Sat. Liq Sat. Vap m 1,sat vap k x k vap/k The total mass is the sum of the masses of the sat'd liquid and sat'd vapor : m k Dr. Baratui - ENGR hw1-sp11.xlsm, /11/2011

18 Part.) We an determine V 2 if we an first determine the final speifi volume. Sine we know both T 2 and P 2, we an look up the final spei volume in the steam tables. Beause T 2 > T sat (600 kpa), we will need to look in the superheatd steam tables. V m Vˆ 2 sat vap 2 Eqn 3 V m 3 /k V m 3 Part d.) Verify : None of the assumptions made in the solution of this problem an be verified based on the iven information. Answers : a.) T o C b.) m 7.39 k.) V m 3 d.) See above. Dr. Baratui - ENGR hw1-sp11.xlsm, /11/2011

19 ENGR Thermodynamis Baratui Problem : R-134a Table Fundamentals - 4 pts 6-Apr-11 A 0.5 m 3 vessel ontains 10 k of R-134a at -20 o C. Determine a.) The pressure in kpa b.) The total internal enery in kj.) The volume oupied by the liquid phase in m 3 Parts (a) and (b) are just a review of how to use thermodynami data tables. The key is to use the iven volume and mass to alulate the overal speifi volume. Part () requires you to USE the liquid speifi volume and vapor speifi volume to determine how muh of the mass is in the liquid and vapor phases. Given : V total 0.5 m 3 T -20 o C m total 10 k Find : a.) P??? kpa b.) U total??? kj.) V liq??? m 3 Assumptions : - The system exists in an equilibrium state. Part a.) We know T, but we need to know the values of 2 intensive variables. So, lets use V and m to alulate the overall speifi volume. V ˆV m total total Eqn 1 V 0.05 m 3 /k Next, we neeed to use the iven T to obtain data from the Saturation Temperature table. This will allow us to ompare the iven value of H to the values of H sat liq and H sat vap in order to determine the phase or phases present in the system at equilibrium. Here is the relevant data : Sine : Vˆ Vˆ Vˆ sat liq sat vap Temp. Pressure V (m 3 /k) U (kj/k) ( o C) (kpa) Sat. Liq Sat. Vap Sat. Liq Sat. Vap E the system ontains a saturated mixture and the pressure must be equal to the saturation pressure. Part b.) P= P sat kpa We need to know the speifi internal enery of the system or vapor-liquid mixture so we an alulat the total internal enery in the system. The key equation is: U m Uˆ total total We ould determine the speifi internal enery usin the followin equation, if we only knew the quality! Uˆ x Uˆ 1x Uˆ sat sat sat mix vap liq We an determine the quality from the speifi volume of the mixture usin : Vˆ Vˆ x Vˆ Vˆ sat vap sat liq satliq Eqn 2 Eqn 3 Eqn 4 Now, we have everythin we need to know to omplete part (b). Pluin values into Eqns 4, 3 & 2, in that order yields : x k vap/k U kj/k U total kj Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

20 Part.) We know the speifi volume of the saturated liquid in the system. So, we ould determine the volume oupied by the saturated liquid usin : Vsatliq msatliq Vˆ satliq So, we need to determine the mass of liquid in the system. This is not diffiult beause we know the quality! The equation we need to use is : m 1x m satliq total Eqn 5 Eqn 6 Verify : Now, we an plu values into Eqns 6 & 5, in that order, to omplete the solution of this problem. m sat liq k V sat liq E-03 m 3 Answers : a.) P kpa b.) U total 904 kj.) V liq L L None of the assumptions made in the solution of this problem an be verified based on the iven information. Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

21 ENGR Thermodynamis Baratui Problem : Inflatin an Automobile Tire - 6 pts 6-Apr-11 The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25 o C, the pressure aue reads 210 kpa. If the volume of the tire is m 3, determine the pressure rise in the tire when the air temperature rises to 50 o C. Also determine the amount of air that must be bled off to restore pressure to its oriinal value at this temperature (in k and in L at P atm and 25 o C). Assume atmospheri pressure is 100 kpa. Assume that the air in the tire behaves as an ideal as, but then hek the validity of this assumption The keys to this problem are that the air is to be treated as an ideal as and that the volume of the air in the tire remains onstant throuhout both proesses: heatin and pressure relief. Given : V m 3 T 2 50 T 1 25 o C P atm 100 kpa P 1,aue 210 kpa R J/mol-K P 1,abs 310 kpa MW 29 /mol Find : a.) P 2 - P 1??? kpa b.) m 2-3??? V amb??? L Assumptions : - The air in the tire is an ideal as throuhout both proesses: heatin and pressure relief. o C Part a.) The first thin we need to do is determine the number of moles of air initially inside the tire. We an use the Ideal Gas EOS diretly beause we were iven the values of P 1, V and T 1 in the problem statement. Be sure to use the absolute pressure! PV nrt PV Eqn 1 n Eqn 2 RT No air leaves the tire durin the heatin proess, so : n 1 = n 2 = moles Now, we know n 2, V, T 2, so we an apply the Ideal Gas EOS to state 2 (at the end of the heatin proess) to determine P 2. n2rt2 P2 V Eqn 3 P kpa The rise in pressure for the heatin proess is: P 2 - P kpa Part b.) Lets start by determinin the mass of air initially inside the tire. We know the number of moles and the MW of air, so this should not be too diffiult. m= n*mw Eqn 4 m Next, lets determine how many moles of air must be in the tire at 50 o C when the pressure is 210 kpa (aue). This is the final state, state 3. We an do this by applyin Eqn 2 to state moles Next, we apply Eqn 4 to determine the mass of air in the tire in state 3. m Now, we an determine the mass of air that must be bled off from the tire and the volume this air would oupy at ambient T and P. n 3 D m = m -m Eqn 5 m Now, we an solve Eqn 1 for V to determine the volume that the air removed from the tire would oupy at 100 kpa and 25 o C. V amb n n RT P 2 3 amb amb Eqn 6 V amb 6.00 L Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

22 Verify : We need to verify the validity of the Ideal Gas EOS by determinin the molar volume of the air in the tire in both the initial and final states. If we onsider air to be a diatomi as, then the riterion for ideal as behavior is : V 20 L/mole Use the fat that the molar volume is V/n and rearrane Eqn 1 to obtain : RT V P Eqn 7 Pluin values into Eqn 7 for states 1 yields : V L/mol Note that the molar volume in state 2 is the same as the molar volume in state! The molar volume is too small. It is not aurate to treat the air in the tire as an ideal as. The error in doin so may exeed 1%. Answers : a.) P 2 - P 1 26 kpa b.) m V amb m 3 Dr. Baratui - ENGR 224 hw1-sp11.xlsm, /11/2011

23 ENGR Thermodynamis Baratui Problem : WB-3 - An Appliation of Equations of State - 8 pts 6-Apr-11 A m 3 tank ontains 1 k of R-134a at 110 o C. Determine the pressure of the refrierant usin: a.) b.).) d.) e.) The Ideal Gas EOS The Compressibility Fator EOS The R-134a Tables The van der Waal EOS The Soave-Redlih-Kwon EOS Not muh to say here. Given : m 1 k T K T 110 o C P 4.067E+06 Pa K V m 3 MW / mol R J/mol-K Part a.) Ideal Gas EOS : PV RT Eqn 1 RT Solve Eqn 1 for pressure : P Eqn 2 V We must determine the molar volume before we an use Eqn 2 to answer the question. V Use the definition of molar volume: V Eqn 3 n Where : m n MW Eqn 4 Pluin values into Eqn 4 yields : n 9.80 mol H 2 O Now, plu values bak into Eqns 3 & 2. V 1.711E-03 m 3 /mol Be areful with the units. P 1.86E+06 Pa P 1861 kpa Part b.) Compressibility EOS : Given T R and the ideal redued molar volume, use the ompressibility harts to evaluate either P R or the ompressibility, Z PV Z RT T Eqn 5 TR Eqn 6 T Pluin values into Eqn 6 yields : T R ideal V Definition of the ideal redued molar volume : V R Eqn 7 RT /P Pluin values into Eqn 7 yields : Read the Generalized Compressibility Chart for P R = 0 to 1 : V R ideal P R 0.39 Z We an use the definition of P R to alulate P : P PR P Pluin values into Eqn 9 yields : P 1586 kpa R P P P Eqn 8 Eqn 9 RT IG Or, we an use Z and its definition to determine P : P Z ZV Eqn 10 V Pluin values into Eqn 10 yields : P 1610 kpa Dr. Baratui - ENGR 224 hw1-sp11.xlsm, WB-3 4/11/2011

24 Part.) The R-134a Tables provide the best available estimate of the pressure in the tank. Beause T > T, the properties of the R-134a in the tank must be obtained from the superheated vapor table, even thouh the water is atually a superritial fluid in this system. At this point we an make use of the fat that we have a pretty ood idea of what the atual pressure is in the tank (from parts a-d) or we an san the superheated vapor tables to determine whih two pressures braket our known value of the speifi volume. In either ase, we bein by onvertin the molar volume into a speifi volume : V ˆV MW Eqn 11 Usin the MW of R-134a yields : v 1.677E-05 m 3 / v 1.677E-02 m 3 /k The Superheated R-134a Table ives us : At P = 1.6 MPa v = m 3 /k Fortunately, no interpolation is required : P 1600 kpa Part d.) van der Waal EOS : RT a P V b V 2 We an determine the values of a and b, whih are onstants that depend only on the hemial speies in the system, from the followin equations. Eqn R T a 64P 2 2 RT Eqn 13 b 8P Eqn 14 a Pa-mol 2 /m 6 b 9.56E-05 m 3 /mol Now, we an plu the onstants a and b into Eqn 12 to determine the pressure. P 1629 kpa RT a Part e.) Soave-Redlih-Kwon EOS : P V b V V b Eqn 15 We an determine the values of a, b and, whih are onstants that depend only on the hemial speies in the system, from the followin equations. R T a P m 1 T R m RT Eqn 16 b P Eqn 17 T Eqn 18 TR T Eqn 19 Eqn 20 Where : Now, plu values into Eqns : T R a Pa-mol 2 /m 6 m b 6.629E-05 m 3 /mol P 1610 kpa Answers : P (kpa) a.) The Ideal Gas EOS 1861 b.) The Compressibility Fator EOS 1586.) The R-134a Tables 1600 d.) The van der Waal EOS 1629 e.) The Soave-Redlih-Kwon EOS 1610 Dr. Baratui - ENGR 224 hw1-sp11.xlsm, WB-3 4/11/2011

25 ENGR Thermodynamis Baratui Problem : WB-4 - Relative Humidity and Foed Glasses - 5 pts 6-Apr-11 On enterin a dwellin maintained at 20 o C from the outdoors where the temperature is 10 o C, a person's eye-lasses are observed not to beome foed. A humidity aue indiates the relative humidity in the dwellin is 55%. Can this readin be orret? Provide supportin alulations. The key to this problem is to assume that the lasses are at 10 o C when they ome in from the outdoors and to reonize that if the dew point of the indoor air is GREATER THAN 10 o C, then when the humid air hits the old lasses, water will ondense onto the lasses and they will "fo". Sine the lasses did not atually fo, we know that the dew point temperature of the humid air in the house is LESS THAN 10 o C. So, we an alulate Tdew of indoor air at h r = 55% and ompare to 10 o C to see if the 55% relative humidity is possibly orret in liht of the fat that the lasses did not fo. Given : T in 20 o C T out 10 o C h r 55% Find : Can the humidity aue be orret if the lasses do not fo? Assumptions : 1- The humid air behaves as an ideal as. 2- The lasses are at a temperature of 10 o C when they enter the house. The water in the humid air in the house will ondense on the lasses if 10 o C is less than or equal to the dew point temperature of the air in the house. So, we an determine whether the humidity aue is orret by omparin the dew point of the humid air in the house to 10 o C. If T dew > 10 o C, then the lasses SHOULD have foed and we an onlude that the humidity aue is not orret. If T dew < 10 o C then we annot determine whether the humidity aue is orret Pw Relative humidity is defined by : hr * P T Eqn 1 We know the relative humidity of the indoor air and we an look up the vapor pressure of water at the indoor T of 20 o C, so we an solve Eqn 1 for the partial pressure of water in the indoor air and evaluate it. P h P T * w r w w Eqn 2 P* w = P sat (20 o C) kpa P w kpa The key to solvin this problem is that the mole fration of water in the humid indoor air remains onstant as it approahes the surfae of the old lasses. The total pressure in the room also remains onstant. So, if the indoor air is an ideal as, then the partial pressure of water in the indoor air also remains onstant beause : P y P w w total Now, let's onsider what happens when the humid indoor air is ooled down to its dew point. At the dew point temperature, the relative humidity is 100%. P P T * w w dew Eqn 3 Eqn 4 Sine the partial pressure of water in the humid indoor air is onstant, we an use Eqn 4 to evaluate the vapor pressure of water at the dew point temperature. P* w (T dew ) kpa Now, we an o bak to the saturation temperature tables and interpolate to detemine the T at whih the vapor pressure is kpa. This T is the dew point! Here is the relevant data : Temp. Pressure ( o C) (kpa) T dew T dew o C Dr. Baratui - ENGR 224 hw1-sp11.xlsm, WB-4 4/11/2011

26 Verify : We need to verify the validity of the Ideal Gas EOS by determinin the molar volume of the humid air in both the indoor and outdoor states. We annot onsider humid air to be a diatomi as, so the riterion for ideal as behavior is : V 20L/mole PV RT RT Eqn 5 V Eqn 6 P Assume atmospheri pressure : P kpa R J/mol-K Pluin values into Eqn 6 yields : V L/mole L/mole Sine the molar volume of both the indoor and outdoor air exeed 20 L/mole, it is aurate to 2 sinifiant fiures to treat the air as an ideal as. V 2 Answers : Sine T out < T dew, the lasses SHOULD have foed when they ame into the house if the humidity were really 55%. However, sine the lasses did not fo, we an onlude that the atual relative humidity was less than 55% and the humidity aue is not orret. Dr. Baratui - ENGR 224 hw1-sp11.xlsm, WB-4 4/11/2011