Name: ________________________ ID: __________________ MAE 241 - Summer 2022 Recitation 10 - Administered 16-June, 2022 Instructions - You should attempt problems 1 and 2 to receive credit. We just want to see your attempt, regardless correct or incorrect. - Provide as much details as you can in your solutions - Both problems are good practice problems for Exam 2. - At the end of the recitation please upload your work to GradeScope; individual submission. Problem 1– Four stroke heat engine modeled as a Carnot Cycle A particular engine is modeled as heat engine that operates as an air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, please answer the following. TH= 1200 K and TL = 350 K a. Sketch the cycle on a T-s diagram. Include the isobars for 350 kPa and 300 kPa and the heat interactions between the cycle and the surroundings according to the processes that exchange heat. Page 1 of 15 MAE 241 – Summer 2022 Recitation 10 b. Using variable specific heats, determine the maximum pressure in the cycle. Hint: if you want to visualize where the maximum pressure occurs and the T-s diagram is not helping you, represent the cycle on the P-v diagram. The maximum pressure in the cycle will be after isentropic compression process (4-1). To visualize it better, P-v diagram of Carnot cycle is shown below, Process 4-1 is an isentropic compression process. So, based on variable specific heats, π4 ππ4 ππ1 ( ) = => π1 = ( ) π4 π1 π =ππππ π‘πππ‘ ππ1 ππ4 From Table A-17 At 350 K , ππ4 = 2.379 and At 1200 K ππ1 = 238.0 238.0 π1 = ( ) 300 πππ = 30,012.61 πππ 2.379 c. Compute the heat transfer to the air in the cycle. Hint: look at your cycle diagram on part (a) Conservation of the Energy Equation for process 1-2 as a closed system is given by πΈππ − πΈππ’π‘ = βπΈπ π¦π π‘ππ (πππ + πππ + πΈπππ π ππ ) − (πππ’π‘ + πππ’π‘ + πΈπππ π ππ’π‘ ) = βπ + βπΎπΈ + βππΈ Let’s make some assumptions that will simplify the equation: 1- The system is stationary, βπΎπΈ = βππΈ = 0 2- Since the piston is a closed system then, there is no mass enter or leave the system, πΈπππ π ππ = πΈπππ π ππ’π‘ = 0 Page 2 of 15 MAE 241 – Summer 2022 Recitation 10 3- No heat leaves the piston-cylinder assembly, πππ’π‘ = 0 4- Air behaves as an ideal gas Then, we will end up with the following equation, πππ + πππ − πππ’π‘ = βπ Or in terms of specific quantities (since mass inside the system is constant always) πππ + π€ππ − π€ππ’π‘ = βπ’ (1) Process 1-2 is an expansion process. So, π€ππ = 0. The boundary work for an isothermal expansion is given by π π€ππ’π‘,12 = π π1 ln π1 (2) 2 Finding π·π From ideal gas law, π£4 = π π4 π4 0.2867 = ππ½ . ππ−πΎ 350πΎ 300 πππ π3 = 0.3344 ππ Process 4-1 is an isentropic process. So from variable specific heats, π£4 π£π4 π£π1 = => π£1 = ( ) π£4 π£1 π£π1 π£π4 From Table A-17 At 350 K , π£π4 = 422.2 and At 1200 K π£π1 = 14.470 π£π1 14.470 π3 π3 π£1 = ( ) π£4 = ( ) 0.3344 = 0.01146 π£π4 422.2 ππ ππ From ideal gas law, π£3 = π π3 π3 = 0.2867 ππ½ . ππ−πΎ 350πΎ 150 πππ π3 = 0.6689 ππ Similarly 2-3 is an isentropic expansion process. So, from variable specific heats, π£2 π£π2 π£π2 = => π£2 = ( ) π£3 π£3 π£π3 π£π3 From Table A-17 At 350 K , π£π3 = 422.2 and At 1200 K π£π2 = 14.470 π£π2 14.470 π3 ) π£3 = ( ) π£3 = 0.02292 π£π3 422.2 ππ Process 2-3 is an isothermal process. So, π1 π£1 = π2 π£2 π£2 = ( π2 = ( π£1 )π = π£2 1 0.01146 π3 ππ π3 0.02292 ππ ) ( Alternatively, π2 can be found as follows, 30,012.61 πππ = 15,001.46 πππ Similarly 2-3 is an isentropic expansion process. So, from variable specific heats, π2 ππ2 ππ2 = => π2 = ( ) π3 π3 ππ3 ππ3 From Table A-17 At 350 K , ππ3 = 2.379 and At 1200 K ππ2 = 238.0 Page 3 of 15 MAE 241 – Summer 2022 Recitation 10 π2 = ( ππ2 238.0 ) π3 = ( ) 150 πππ = 15,006.30 πππ ππ3 2.379 So, from (2) π€ππ’π‘ = π π1 ln π1 ππ½ 30,012.61 πππ ππ½ = 0.2867 . 1200 πΎ ln = 238.47 π2 ππ − πΎ 15,006.30 πππ ππ From (1) πππ = π€ππ’π‘ + βπ’ = π€ππ’π‘,12 + π’2 − π’1 π’2 = π’1 since both states are at same temperature. ππ½ πππ = π€ππ’π‘,12 = 238.47 ππ d. Compute the entropy change per unit mass, in kJ/kg-K, for the high temperature process. Entropy balance for a closed system and the process 1-2 is given by, π π βπ12 = ∑ − ∑ + ππππ π π ππ ππ’π‘ In a Carnot cycle, all the processes are reversible. So, ππππ = 0. In the high temperature process, there is only heat addition. So, πππ’π‘ = 0. βπ12 = πππ ππ» In terms of specific quantities, ππ½ πππ 238.47 ππ ππ½ βπ 12 = = = 0.1987 ππ» 1200 πΎ ππ − πΎ e. Determine the mass of air in the cycle, in kg, with three significant digits. Energy balance for the entire cycle Conservation of the Energy Equation is given by πΈππ − πΈππ’π‘ = βπΈπ π¦π π‘ππ (πππ + πππ + πΈπππ π ππ ) − (πππ’π‘ + πππ’π‘ + πΈπππ π ππ’π‘ ) = βπ + βπΎπΈ + βππΈ Let’s make some assumptions that will simplify the equation: 1- The system is stationary, βπΎπΈ = βππΈ = 0 2- Since the piston is a closed system then, there is no mass enter or leave the system, πΈπππ π ππ = πΈπππ π ππ’π‘ = 0 For a cycle, βπ = 0 we will end up with the following equation, πππ’π‘ − πππ = ππππ‘ = πππ − πππ’π‘ = π(πππ − πππ’π‘ ) π=π ππππ‘ ππ−πππ’π‘ (3) Page 4 of 15 MAE 241 – Summer 2022 Recitation 10 Conservation of the Energy Equation for process 3-4 is given by πΈππ − πΈππ’π‘ = βπΈπ π¦π π‘ππ (πππ + πππ + πΈπππ π ππ ) − (πππ’π‘ + πππ’π‘ + πΈπππ π ππ’π‘ ) = βπ + βπΎπΈ + βππΈ Let’s make some assumptions that will simplify the equation: 1- The system is stationary, βπΎπΈ = βππΈ = 0 2- Since the piston is a closed system then, there is no mass enter or leave the system, πΈπππ π ππ = πΈπππ π ππ’π‘ = 0 3- No heat is added to the piston, πππ’π‘ = 0 Then, we will end up with the following equation, πππ’π‘ = πππ − πππ’π‘ + βπ Or in terms of specific quantities (since mass inside the system is constant always) πππ’π‘ = π€ππ − π€ππ’π‘ + βπ’ (4) The term π€ππ − π€ππ’π‘ will be the work done during isothermal process. Calculating π€ππ’π‘ and identifying its sign will help us identify if the work is transferred out or into the system π ππ½ 150 ππ½ π€ππ’π‘ = π π ln π3 = 0.2867 ππ−πΎ 350 πΎ ln 300 = −69.55 ππ 4 ππ½ ππ From (4) after enforcing βπ’ = 0, since both 2 and 3 are at same temperatures ππ½ πππ’π‘ = 69.55 ππ From (3) 0.5 ππ½ π= = 2.960 × 10−3 ππ ππ½ ππ½ 238.47 ππ − 69.55 ππ • πππ − πππ’π‘ is the area of rectangle in the T-s diagram which will give the net work. ππ, π€ππ = 69.55 Page 5 of 15 MAE 241 – Summer 2022 Recitation 10 Problem 2 (Problem 9.52 ideal Diesel cycle) An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process, air is at 95 kPa and 27°C. Accounting for the variation of specific heats with temperature, determine (a) the temperature after the heat-addition process, (b) the thermal efficiency, and (c) the mean effective pressure. Air-standard assumptions for diesel cycle From textbook Page 480 (heat addition at constant pressure for diesel) State 1: π1 = 95 πππ, π1 = 27 β From ideal gas law, we can calculate π£1 = π π1 π1 = (0.2867 ππ½ )(273+27)πΎ ππ−πΎ 95 πππ π3 = 0.9063 ππ Page 6 of 15 MAE 241 – Summer 2022 Recitation 10 State 2: Compression ratio π = π£ So, π£2 = 161 = π3 0.9063 ππ 16 π£πππ₯ π£πππ π£ = π£1 = 16 2 = 0.0566 π3 ππ Also, Process 1-2 is an isentropic compression process. Assuming variable specific heats, π£2 π£π2 ( ) = = 16 π£1 π =ππππ π‘πππ‘ π£π1 From Table A-17 at 300 K we get π£π1 = 621.2. So, π£π2 = π£π1 16 = (621.2) 16 = 38.825. Again, from Table A-17, we get π2 = 862.35 πΎ after linear interpolation So, from ideal gas law, ππ½ π π2 (0.2867 ππ − πΎ ) (862.35 )πΎ π2 = = = 4368.12 πππ π3 π£2 0.0566 ππ State 3: π2 = π3 = 4368.12 πππ π£ Cut off ratio ππ = π£3 = 2. So, π£3 = 2. π£2 = 2 (0.0566 2 π3 = π3 π£3 = π π3 π3 ) = 0.1132 ππ ππ (4368.12 πππ) (0.1132 ππ½ 0.2867 ππ − πΎ π3 ππ ) = 1724.69 πΎ State 4: π3 π£4 = π£1 = 0.9063 ππ Process 3-4 is isentropic expansion. So, assuming variable specific heats π£4 π£π4 = π£3 π£π3 π£4 π£π4 = ( ) π£π3 π£3 From Table A-17, at π3 = 1724.69 πΎ, we get π£π3 = 4.5380 π3 0.9063 π£4 ππ π£π4 = ( ) π£π3 = π£π4 = ( ) 4.5380 = 36.332 π3 π£3 0.1132 ππ From Table A-17, we get π4 = 882.41 πΎ From ideal gas law Page 7 of 15 MAE 241 – Summer 2022 Recitation 10 ππ½ π π4 (0.2867 ππ − πΎ ) 882.41 πΎ π4 = = = 279.14 πππ π3 π£4 0.9063 ππ a. the temperature after the heat-addition process, π»π = ππππ. ππ π² b. the thermal efficiency, The thermal efficiency of the cycle is given by, ππ‘β = 1 − ( πππ’π‘ ) πππ The heat addition (Process 2-3): Noting that the Diesel cycle is executed in a piston–cylinder device, which forms a closed system. Conservation of the Energy Equation is given by πΈππ − πΈππ’π‘ = βπΈπ π¦π π‘ππ (πππ + πππ + πΈπππ π ππ ) − (πππ’π‘ + πππ’π‘ + πΈπππ π ππ’π‘ ) = βπ + βπΎπΈ + βππΈ Let’s make some assumptions that will simplify the equation: 1. The system is stationary, βπΎπΈ = βππΈ = 0 2. Since the piston is a closed system then, there is no mass enter or leave the system, πΈπππ π ππ = πΈπππ π ππ’π‘ = 0 3. No heat leaves the piston, πππ’π‘ = 0 4. Only boundary work is present and is assumed to be transferred to the surrounding at a constant pressure so, πππ = 0 Then, we will end up with the following from Eq3: πππ − πππ’π‘ = βπ πππ = π(π2 (π£3 − π£2 ) + (π’3 − π’2 )) = π(β3 − β2 ) ππ πππ = πππ π = (β3 − β2 ) (1) The heat rejection (Process 1-4) Conservation of the Energy Equation is given by πΈππ − πΈππ’π‘ = βπΈπ π¦π π‘ππ (πππ + πππ + πΈπππ π ππ ) − (πππ’π‘ + πππ’π‘ + πΈπππ π ππ’π‘ ) = βπ + βπΎπΈ + βππΈ Let’s make some assumptions that will simplify the equation: 1- The system is stationary , βπΎπΈ = βππΈ = 0 Page 8 of 15 MAE 241 – Summer 2022 Recitation 10 2- Since the piston is a closed system then, there is no mass enter or leave the system, πΈπππ π ππ = πΈπππ π ππ’π‘ = 0 3- No heat enters the piston, πππ = 0 4- No work interactions , πππ = 0 and πππ’π‘ = 0 Then, we will end up with the following from Eq3: −πππ’π‘ = βπ πππ’π‘ = π(π’4 − π’1 ) πππ’π‘ = πππ’π‘ π or = π’4 − π’1 (2) So, ππ‘β = 1 − ( πππ’π‘ π’4 − π’1 )=1−( ) πππ β3 − β2 From Table A-17, State 1: at 300 K, π’1 = 214.07 ππ½ ππ ππ½ State 2: at 862.35 πΎ, β2 = 890.89 ππ ππ½ State 3: at 1724.6 πΎ β3 = 1910.36 ππ State 4: at 882.41 πΎ π’4 = 659.95 ππ½ ππ ππ½ ππ½ 659.95 ππ − 214.07 ππ π’4 − π’1 ππ‘β = 1 − ( )= 1−( ) = 56.26 % ππ½ ππ½ β3 − β2 1910.36 ππ − 890.89 ππ c. Mean effective pressure π€πππ‘ πππ − πππ’π‘ (β3 − β2 ) − (π’4 − π’1 ) = = π£πππ₯ − π£πππ π£1 − π£2 π£1 − π£2 ππ½ ππ½ ππ½ ππ½ (1910.36 ππ − 890.89 ππ) − (659.95 ππ − 214.07 ππ) = = 675.05 πππ π3 π3 0.9063 ππ − 0.0566 ππ ππππ ππππππ‘ππ£π ππππ π π’ππ = Answers in textbook: (a) 1725 K, (b) 56.3 percent, (c) 675.9 kPa Page 9 of 15 MAE 241 – Summer 2022 Recitation 10 Problem 3 – Ideal steam power cycle (10-95) Steam enters the turbine of a steam power plant that operates on a simple ideal Rankine cycle at a pressure of 6 MPa, and it leaves as a saturated vapor at 7.5 kPa. Heat is transferred to the steam in the boiler at a rate of 40,000 kJ/s. Steam is cooled in the condenser by the cooling water from a nearby river, which enters the condenser at 15°C. Show the cycle on a T-s diagram with respect to saturation lines, and determine (a) the turbine inlet temperature, (b) the net power output and thermal efficiency, and (c) the minimum mass flow rate of the cooling water required. a) the turbine inlet temperature (T3) ππ½ State 4: β4 = βπ@7.5πππ = 2574.0 ππ (A-5) ππ½ Process 3-4 is an isentropic process. So, π 3 = π 4 = π π@7.5πππ = 8.2501 ππ−πΎ (A-5) At 6 MPa, value of π 3 > π π@6πππ . So, state 3 should be in superheated state. (A-5) From Table A-6, under 6 MPa, π3 = 1089.18 πΎ after linear interpolation. b) the net power output and thermal efficiency • To calculate the net power output, mass flow rate is needed. It can be calculated from heat input to the boiler which is already given. Net power output: Analyzing the boiler as a control volume: Starting with the general mass balance: Page 10 of 15 MAE 241 – Summer 2022 Recitation 10 ∑ππ πΜ − ∑ππ’π‘ πΜ = πππΆπ Eq1 ππ‘ The energy balance equation: πΈΜππ − πΈΜππ’π‘ = ππΈπ π¦π π‘ππ /ππ‘ We can expand it knowing that the boiler is under steady state condition as follows: πΜππ + πΜππ + ∑ πΜ π = πΜππ’π‘ + πΜππ’π‘ + ∑ πΜ π ππ ππ’π‘ Then it can be written as: 2 2 π π πΜππ + πΜππ + ∑ππ πΜ (β + + ππ§) = πΜππ’π‘ + πΜππ’π‘ + ∑ππ’π‘ πΜ (β + + ππ§) 2 2 Eq2 The assumptions: 1- This is a steady-flow process since there is no change with time, πππΆπ ππ‘ = 0. 2- Potential energy and kinetic energy changes are negligible, so small compared with (ββ β« βπΎπΈ & ββ β« βππΈ) so, βπΎπΈ = 0, & βππΈ = 0. 3- There is one inlet and one outlet. 4- There is no heat out from the boiler, πΜππ’π‘ = 0. 5- No work interactions between the surroundings and the system, πΜππ = 0, πΜππ = 0 Using the assumptions, Eq10 become as follows to represent the total mass of the boiler (πΜ ): πΜππ = πΜππ’π‘ = πΜ Eq3 Using the assumptions, Eq11 becomes as follows: πΜππ + πΜππ + ∑ πΜ (β + ππ π2 π2 Μ Μ + ππ§) = πππ’π‘ + πππ’π‘ + ∑ πΜ (β + + ππ§) 2 2 ππ’π‘ Which will lead to: πΜππ = πΜ [β3 − β2 ] Eq4 Let’s analyze the pump as control volume as follows: Starting with the general mass balance: ∑ππ πΜ − ∑ππ’π‘ πΜ = πππΆπ ππ‘ Eq5 The energy balance equation: πΈΜππ − πΈΜππ’π‘ = ππΈπ π¦π π‘ππ /ππ‘ We can expand it knowing that the pump is under steady state condition as follows: πΜππ + πΜππ + ∑ πΜ π = πΜππ’π‘ + πΜππ’π‘ + ∑ πΜ π ππ ππ’π‘ Page 11 of 15 MAE 241 – Summer 2022 Recitation 10 Then it can be written as: 2 2 π π πΜππ + πΜππ’ππ,ππ + ∑ππ πΜ (β + + ππ§) = πΜππ’π‘ + πΜππ’π‘ + ∑ππ’π‘ πΜ (β + + ππ§) 2 2 π€ππ’ππ,ππ = πΜππ’ππ,ππ πΜ Eq6 = β2 − β1 Eq7 Since this is an ideal Rankine cycle, the pump is adiabatic and reversible, i.e. isentropic. For open flow devices, the reversible specific work input is 2 π€rev,in = ∫ π£ ππ + Δππ + Δππ 1 2 π€rev,in = ∫ π£ ππ 1 where we have utilized the above assumptions that Δππ and Δππ are both small compared to any other relevant energy terms. The inlet to the pump (state 1) is a saturated liquid, and the outlet is a compressed liquid, so we can treat the water as an incompressible (constant π£) fluid as it flows through the pump. The specific work in the pump then becomes 2 π€ππ’ππ,ππ = π€rev,in = ∫ π£ ππ = π£1 (π2 − π1 ). 1 m3 State 1; π£1 = π£π@7.5πππ = 0.001008 ππ π€ππ’ππ,ππ = π£1 (π2 − π1 ) = 0.001008 m3 ππ½ (6000 πππ − 7.5 πππ ) = 6.04 ππ ππ π€ππ’ππ,ππ = β2 − β1 = 6.04 State 1: β1 = βπ@7.5πππ = 168.75 ππ½ ππ ππ½ ππ β2 = β1 + 6.04 ππ½ ππ½ = 174.79 ππ ππ So from Eq4, πΜππ = πΜ [β3 − β2 ] ππ½ At state 3: β3 = 4852.18 ππ (Table A-6) ππ½ 40,000 πΜππ ππ π πΜ = = = 8.551 ππ½ ππ½ β3 − β2 4852.18 π ππ − 174.79 ππ Page 12 of 15 MAE 241 – Summer 2022 Recitation 10 πΜππ’ππ,ππ = πΜ π€ππ’ππ,ππ = 8.551 ππ ππ½ ππ½ . 6.04 = 51.65 π ππ π Now analyzing the turbine as a control volume: Starting with the general mass balance: ∑ππ πΜ − ∑ππ’π‘ πΜ = πππΆπ Eq8 ππ‘ The energy balance equation: πΈΜππ − πΈΜππ’π‘ = ππΈπ π¦π π‘ππ /ππ‘ We can expand it knowing that the turbine is under steady state condition as follows: πΜππ + πΜππ + ∑ πΜ π = πΜππ’π‘ + πΜππ’π‘ + ∑ πΜ π ππ ππ’π‘ Then it can be written as: 2 2 π π πΜππ + πΜππ + ∑ππ πΜ (β + + ππ§) = πΜππ’π‘ + πΜππ’π‘ + ∑ππ’π‘ πΜ (β + + ππ§) 2 2 Eq9 The assumptions: 1- This is a steady-flow process since there is no change with time, πππΆπ ππ‘ = 0. 2- Potential energy and kinetic energy changes are negligible, so small compared with (ββ β« βπΎπΈ & ββ β« βππΈ) so, βπΎπΈ = 0, & βππΈ = 0. 3- There is one inlet and one outlet. 4- The turbine is adiabatic , πΜππ = 0 πππ πΜππ’π‘ = 0. 5- No work applied from the surroundings to the system, πΜππ = 0 Using the assumptions, Eq6 become as follows to represent the total mass of the turbine (πΜ ): πΜππ = πΜππ’π‘ = πΜ Eq10 Using the assumptions, Eq7 becomes as follows: πΜππ + πΜππ + ∑ πΜ (β + ππ π2 π2 + ππ§) = πΜππ’π‘ + πΜπ‘π’ππ,ππ’π‘ + ∑ πΜ (β + + ππ§) 2 2 ππ’π‘ Which will lead to: ππ ππ½ ππ½ ππ½ πΜπ‘π’ππ,ππ’π‘ = πΜ [β3 − β4 ] = 8.551 π (4852.18 ππ − 2574.0 ππ) = 19,480.71 π Eq11 πΜπππ‘ = πΜπ‘π’ππ,ππ’π‘ − πΜππ’ππ,ππ = 19,480.71 ππ½ ππ½ ππ½ − 51.65 = 19,429.06 π π π Page 13 of 15 MAE 241 – Summer 2022 Recitation 10 ππ‘β ππ½ πΜπππ‘ 19,429.06 π = = = 48.57 % ππ½ πΜππ 40,000 π c) the minimum mass flow rate of the cooling water required. Assumptions ππΈππ£ ππππ£ • Condenser is operating under steady flow conditions. • • • No mixing happens inside the Condenser. Pressure drop is negligible for steam and cooling water. Change in kinetic and potential energies are small compared to change in other energies. βπΎπΈ = βππΈ ≈ 0 ππ‘ = 0, ππ‘ =0 The cooling water should be raised to steam temperature so that steam completely cools down. So, the temperature of the cooling water will be saturation temperature at 7.5 KPa. So, Tout=40.29 β. Writing energy balance for the system chosen in this part and applying the assumptions listed above, ππΈ πΈΜππ − πΈΜππ’π‘ = ππ‘ππ£ Eq11 (πΜππ + πΜππ + ∑ πΜ (β + ππ π2 π2 ππΈππ£ Μ Μ ) (π + ππ§) − ππ’π‘ + πππ’π‘ + ∑ πΜ (β + + ππ§)) = 2 2 ππ‘ ππ’π‘ πΜ π (β4 ) + ππ€ Μ (βππ ) − πΜ π (β1 ) − ππ€ Μ (βππ’π‘ ) = 0 πΜ π (β4 − β1 ) = ππ€ Μ (βππ’π‘ − βππ ) πΜ (β4−β1 ) ππ’π‘ −βππ ) πΜπ€ = (βπ Eq12 kJ At inlet of the cooling water, βππ = βπ@15β = 62.982 kg(A-4) ππ½ So, βππ’π‘ = βπ@40.29β = 168.74 ππ(A-4) Page 14 of 15 MAE 241 – Summer 2022 Recitation 10 ππ ππ½ ππ½ 8.551 π (2574.0 ππ − 168.75 ππ) ππ πΜπ€ = = 194.47 ππ½ kJ π (168.74 ππ − 62.982 kg) Page 15 of 15
