ENSY 5000 Fundamentals of Energy System Integration 2022 Fall, Homework Problem Set 5NOTE: To solve this problem set as well as any other problem set, it will help if you draw your system and...

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In all of your solutions,



  • clearly show your system definition with proper sketches indicating related energy flows

  • clearly show how you write your equations (e.g. Mass, Energy balance, etc.)

  • identify relevant energy flows

  • use of definitions of power-energy relationships and efficiencies as needed

  • use of proper equation of state, or property relations as needed

  • clearly show your steps, assumptions and approximations.




ENSY 5000 Fundamentals of Energy System Integration 2022 Fall, Homework Problem Set 5 NOTE: To solve this problem set as well as any other problem set, it will help if you draw your system and identifying the energy component (property) changes and the energy interactions at the boundaries. In some cases, you may need a sub system analysis to identify the energy interaction, especially the chemical energy changes and heat flows associated with the inefficiencies of the system. Also, there is no external energy input until the end of the driving cycle for the hybrid vehicle, all the energies for the driving processes are supplied by the battery. For the nonhybrid vehicle, there is an external energy input from the fuel supplied to the engine for driving process. In all of your solutions, • clearly show your system definition with proper sketches indicating related energy flows • clearly show how you write your equations (e.g., Mass, Energy balance, etc.) • identify relevant energy flows • use of definitions of power-energy relationships and efficiencies as needed • use of proper equation of state, or property relations as needed • clearly show your steps, assumptions, and approximations. READING ASSIGNMENT: Read Chapter 2, section 2.3 to 2.6 (to end of Chapter 2) of the Textbook. Q1: A piston cylinder device has a volume of 0.04 m3 and initially contains air at 400 K and 2 bar. This device is used to perform a cycle in which the gas is heated at a constant volume until the temperature reaches 1200 K. The air is allowed to expand following an isothermal process until the volume is two times the original volume. It is then cooled at a constant volume process. The last process completes the cycle going back to the initial state such that PVn = constant, where n=1.3633. For the air, the constant volume specific heat capacity is cv =0.790 kJ/(kg·K) and ideal gas constant is Rg=0.287 kJ/(kg·K). A) Draw pressure-volume and temperature-volume diagrams indicating all 4 states and process lines between those states. B) Determine the energy interactions in the form of work for all processes. Explicitly state direction of energy flows (into the air in the cylinder or out of air). C) Determine the energy interactions in the form of heat for all processes. Explicitly state direction of energy flows (into the air in the cylinder or out of air). D) Calculate the net work produced by this cycle, the required heat input to the cycle and its thermal efficiency. E) Assume all heat input to this cycle comes from a heat source at 1800 K, and all heat discharged goes to a heat sink at 290 K. Calculate entropy production for each process. Find entropy produced in entire cycle. Which process has the largest irreversibility? Q2: Two heavily insulated beakers contain an incompressible liquid at 1 bar atmospheric pressure, but at different temperatures. Temperature of the hot liquid is 100oC and its mass is mH=6 kg. Temperature of the cold liquid is 20oC and its mass is mC=10 kg. Specific heat of liquid is constant and equal to c = 4 kJ/(kg·K). Density of the liquid is constant and equal to ρ=800 kg/m3. Both liquids are allowed to exchange heat through a part of their boundary (see Figure 1). Assume: FIGURE 1 • the liquids reach thermodynamic equilibrium at the end of this process. • there is no energy transfer between the liquids and the outside, i.e. the only energy transfer is between the fluids. • there is no significant kinetic or potential energy change. • Temperature of the liquids are spatially uniform in the beakers but changes overtime. Determine A. the final equilibrium temperature of the liquids by applying 1st law of thermodynamics (energy balance). HINT: you can combine both liquids within a single system boundary, or you can treat them two separate system such that energy coming from one goes to the other. Either way you should be able find the same final temperature. B. amount of energy transferred (heat) from the hot liquid to cold liquid. C. change in entropy separately of cold liquid and hot liquid. D. the entropy production associated with this process by applying entropy balance equation for a system that combines both liquids. HINT: Although you can find entropy production in each liquid separately and add them up to find total entropy production, this approach requires knowing boundary temperature for heat during the process which is not constant. This will be our next question. Q3: Consider the same configuration in Q2. Rate of heat going from hot liquid to cold liquid can be determined from the thermal resistance of contact medium (see figure 2, where contact region is exaggerated for clarity), as �̇�? = (???? − ????)/???? . Assume for this problem ???? = 10 K/W. Choose a small-time increment Δ?? where variation of temperatures in the cold liquid and hot liquid are small, and incremental heat within Δ?? can be approximated from ???? = �̇�?Δt. Since mass of liquids are different change in temperatures in hot (Δ????) and cold (Δ????) liquids will be different. But Δ???? and cold Δ???? can be determined from the energy balance equation using known ????. Develop a multi-step solution for transient behavior that starts from given initial temperatures for both liquids, then 1. evaluate ???? 2. evaluate Δ???? and Δ???? 3. update hot and cold liquid temperatures for the next step, new ???? → old TH − Δ???? and new ???? → old TC − Δ???? 4. if TH>TC go back to step 1 5. Stop. equilibrium is reached Please note that you can freely choose time increment Δ??. However, if it is too large you will have larger approximation errors. If it is too small, you will need to many steps above. I would recommend use a value which gives 0.1 < δ????="">< 0.5. if you prefer you can choose a value for δ???? and find corresponding δ?? for each step, and do the rest of the calculations as above. figure 2 hints: we are looking for a transient solution here. this problem can be solved analytically like the transient solutions we have done before. however, derivations would be longer. if the analytical solutions get too complex, we can use numerical techniques. you will learn how to do it in this exercise. do not get panic. follow the steps. you can use an excel sheet, a computer program written in any language you prefer (c, fortran, basic, etc), or mathematical tool like matlab, mathematica, etc. a) find total heat until equilibrium reached by adding ????′s in all iterations above and compare it to the one you found in q2. b) make a plot of temperature of liquids versus time. please note that if you set δ???? , δ?? for each iteration will be different and you need to keep a track of time for each iteration until equilibrium is reached. but solution will be the same regardless of your choice. 0.5.="" if="" you="" prefer="" you="" can="" choose="" a="" value="" for="" δ????="" and="" find="" corresponding="" δ??="" for="" each="" step,="" and="" do="" the="" rest="" of="" the="" calculations="" as="" above.="" figure="" 2="" hints:="" we="" are="" looking="" for="" a="" transient="" solution="" here.="" this="" problem="" can="" be="" solved="" analytically="" like="" the="" transient="" solutions="" we="" have="" done="" before.="" however,="" derivations="" would="" be="" longer.="" if="" the="" analytical="" solutions="" get="" too="" complex,="" we="" can="" use="" numerical="" techniques.="" you="" will="" learn="" how="" to="" do="" it="" in="" this="" exercise.="" do="" not="" get="" panic.="" follow="" the="" steps.="" you="" can="" use="" an="" excel="" sheet,="" a="" computer="" program="" written="" in="" any="" language="" you="" prefer="" (c,="" fortran,="" basic,="" etc),="" or="" mathematical="" tool="" like="" matlab,="" mathematica,="" etc.="" a)="" find="" total="" heat="" until="" equilibrium="" reached="" by="" adding="" ′s="" in="" all="" iterations="" above="" and="" compare="" it="" to="" the="" one="" you="" found="" in="" q2.="" b)="" make="" a="" plot="" of="" temperature="" of="" liquids="" versus="" time.="" please="" note="" that="" if="" you="" set="" δ????="" ,="" δ??="" for="" each="" iteration="" will="" be="" different="" and="" you="" need="" to="" keep="" a="" track="" of="" time="" for="" each="" iteration="" until="" equilibrium="" is="" reached.="" but="" solution="" will="" be="" the="" same="" regardless="" of="" your="">
Answered 1 days AfterOct 14, 2022

Answer To: ENSY 5000 Fundamentals of Energy System Integration 2022 Fall, Homework Problem Set 5NOTE: To...

Sudarshan K answered on Oct 15 2022
40 Votes
Thermodynamics – Energy System Integration
Question:
B) Work done in each step: Work done is calculated by change in volume over the process.
Process 1: Heating of gas at constant volume. So no work done as volume is constant. However, one can figure out mass of gas from initial condition.
PV = m
RT => 200*0.04= m *0.287 * 400.
Therefore, m=0.07kg
Process 2: expansion of gas at isothermal condition. Since the volume of gas is increasing the work is done by gas on surrounding. However the gas is expanding at constant temperature, so change in internal energy is zero. Therefore work done by gas on surrounding = heat added to system = mRT ln(V2/V1) = 0.07*0.287*1200 *ln2 = 16.7 kJ
Process 3: Cooling at constant volume means again no work is done.
Process 4: return to initial condition via PV process.
200*0.041.3633= P4*0.081.3633
P4 = 200*(1/2)1.3633
P4 =77.74 kPa
PV=mRT
77.74*0.08 = 0.07*0.287*T
T=309.5 K.
Work done = P4V4 – P1V1 /(n-1) = (77.74*0.08 – 200*0.04)/0.3633 = -4.9 kJ.
This is the work done on the gas as gas is being compressed.
C) Heat addition to the system or removal from the system.
Process 1: Constant volume or isochoric process. No work done.
Heat added is equal to change in internal energy.
Q1 = mCv *T = 0.07*0.790 *(1200-400) = 44.24 kJ
Process 2: Isothermal process. So heat added to the system is equal to work done. Q2 = 16.7 kJ
Process 3: Constant Volume Process. No work done. Heat removed is equal to change in internal energy.
Q3 = mCv *T = 0.07*0.790 *(309.5-1200) = -49.24 kJ. This is heat removed from the system
Process 4: Polytropic process:
Q4 = U +W = 0.07*0.790*(400-309.5) +(-4.9) = 0.1 kJ. Heat added to the system.
D. Net work and Efficiency.
Net work = work done by air – work done by surrounding. = 16.7-4.9 = 11.8 kJ
E. Total Heat input = all the heat added = 44.24+16.7+0.1 = 61.04 kJ
Efficiency = Net work done /Heat added = 11.8/61.04 = 19.3 %
E. Entropy Calculations.
Process 1: S = q/T =44.24/1800 = 0.0245 kJ/K
Process 2: S = q/T = 16.7/1800 = 0.00927 kJ/K
Process 3: S = q/T = -49.24/290 =-0.17 kJ/K
Process 4: S = q/T = 0.1/1800 = 0.00005 kJ/K
Process 3 has the largest irreversibility
A. Final Temperature Calculations.
Since heat removed from first is heat added to second and n0o work is done.
Qh = Qc => mhCT = mc CT =>6*(100 –Tf) = 10*(Tf-20)
600 -6Tf = 10 Tf -200 => 800 = 16 Tf => Tf = 50 degree Celsius
B. Amount of Energy Calculations
Energy transferred = mhCT = 6*4*(373.15-323.15) = 24*50 = 1200 kJ
C. Change in Entropy of the cold and Hot Liquid
Entropy transfer from hot fluid = -1200/373.15 = -3.215 kJ/K
Entropy transfer from cold fluid = 1200/293.15 = 4.093 kJ/K
D. Entropy production of the system
Entropy Calculations = 4.093-3.215 = 0.8785 kJ/K
A) Total Heat loss analysis
Total heat loss from transient analysis is 1199.89 kJ which is approximately equal to one calculated above i.e. 1200 kJ
B) Plot in Excel Sheet
Temperture Change of Hot and Cold Liquid
Initial Temp...
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