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Cengel & Boles: Ch 7:
7.14 - The Increase of Entropy Principle - 2 pts
7.15 - The Increase of Entropy Principle - 2 pts
7.18 - The Increase of Entropy Principle - 2 pts
7.28 - ΔSSys, ΔSRes, and ΔSUniv, for a H.T. Process - 4 pts
7.29 - Entropy Change in the Evaporator of a Refrigerator - 6 pts
7.66 - ΔSUniv Upon Quenching an Iron Block - 6 pts
| WB-1 - |
Thermal Efficiency of an Internally Reversible HE with Multiple Heat Transfers - 4 pts |
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Problem
Statement : |
A system executes a power cycle while receiving 750 kJ by heat transfer at its boundary where the temperature is 1500 K and discharges 100 kJ by heat transfer at another portion of its boundary where its temperature is 500 K. Another heat transfer from the system occurs at a portion of the system boundary where the temperature of the system is 1000 K. No other heat transfer crosses the boundary of the system. If no internal irreversibilities are present, determine the thermal efficiency of the power cycle. |
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Hints : |
The given temperatures are the temperatures within the system at which heat is exchanged with the surroundings. Use the 1st Law and the definition of ηth. Sgen = 0 within the system. ΔScycle = ?
Ans.: ηth ≈ 47% |
| WB-2 - |
Efficiency and Reservoir Temperatures for Reversible and Irreversible Cycles - 6 pts |
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Problem
Statement : |
Complete the following involving reversible and irreversible cycles.
a.) Reversible and irreversible power cycles each discharge QC to a cold reservoir at TCand receive energy QH from hot reservoirs at TH and T'H, respectively. There are no other heat transfers involved. Show that T'H> TH.
b.) Reversible and irreversible heat pumpcycles each receive QC from a cold reservoir at TCand discharge QH to hot reservoirs at TH and T'H, respectively. There are no other heat transfers involved. Show that T'H< TH. |
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Hints : |
Each part of this problem involves the algebraic manipulation of the defining equation for entropy generation for both reversible and irreversible cycles. |
| WB-3 - |
Specific Entropy Change Using Tabluar Data - 4 pts |
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Problem
Statement : |
Determine the change in specific entropy in kJ/kg-K for :
a.) Water: P1 = 10 MPa, T1 = 400oC and P2 = 10 MPa, T2 = 100oC. ( Use the NIST Webbook )
b.) R-134a: H1 = 211.44 kJ/kg, T1 = - 40oC and P2 = 5 bar, x2 = 1.0. (Use the NIST Webbook with the default reference state: U = 0 and S = 0 for saturated liquid at 0.01oC)
c.) Air (IG): T1 = 27oC, P1 = 2 bar and T2 = 327oC, P2 = 1 bar.
d.) Hydrogen (H2, IG): T1 = 727oC, P1 = 1 bar and T2 = 25oC, P2 = 3 bar. |
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Hints : |
a & b.) Exercises in the use of thermodynamic property tables to evaluate ΔS.
c & d.) exercises in the use of the ideal gas property tables and Gibbs' 2nd Equation to evaluate ΔS for ideal gases.
Ans.: b.) ΔS ~ 0.65 kJ/kg-K , c.)ΔS ~ 0.91 kJ/kg-K |
| WB-4 - |
"Show That" for a Cycle Interacting with Three Reservoirs - 4 pts |
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Problem
Statement : |
The system shown in the figure undergoes a cycle while receiving energy at the rate Qsurr from the surroundings at temperature Tsurr, receiving QH from a heat source at temperature TH, and rejecting QC to a thermal reservoir at TC. Derive an expression for the maximum theoretical value of QC in terms of QH and T0, TS and TU only. There is no work produced by this system. This is an absorption heat pump system. The heat source might be a propane flame and the heat sink might be the air inside your camper. So, it is important to note that TS > TU > Tsurr. Hint: TH, Tsurr and TC are not important in the solution of this problem ! |
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Hints : |
The key here is to recognize that TH, TC and Tsurr do not directly effect the solution of this problem and then determine how entropy generation affects the heat output of this heat pump, QC. Because Sgen > 0 we will be able to determine a maximum value for QC as a function of QH and the system temperatures at which heat is transferred. |
| WB-5 - |
Three-Step, Ideal Gas Cycle Analysis - 8 pts |
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Problem
Statement : |
A quantity of air undergoes a thermodynamic cycle consisting of three internally reversible processes in series. Assume that the air behaves as an ideal gas. This may not be a good assumption, but let's work with it here anyway.
Step 1-2 : Isothermal expansion at 350 K from 4.75 bar to 1.0 bar.
Step 2-3 : Adiabatic compression to 4.75 bar.
Step 3-1 : Isobaric cooling.
a.) Sketch the cycle on a PV diagram.
b.) Sketch the cycle on a TS diagram.
c.) Determine T3 in Kelvin
d.) If the cycle is a power cycle, determine its thermal efficiency. If the cycle is a refrigeration cycle, determine its COP. |
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Hints : |
The key to determining T3 is to recognize that since step 2-3 is both adiabatic and internally reversible, it is also isentropic. |
| WB-6 - |
Maximum Work output from an Adiabatic Turbine - 5 pts |
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Problem
Statement : |
Steam enters an adiabatic turbine at 800 psia and 900°F and leaves at a pressure of 40 psia. Determine the maximum amount of work that
can be delivered by this turbine. |
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Hints : |
Make the usual assumptions about the turbine. The maximum work would be delivered by a reversible turbine.
Ans.: Wsh ~ 310 Btu/lbm |
| WB-7 - |
ΔS for Heat Transfer to R-134a in a Rigid Tank - 6 pts |
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Problem
Statement : |
A 0.5 m3 rigid tank contains R-134a initially at 200 kPa and 40% quality. Heat is transferred to the refrigerant from a source at 35oC until the pressure rises to 400 kPa. Determine…
a.) The entropy change of the R-134a.
b.) The entropy change of the heat source.
c.) The entropy change of the universe for this process. |
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Hints : |
The keys to this problem are that the specific volume is the same in states 1 and 2 and that the heat source can be treated as a true thermal reservoir (its ΔS can be determined from the definition of entropy).
Ans.: a.) ΔSR-134a ~ 3.9 kJ/K , c.) ΔS ~-3.5 kJ/K |
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