CHE 33, October 8, 014. Cemical Engineering Termodynamics. Tutorial problem 5. In a simple compression refrigeration process, an adiabatic reversible compressor is used to compress propane, used as a refrigerant. Te compressed gas delivered by te compressor is fed to a condenser tat produces saturated liquid at 40 C. Te expansion valve produces a mixture of vapour and liquid at -30 C witin te evaporator. Te evaporator produces a saturated vapour at -30 C tat is fed to te compressor. Te refrigeration unit is designed as a capacity of tons (4000 BTU/r or 7.4kW, QL = 7.4 kw). Determine te molar flow rate of propane, te power of te compressor, te CO of te refrigerator, te duty of te condenser, te total entropy of te system, and te total entropy of te process, considering tat eat is removed from an environment at -10 C, and tat te eat of te condenser is released to te surroundings at 30 C. Find below a simplified diagram of te compression refrigeration cycle: 4 Compressor 1 Evaporator 3 Condenser Expansion valve To calculate tis cycle, assume tat wen te fluid is in vapour pase, it beaves as an ideal gas. Te following are te properties of te fluid. C3H8 MW= 44g/mol Tc = 369.8K c = 4.49 Ma ω=0.15 Cp gas = 68 J/mol-K (assume to be constant) sat, ka = exp(13.84-191.3/(t C+50.4)) ΔHvap /(R*Tc) = 7*(1-Tr) 0.354 +11*ω*(1-Tr) 0.456 To calculate your entalpies and entropies use te condition of stream (saturated liquid at 40 C) as reference, in oter words, make = 0 J/mol, s = 0 J/mol-K
Solution: Basic approac to follow: Wat data is given? Wic are te assumptions? Make drawings and process diagrams. Wat is appening? Wic are te equations required? And Units? Start solving te easiest first Items tat ave te more information available IMORTANT: Read ALL te text twice, not just look for te numbers in between te letters. - Wat data is given? Fluid: ropane. Formulas and values for properties given. Cp, Hvap, sat, MW, Tc, etc. Compressor: Adiabatic + Reversible Isentropic Condenser: Outlet is Sat.Liq @ 40C Evaporator: Outlet is Sat.Vap @ -30C Valve: Expands fluid instantly. Tis case is from Sat.Liq to Liq.Vap.Mix. System designed for: QL= tonref = 7.4kW Surroundings: Hig: TH= 30C ; Low: TL=-10C - Wic are te assumptions? No pressure drops in Evaporator nor Condenser, terefore: = 1 ; 4 = 3. But 1 > 4. Valve: is Adiabatic and no Work, ten: 3 = (proved troug te energy balance) Vapour ase ONLY: Assume Ideal Gas. Valid for Sat.Vap and Super Heated Vapour ONLY. Reference State: Stream (Sat.Liq @ 40C). Take = 0 and s = 0.
- Drawings and Diagrams: TL Evaporator QL Compressor 4 1 3 Expansion valve WC QH TH Condenser 1 4 QH 40 C 1 3-30 C 4 QL WC - Basic Equations required: - Energy Balance (See Inside of Front Cover of your Book) Write te equation for eac equipment - Entropy Cange for System and Surroundings - Isentropic Relations for Ideal Gases - Units: Remember work in Kelvins and te conversions of bar to ka, m 3 to liters, etc. Start Solving: - Easiest calculations first: We can identify te two levels of pressure of te refrigeration cycle by calculating te vapour pressures at 40 C (te condenser produces saturated liquid), and at -30 C (te evaporator produces saturated vapour). sat, @40 C = 1== exp(13.84-191.3/(0+50.4)) = 1366.9 ka. sat, @-30 C = 3=4= exp(13.84-191.3/(-30+50.4)) = 166.7 ka. Now we start writing te Energy Balance equations for eac equipment and see wat we know for eac one (Careful wit sign conventions) - Compressor: WC = -n*(1 4) Units: W [J/s] WC (+) since is consumed ; n [mol/s] ; [J/mol] Now we need n, 1 and 4 (wait, still oter equations to write ) From assumptions: Compressor is Isentropic and Streams 1 and 4 are vapours, terefore, we can take tem as Ideal Gases.
Since te compression process is adiabatic reversible, ten te process is isentropic. ΔS41 = Cp*dT/T R*ln(1/4) = 0 T4 =-30 C = 43K. Use te equation above to find T1. ΔS41 =68*ln(T1/43) -8.314*ln(1366.9/166.7) = 0 Ten T1 = 43*exp(8.314*ln(1366.9/166.7)/68) K = 314.3 K (~ 41. C) 0 ln Tis is te derivation of one of te isentropic relationsips for Ideal Gases To find 1, we need to add te terms necessary from te reference point (), see diagram: 1 1 40 C 41. C To find 1, ten 1= +ΔHvap@40 C (to saturated vapour@40 C) + Cpgas*(41.-40)(to supereated vapour@ 41. C) Hvap Cp* T From given Correlation for HVap: ΔHvap@40 C =8.314*369.8*(7*(1-313/369.8)^0.354+11*0.15*(1-313/369.8)^0.456) J/mol ΔHvap@40 C =1376 J/mol Finally 1 = 0+1376+68*(41.4-40) J/mol = 13371 J/mol To calculate 4, we follow a similar approac, we can calculate te entalpy of te saturated vapour at 40 C, and ten reduce te pressure to 166.7 ka and 40 C, ten reduce te temperature to -30 C. Note: for an Ideal Gas, te entalpy does not depend on pressure. Remember, =u+v => = u+rt, were u is only a function of temperature. Finally: 1 4 40 C 1-30 C Hvap 4 Cp* T ressure Effect 4= + ΔHvap@40 C (to saturated vapour@40 C) + 0 (effect of cange of pressure to 166.7ka)+ Cp gas*(-30 C-40 C) 4= 0+ 1376+0+68*(-30-40) J/mol = 8516 J/mol We still need n and te oter entalpies, so we continue to write te equations for te following equipments. - Condenser: QH = n*(-1) Units: Q [J/s] QH (-) since is rejected to surroundings
We ave already 1= 13371 J/mol and =0 (Reference). Still need n for te refrigerant - Valve: Because te expansion valve is isentalpic, 3==0 J/mol. - Evaporator: QL = n*(4-3) Units: Q [J/s] QL (+) since is taken from surroundings At tis point, we know QL (given), 3 and 4 (calculated above). Applying te first law for open systems around te evaporator, neglecting canges in kinetic and potential energy, and at steady state: H4- H3 = QL since no work in te evaporator. nrefrigerant*(4-3) = QL = 7.4 kw. Ten, nrefrigerant = 7.4*10^3 J/s/(8516-0)J/mol = 0.869 mol/s Now, since we got te value for n and we already ave all te entalpies, we can go back to energy balances for te oter equipments and solve for te missing variables, WC, QH. Compressor: Applying te first law for open systems around te compressor, neglecting canges in kinetic and potential energy, and at steady state: H1- H4 = W since te compressor is adiabatic. nrefrigerant*(1-4) = WC = 0.869 mol/s *(13371-8516)J/mol = 419 W = 4.kW (Consumed) Condenser: Applying te first law for open systems around te condenser, neglecting canges in kinetic and potential energy, and at steady state: H- H1 = QH since no work in te condenser. mrefrigerant*(-1) = QH =0.869 mol/s *(0-13371)J/mol = -11619W = -11.6kW We can calculate te CO of te process = QL/ W = 7.44kW/4.kW = 1.76 Compare tis number to te CO for te Carnot cycle in between te same TH and TL. 6.575 See tat CO < COCarnot.
Entropy of te system: For simplicity, we calculate only te entropy canges for eac process, and for te entire system we get: ΔSsystem= ΔS41 + ΔS1 + ΔS3 + ΔS34 were: ΔSi-o= nrefrigerant*(si-so) NOTE: Te use of SGen term explained in Aurelio s tutorial will NOT be evaluated ere for simplicity. But first, we need to calculate all te entropies of eac stream. Reference => s = 0 J/mol-K saturated liquid at 40 C ( =1366.9 ka) Remember te definition of Entropy: % &' ()* For determining s1 we follow a similar approac as we did for 1. We can approximate te entropy cange due to te pase cange to Hvap/Tsat, assuming tat te pat is reversible and pressure is constant. 1 1 Te contribution of te vapour eating section can be calculated troug te expression for Ideal Gas: Hvap/T Temp+ress cange Ten: s1 = s + ΔHvap@40 C/313K (to Sat.Vap @ 40 C) + Cp*ln(314.3/313) (to Sup.Heated.Vap @ 1366ka & 314.3K) Finally, s1 = 0 + 1376/313+68*ln(314.3/313) J/mol-K = 4.7 J/mol-K Since te compressor is isentropic, ten s1=s4 = 4.7 J/mol-K We can calculate te entropy s3 starting from s4 (saturated vapour at -30 C) and considering 4 and 3 (evaporation at -30 C), we can approximate s3 as in te diagram 1 40 C 1 Hvap@40C/T s3 = s4 +(3-4)/(73-30)K = 4.7 J/mol-K + (0-8516)J/mol/43K = 7.65 J/mol-K 4 3-30 C 4 (3-4)/T4
Finally, it is possible to calculate te entropy cange for eac equipment: Entropy of te compressor ΔS41= 0 W/K (te compressor is adiabatic+reversible = isentropic) Entropy of te condenser ΔS1= nrefrigerant*(s-s1) = 0.869 mol/s*(0-4.7)j/mol-k = -37.1 W/K Entropy of te expansion valve ΔS3= nrefrigerant*(s3-s) = 0.869 mol/s*(7.65-0)j/mol-k = 6.65 W/K Entropy of te evaporator ΔS34= nrefrigerant*(s4-s3) = 0.869 mol/s*(4.7-7.65)j/mol-k = 30.45 W/K ΔSsystem= ΔS41 + ΔS1 + ΔS3 + ΔS34 = 0-37.1+6.65+30.45 W/K = 0 W/K Tis makes sense because being a cycle, ten te cange in entropy around te cycle sould be zero (remember, entropy is a state function and if you return to te point of origin of te cycle, ten te cange of entropy sould be zero). - Entropy of te surroundings = + QH /Toutside air - QL /Trefrigerated In tis case te term + QH is because tis eat is absorbed by te outside air, and - QL is eat released from te refrigerated environment. ΔSsurroundings = +11.6kW/(30+73)K - 7.4 kw/(-10+73)k =0.010 kw/k = +10 W/K Ten, ΔStotal = ΔSsystem + ΔSsurroundings = 0 + 10 W/K = +10 W/K Because te total entropy of te system is positive, ten te process complies wit te second law of termodynamic ( ΔStotal 0), and is feasible.