Tarik Al-Shemmeri. Engineering Thermodynamics (776123), страница 8

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The condenser is cooled bywater which enters at a temperature of 5 C and leaves at a temperature of 25 C. Calculate themass flow rate of cooling water required if all changes in kinetic and potential energy may beneglected. Assume Cp = 4.2 kJ/kgK, and the enthalpy at 100 kPa & 30oC = 125 kJ/kg.ptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)Download free ebooks at bookboon.com783. Laws of ThermodynamicsEngineering Thermodynamics10099.630.001041.694417.462,6761.30267.3594Solution:At 100 kPah1 = 340.49 + 0.8x ( 2676 − 417.46) = 2224.3 kJ / kgat 100 kPa & 30 oCh 2 = 125 kJ/kgPlease click the advertdongenergy.com/jobdDear highly educated engineeringand finance students,if you are driven, ambitious, open-mindedand focused - we have a challenge for you.Actually, the greatest challenge in theworld.

Curious? Visit dongenergy.com/jobBest wishesDONG EnergyDownload free ebooks at bookboon.com793. Laws of ThermodynamicsEngineering ThermodynamicsEnergy Balancem s (h1 − h2 ) = m w C pw (t w 2 − t w1 )3(2224.3 − 125) = mw x 4.2(25 − 5)∴ mw = 75 kg / sNote that this represents a ratio of nearly 25 kg of water to every kg of steamWorked Example 3.18A reciprocating compressor delivers 0.1 kg/s of air at a pressure of 12 bar. The air enters thecompressor at a pressure of 1 bar and a temperature of 15  C. Calculate the delivery temperature ofthe air, the work transfer rate and the heat transfer rate in the compression process for:reversible polytropic compression, PV 1.2 = constant;reversible adiabatic compression;reversible isothermal compression.i.ii.iii.Air - R = 0.287 kJ/kgK, C p = 1.005 kJ/kg K 2 C v = 0.718 kJ/kgK and n = 1.4.Solution:i.P T2 = T1  2  P1 W =n −ln0.2 12  1.2= (288)  = 435.7 K1mR(T1 − T2 ) 0.1x 287(288 − 435.7 )== −21.2kWn −10.2∆U = mC v (T2 − T1 ) = 0.1x0.718(435.7 − 288) = 10.6kWQ − W = ∆U ∴ Q = ∆U + W = 10.6 − 21.2 = −10.6kW0.4ii. 12  1.4T2 = 288  = 585.8 K1Download free ebooks at bookboon.com803.

Laws of ThermodynamicsEngineering ThermodynamicsW =0.1x0.287(288 − 585.8)= −21.36kW0.4∆U = 0.1x0.718(585.8 − 288) = 21.36kW∴ Q = ∆U + W = 21.36 − 21.36 = 0iii.W = mRT1 lnP11= 0.1x0.287 x 288 ln = −20.5kWP212∆U = mC v ∆T = 0∴ Q = W = −20.5kWie lossWorked Example 3.19A reciprocating internal combustion engine has a clearance volume of 0.0001m3 and a compressionratio (volume ratio) of 10. The pressure and temperature of the combustion gases when the pistonis at top dead centre are 4000 kN/m2 and 1800ºC respectively.Assuming that the expansion process follows PV1.3 = constant, calculate:a) the work transfer in this process, andb) the temperature of the gases at the end of the process.Solution:a) since V1 = 0.0001m3x1PV2= 10V1PV1.3 = c∴V2 = 0.001m 3VP2 = P1  1 V2x2nVDownload free ebooks at bookboon.com813.

Laws of ThermodynamicsEngineering Thermodynamics1= 4000  10 W =1.3= 200kN / m 2P1V1 − P2V2 4000 x10 3 x 0.0001 − 200 x10 3 x 0.001= 66.7 J=n −11.3 − 1Vb) T2 = T1  1 V2n −11= (1800 + 273)  10 0.3= 1039 KPlease click the advertdongenergy.com/jobMay we offer you oneof the world’s greatestchallenges?In all humility.We are looking for highlyeducated engineers andfinance students.Join us at dongenergy.com/jobDownload free ebooks at bookboon.com823. Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.20A reciprocating steam motor is supplied with dry saturated steam at a pressure of 1.6MPa ( specificvolume = 0.1238 m3/kg). The stroke of the motor is 0.8m and the bore is 0.3m.

The clearancevolume is negligible. The steam enters the cylinder, expands at constant pressure for ¼ of thestroke and then expands reversibly according to a law PV = constant, til the end of the stroke.Calculatea) the mass of the steam,b) the work transfer and the heat transfer in the process?Solution:Pa) V 2 =π4x0.3 2 x0.8 = 0.0565m 3 ;0VV1 = 2 = 0.0141m 34VP2 = P1  1 V2121 = 16 x = 0.4MPa4∴ m = V1 / v1 =V0.0141= 0.114kg0.1238b) The work Transfer isW = P1V11nV2= 1.6 x10 6 x0.01411n4 = 31.3kJV1This system can be considered as a Closed system for which the NFEE appliesQ − W = ∆Ubut∆U = 0Q = W = 31.3kJsince isothermal process, hencei.e.

gainDownload free ebooks at bookboon.com833. Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.21A piston and cylinder mechanism has its piston fixed so that the volume contained is 0.0025m3.The mechanism is filled with wet steam at a pressure of 2 bar. The steam is heated until it reachesthe critical point. The piston is released and the steam expands adiabatically to a pressure of 2 barand a volume of 0.5m3. Calculate:a) the mass of steam in the mechanism,b) the dryness fraction of the steam after expansion.ptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)200120.230.001060.8857504.72,7071.53017.127122120374.150.003170.003172,0842,0844.434.43Solution:a)P1 = 2bar , V1 = 0.0025m 3P2 = 221bar , V2 = 0.0025m 3x2PPV1.25 = c1x-----------------------x 3v 2 = vc = 0.00317 m 3 / kgas it is in a critical stateVP3 = 2bar , V3 = 0.5m 3m=V20.0025== 0.788kgv 2 0.00317b) v 3 =V30.5== 0.634m 3 / kgm 0.788v = vf + X.(vg - vf)0.634 = 1.0605 x10 −3 + x 3 (0.8857 − 0.00106)∴ x 3 = 0.715Download free ebooks at bookboon.com843.

Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.22A mass of gas occupying 0.08m3 at 6kN/m2 and 80°C is expanded reversibly in a non-flow processaccording to a law PV1.2 = constant. The pressure at the end of expansion is 0.7kN/m2. The gas isthen heated at constant pressure to the original temperature. The specific heat capacities atconstant pressure and constant volume are 1.00 and 0.74 kJ/kg K respectively. Determine:a)b)c)d)the work transfer in the expansion process;the heat transfer in the expansion process;the volume at the end of the heating process;the change in internal energy during the heating process.Solutiona)PR = C p − C v = 1.00 − 0.74 = 0.26kJ / kgK∴m =P1V26 x10 3 x0.08== 5.23 x10 −3 kgRT1 260 x(80 + 273)x1PV1.2 = c2_______3VPlease click the advertdongenergy.com/jobAmbitiousand curious?We are looking for young andhighly educated engineersand finance students.Join us at dongenergy.com/jobDownload free ebooks at bookboon.com853.

Laws of ThermodynamicsEngineering ThermodynamicsP V2 = V1  1  P2 1/ n 6 = 0.08 0.7 1 / 1.2= 0.479m333 P V − P2V2   6 x10 x0.08 − 0.7 x10 x0.479  = +0.723kJ=W = 1 11.2 − 1n − 1  P b) T2 = T1  2  P1 n −1n0.2 0.7  1.2= (80 + 273) = 246.7 K 6 ∆U = mC v (T2 − T1 ) = 5.23 x10 −3 x740(−26 x80) = −0.411kJ∆U = Q − W∴ Q = ∆U + W = −0.411 + 0.723 = +0.312kJc) V3 = V2T3T2∴V3 = 0.479 xi.e. heat gain353= 0.685m 3246.7d) ∆U 23 = mC v (T3 − T2 )= 5.23 x10 −3 x0.740(246.7 − 353) = −0.411kJWorked Example 3.23An ideal centrifugal air compressor takes in air at 1 bar, 15ºC and compresses it reversibly andadiabatically to a pressure of 4 bar.a) (Calculate the delivery temperature of the gas.b) (If kinetic energy and potential energy changes are negligible calculate the specific worktransfer in the compression process.Air may be assumed to be a perfect gas with specific heat capacity at constantpressure Cp = 1.005kJ/kg K and n = 1.4.Download free ebooks at bookboon.com863.

Laws of ThermodynamicsEngineering ThermodynamicsSolution:P a) T2 = T1  2  P1 n −1n0.4 4  1.4= (15 + 273)  = 428 K1b) For the present system, the SFEE appliesV 2 2 − V 21+ g ( Z 2 − Z 1 )Q − W = m C p (T2 − T1 ) +2Q = 0 adiabatic∆PE = 0 neglected (assumed)∆KE = 0 neglected (assumed)hence:W = mC p (T1 − T2 )= 1 x 1.005 (288 - 428) = -140.7 kJ/kgWorked Example 3.24A reversible adiabatic air turbine drives a small generator which requires a power of 2kW. The airsupply for the turbine is provided by a reservoir and the pressure and temperature at turbine entrymay be considered constant at 9 bar, 20ºC respectively.

The velocity of the air at inlet to theturbine is small and may be neglected but at exit the velocity is 55m/s. The exit pressure is 1.2 bar.Calculate:a) the air temperature at exit from the turbine, andb) the mass flow rate of air stating any assumptions made.Air may be considered a perfect as for which the specific heat capacity at constant pressure Cp =1.005 kJ/kg K and n = 1.4.Download free ebooks at bookboon.com873. Laws of ThermodynamicsEngineering ThermodynamicsSolution:a) The situation is an open system for which the SFEE applies:V 2 2 − V 21+ g ( Z 2 − Z 1 )Q − W = m C p (T2 − T1 ) +2Q = 0 adiabaticg ( Z 2 − Z 1 ) = 0 (assumed)P T2 = T1  2  P1 n −1n0.4 1.2  1.4= 293  = 164.76 K 9 Please click the advertdongenergy.com/jobEverybodyis talking......about future energy supply.

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