Tarik Al-Shemmeri. Engineering Thermodynamics (776123), страница 7

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Laws of ThermodynamicsEngineering ThermodynamicsSolution:SFEE neglecting ∆KE and ∆PE reduces to:Q - W = ∆H + ∆KE + ∆PENeglecting changes in velocity and elevation between inlet and exit to the compressor, change inenthalpy is equal to∆H = Q - WHeat loss to cooling water isQ = - mw Cpw (Tw2 - Tw1)= - (6/60) x 4.186 x ( 20 – 10 )= - 4.186 kWActual work taking efficiency into account isW = - 10 / 0.80 = -12.5 kW∆H = ma Cpair(Ta2 – Ta1)Hence back to the SFEE :∆H = Q - W-4.186 + 12.5 = 0.5 x 1.005 ( Ta2 -27)solving to findTa2 = 43.5oC.Download free ebooks at bookboon.com663.

Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.8Air at 27oC receives heat at constant volume until its temperature reaches 927oC. Determine theheat added per kilogram? Assume for air CV = 0.718 kJ/kgK.Solution:Closed system for which the first law of Thermodynamics applies,Q − W = ∆UPlease 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.com673. Laws of ThermodynamicsEngineering ThermodynamicsW=0No work transfer at constant volume process.∴Q = ∆U∆U = Q = m.C v .(T2 − T1 )= 1 x 0.718 (927 - 27)= 646.2 kJ/kghenceQ = ∆U = 646.2 kJ/kg.Note that the Cv used is an average value.Worked Example 3.9An insulated, constant-volume system containing 1.36 kg of air receives 53 kJ of paddle work. Theinitial temperature is 27oC. Determinea) the change of internal energy.b) the final temperature.Assume a mean value Cv = 0.718 kJ/kgK.Solution:a) Q - W = ∆UQ = 0 ( insulated system )W = -53 kJ ( externally inputted work )The change in internal energy ∆U is∆U = -W = +53 kJSince Q = 0b) ∆U = m Cv ∆TDownload free ebooks at bookboon.com683. Laws of ThermodynamicsEngineering Thermodynamics∴ 53 = 1.36 x 0.718 (T2 - 27)T2 = 27 +53= 81.3O C1.36 x 0.718Worked Example 3.10An ideal gas occupies a volume of 0.5 m3 at a temperature of 340 K and a given pressure.

The gasundergoes a constant pressure process until the temperature decreases to 290 K. Determinea) the final volume,b) the work if the pressure is 120 kPaSolution :a) Since P = constantV1 V2=T1 T2V2 = V1 xT2290= 0.5 x= 0.426 m 3T1340∫b) W = P.dVfor a constant pressure process,Download free ebooks at bookboon.com693. Laws of ThermodynamicsEngineering ThermodynamicsW= p (V2 - V1)= 120 (0.426 - 0.5)= - 8.88 kJthe negative sign indicates that work is imported from an external source on the systemWorked Example 3.1130 kg/s steam at 3 MPa, 300oC expands isentropically in a turbine to a pressure of 100 kPa. If theheat transfer from the casing to surrounding air represents 1 per cent of the overall change ofenthalpy of the steam, calculate the power output of the turbine.

Assume exit is 2 m above entryand that initial velocity of steam is 10 m/s whereas exit velocity is 1 m/s.p = 3.00 MPa (233.90 C)T300v0.08114uh2750.12993.5s6.5390Please 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.com703. Laws of ThermodynamicsEngineering Thermodynamicsptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)10099.630.001041.694417.462,6761.30267.3594Solution:At 3 MPa, 300 oCh1 = 2993.5 kJ/kgs1 = 6.539 kJ/kgKat 100 kPa,sf = 1.3026, sg = 7.3594 kJ/kgKhf = 417.46, hg = 2676 kJ/kgx2 =6.539 − 1.3026= 0.8647.3594 − 1.3026h2 = 417.46 + 0.864 x(2676-417.46) = 2370 kJ/kgThe Steady Flow Energy Equation applies to this ssituation:Q − W = m[(h 2 −h1 ) +with Q = - 0.01(h2 - h1)V22 − V12+ g ( z 2 − z1 )]2Heat loss ( negative sign )V22 − V12+ g ( z 2 − z1 )]W = −m[1.01x (h 2 −h1 ) +212 − 10 2= −30[( 2993.5 − 2370) ++ 9.81x 2 / 1000] = +19 kW2 x1000+ indicate useful power outputDownload free ebooks at bookboon.com713.

Laws of ThermodynamicsEngineering ThermodynamicsWorked Example 3.12A piston and cylinder mechanism contains 2 kg of a perfect gas. The gas expands reversibly andisothermally from a pressure of 10 bar and a temperature of 327  C to a pressure of 1.8 bar.Calculate:a) the work transfer,b) the heat transfer; andc) the specific change in enthalpy of the gas.Take R=0.3 kJ/kg K and n=1.4Solution:a) The work done is given byW = ∫ PdVPV = c or P = c/VPressureVcW = ∫ dV = ln 2vV1P1V1 = P2V2 ∴V2 P1=V2 P2∴ W = mRT1 lnvolumeP1P2= 2 x 0.3x 600 x ln(10) = 617 kJ1.8b) This is a closed system, hence Non-flow Energy Equation appliesQ − W = ∆U∆U = Cv .∆TDownload free ebooks at bookboon.com723.

Laws of ThermodynamicsEngineering Thermodynamicssince for isothermal process isotherma, ie temperature difference is zero, then the internalenergy = 0, and hence Q = W = 617 kJc)∆h = C p .∆T = 0 since for isothermal process .Worked Example 3.13The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 50bar and an initial temperature of 1623  C. The initial volume is 50000 mm 3 and the gas expandsthrough a volume ratio of 20 according to the law pV 1.25 = constant. Calculatea) the work transfer andb) heat transfer in the expansion process.Take R = 270 J/Kg K and C v = 800 J/Kg K.Please 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.com733.

Laws of ThermodynamicsEngineering ThermodynamicsSolution:x1VP2 = P1  1 V2P T2 = T  2  P1 W ==1.25n −ln 1 = 50  20 1.25= 1.182 barPV1.25 = cP 1.182 = (1623 + 273) 50 0.251.25x2= 896.6 KVP1V1 − P2V2n−l50 x10 5 x 50,000 x10 −9 − 1.182 x 50,000 x10 −9 x 20= 527 J1.25 − 1.0m=P1V1 50 x10 5 x5000 x10 −9== 4.88 x10 − 4 kgRT1270 x(1623 + 273)∆U = mC v ∆T = 4.88 x10 −4 x800(1623 + 273 − 896.6 ) = −1390 JQ − W = ∆U∴ Q = ∆U + W = −390 + 527 = 136 JWorked Example 3.14A reciprocating steam engine cylinder contains 2kg of steam at a pressure of 30 bar and atemperature of 300  C. The steam expands reversibly to a final pressure of 2 bar, according to thelaw pv 1.2 = c Calculatea) the final state of the stream,b) the work transfer andc) the heat transfer in the process.p = 3.00 MPa (233.90 C)T300Vu0.08114h2750.1s2993.56.5390Download free ebooks at bookboon.com743.

Laws of ThermodynamicsEngineering Thermodynamicsptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)200120.230.001060.8857504.72,7071.53017.1271Solution:a) V 1 =m.v1 =2 x 0.08114 = 0.16228m 311 P  1.2 30  1.2V2 = V1  1  = 0.16228  = 1.55m 3 2 P2 v2 =x1V2 155== 0.775m 3 / kgm2v 2 = v f 2 + x 2 v fg 2PPV1.2 = cx2∴ x2 =0.775 − 0.0010605= 0.8750.8857 − 0.0010605P V − P2V2 30 x10 5 x0.16228 − 2 x10 5 x1.55b) W = 1 1== 884kJn−l1.2 − 1Vc) u 2 = 504.49 + 0.875 x 2025 = 2276.4kJ / kg∆u = 2276.4 − 2750.1 = −473.7 kJ / kg = −947.4kJ ,∴ Q = ∆U + W = −947.4 + 884 = −63.4kJ / kgWorked Example 3.15Steam at a pressure of 6 MPa and a temperature of 500  C enters an adiabatic turbine with avelocity of 20 m/s and expands to a pressure of 50 kPa, and a dryness fraction of 0.98. The steamleaves with a velocity of 200 m/s. The turbine is required to develop 1MW.

Determine:Download free ebooks at bookboon.com753. Laws of ThermodynamicsEngineering Thermodynamicsa) the mass flow rate of steam required, when KE is neglected, andb) What is the effect of KE on the answer?p = 6.0 MPa (257.64 deg-C)Tv0.05665500U3082.2h3422.2s6.8803ptsvfvghfhgsfsg(kPa)(oC)(m3/kg)(m3/kg)(kJ/kg)(kJ/kg)(kJ/kg.K)(kJ/kg.K)10099.630.001041.694417.462,6761.30267.3594Solution:a)h1 = 3422.2kJ / kgh2 = 340.49 + 0.98x ( 2676 − 417.46) = 2599.8kJ / kgPlease click the advertdongenergy.com/jobEverybodyis talking......about future energy supply.

We are not.Stop talking and make a careermoving energy forward.Ambitious engineers andfinance students go todongenergy.com/jobDownload free ebooks at bookboon.com763. Laws of ThermodynamicsEngineering ThermodynamicsThis situation is governed by the Steady Flow Energy EquationV 2 − V12+ g (Z 2 = Z 1 )Q − W = m (h2 − h1 ) + 22Neglecting the changes in Kinetic and Potential energies, Q = 0 for an adiabatic processApproximately W = (h1 − h2 )(b)1x10 3 = m(342.2 − 2599.8)hence m = 1.216 kg/staking KE into consideration:V 2 − V12 W = m (h2 − h1 ) + 22 200 2 − 20 2 1x10 3 = m 3422.2 − 2599.8 +2m = 1.187 kg/s or an error of 2.4%Worked Example 3.16Air, which may be considered a perfect gas, enters an adiabatic nozzle with negligible velocity.The entry pressure is 6 bar and the exit pressure is 1 bar; the entry temperature is 760 K.

The flowthroughout the nozzle is reversible and the mass flow rate is 2 kg/s. Calculate the exit velocity.Take Cp = 1004.5 J/kg K and n = 1.4Download free ebooks at bookboon.com773. Laws of ThermodynamicsEngineering ThermodynamicsSolution:This situation is an open system for which the SFEE appliesV 2 − V12+ g (Z 2 − Z 1 )Q − W = m (h2 − h1 ) + 22Q = 0 adiabaticW = 0 no work transfer in the systemFor a perfect gash2 − h1 = C p (T2 − T1 )g (Z 2 − Z1 ) = 0 negligableP T2 = T1  2  P2 n −1n0.4 1  1.4= 760  = 455.5 K6The SFEE reduces to (dividing by the mass)∴ 0 = C p (T2 − T1 ) +V22 − V1220 = 1004.5(455.5 − 760 ) +V222∴V2 = 782m / sWorked Example 3.173 kg/s of steam enters an adiabatic condenser at a pressure of 100 kPa with dryness fraction 0.80,and the condensate leaves the condenser at a temperature of 30 C.

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