Thermodynamics worked examples



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An Introduction to Mechanical Engineering Part hermodynamics worked examles. What is the absolute ressure, in SI units, of a fluid at a gauge ressure of. bar if atmosheric ressure is.0 bar? Absolute ressure g + a.0 +.0. bar kpa. Conert - o C to a temerature in degrees Kelin. [K] [ o C] + 7. -+7. 48.K. Calculate the following: (i) the kinetic energy of a body which has a mass of kg and a elocity of 0m/s; (ii) the change in otential energy of a mass of kg when it is raised a height of m; (iii) the strain energy stored in a sring comressed by 8mm from its free length if the sring constant is.0 MN/m; (i) the increase in internal energy of a gas in a closed system during a rocess in which -00 J of heat transfer and 400 J of work transfer take lace. (i) Kinetic energy mc 00 (ii) Change in otential energy mg z 0J 9.8 47J (iii) Strain energy stored kx 6. 0 08 4J (i) Increase in internal energy U U First Law: W+Q U U 400-00 00J Hodder Education 009

An Introduction to Mechanical Engineering Part 4. he iston in Figure 4. has no mass and is free to moe. he oint force F acting on the iston is alied gently so that the iston moes slowly to the left. As the gas in the cylinder is comressed, the roduct of ressure and system olume, V, remains constant. he iston area is 0 m and when x is x the system olume is 006m. Atmosheric ressure is bar. Calculate the distance (x -x ) moed by the iston as F increases from 0 to 70N. Equilibrium of forces on iston: A 70 + aa 70 + 0 0.8 x 0 Pa Gien V constant V V V 0 006. 8 0 0 004m. 8 0 V. x x Distance moed by iston: l x x l l l V V V A 66 084m or 8.4mm. Calculate the work inut to the closed system undergoing the cycle shown in Figure 4. if the ressure during rocess () to () is raised from 000 kpa to 400 kpa. 400 kpa P 00 kpa 4 m.0m Hodder Education 009

An Introduction to Mechanical Engineering Part W net W + W + W 4 + W 4 0 + 4 x 0 (.0) + 0 + x 0 (.0 ) -7 x 0 + x 0-600kJ 6. (i) (ii) Calculate the efficiency of a reersible heat engine oerating between a hot reseroir at 900K and a cold reseroir at 00K. he temerature of one of the heat reseroirs can be changed by 00 degrees kelin u or down. What is the highest efficiency that can be achieed by making this temerature change? 00 (i) η 44. 44% 900 (ii) he largest reduction in lower temerature. 400 η.% 900 is roduced by subtracting the temerature change from the 7. A erfect gas at a ressure of 8 bar and a temerature of 40K has a density of 0 kg/m. he ratio of secific heats γ is.48. (i) Calculate the alues of molar mass m ~, secific heat at constant ressure c and secific heat at constant olume c. (ii) Calculate the change in secific entroy of the gas if the ressure is raised to 00 bar and the temerature is lowered to 400K. (i) Perfect gas, R ρ 8 0 R 7. 77J / kgk ρ 0 40 R ~ 8. 4 0 m ~. kg / kmol R 7. 77 γ R. 48 c 7. 77 794. 79J / kgk γ 48 Hodder Education 009

An Introduction to Mechanical Engineering Part R 7. 77 c 7. 0J / kgk γ 48 (ii) Secific entroy change s s cln Rln 400 00 794. 8l n 7. 8l n 40 8 4.0J/kgK 8. Calculate the dryness fraction x of saturated steam if it has a secific enthaly of 800 kj/kg and at the same ressure and temerature, the secific enthaly of saturated liquid is 0 kj/kg and the latent heat of aorization is 0 kj/kg. h h f + xh fg h hf 800 0 x 4 h 0 fg 9. A erfect gas with a γ alue of.4 undergoes an exansion rocess from a ressure of 600 kpa. he ratio of secific olumes / is. Calculate the ressure at the end of the rocess if this is (i) olytroic with an index n of.6, (ii) isothermal, (iii) isentroic. (i) n ( ). 6 0 4kPa 600 0. (ii) R constant 600 0 00kPa (iii) γ ( ). 4 8 9kPa 600 0 x. Hodder Education 009

An Introduction to Mechanical Engineering Part A erfect gas with a γ alue of. can be comressed in either a non-flow or a steady flow rocess between the same initial and final states. Neglecting any change in kinetic and otential energy, what is the ratio of the secific work done is the two cases if the rocess is (i) olytroic with an index n of.4; (ii) (iii) isothermal; isentroic? Refer to results gien in able 4.8 w Ratio w (i).4 (ii) (iii). flow non flow is:. Calculate the efficiency of the following ideal cycles when undergone by a erfect gas with a γ alue of.4: (i) a Stirling cycle oerating between a hot reseroir at 600K and a cold reseroir at 00K; (ii) a Brayton cycle with a ressure ratio of 8; (iii) an Otto cycle with a comression ratio of 8; (i) a Diesel cycle with a comression ratio of and a cut-off ratio of. (i) η ηcarnot 0% (ii) η ( ) γ γ 4 /. 4 r 8 44.8% (iii) η ( ) γ 4 c 6.% r 8 (i) η c γ c0 γ r ( ) ( ) γ c0 Hodder Education 009

An Introduction to Mechanical Engineering Part. 4 ( ) 4 6.7% (. 4 ). A erfectly insulated, rigid tank with a olume of m contains a erfect gas which has a molar mass of 8 kg/mol and a ration of secific heats of.4. Initially the ressure and temerature in the tank are 9 bar and 0 K resectiely. A fan inside the tank is sun at 600 re/min for 0 seconds. he torque required to turn the fan is 0 Nm. Calculate the following: (a) he R, c and c alues of the gas and the mass of gas in the tank. (b) he work inut (gien by torque x angular rotation in radians) to the gas from the fan. (c) he final temerature of the gas. (Exlain briefly why the temerature continues to rise for a short time after the fan has stoed rotating.) (d) he increase in entroy of the gas. V m γ.4 m ~ 8kg / 0K P 9bar kmol orque 0 Nm 600 N 600 re / min 60re / s 60 (a) ~ R 84. J / kmolk R 46J / kgk m ~ 8kg / kmol R C C C C C γ 4 C 06. 7J / kg / K C γc 488. 7kg / K Hodder Education 009

An Introduction to Mechanical Engineering Part P V 9 0 m. kg R 46. 0 (b) Work inut: W fan θ ( N 0 π) 0 60 0 π 6. kj (c) No heat transfer and constant olume roes so. st Law: U + U Q+ W W Wfan where Q0 and W has constant olume, 0 mc ( ) W fan W mc fan + 6. 0 + 0. 06. 7 8 6 + 0 0 6K (d) V S S mr ln mc ln V + 0 6 0 +. 06. 7 ln 0 6 J / K emerature will continue to increase after the fan stos because the KE generated by the fan takes a (short) time to dissiate from KE into internal energy.. A closed system containing argon undergoes a reersible isothermal rocess from an initial state () where 0 bar, V 0m and 40 K to state (). he work done during the rocess is -00 kj. he system is then heated reersibly at constant olume to final state (). he total heat transferred during the two rocesses is 70 kj. reat argon as a erfect gas with c 0 J/KgK and molar mass m ~ 40 kg/kmol. Calculate the following: Hodder Education 009

An Introduction to Mechanical Engineering Part (a) he mass of argon in the system. (b) he heat transferred during the constant olume rocess () (). (c) he final temerature,. (d) Sketch the rocesses on a ressure olume diagram. Argon m ~ 40kg / kmol C 0 J / KgK ~ R m ~R 8. 4 0 40 07. 9KJ / kgk (a) Mass V m R 0 0 0 07. 9 40. 60kg (b) Isothermal () () Q Q + W W U U 00kJ 0 But Q + Q 70kJ Q 70 00 70kJ (c) Constant olume () () Q + W U U where W 0 U U 70kJ mc ( ) Hodder Education 009

An Introduction to Mechanical Engineering Part and C 40K (() () Isothermal) 70 0 40. 60 C C R 0 07. 9. J / kgk 70, 000 40 +. 60. 89. 9K (d) isothermal P () () () Constant olume V 4. A closed system containing steam undergoes a reersible constant ressure rocess during which 400 kj/kg of heat transfer takes lace. Initially the steam has a dryness fraction, x, of.0 and a temerature of 7ºC. Using the tables, and using linear interolation where necessary, determine: (a) he secific enthaly, temerature and secific internal energy of the steam at the end of the rocess. (b) he secific work transfer. (c) State whether the rocess is an exansion or comression rocess and gie two reasons to suort your choice. (a) Initial conditions dry saturated (x.0) at 7 C u h g 00bar 7kJ / kg h g 0kJ / kg 007m / kg Constant ressure rocess (reersible) st Law: q+ w u u q ( h ) + u - u 400 + 0 90 KJ / kg h h emerature interolate at 80 bar and 90kJ Hodder Education 009

An Introduction to Mechanical Engineering Part u (90 888) 400 + (4 400) (004 888) 400 + 897 404. 74 C h 0. 9+ (. 8. 9) 897. 9m / kg u 09m 90 696kJ/kg / kg ( 80 0 09) 0 kj / kg (b) st Law: q+ w u u w 400 +. 6 84. 4kJ / kg. 6kJ / kg (c) Exansion. (i) w is negatie (ii) >. Air flows through an oen system at the steady mass flow rate of kg/s. At inlet the air elocity is negligible, the ressure is bar and the temerature is ºC. he air flow is comressed isentroically and leaes the system at a ressure of bar through a ie with an internal diameter of 80 mm. Calculate: (a) the temerature of the air as it leaes the system; (b) the elocity in the exit ie; (c) the ower inut to the air; (d) Exlain why the ower inut is not gien by Assume for air, R87 J/kgK and γ.4. W ~ m& d Hodder Education 009

An Introduction to Mechanical Engineering Part m& W & Q & 0 () () C o φ 80mm bar bar 7 + 88K (a) Isentroic comression P P γ γ 88. 86 46K (b) mass continuity m& ρ Ac m& c ρ A ρ R 0. 8kg / m 87 46 πd π A 08. 0 0 4 4 000 c 60m / s. 8. 0 m (c) S.F.E.E Q& + W& m(h & 60 W& (c( 46 88) + ) ( 0 68 + 800 ) x 6kW 0 kw c h+ ) (d) W & m ~ c d neglects changes in And otential energy between () and (). Would underestimate by.8 kw in this examle. Hodder Education 009

An Introduction to Mechanical Engineering Part 6. (a) Determine the rate of heat rejection from a reersible heat engine oerating between a hot reseroir at 900K and a cold reseroir at 400K if the engine roduces a ower outut of 400 kw. (b) A ie with an inside diameter of 80 mm, a wall thickness of 0 mm and a thermal conductiity of 09 W/mK is lagged with a 0 mm thick layer of lagging with a thermal conductiity of 0 W/mK. he ambient temerature is 4ºC and the inner surface of the ie wall is maintained at 80ºC. he heat transfer coefficient at the outer surface of the lagging is 8 W/mK. Calculate: (i) the rate of heat transfer er unit length of ie; (ii) the temerature at the inner surface of the lagging. (a) Reersible, and two reseroirs so: η η carnot 400 900. % 900k Q R? 400k W& Q& S Q& s 40 7kW First law: Q & R Q& S W& 7 40 kw (b) 0 k0 W/mk 0 () 80 o C? () h8 W/m k Hodder Education 009

An Introduction to Mechanical Engineering Part q& ΣR H 80 4 70 90 ln ln 40 70 000 + + π. 09 π. 0 π. 9 8 6 6 990 +. 00 + 098. 09 W / m 80 q & 70 990 ln 40 π 09 80 99 0 C Hodder Education 009