0. (a) Sol: Section A A refrigerator macine uses R- as te working fluid. Te temperature of R- in te evaporator coil is 5C, and te gas leaves te compressor as dry saturated at a temperature of 40C. Te mean specific eat of liquid R- between te above temperatures is 0.963 kj/kgk. Te entalpy of evaporation at 40C is 03. kj/kg. Neglecting losses, find te COP. ( 3 ) = 03. kj/kg (s s 3 ) = 03. 33 s = s, 3 = 4 (s s 3 ) = 0.649 kj/kgk = 0.649 kj/kgk 33 (s 3 s 4f ) = 0.963 ln = 0.495 kj/kg 68 (s s 4f ) = (s s 3 ) + (s 3 s 4f ) = 0.649 + 0.495 = 0.7987 kj/kgk ( 4f ) = 68 (0.7987) = 4.05 kj/kg 40C 5C T 4f 3 4 s ( M) Net refrigerating effect, ( 4 ) = ( 4f ) ( 4 4f ) = 4.05 ( 3 4f ) ( 3 4f ) = c p (T 3 T 4f ) = 0.93 (40 ( 5)) ( 3 4f ) = 0.963 (45) = 43.335 ( 4 ) = 4.05 43.335 = 70.7 kj/kg Work of compression, ( ) = ( 3 ) ( 3 ) = 03. ( 4 ) = 03. 70.7 = 3.485 kj/kg COP 4 = net refrigerant effect kj / kg work of compression kj / kg 70.7 3.5 5.5
(b). Compare te knocking penomena in SI and CI engines. Explain clearly te factors wic tend to prevent knock in SI engines in fact promote knock in CI engines. ( M) (c). A laboratory wind tunnel as a test section tat is 305mm square. Boundary layer velocity profiles are measured at two cross-sections and displacement ticknesses are evaluated from te measured profiles. At section, were te free stream speed is u = 6m/s, te displacement tickness is.5mm. At section, te displacement tickness is.mm. Calculate te cange in static pressure between sections and as a fraction of te free stream dynamic pressure at section. ( M) Sol: * * W W - * W - * () () W - * W - * W Te displacement boundary layer can be considered as an imaginary growt in tickness of wall and te flow in core region can be treated as uniform flow. By continuity equation, A V = A V i.e, (w - *) V = V (w - *) V 6 305.5 305. V = 6. m/s Applying Bernoulli s equation between () & ()
V P P V P P V V P P V V V = 0.06 =.6 % (d). A drying oven consists of a long semicircular duct of diameter D = m as sown in Fig. below. Materials are to be dried over te base of te oven wile te wall is maintained at 00 K. Wat is te drying rate per unit lengt of te oven if a water-coated layer of material is maintained at 35 K during te drying process? Blackbody beavior may be assumed for te water surface and te oven wall. T = 00 K D= m Fig. T = 35 K ( M) Sol: Assumptions: Neglecting convection effect Surfaces are black L = m Given, D = m, T = 00 K, T = 35 K T = 00 K D= m T = 35 K
Net eat transfer rate : Q net = A F - (T 4 T 4 ) = D L 5.67 0 8 (35 4 00 4 ) = 5.67 0 8 (35 4 00 4 ) = 6940.53 W/m Net eat transfer rate : = Q net = 6.94 kw/m Net eat transfer to water = m C T 6.94 m 4.8 373 35 57 pw fg m = 0.04758 kg/sec m = 7.9 kg/r Drying rate per meter lengt = 7.9 kg/r ( fg of water at 00C and atmosperic pressure = 57 kj/kg) (e). Explain te desirable properties of refrigerants. List all te possible alternative refrigerants to CFCs and HCFCs. ( M) 0. (a). In aircraft refrigerating unit using air cycle, 50 kg/min of air at 80 cm Hg gauge and 05C are bled off te air compressor serving te jet engine of an airplane. Te bled air is passed troug a eat excanger leaving at 75 cm Hg gauge and 75C. At tis point, it is expanded troug a small cooling turbine to 0 cm Hg vacuum and 0C. Te air exausted out of te plane is at 5C. Assume C p =.0 kj/kg K. (i) Find te cooling in ton (refrigeration). (ii) If te compressor receives air at stagnation state of cm Hg gauge and 50C and if te small air-cooling turbine output serves te centrifugal fan for passing coolant air troug te eat excanger, determine te input power for te refrigerant plant. (iii) Wat is te COP based on input power to bled off air? (0 M)
Sol: Compressed air m a () Heat excanger () Fan Cooling turbine T 3 4 0 (0) Cooling air (3) T 3 To Cabin s Massflow rate of air troug te compressor Pressure at te end of compression, m a = 50 kg/min, 56 P = P atm + P g = 76 + 80 = 56 cms of Hg 0. 35 = 34.3 kpa 76 Temperature at end of compression T = 73 + 05 = 478 K Pressure at end of cooling 5 P = P atm + P g = 76 + 75 = 5 cms of Hg 0. 35 = 334.64 kpa 76 Temperature at end of cooling T = 73 + 75 = 348 K Pressure at end of expansion from turbine, 56 P 3 = P atm P vac = 76 0 = 56 cm of Hg 0. 35 = 74.66 kpa 76 Temperature at end of expansion from turbine, T 3 = 73 0 = 63 K T 4 = cabin temperature = 5C = 98 K Cooling load = m C T T p 4 3 50 98 63 = 9.66 kw 60 9.66 8.93T R 3.57
Pressure at inlet to te compressor, 78 P 0 = P atm + P g = 76 + = 78 cm of Hg 0. 35 = 03.99 kpa 76 Temperature at inlet to te compressor, T 0 = 73 + 50 = 33 K Compressor work = W C = m C T T 478 33 a Turbine work = W T = m C T T = 348 63 a p p 3 0 50 = 9.6 kw 60 50 = 70.83 kw 60 Power output of cooling turbine = power input to fan = 70.83 kw Net power input to system = W C = 9.6 kw Cooling load kw 9.66 COP 0. 58 W kw 9.6 C (b). Te pressure in an automobile tire depends on te temperature of te air in te tire. Wen te air temperature is 5C, te pressure gauge reads 0 kpa. If te volume of te tire is 0.65 m 3, determine te pressure rise in te tire wen te air temperature in te tire rises to 50C. Also determine te amount of air tat must be bled off to restore pressure to its original value at tis temperature. Assume atmosperic pressure to be 00 kpa and R = 0.87 kj/kg K. (0 M) Sol: Initial pressure of air, P = P atm + P g = 00 + 0 = 30 kpa Initial volume of air, V = 0.65 m 3 Initial temperature of air, T = 73 + 5 = 98 K, Final temperature of air on eating, T = 33 K R = 0.87 kj/kgk Mass of air, V = C P T P T P v m RT 30 0.65 =.356 kgs 0.87 98
T 33 Final pressure of air, P P 30 = 336 kpa T 98 Pressure rise in tire = P P = 336 30 = 6 Let m be te mass at T wic gives a pressure 30 kpa P v m RT 30 0.65 =.736 kg 0.87 33 Amount of mass to be removed = m m =.356.736 = 0.84 kgs (c). Te temperature distributions witin a series of one-dimensional plane walls at an initial time (t =0), at steady state (t = ) and at several intermediate times are as sown in Fig. below (Case A and Case B). For eac case, write te appropriate form of diffusion equation. Also write te equations for te initial condition and te boundary condition tat are applied at x = 0 and x = L. If te volumetric generation occurs, it is uniform trougout te wall. Te properties are constant. t= t=0 T=T T i Increasing time t=0 Increasing time t= L T=T i L x Case - A Fig. x Case - B (0 M) Sol: Case (A) : -D, eat diffusion equation T q T x k t t= T=T Increasing time Boundary condition, t=0 T is a function of distance x and time t T= T(x,t) T(x,0) = T i L x Case - A T=T i
T(L,t) = T T x 0,t 0 Case (B) : -D eat diffusion equation T q T x k t Boundary condition T is a function of distance x and time t T= T(x,t) T(x,0) = T i L x Case - B t=0 T i Increasing time t= T(L,t) = T i T x 0,t 0 Were, q = uniform volumetric termal energy generation = termal diffusivity k = termal conductivity of te wall 03. (a). Determine te cange of air-fuel ratio of an airplane engine carburetor wen it takes off from sea level to a eigt of 5000 m. Te carburetor is adjusted for 5: ratio at sea level, were te air temperature is 7C and pressure bar. Assume te variation of temperature of air wit altitude as t = t s 0.0065, were is in meter and t is in C. Te air pressure decreases wit altitude as per te relation = 900 log 0 (/p), were p is in bar. Evaluate te variation of air-fuel ratio wit respect to altitude in steps of 000 m on te trend. Sow te variation on a grap and discuss. (0 M)
Sol: Altitude = 5000 m, Air fuel ratio at sea level = (AFR) SL = 5: Temperature at sea level, T 0 = 7C = 300 K Pressure at sea level, P 0 = 00 kpa Density of air at sea level o P0 RT 0 = 900 log 0 P 0 P P = 900 900 0 t = t s 0.0065 00.64 kg / 0.87 300 3 m Parameters Formula Heigt Temperature t = t s Pressure 0.0065 (C) P = 0900 (bar) 000 m 000 m 3000 m 4000 m 5000 m =7 0.0065 = 7 0.0065 = 7 0.0065 = 7 0.0065 = 7 0.0065 000 = 0.5C 000 = 4C 3000 = 7.5C 4000 = C 5000 = 5.5C 000 0900 = 0.887 bar = 88.7 kpa 000 0900 = 0.7867 bar = 78.67 kpa 3000 0900 = 0.6978 bar = 69.78 kpa 4000 0900 = 0.688 bar = 6.88 kpa 5000 0900 = 0.549 bar = 54.9kPa Density Air Fuel Ratio at altitude P 88.7 RT 0.87 93.5 (kg/m 3 ) =.053 kg/m 3 (AFR) SL.053 5. 64 = 4.8 o 78.67 69.78 6.88 54.9 = 0.87 87 0.87 805 0.87 74 0.87 67. 5 = 0.955 kg/m 3 = 0.867 kg/m 3 = 0.7869 kg/m 3 = 0.75 kg/m 3 0.955 5.64 3.6 0.867 5.64.96 0.7869 5.64.34 5 =.77 0.75.64 Cange of air fuel ratio at 5000 m altitude = (AFR) 5000m (AFR) SL =.77 5 = 3.3
Variation of air fuel ratio wit altitude : AFR 5 4 3 0 SL 000 000 3000 4000 Altitude 5000 Discuss: As altitude increases ricness of air fuel ratio increases. Hence an altitude compensating device must be used for proper combustion of fuel. (b). Consider te laminar flow of a fluid layer falling down on a plane at an angle wit te orizontal. If is te tickness of te layer in te fully developed state, ten (i) sow tat te velocity distribution is U = g sin( y )/v, were v is te kinematic viscosity (te x-axis points along te free surface and te y-axis points towards te plane); (ii) develop te expression for volume flow rate per unit widt; (iii) develop te expression for frictional stress on te wall. (0 M) Sol: Du dt p u y g x x y x d u dy g sin y x Integrating w.r.t y gsin du dy g sin y c gcos g y u sin c y c g
du At y = 0, 0 [c =0] dy At y =, u = 0 c u Q g sin gsin y ------ () 0 udy gsin 0 y 3 g sin y y 3 g sin Q 3 3 0 w du dy y g sin u g sin = g sin (c). Derive te Euler s equation for turbo-macines and sow tat for single-stage axial impulse turbine, work done can be represented as W V V were V and V are absolute velocities at inlet and exit of rotor blades. (0 M) 04. (a). Explain te working of electrostatic precipitator and discuss variation of its collection efficiency wit operating parameters like collector area, migration velocity and mass flow rate. (0 M)
(b). Wat are te tree different types of fuel cell reactions? Give te ydrogen-oxygen, carbonoxygen and metane-oxygen fuel cell reactions. (0 M) (c). Steam enters te condenser of a steam power plant at 0 kpa and a quality of 95% wit a mass flow rate of 0000 kg/. It is circulating te water troug te tubes witin te condenser. To prevent termal pollution, te river is not allowed to experience a temperature rise above 0C. If te steam is to leave te condenser as saturated liquid at 0 kpa, determine te mass flow rate of te cooling water required. Data from steam table: At 0 kpa, f = 5.4 kj/kg; fg = 358.3 kj/kg; Specific eat of water = 4.8 kj/kg C. (0 M) Sol: x = 0.95, T = ( f ) 0 kpa = 5.4 kj/kg = f + x ( fg ) = 5.4 + 0.95 358.3 = 49.79 kj/kg Mass flow rate of steam = m s = 0000 kg/r 0 kpa Mass flow rate of cooling water = m w =? Specific eat of water = C pw = 4.8 kj/kgk Rise in temperature of water = (T) w = 0C Heat gained by water = eat lost by steam m m w w kg / sec C kj / kgkt m kg / sec ms C 0000 3600 pw T pw w 49.79 5.4 4.80 = 97.77 kg/sec w s s
Section B 05. (a). A 30 kg iron block and a 0 kg copper block bot initially at 80 o C are dropped into a large lake at 0C. Termal equilibrium is establised after a wile as a result of eat transfer between te blocks and te lake water. Determine te total entropy cange for tis process. For copper and iron, specific eats are respectively 0.386 kj/kg K and 0.46 kj/kg.k. ( M) Sol: Mass of iron block = m i = 30 kg, Specific eat of iron = C pi = 0.46 kj/kgk Mass of copper block = m c = 0 kg Specific eat of copper block = C pc = 0.386 kj/kg Initial temperature of bot copper and iron = T = 73 + 80 = 353 K Final temperature of bot copper and iron blocks = T f = 73 + 0 = 93 K Heat lost by copper and iron to lake = Q = m C m C (T T f ) i pi c pc Q = (30 0.46 + 0 0.386) (353 93) = (3.8 + 7.7) (60) = 9. kj Entropy cange of iron = Tf m i Cpi n T 93 ds i =.5708 k/k 353 30 0.46 n Entropy cange of copper = Tf m c Cpc n T 93 ds c =.438 kj/k 353 0 0.386 n Entropy cange of lake (ds) lake = Q T f 9. = 4.4068 kj/k 93 Entropy cange of process = (ds) i + (ds) c + (ds) lake = (.5708) + (.438) + 4.4068 = 0.3978 kj/k > 0
(b). Te wind speed V at a location is 4.47 m/s, te speed at turbine rotor is 60% of tis value and te speed at te exit is 30% of V. Te rotor diameter is 9 m, density =.93 kg/m 3. Calculate (i) te power available in te undisturbed wind at te turbine rotor, (ii) te power in te wind at outlet, (iii) te power developed by te turbine and (iv) te coefficient of performance. ( M) Sol: speed () () (3) (4) Pressure P atm 3 (i) Power available = AV 3 =.93 9 4.47 = 3678.3 W 4 (ii) Power developed = m V V AV (iii) Power in te wind at outlet = 4 V V4 4 0.6 4.47 4.47 0.3 4.47.93 9 = 007.3 W m V 4 AV V 4.93 9 0.6 4.47 0.3 4. 47 = 98.63 W 4 (iv) Coefficient of Performance (c p ) : Power developed c p power available 007.3 = 54.57 % 3678.3
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