Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

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0-1-1[US]Two thermodynamics books, each with a mass of 1 kg, are stacked one on topof another. Neglecting the presence of atmosphere, draw the free body diagram of thebook at the bottom to determine the vertical force on its (a) top and (b) bottom faces inkN.SOLUTION(a)From the free body diagram of the book at the top, avertical force balance produces: 121m kNkgk0.00981 kNN1000sN19.811000mgFF(b) Using the free body diagram of the book at the bottom, 212kN100019.810.0098110000.01962 kNmgFFF1Fmg2Fmg1F

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0-1-2[UA]Determine (a) the pressure felt on your palm to hold a textbook of mass 1 kgin equilibrium. Assume the distribution of pressure over the palm to be uniform and thearea of contact to be 25 cm2. (b)What-if Scenario:How would a change in atmosphericpressure affect your answer(0: No change; 1: increase;-1 decrease)?SOLUTION(a)2kN;=kPamFpA 19.81;;100010000.0025mgppA3.924 kPap(b)A change in atmospheric pressure would not affect the answer because the net forcecreated by a uniform pressure around any object is zero. Answer: 0

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0-1-3[UH]The lift-off mass of a Space Shuttle is 2 million kg. If the lift off thrust (thenet force upward) is 10% greater than the minimum amount required for a lift-off,determine the acceleration.SOLUTIONthrustgravity;1.1;0.1 ;FFmamgmgmaag0.19.81 ;a2m0.981 sa

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0-1-4[UF]A body weighs 0.05 kN on earth whereg= 9.81 m/s2. Determine its weighton (a) the moon, and (b) on mars withg= 1.67 m/s2andg= 3.92 m/s2, respectively.SOLUTION0.051000;;5.097 kg;9.81wmmmg(a)On the moon:5.0971.67;100.0008 12 k05Nww(b)On Mars:5.0973.92;0.01998N00k10ww

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0-1-5[UD]Calculate the weight of an object of mass 50 kg at the bottom and top of amountain with (a)g= 9.8 m/s2and (b)g= 9.78 m/s2respectively.SOLUTION(a)509.8;;10001000.49 k0Nmgwww(b)509.78;0.489 k1000Nww

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0-1-6[UM]According to Newton's law of gravity, the value ofgat a given location isinversely proportional to the square of the distance of the location from the center of theearth. Determine the weight of a textbook of mass 1 kg at (a) sea level and (b) in anairplane cruising at an altitude of 45,000 ft. Assume earth to be a sphere of diameter12,756 km.SOLUTION(a)At sea level002;GmMwg mR 09.811 ;w09.81 Nw(b)3kmAt45,000 ft0.3048 1013.716 kmfth2202222;;;;GmMGmMRRwwwgmwgmRRhRhRh 2222m9.81(6,378 km)ms;9.76;(6,378 km13.716 km)s9.7671 ;ggw9.76 Nw

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0-1-7[UJ]ThefrictionalforceonablockofmassmArestingonatable(seeaccompanying figure) is given asF= μN, whereNis the normal reaction force from thetable. Determine the maximum value formBthat can be supported by friction. Assumethe pulley to be frictionless.SOLUTIONThe maximum value ofBmthat can be supported by the friction:;kN1000;kN1000;AABBABm gFm gFFF;ABm gm gBAmm

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0-1-8[UW]If the block A in problem 0-1-7 [UJ] sits on a wedge with an angleθwith thehorizontal, how would the answer change?SOLUTIONThe normal force on the block is:cos;kN1000Am gNA balance between the tension in the string and friction produces:The normal force on tcos;10001000ABm gm g cos;ABm gm gcosABmm

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0-1-9[XR]A block with a mass of 10kg is at rest on a plane inclined at 25oto thehorizontal. Ifμs= 0.6, determine the range of the horizontal push forceFif the block is(a) about to slide down, and (b) about to slide up.SOLUTION(a)As the block is about to slide down, friction acts upward and a forcebalance along the slope yields:sincoscossin;1000smgFmgF(sincos)cossin;1000(sincos) ;1000 cossin109.81 (sin 250.6 cos 25 ) ;1000cos 250.6sin 25ssssmgFmgFF  0.010248 kNF The negative sign indicates that a slight pull force is necessary to overcome frictionfor the block to slide down.(b)As the block is about to slide up, friction acts downward and a force balance alongthe slope yields:cossinsincos;1000smgFmgF(cossin)(cossin)cossin;10001000 cossin109.81 (0.6cos 25sin 25 ) ;1000cos 250.6sin 25ssssmgmgFFF  0.1452 kNF

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0-1-10[XO]A vertical piston cylinder device contains a gas at an unknown pressure. Ifthe outside pressure is 100 kPa, determine (a)the pressure of the gas if the piston has anarea of 0.2 m2and a mass of 20 kg. Assumeg= 9.81 m/s2. (b)What-if Scenario:Whatwould the pressure be if the orientation of the device were changed and it were nowupside down?SOLUTION(a)piston2piston0piston2m kN;kPa mkg=kN1000sNimgp Ap Apiston0222pistonkNm kN1;kPa=kg1000msN m209.81100;10000.2iimgppAp100.981 kPaip(b)piston2piston0piston2m kN;kPa mkg=kN1000sNimgp Ap Apiston0piston;1000209.81100;10000.2iimgppAp99.019 kPaippopimg

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0-1-11[XB]Determine the mass of the weight necessary to increase the pressure of theliquid trapped inside a piston-cylinder device to 120 kPa. Assume the piston to beweightless with an area of 0.1 m2, the outside pressure to be 100 kPa andg= 9.81 m/s2.SOLUTION2piston0piston2m kN;kPa mkg=kN1000sNwim gp Ap A0piston0piston;kPa10001000 ;kg1201000.11000 ;9.81wiiwwm gppAppAmgm203.873 kgwm

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0-1-12[XS]A mass of 100 kg is placed on the piston of a vertical piston-cylinder devicecontaining nitrogen. The piston is weightless and has an area of 1 m2. The outsidepressure is 100 kPa. Determine (a) the pressure inside the cylinder. The mass placed onthe piston is now doubled to 200 kg. Also, additional nitrogen is injected into the cylinderto double the mass of nitrogen. (b) Determine the pressure inside under these changedconditions. Assume temperature to remain unchanged at 300 K and R for nitrogento be0.296 kJ/kg-K.SOLUTION(a)A vertical force balance on the piston yields:2piston0piston2m kN;kPa mkg=kN1000sNwim gp Ap A 0piston;kPa10001009.8110100.98 k0;1000a1Pwiiim gppApp(b)As the weight is doubled, the force balance on the piston yields: 2009.81=100;11000ip101.96 kPa=ipNote the amount of nitrogen (or the identity of the gas) is not relevant.

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0-1-13[XN]A piston with a diameter of 50 cm and a thickness of 5 cm is made of acomposite material with a density (ρ) of 4000 kg/m3. (a) If the outside pressure is 101kPa, determine the pressure inside the piston-cylinder assembly if the cylinder containsair. (b)What-if Scenario:What would the inside pressure be if the piston diameter were100 cm instead? (c) Would the answers change if the cylinder contains liquid waterinstead? (1: yes; 2: no)SOLUTION(a)2220.5;;0.1963 m ;44dAAApiston;kgmAt2pistonpiston0.540000.05;39.25 kg;4mmpiston0piston;kPa1000imgppA39.259.81101;10000.1102.966kPa9 3iipp(b) 2pistonpiston140000.05 ;157 kg;4mmpiston0piston;1000imgppA1579.81101;10000.785102.96 kPa4iipp(c)No, the force balance on the piston does not depend on the identity of the fluidsexerting pressures. Answer: 2popimg

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0-1-14[XA]A piston-cylinder device contains 0.02 m3of hydrogen at 300 K. It has adiameter of 10 cm. The piston (assumedweightless) is pulled by a connecting rodperpendicular to the piston surface. If the outside conditions are 100 kPa and 300 K, (a)determine the pull force necessary in kN to create a pressure of 50 kPa inside. (b) Thepiston is now released; as it oscillates back and forth and finally comes to equilibrium,the temperature inside is measured as 600 K (reasons unknown). What is the pressure ofhydrogen at equilibrium? (c)What-if Scenario:What would be the answer in part (a) ifthe gas were oxygen instead?SOLUTION(a)A vertical force balance on the piston yields:0;kNiFp Ap A02;100500.393 kN0.1;4iFppAFF(b)With the external force absent, a new force balance on the piston after it comes toequilibrium yields:0;kNip Ap A0;100 kPaiippp(c)The identity of the gas has nothing to do with the force balance. Therefore, theanswer does not change.F= 0.393 kN.

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0-1-15[XH]Air in the accompanying piston-cylinder device is in equilibrium at 200oC.If the mass of the hanging weight is 10 kg, atmospheric pressure is 100 kPa, and thepiston diameter is 10 cm, (a) determine the pressure of air inside. Assumeg= 9.81 m/s2.(b)What-if Scenario:What would the pressure be if the gas were hydrogen instead?Molar mass of air is 29 kg/kmol and that of hydrogen is 2 kg/kmol. Neglect piston massand friction.SOLUTION(a)A vertical force balance on the piston yields:0;kN1000ippmgp Ap A02;kPa1000109.814100;(10012.49);1000(0.1)ipiimgppApp87.51 kPainp(b)The force balance does not depend on the identity of the gas. Hence,87.51 kPaip

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