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me104hw456 - we 2-1C Initially the rock possesses potential...

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Unformatted text preview: we / 2-1C Initially the rock ' . , possesses potential energy relative t th b potential ener _ . _ . .o e ottom of the sea. As the roc ' as a resmt of tgryciisocgiliyleergiénég(kinetic energy. Part of this kinetic energy is converted to theli'nfijllseriiligy 0 air resrstance, which is transferred to the ' an and the rock. Same thing h pp [15 In ater ASsumln ea 1“ )ttOIIl ls llegllg [I’le [he 6] [Inc a e W . g the Impact elOCIty 0 the 06k at th , V 1 I e S potentlal ellelgy 01 the rock 15 converted to themal energy In water and alr /2-8 The specific kinetic energy of a mass whose velocity is given is to be determined. Analysis Substitution of the given data into the expression for the specific kinetic energy gives 2 2 V _ (30 m/S) (452%?) = 0.45 lekg 2 1000m /S 2-18 A person with his suitcase goes up to the 10'11 floor in an elevator. The part of the energy of the elevator stored in the suitcase is to be determined. Assumptions 1 The vibrational effects in the elevator are negligible. Analysis The energy stored in the suitcase is stored in the form of potential energy, which is mgz. Therefore, w?)=1o_3 kJ 1000m /s y compared to an identical suitcase on the lobby AEsmse = APE = mgAz = (30 kg)(9.81m/s2 )(35 m)( Therefore, the suitcase on 10th floor has 10.3 k] more energ level. Discussion Noting that 1 kWh = 3600 k], the energy transferred to the suitcase is 10.3/3600 = kWh, which is very small. 0.0029 2-25C When the length of the spring is changed by applying a force to it, the interaction is a work interaction since it involves a force acting through a displacement. A heat interaction is required to change the temperature (and, hence, length) of the spring. 2-29E The power produced by a model aircraft engine is to be expressed in different units. Analysis Using appropriate conversion factors, we obtain . 1 Btu/s 778.1691bf-ft/s = w = 7.38 Ibf -ftls (a) W (10 {1055.056 WI lBtu/s ) . lhp = ———— = 0.0134 h (17) W (10 W{745.7 W) p 2-33 A man is pushing a cart with its contents up a ramp that is inclined at an angle of 20° from the horizontal. The work needed to move along this ramp is to be determined considering (a) the man and (b) the cart and its contents as the system. Analysis (a) Considering the man as the system, letting I be the displacement along the ramp, and letting 0 be the inclination angle of the ramp, lkJ/kg _ = 67.0 kJ 1000m2/52) W = Fl sin 19 = mg] sin 6 = (100 +100 kg)(9.8 n1/52)(100 m)sin(20)( This is work that the man must do to raise the weight of the cart and contents, plus his own weight, a distance of lsina (b) Applying the same logic to the cart and its contents gives 1 kJ/kg — = 33.5 kJ 1000m2/52) W = F 1 sin 19 = mg] sin 6 = (100 kg)(9.8 m/s2 )(100 m)sin(20)[ different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power re uired for ea ' ' ' energies_ That is, q ch case is the sum of the rates of changes in potential and kinetic Wtotal = Wu + Wg (a) Zero. ([7) W, = 0 . Thus, _ ' _ A2 onmi —Wg —mg(zz ‘Zi)/At=ng=msz =mngin30° = (1200 kg)(9.8lm/52) M AWL (0 5) =81 7 kW 3600 s 1000 mZ/s.2 ' ' (c) W8 =0 . Thus, J/(lz s) =31.3 kW . . 1 2 Wtotal = We = 3 "1(sz - V12)/At = i (1200 kg) L’OOO m —o “(J/kg 2 3600 s 1000 m2/52 2-44C Energy can be transferred to or from-a control volume as heat, various forms of work, and by mass transport. 2-48E A water pump increases water pressure. The power input is to be determined. Analysis The power input is determined from W=V(P2 —P1) 50 psia <— =(1.2ft3/s)(50—10)psia ——“3t“__ $] 5.404 psia - 03 0.7068 Btu/s =12.6hp The water temperature at the inlet does not have any significant effect on the required power. 2-54 A fan is to accelerate quiescent air to a specified velocity at a specified flow rate. The minimum power that must be supplied to the fan is to be determined. Assumptions The fan operates steadily. Properties The density of air is given to be p = 1.18 ’kg/ma. Analysis A fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan, the energy balance can be written as E,” — E = dE 011! system / dt <90 (steady) = 0 _) Em = E out h’w—d Rate of net energy transfer Rate of change in internal, kinetic, by heat, WOIk. and mass potential, etc. energies 2 . V _ . _ - out Wsh, in " mairkeout _ mair 2 where Air ma, = pi) = (1.18 kg/m3 )(4 m3/s) = 4.72 kg/s Substituting, the minimum power input required is determined (lOm/s)2( lJ/kg 2 lmz/s2 . V2 Wsh, in = Mair % = (4'72 kg/S) j: 236 J/s =236W Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of mechanical shaft energy to kinetic energy of air. 3-1C Yes, since the chemical composition throughout the tank remain the same. ...
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me104hw456 - we 2-1C Initially the rock possesses potential...

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