ME231C S09 ME231C Engineering Thermodynamics I Name: HW#2 due Friday Jan 30 in class Group: Problem 2.7: A 30-seat turboprop airliner whose mass is 14,000 kg takes off from an airport and eventually achieves its cruising speed of 620 km/h at an altitude of 10,000 m. For g = 9.78 m/s2, determine the change in kinetic energy and the change in gravitational potential energy of the airliner, each in kJ. Known; 14000m kg= 1 0 /V m s= 2 1000 1620 172.22 / 3600 m hkmV m s h km s = = 2 1 10000Z Z m− = 9.81 /g m s= Find and PEKE∆ ∆ ( ) ( )( )22 22 1 2 14000 172.22 / 1 1 2 2 1 . / 1000 . m V V kg m s N kJKE kg m s N m −∆ = = 2.076 5KE E kJ∆ = ( ) ( )( )( )22 1 21 114000 9.81 / 10000 1 . / 1000 .N kJPE mg Z Z kg m s m kg m s N m∆ = − = 1.369 6PE E kJ∆ = ME231C S09 Problem 2.23: Measured data for pressure verses volume during the compression of a refrigerant within the cylinder of a refrigeration compressor are given in the table below. Using data from the table, complete the following: a) Determine a value of n such that the data are fit by an equation of the form b) Evaluate analytically the work done on the refrigerant, in Btu, using Eq. 2.17 along with the result of part (a) c) Using graphical or numerical integration of the data, evaluate the work done on the refrigerant, in Btu. d) Compare the different methods for estimating the work used in parts (b) and (c). Why are they estimates? Data Point P (lbf/in2) V (in3) 1 112 13.0 2 131 11.0 3 157 9.0 4 197 7.0 5 270 5.0 6 424 3.0 a) npV Constant= Taking natural log on both sides of the equation yields ( ) ( )ln ln lnp n V Constant= − + This equation can be compared with an equation of a straight line Y mX C= + , where ( )n− corresponds to the slope ( )m of a plot of ( )ln p verses ( )lnV . Using a spreadsheet program to obtain the plot and the lease square best fit line: Thus, 0.90890.9089m nn = − = −= Data Point p (psi) v (in3) ln(P) ln(V) 1 112 13 4.718 2.565 2 131 11 4.875 2.398 3 157 9 5.056 2.197 4 197 7 5.283 1.946 5 270 5 5.598 1.609 6 424 3 6.050 1.099 ME231C S09 b) Using the value of n from part (a) and eq. 2.17 ( )2 1 2 2 1 1 1 1 V V p V p VW p dV Polytropic process, n n −= = ≠ −∫ ( )( ) ( )( ) 32424 3 112 13 1 11 0.9089 12 778.2 .ft Btulbf inin in ft lbf− = − 0.2163W Btu= − c) Using graphical integration: Create a p-V diagram using excel. Word done on the refrigerant is the area under the process curve on p-V diagram. Each rectangle under the curv corresponds to: ( )32 1 110 0.2 2.14 412 778.2 .ft Btulbf in E Btuin in ft lbf − = − − The work is negative, because V is decreasing. Total number of rectangles ≈ 1002, thus ( )( )1002 2.14 4 0.2144W E Btu= − − = − npV Constant= W Area= 12 34 5 6 ME231C S09 d) The results obtained in parts (b) and (c) are in good agreement. The only measured data are the tabulated data shown as filled squares in the plot. The smooth curve does not necessarily represent the actual pressure for the corresponding volume. Hence the work evaluated in part (c) is an estimate. Problem 2.44: As shown in Fig. P2.44, the 6-in thick exterior wall of a building has an average thermal conductivity of 0.32 Btu/h · ft · ˚R. At steady state, the temperature of the wall decreases linearly from T1 = 70˚F on the inner surface to T2 on the outer surface. The outside ambient air temperature is T0 = 25˚F and the convective heat transfer coefficient is 5.1 Btu/h · ft2 · ˚R. Determine (a) the temperature T2, in ˚F, and (b) the rate of heat transfer through the wall, in Btu/h per ft2 of surface area. 26 0.5 0.32 / . .°L in ft k Btu h ft R= = = 21 070° T =25°F 5.1 / . .°T F h Btu hr ft R= = ( )Conv 2 0 QCond dTQ kA hA T Tdx= − = −dotnosp dotnosp At steady state ConvQCondQ =dotnosp dotnosp ( )2 1 2 0T TkA hA T TL− − = − a) ( )( )( ) ( )( )( )( ) ( ) 2 0 1 2 2 5.1 / . .° 0.5 25° 0.32 / . .° 70° 30° 5.1 / . .° 0.5 0.32 / . .° Btu h ft R ft F Btu h ft R Fh L T k TT F h L k Btu h ft R ft Btu h ft R +−= = = + + b) 2 1 230 70 °0.32 25.6. .° 0.5 .CondQ T T Btu R Btuq kA L h ft R ft h ft − − = = − = − = dotnospdotnosp Fig. P2.44 ME231C S09 Problem 2.71: The following table gives data, in Btu, for a system undergoing a thermodynamic cycle consisting of four processes in series. Determine a) The missing table entries, each in Btu b) Whether the cycle is a power cycle or a refrigeration cycle Process ∆U ∆KE ∆PE ∆E Q W 1-2 72 -5 -6 d = 61 0 j = -61 2-3 64 0 c = 3 e = 67 90 k = 23 3-4 -97 b = 5 0 f = -92 h = 0 92 4-1 a = -39 0 3 g = -36 i = -36 0 For each process (1-2, 2-3, 3-4, and 4-1) conservation of energy equation yields: E KE PE U Q W∆ = ∆ +∆ +∆ = − For a complete cycle 0 0 0U KE PE∆ = ∆ = ∆ =∑ ∑ ∑ 0 39U a Btu∆ = ⇒ = −∑ 0 5KE b Btu∆ = ⇒ =∑ 0 3PE c Btu∆ = ⇒ =∑ E= PE 61 , 67 , 92 , 36U KE d Btu e Btu f Btu and g Btu∆ ∆ +∆ +∆ ⇒ = = = − = − 0 , 36 , 61 , 23E Q W h Btu i Btu j Btu k Btu∆ = − ⇒ = = − = − = Net power output of the cycle: 1 2 2 3 3 4 4 1CycleW W W W W− − − −= + + + 54CycleW Btu= 0CycleW > , so it is a power cycle ME231C S09 Problem 2.76: A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2: Constant volume, V = 0.028 m3, U2 - U1 = 26.4 kJ Process 2-3: Expansion with pV = constant, U3 = U2 Process 3-1: Constant pressure, p = 1.4 bar, W31 = -10.5 kJ There are no significant changes in kinetic or potential energy. a) Sketch the cycle on a p-V diagram. b) Calculate the net work for the cycle, in kJ. c) Calculate the heat transfer for process 2-3, in kJ d) Calculate the heat transfer for process 3-1, in kJ Is this a power cycle or a refrigeration cycle? During the cycle the substance is always in gas phase. So it is not necessary to show vapor dome in p-V diagram. a) b) 1 2 2 3 3 1CycleW W W W− − −= + + 1 2 0W − = (Constant volume process, V=constant) 3 32 3 2 2 3 3 2 2 ln lnV VW p V p VV V− = = in this we know p3 (1.4 bar) and V2 (V2=V1). We can calculate V3 from W3-1 ( ) 3 13 1 3 1 3 3 1 3 WW p V V V V p − − = − ⇒ = − ( )3 33 210.5 1 1000 .0.028 0.1031.4 1 5 / 1kJ bar N mV m mbar E N m kJ−= − = 1 3 2 V=Constant P=Constan pV=Constant 1 1 V p ME231C S09 Thus, ( )( ) 2332 3 3 3 2 1 5 / 10.103ln 1.4 0.103 ln 18.78 0.028 1 1000 , V E N m kJW p V bar m kJ V bar N m− = = = And 1 2 2 3 3 1 0 18.78 ( 10.5) 8.28CycleW W W W kJ− − −= + + = + + − = 0CycleW > so it is a power cycle c) Energy balance of process 2-3 2 3 2 3 2 3 2 3 2 3U KE PE Q W− − − − −∆ +∆ +∆ = − where, 2 3 2 3 2 30, 0, 0U KE and PE− − −∆ = ∆ = ∆ = So 2 3 2 3 18.78Q W kJ− −= = d) For a cycle ( ) ( ) ( )2 1 3 2 1 3 0U U U U U U− + − + − = ( ) ( ) ( )1 3 2 1 3 2 26.4 0 26.4U U U U U U kJ− = − − − − = − − = − For process 3-1: ( )1 3 3 1 3 1U U Q W− −− = − 3 1 36.4 ( 105.) 36.9Q kJ− = − + − = − meisu Microsoft Word - ME231D_HW_2_Solution.doc