resoluções van ness 7 edição, Exercícios de Termodinâmica

Baixe resoluções van ness 7 edição e outras Exercícios em PDF para Termodinâmica, somente na Docsity! SOLUTIONS MANUAL for Introduction to Chemical Engineering Thermodynamics 8th Edition by Smith IBSN 1259696529 full download: http://downloadlink.org/p/solutions-manual-for-introduction-to- chemical-engineering-thermodynamics-8th-edition-by-smith-ibsn-1259696529/ b) The internal energy change (1715 J) is equal to the total heat capacity times the temperature change ( U'=mC, Tor 1715 J=30kg-4180] kg' K: TK, from which T= 1715/(30-4180) = 0.0137 K should be some amount smaller than the electrical work provided to the mixer, because the mixer should have some inherent irreversibilities that prevent it from fully converting the incoming electrical work into mechanical work done on the dough. From the given data, we cannot evaluate all terms in the energy balance for either the dough alone or the mixer and dough together as the system. 2.6 (7" edition Prob. 2.5) For this problem, we need to use two principles: U=Q+W, which gives us the 3º number in p Pp. Pp g any row in which we already have 2 numbers, and Lis a state function, so its value doesn"t change in a cyclic process (if we end up at the same conditions as where we started, U' will return to its initial value). Step Ui) o() W(J)) 12 —200 ? —6000 23 ? —3800 ? 34 ? —800 300 41 4700 ? ? 12341 ? ? —1400 So, we just go through and fill in the numbers by adding and subtracting things. Q for the first step hastobeQ= U'-W= 5800]. Likewise, whole cycle is 0, as stated above. Filling these in gives U=0Q + W= —500 or the third step. Also, Step Ui) 20) W0]) 12 —200 5800 —6000 U'for the  23 ? —3800 ? 34 | —-500 | -800 | 300 41 4700 ? ? 12341 0 ? —1400 Now, we can fill in Q = 1400J for the overall cycle (so OQ + W= U). Then we know all but one number in the columns for U'and Q, and the sum of the 4 steps in cach case has to add up to the value for the overall cycle, so we have U.. = O + 200 + 500 — 4700 = —4000 J and Q,, = 1400-5800+3800+800 = 200. Filling these in gives: Step vo, om | wo) 12 —200 | 5800 | —6000 23 | —4000 | —3800 ? 34 —500 | —800 | 300 41 4700 | 200 ? 12341 0 1400 | —1400 Finally, we now know two of the three entries in the 2 and 4" rows, so we can fill in these last two numbers. In cach case, wc have W= U-— Q, which gives W = —200 J for step 23 and W= 4500 J tor step 41. Filling these in completes the table:  Step | UM] OM | WD) 12 —200 | 5800 | —6000 23 | —4000 | 3800 | —200 34 —500 | —800 | 300 41 4700 | 200 | 4500 12341 0 1400 | —1400 As a check, we can add up the work values for the four steps and make sure the sum comes out to be —1400 (it does). 2.7 (7" edition Prob. 2.6) If the refrigerator is inside the kitchen, then the electrical energy entering the refrigerator (from outside the kitchen) must inevitably appear in the kitchen. The only mechanism is by heat transfer (from the condenser of the refrigerator, usually located behind the unit or in its walls). This raises, rather than lowers, the temperature of the kitchen. The only way to make the refrigerator double as an air conditioner is to place the condenser of the refrigerator outside the kitchen (outdoors). If, for example, one mounted the refrigerator in an exterior doorway and left the refrigerator door open, once would have the (bulky, expensive, and relatively incfficient) equivalent of a window air conditioner. 2.8 (7" edition Prob. 2.11) The total hcat capacity of the water is 20 kg-4.18 kJ/(kg- C) = 83.6 kJ/ € = 83600 J/ €. So, to increase the temperature by 10 C requires an energy input of 836 kJ. If we are doing work on the water at a rate of 0.25 KW = 0.25 kJ/s, then we will have to do so for 836/0.25 = 3344 s = 55.7 minutes = 0,029 hr. Electrical and mechanical irreversibilities cause an increase in the internal energy, and therefore mechanical energy. The amount of energy required to heat a quart of water (1 kg of water is about 1 liter or 1 quart) from room temperature to its boiling point (increasing the temperature by about heat transfer from the motor to the surroundings is equal to the rate at which heat is produced within the motor by these irreversibilitics. Insulating the motor does nothing to decrease the irreversibilitics in the motor and mercly causes the temperature of the motor to rise until steady- state heat-transfer is reestablished with the surroundings. Insulating the motor will just make it run hotter and probably damage it. 2.14 (7º edition Prob. 2.17) A general encrgy balance on the hydroturbinc can be written as ., dude + ((u+ 1 2 + cem) =0+W The system can reasonably be assumed to operate at steady-state, so the first term is zero. The change in kinctic energy will also be negligible, because the inlet and outlet pipes have the same diameter, and therefore the water will have the same velocity at the inlet and outlet. A somewhat stronger assumption is that the change in enthalpy of the water is negligible. All real processes, including hydroturbines, have some degree of irreversibility. However, it turns out that a well- designed hydroturbine can be quite efficient, and can convert more than 90% of the potential energy of the water into mechanical energy (to be used to drive a gencrator, which will have its own inefficiencies, so that the final electrical energy output will be smaller). Thus, for the hydroturbine, it is reasonable to neglect the change in enthalpy of the water, as well as heat flow to or from the system. Doing so gives simply W= Am+ SmgAz This will be an upper limit for the mechanical power (rate of work) produced by the turbine. To compute it, wc need to know the mass flow rate of the water through the turbine (which is the same at the inlet and outlet, by a trivial mass balance). The cross-sectional area of the 2 m diameter inlets and outlets is m-(1 m)” = 3.14 m”, Multiplying this by the water velocity of 5 ms” gives a volumetric flow rate of 15.7 m'/s. Multiplying this by the density of water (1000 kg/m”) gives the mass flow rate as 15,700 kg/s. Thus, the mechanical work is estimated as W= 15700 kg/s + 9.8 m/s? e 50 m = 7.70 10º kg m? 5? = 7.7 MW A crude estimate is that the average power consumption rate of a U.S. houschold is around 1 KW. Allowing for an overall efficiency (turbine plus generator) of 85%, this is still enough to power about 6500 households. 2.15 (New) If wc take a cylinder of 40 m diameter and wind speed of 8 m/s, then the volumetric flow rate of the air impinging on the turbine is 8 m/s-(40 m)/4-7 = 10050 m/s. The corresponding mass flow rate is 10050 m'/s-1.2 kg/m” = 12060 kg/s. The Kinetic energy per unit mass is 8/2 = 32 m/s” = 32 J/kg. Thus, the total kinctic energy per time is 12060 kg/s-32 J/kg = 386000 J/s = 386 kW. The fraction of this converted to electrical power is 90/368 = 0.233. That is, 23.3% of the kinctic energy of the oncoming wind is captured. 2.16 (New) insulated) the heat is also zero. the change in-internal energy of the fluid is also zero. The internal discharge rate is 56/4 = 14 W, or 14 J/s. Remembering that power is current times voltage, the average current drawn by the laptop is I4W/LI V= 1.264. Taking the computer (not including the battery) as the system, and neglecting any change in internal energy of the computer, all of the electrical work supplied to the computer by the battery (from the condenser of the refrigerator, usually located behind the unit or in its walls). This raises, as au - W=0 dt 2+ TF U remains constant, then O = —W = —14 W. When the computer is at stcady state (not changing temperature) 14 W must be dissipated into the surroundings as heat. If we instead considered a system that includes the battery as well as the rest of the computer, then the work would be zero (the battery is now part of the system. rather than the surroundings). We could write the electrical potential energy of the battery (when fully charged) as Es = 56 W-hr Thus, the total energy of the system would include this as well as the internal energy of the computer and we could write the first law as MU +ED=0+W Taking W=0, AU'= 0, and AE;p = —56 W-hr (when the battery is fully discharged, Esp = 0), we have —56 W-hr=Q The total heat removal is 56 W-hr over a period of 4 hours, so the average rate of heat flow is Q=-56/4=-14W 2.17 (New) If we take the laptop and bag as the system, then there is no exchange of heat or work between the system and surroundings, and the first law tells us that AU” = 0. The internal cnergy does not change, but the chemical energy stored in the battery is released as sensible heat, which raises the temperature of the computer. The total amount of energy stored in the battery is 56 W-h = 56 J/s-h-3600 s/h = 201600 J = 201.6 kJ. The total heat capacity of the laptop is 2.3-0.8 = 1.84 KJ/K. Thus, encrgy release of 201.6 kJ would increase the temperature by 201.6 kJ/1.84 kJ/K = 110 K = 110ºC (because this is a temperature DIFFERENCE, and not an absolute temperature, units of K and ºC are the same). If the computer started off at a cool 20ºC, the estimated final temperature is 130ºC (about 270ºF). At that point, the foam briefcase would probably start melting onto the surface of the laptop. Fortunately, in real life, this is unlikely to happen, and even if it did, the bag would not be perfectly insulating. Nonetheless, a typical laptop battery does store sufficient energy to cook the laptop in which it is installed if no heat could be removed. o dt Tuy O -r(E dr ( e Sa = al- + SI No “TT T T dT = Kdt Had (d+ 1 ) 1 ( Cy o Cy Integrating from t = 0 (where T = Tjtor gives: TC + CyTua= (+ cult RR exp (= +CW) ) CC, o CC, =K(C' + Cut TCA Cylwo (+ CylT= Cy To To exp (e) “Cy é =K(C + Cdr BC + Chiyo — Cy Too” To exp a 2) p= CC 2.21 (7 edition Prob. 2.20) The general equation applicable here is Eg. (2.29): alH+l e? + zm], =0+W, (a) We write this for the single stream flowing within the pipe, neglect potential and kinetic encrgy changes, and set the work term equal to zero. This yiclds: AmzgmmAd=m HH, =Q In the second and third versions of the LHS, wc have used the fact that there is a single inlet and outlet, and that mass conservation requires that the mass flow rate at the inlet and outlet are equal. (b) The equation is here written for the two streams (1 and 2) flowing in the two pipes, again neglecting any potential- and kinetic-energy changes. There is no work, and the heat transfer is internal to the system, between the two streams, making OQ = O. Thus, Amzg mmAH, +mAH —0O (c) For a pump operating on a single liquid stream, the assumption of negligible potential- and the mechanical work done by the system (negative). Thinking way, way, back to first year surroundings. Whence, Amzgmmid=m HH =W (d) For a properly designed gas compressor the result is the same as in Part (c). Amzgmmid=m HH =W (e) For a properly designed turbine the result is the same as in Part (c). AmzgmmAH=m Ha Ha -W (f) The purpose of a throttle is to reduce the pressure on a flowing stream. One usually assumes adiabatic operation with negligible potential- and kinctic-energy changes. Since there is no wotk, the equation is: AmzgmmAH=m Hoau-H =0 £) The whole purpose of a nozzle is to produce a stream of high velocity. The kinctic-cnergy change must therefore be taken into account. However, one usually assumes negligible potential-cnergy change. Then, for a single stream, adiabatic operation, and no work: 1. — al(H + au )m| =0 ora(H+1,4)=0 The usual case is for a negligible inlet velocity. The equation then reduces to: 2 AH=-1,6 wherc u, is the exit velocity. 2.22 (7º edition Prob. 2.21) The mass flowrate, m, is related to the density, diameter, and average velocity by m= puz 4 Thus, puD = dm x And e= puD Am H a Du Thus, the Reynolds number is directly proportional to the mass flow rate. dU dm di gm HW m dt + H m We can use the mass balance to climinate 11" from this equation to get mt gim = dm pp dt di dt Rearranging this gives dU dm,w MU dm y go aa , From which we can climinate time as a variable and get the desired relationship dU dm H-U m If conditions within the tank arc uniform (no gradients in pressure, temperature, etc. between the main part of the tank and the tank exit) then H' = H. 2.26 (7º edition Prob. 2,25) Mass balance at steady state requires (hat the mass flow (and therefore volumetric [low for water with nearly constant density) be the same before and after the diameter change. The mass flowrate in the 2.5 cm diameter pipe is m=pymoke mS0ms!. 0,025m?/4 and the mass flow rate after the expansion is 3 . m= prokgm” ms daum?/4 Vou our Setting these equal shows that the velocity in the outlet pipe is Vau 14 m/s: (2.5 cm/dacm) So, if du = 3.8 em, then v,, = 6.06 m/s The change in kinetic energy per unit mass is then Edm =4 (6.06 - 14) = 79.6m's” = 79.6] kg! Since no heat is exchanged with the surroundings, there is no change in elevation (no change in gravitational potential energy) and no non-flow work is done on the surroundings, the energy balance is simply H+ E =0 From which H=79.6]kg” This enthalpy change corresponds to a temperature increase given by H=C, T=4180]kg'K' T=79.6]kg' from which T=0.019K. If the downstream diameter were 7.5 cm, then we would have v,; = 1.556 m/s, and the change in kinetic energy per unit mass would be —96.8 J kg”. The corresponding temperature increase would be 96.8/4180 = 0.0232 K. The maximum temperature increase would occur if the downstream velocity went to zero, so that all of the kinetic energy of the flow were converted to internal energy. In this case, we would have Edm = Ya (0 — 14) = —98 J kg”, Icading to a temperature increase of 98/4180 = 0.0234 K. For the downstream diameter of 7.5 cm, wc had very nearly reached this limit. 2.27 (7” edition Prob. 2.26 The energy balance, including kinctic energy contributions, for a steady flow system gives us Ls Ls m (Ho — Hi + tou 58) =0+W The mass flow rate (in and out) is 50 kmol hr'-29 kg/kmol = 1450 kg/hr = 0.40278 kg/s The enthalpy change is 7 - 3:5º8314K kmol' E! H = in 29 kg kmol-! Ha=C 7, out — Ein p Pout — The kinctic energy change is 2 4ou — 58in So, the energy balance becomes 520K — 300K = 220.75 kJ kg”! LL =05(1225m252-100nê 52) =-439m? 52 = 00439] kg”! so often the case, the change in kinetic energy is negligible. 040278 kgs! (220.71 kl kg!) =88.9k]s! =889kW=0+W=0Q+88.9kW from which Q = -9,9 KW. That is, 9.0 kW of heat flows from the surroundings, or 9.9 KW of the input mechanical energy Icaves the compressor as heat. 2.28 (7º edition Prob. 2,27) system (the air) to the Because the pipe is insulated, we can assume Q is zero, and because the pipe and valve presumably have no moving parts, W is also zero. Furthermore, the pipe is horizontal, so there is no change in gravitational potential energy between the inlet and the outlet streams. Thus, we can write the energy balance for an open system with one inlet and one outlet like it is written in equation 2.31. Note that this equation, as written, is per unit mass. AR aH4L BS Az=0+W 28. Be AZ aH4 DL = 28, The mass flow rates in and out must be equal (at steady state, with no chemical reactions), m, = Mas And the mass flow rate is the velocity times the cross-sectional area divided by the specific volume (volumetric flow rate divided by specific volume): m = uA4/V. So, we have UjnAin — Hour Á out andu — UinAin Vout = r = V. V. out V. in out Au in A. and A,, arc the same, and because we have PV/T = a constant, we can write Vou — Tou Pim Va TinPou So, we can then write o tinPinho inf ow Cp T ourt in 4H, Putting in the numbers, we have P,/P,, = 5.385 andu,=20ms',7,=45 C=318.15K,so our ia in m = *20018.15+AD + Umu(mesT)= E =322.0488.9EWAT Where ATisinKor*C. Also, wc have C, = 3.5-R = 3.5-8.314 J/(mol K) = 29.10 J/(mol K). As given, this is the molar heat capacity. To get the specific heat, we divide by the average molecular weight of nitrogen (0.02897 kg/mol) to ger €, = 1004.5 J/(kg K). Putting this all together, we see that 107.69 88.9 kW AT? — 20? 3204 DS = where both terms have units of J/kXg, and Thas units of K or ºC. Multiplying things out gives 77MW kg 7 + 322. 04AT+5398.8=0 Applying the quadratic formula to this gives T=-5.19 K = -5.19C, so the downstream temperature is 39.8 €C. 2.30 (7” edition Prob. 2.28) We can write the energy balance (per unit mass) as AH+AER=0+W, From the problem statement, we have H'=2726.5 - 334.9 = 2391.6 kJ kg”, and no shaft work is done. Since there is a large change in velocity from inlet to outlet, we will take into account the change in kinetic energy, which is Edm= (200 -3)m's”= 19996] kg” =20.0kJ kg”. So. wc have OQ =AH+AEp = 322.044200=220.75k kg”! Even for this huge change in velocity, the change in kinctic cnergy is a very small part of the overall heat requirement. 2.31 (7º edition Prob. 2.26) Because the nozzle is insulated, Q is zero, and because it is a nozzle (with no moving parts) the nonflow work, W. is zero. so from the stcady-state energy balance we have 1 15º m (Ho — Hi +5 ou 58) =0 or H H, =1 tg Bu) our O fZin S 5 Nin O Hour trom which H, H, =1 tg Bu) out — HHin 2 Min — “our “ou =2 Ha Hou + = 2(3112500 ] kg”! — 2945700 kg!) + 900 T kg”! ud = 334500 ] kg”! = 334500 m? s? tou = 578ms” The mass balance simply tells us that the mass out equals the mass in, or Aintin V; in A — Poutttowt V, out trom which V, 667.75 em? g1+30ms! H; out — Pouttiin + = 322.04 Therefore dom = Ay = 322.04 = 0,299 Tim Ain so du = 0.299-5 cm = 1,49 cm out Taking the alternative approach, we could substitute for P in terms of V, writing RT V-b r w=- 8Lavy= rm (bo JV-b h-b K Substitution for V, andV, in terms of P, and P, shows that this result is identical to the first one. 2.35 (7” edition Prob. 2.33) The volumetric flow rate into the pipe is the velocity (u = 10 fts! = 3.048 ms) times the cross- sectional area (A = (0.25 ft) /4 = 0.04909 ft), so the volumetric flowrate is 0.4909 ft" 5”, Because the specific volume at these conditions is 3.058 ft Ib,, the mass flowrate in is mi, = 0.4909 tt s! 73.058 ft Ib,] = 0.1605 lb, s! = 0.07280 kg s', At steady-state, the mass flow rate out must be the same as the mass flowratc in. So, if the specific volume at the outlet conditions is 78.14 ft 1b,;, then the volumetric flowrate out is 78.14 tt 1b,-0.1605 1b, s! = 12.543 ft s!, The cross-sectional area of the 10-inch diameter exit pipe is (10/12 ft)/4 = 0.5454 ft, so the velocity is tu = 12.543 ts! / (0.5454 fÉ = 23.00 fts! = 7.0104 m s!, Now, we can use this in the energy balance for an open system as written in equation 2.30. AZ (aut L ssa )n= 04 We will assume that the heat loss from the turbine is negligible (Q = 0) and that the change in clevation from inlet to outlet is negligible ( z = 0), so the work output of the turbine is given by: 2 W= (am+ A ) m 28. (7.0104 ms? — (3.048 ms) 5 0.07280 kg s”! W=| 322.04- 1322.6 Btu lbj;! + W= (174.0 Btu bz! + 19.93 1 kg”!)0.07280 kg s! W= (334500 J kg”! — 19.93 1kg"!)0.07280kg 5! = —29462 197! W=-29.46k]s! =-29,46 kW = -27.92 Btu s!. Our sign convention for W is that it is work done on the fluid, so the work output from the system is 27.92 Btu s! = 39.5 hp. 2.18 (New) The volumetric flow rate into the pipe is the mass flow rate times the specific volume (0.1 kg s!- 0.20024 m” kg! = 0.020024 ms). The cross = 0.005027 m3, so the average flow velocity is 0.020024/0.005027 = 3.984 ms At steady-state, the mass flow rate out must be the same as the mass flowrate in. So, if the specific volume at the sectional area of the inlct pipe is 4= (0.08 m)/4 outlet conditions is 3.4181 m” kg”, then the volumetric flowrate out is 3.418] m' kg-0.L kgs! = 0.34181 m? s!. The cross-sectional area of the 25 em diameter exit pipe is (0.25 mj/4 = 0.04909 m”, so the velocity is tí, = 0.34181 m s'/ 004909 m” = 6.963 m s!, Now, we can use this in the energy balance for an open system as written in equation 2.30. AZ (aut L ssa )n= 04 We will assume that the heat loss from the turbine is negligible (Q = 0) and that the change in clevation from inlet to outlet is negligible ( z = 0), so the work output of the turbine is given by: 2 W= (a 10) m 28. (0.025 m Ss y — (3.984 m Ss y W= [ (322.04 - 220.71 K kg”! + > JO ke s! W= (-468.1K] kg”! + 19.93]kg !)0.1 kgs! = 46.81 kJ 5! W= 46.81 ks! = 46.81 KW. Our sign convention for W is that it is work done on the fluid, so the work output from the system is 46.8 KW. 2.37 (New) First, we need to determine the velocity of the discharge, «,. To do this the mass flow rate must be determined by using Eu emêemol a mol m="—=040278. or 335479 —. Using this m, = 7.0104 mys. m=m Now that « has been determined, using Egn. 2.31, - mou W gsm — tê. as O=H+Hi ti Sam = 220 The cross scctional arcá of the 2 m 1 kJ/kg = 1000 m/s” you obtain Q = -127.064 KJ/mol. To get the heat transfer rate, 0=Qem= —19053kJ/s. 2.38 (7 edition Prob. 2.34) First, we need to determine the velocity of the discharge, «,. To do this, the mass flow rate must be determined by using  E 20 3 m 3 mÉkPa, 293.16k 50 pro 13 =Rà O 83140105 tb, H=Rg= 854 10 qui * TOO kPa mol T 3 p=RZ=VaP- AD P, mol Using the intermediate temperature to determine the temperature change in both steps wc have AT,=T'-T|=-263.844kandAT,=T;—T' = 303.844k. Using=m = E Tr and plugging the temperatures into Eqn. 2.16 and Eqn. 2.20, wc have for Step À: As = AL AV, P— Rj = ARO A pagto ago kra = 1000 kra = 548310 mol mol mol AU,=CAT,= ó . ga4— «263.844 k = —0.40278 A 4 esa mol k mol And for Step B: AH, =CAT,= Z, ga * —303.844 k = 0,40278 + P 2 mol E mol ses »m am 4 AL, =AM— B V—W = 884156 — 100 kPa (29770 +10 — 243710 — ) = 548310 . te mol mol mol mol Now using these to determine the total AU and AH AU = AU,+AU,= 832.16 SL mol AH = AH, +AH, = 0.40278 Em mol 2.42 (New) For this problem, we need to find the specific heat or the heat capacity over a range of temperatures. From the text, it is given that Q= AH="CmaAaT Given the power supplied at the Q values and the flow rate, solving for the heat capacity gives: c-L “mAT Given this the specific heat from 0º€C to 10ºC is em — 5.5 Jfsce -m=-L “ 0333g/sec 10k 77 And going from 90º€ to 100ºC is C=mn=— J 71 The key here is that the initial temperature for cach step is O C and that the AT for cach is the difference between 0ºC and the new temperature. Over the whole range the “C values are. hr 10 20 30 40 50 60 70 80 90 100 5.5 1.0 | 16.6 | 223 | 280 | 337 | 396 | 454 | 513 | 573 o 1.650 | 1.650 | 1.660 | 1.673 | 1.680 | 1.685 | 1.697 | 1.703 | 1.710 | 1.719 Jg k) The average is ºC = 1.683 Jg k). 2.43 (New) The relevant steady-state energy balance is AH=0+W Whether wc regard the energy input as heat or work depends on where we draw the system boundary. If we draw the s stem boundary to only contain the fluid, then W= 0, and AF = O, where Q is the flow of heat from the heating element to the water. The total change in enthalpy of the 237 g of water is AH = mC,AT = 237 g-4.18] g'ºC'!.66 ºC = 65384] = 65.4 kJ. This is also the total heat transferred to the water in a period of 60 s. The rate of heat transfer is Lherefore 65.4 kJ/60 s = 1.09 kJ/s = 1.09 kW. We can reasonably assume that the heating element fully converts electrical energy to heat, so at steady state, this would also be the power requirement for the heater. The actual power requirement might be slightly higher, because of heat losses to the surroundings (not all of the heat generated in the heating element actually enters the water). If the coffee maker mechanical energy. The amount of energy required to heat a quart of water (1 kg of water is about 1 liter or 1 quart) from room temperature to its boiling point (increasing the temperature by about amp houschold circuit. 2.44 (7" edition Prob. 2.38) (a) Based on Equ. 2.234: m = u A p,if'the p is constant, the À is constant, and the flow is at steady state, which means it remains unchangec “ A, then that would mean q is constant as well. (b) Due to the law of mass conservation, m must be constant. n will change due to reaction in the stream; the moles of A will go towards Q as it travels to the end of the pipe. Due to the temperature and pressure changes in the pipe both q and u will also vary. 2.45 (7" edition Prob. 2.39) (a) First, the Reynolds number must be determined: kg m *8.3140 9965 0.55 Re-Pets TO Cm TS oa H co E 7710 ms Next, using the equation for the fanning friction factor given, wc get