Solution manual for heat and mass transfer 2nd edition by rolle

Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Heat and Mass Transfer Solutions Manual Second Edition This solutions manual sets down the answers and solutions for the Discussion Questions, Class Quiz Questions, and Practice Problems There will likely be variations of answers to the discussion questions as well as the class quiz questions For the practice problems there will likely be some divergence of solutions, depending on the interpretation of the processes, material behaviors, and rigor in the mathematics It is the author’s responsibility to provide accurate and clear answers If you find errors please let the author know of them at rolle@uwplatt.edu Chapter Discussion Questions Section 1-3 What is meant by a system? By a body? A volume in space having a boundary surface separating it from its surroundings A body is a system that typically has a mass and in the study of mechanics, is subjected to external forces and body forces What is the importance of a boundary in heat and mass transfer? A boundary is the surface across which heat, energy, and mass flow and is the primary location for predicting these transfers What is the difference between a boundary and a surface? A surface is a three dimensional geometric construct but which has no thickness A boundary is a surface but a surface may not necessarily be a boundary What is meant by system properties? These are the quantities that describe the action or reaction of a system to its surroundings Examples of properties are pressure, temperature, volume, mass, and energy What is steady state? Steady State is the situation when a system does not change with elapsing time What is a control volume? A boundary containing a system and defining the system volume A control volume allows for mass to pass through the boundary in both ways What is heat transfer? Heat transfer is rate at which heat or energy flows across a systems boundary © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ What is meant by reversible heat transfer? Reversible heat transfer is an approximation to allow for a predictable answer to heat transfer Rigorously, reversible heat transfer must only occur over a differential temperature or differential temperature gradient and must proceed over an infinitesimally small time interval so that it needs to be fast Section 1-4 What are the three modes of heat transfer? The three modes are conduction, convection, and radiation 10 What is significant about the temperature difference when considering radiation heat transfer? The temperature difference is a difference of temperature, each to the fourth power 11 What mode of heat transfer does not satisfy the thermodynamic definition of heat transfer and why? Convection heat transfer does not rigorously satisfy the thermodynamic concept of heat or heat transfer because there is not a precise boundary across which heat can flow and there is mass transfer mingled with the process Section 1-5 12 What are causes of diffusional mass transfer? Diffusional mass transfer occurs due to a pressure difference or a pressure gradient, due to a concentration or density difference or gradient, due to a temperature difference or gradient, due to body forces on the mixture components, due to forced or free convection currents, due to turbulent flow, or due to phase change phenomena 13 What is the difference between a concentration gradient and a density gradient? Concentration is the number of moles in a prescribed volume while density is the amount of mass in a prescribed volume The gradients just reflect these two definitions Class Quiz Questions What is meant by an ideal or perfect gas? A substance or material which behaves according to the equation ‫ܴܶ݉ = ܸ݌‬ What is meant by an incompressible material? A material which does not change its volume when subjected to different pressures or temperatures is called an incompressible material Define specific heat at constant volume డ௨ Specific heat at constant volume, ܿ௏ is ܿ௏ = ቀడ் ቁ ௏ â 2016 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Define specific heat at constant pressure డ௛௡ Specific heat at constant pressure, ܿ௉ , is ܿ௉ = ቀ డ் ቁ ௉ What is the first law of thermodynamics for a system? The first law of thermodynamics, in words, is heat minus work equals energy change of the system What is flow work? Flow work is the work done against a pressure when pushing a fluid, causing it to move or flow As an equation flow work is ‫ܸ݌‬ What is thermal conductivity? The defining equation for thermal conductivity ߢ is Fourier’s law of conduction; ‫ݍ‬ሶ ௖௢௡ௗ௨௖௧௜௢௡,௫ = ߢ డ் డ௫ What is the convective heat transfer coefficient? Newton’s law of heating or cooling is the defining equation for the convective heat transfer coefficient ℎ; ܳሶ௖௢௡௩௘௖௧௜௢௡ = ℎ ‫ܶ|ܣ‬ஶ − ܶ௕ | What is the relative humidity of air when it is at its dew point? The relative humidity is 100 % Practice Problems Section 1-3 Ten lbm of carbon dioxide gas are cooled from 1500F to 800F The specific heat at constant volume is ܿ௏ = 0.323 − ଵସ଼ ଷଶ଴ସହ + ் మ where ் T is in Rankine degrees and ܿ௏ is in Btu/lbm· F Determine the internal energy change for the cooling process Solution The internal energy change for the cooling process is T2 T2 540 148 32045   ∆U = ∫ mcV dT = m ∫ cV dT = m ∫ 0.323 − + dT T T  T1 T1 610  Using some integral calculus we have 540 32045   = m 0.323T − 148 ln T − T  610   540   = (10lbm ) 0.323 ( −700 F ) − 148ln − 32045  −  610  540 610    = 113.8 Btu â 2016 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Two kg of water are heated from 150C to 3000C at 101 kPa Determine the amount of heat added during the process Assume cP is 4.18 kJ/kg·K for liquid water and 1.86 kJ/kg·K for steam Assume the heat of vaporization is 2258 kJ/kg at 101 kPa Solution For the heat we have Q = ∆Hn = m∆hn and then = m cP , water (T2 − T1 ) + hn fg + cP , steam  = ( 2kg )  4.18 (100 − 15 ) + 2258 + 1.86 ( 300 − 100 )  ∆Hn = 5970.6kJ One hundred lbm of air in a pressure tank is cooled from 1800F and 160 psia to 800F when the surroundings are at 700F Determine the tank pressure at 800F Solution Assume the tank is rigid so that the volume is constant and that the air behaves as an ideal gas Then p1 p2 = T1 T2 and T p2 = p1   T1    5400 R   = (160 psia )   = 135 psia 640 R    Steam is heated from MPa, 4800C to MPa, 6400C Determine the heat added to the steam per unit mass Solution The heat added will be the increase in the steam’s enthalpy, q = hn2 − hn1 The enthalpies may be read from a superheat steam table from a thermodynamics book or approximated from Appendix Chart C-2 hn1 = 3390kJ / kg hn2 = 3766kJ / kg â 2016 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ and then q = 3766kJ / kg − 3390kJ / kg = 376kJ / kg Three hundred kg of water are heated from 200C to 800C and at constant pressure Determine the enthalpy change and the heat Solution The enthalpy change is the heat so Q = m∆hn = mcP ( T2 − T1 ) and then = ( 300kg )( 4.18kJ / kg − K )( 80 − 20 K ) = 75, 240kJ Mercury is heated from 1000F to vapor at 8000F Assume hnfg is 122 Btu/lbm at 6760F and use a cP of 0.032 Btu/lbm ·0F for liquid mercury and 0.015 for the vapor Determine the enthalpy change per unit mass and the heat per unit mass Solution The enthalpy change will be the heat so that ∆hn = q = cP (Tsat − T1 ) + hn fg + cP.vapor (T2 − Tsat ) And then ∆hn = q = ( 0.032 Btu / lbm0 F )( 676 − 1000 F ) + 122 Btu / lbm + ( 0.015 )( 800 − 676 ) = 142.292 Btu / lbm A refrigerator condenser is a heat exchanger that converts a refrigerant from a vapor to a liquid at constant pressure and approximately constant temperature If ammonia is used as a refrigerant and 10 kg/s is condensed at 250C when the surrounding temperature is 200C, determine the heat transfer from the refrigerant using data from the Appendix Chart C-7 Solution The heat transfer is determined from the equation © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ i i i i Q = Hn = m ∆hn = m hn fg The enthalpies may be approximated from Chart C-7 or read from a thermodynamic table of saturation properties of ammonia hn fg = hng − hn f = 517kJ / kg − ( −650kJ / kg ) = 1167 kJ / kg and then i Q = 11, 670kJ / s = 11.67 MW Using the psychrometric chart Appendix Chart C-1, determine the vapor pressure of water in air at 800F, 40% relative humidity Also determine the humidity ratio, the enthalpy, the dew point temperature and the wet bulb temperature Solution For the air-water vapor mixture at 800F, 40% relative humidity, the dry bulb temperature is 800F Then, reading from the Chart C-1E pv ≈ 0.2 psia ϖ ≈ 60 grains / lbmdryair hn ≈ 29.2 Btu / lbmdryair Tdp ≈ 540 F Twb ≈ 640 F Ten m3/s of air at 00C, 80% relative humidity is heated to 250C Determine the amount of heat transfer required and the final relative humidity of the air Determine the amount of water required to increase the final relative humidity to 50% at 250C How much additional heat is then required to accomplish this humidification? Solution From the psychrometric Chart C-1 â 2016 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ ϖ = ϖ ≈ 3.2 g / kgdryair hn1 ≈ 10kJ / kgdryair hn2 ≈ 35kJ / kgdryair β ≈ 17% q = hn2 − hn1 ≈ 25kJ / kgdryair i i Q = qm The mass flow is just i i i m =V ρ =V / v And the volume flow rate, ܸሶ, is 10 m3/s The specific volume is read from Chart C-1 as about 0.775 m3/kg so that the heat transfer is ∆hn i ( 25kJ / kg ) (10m / s ) Q= V≈ = 322.6kJ / s v 0.775m3 / kg i At 50% relative humidity and 250C the humidity ratio is about 10 g/kg from Chart C-1 so that the amount of water needed to be added to the air is  i V mw =  v  i   (ϖ − ϖ ) =  10  (10 − 3.2 ) = 87.7 g / s   0.775   And the heat required for the humidification is Q = m ( hn3 − hn2 ) ≈ (12.9kg / s )( 51.5kJ / kg − 35kJ / kg ) = 213kW i i 10 Air at 1000F, 70% relative humidity is to be conditioned to 750F and 60% relative humidity Determine the partial pressure of the water vapor in the air at both states, the amount of water removed per lbm of dry air, and the lowest temperature to which the air must be cooled to accomplish this conditioning process Solution From the psychrometric Chart C-1E the following properties are read: © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ ω1 ≈ 205 grains / lbmdryair hn1 ≈ 56.2 Btu / lbmdryair ω2 ≈ 78 grains / lbmdryair pv1 ≈ 0.64 psia pv ≈ 0.26 psia The amount of water removed is the difference in humidity ratios, or 205 - 78 = 127 grains/lbm dry air, or 0.18 lbm/lbm dry air The lowest temperature to which the air must be cooled is the dew point temperature, namely about 610F 11 One hundred kg of dry air is humidified by mixing with kg steam at 300C Determine the partial pressure of the dry air and the steam if the total pressure is 100 kPa Solution The humidity ratio is ω = 1kg 100kg = 0.01kg = 0.622 pv pv = 0.622 pda p − pv Solving for the partial pressures, Pv = 1.6 kPa and pda= 98.4 kPa 12 For a real substance the enthalpy is function of temperature and pressure Write an integral equation that expresses the enthalpy change per unit mass of the substance when the temperature and pressure change from T1 to T2 and p1 to p2 Solution We have hn = f ( T , p ) dhn = ∂hn ∂hn dT + dp ∂T ∂p And T2 ∂hn dT + ∂ T T1 ∆hn = ∫ p2 ∂hn dp ∂ p p1 ∫ © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Section 1-4 13 A Styrofoam ice chest is cm thick If the inside of the chest is at 00C and outside the chest it is 250C, estimate the heat transfer by conduction through the Styrofoam per unit area Solution i qA = κ ∆T  W  =  0.029  ∆x  mi K  ( 250 C − 0.02 m ) = 36.25 mW Where the thermal conductivity for Styrofoam is read from Table B.2 14 A concrete nuclear reactor containment wall has a temperature distribution given by the equation T ( x ) = 800 F − 400 x 0 F ft where x is in feet Determine the heat transfer through the wall per unit area due to conduction at the center of the wall, where x = 0.5 ft Solution Using Fourier’s law of conduction i q A = −κ ∂T ∂ = −κ ( 8000 F − 400 x ) = −κ ( −800 x ) ∂x ∂x Reading the thermal conductivity for reinforced concrete from Table B-2E and at x = 0.5 ft we find i  Btu   F Btu q A =  0.925 400  = 370  hr ⋅ ft ⋅ F   ft  hr i ft  15 A large ft by ft thermopane glass window loses 300 Btu/hr of heat when the inside temperature is 700F If the thermopane has an average thermal conductivity of 0.032 Btu/hr·ft·0F and is ½ inch thick, estimate the outside temperature of the window Solution Using Fourier’s law of conduction we have © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ ∆T Btu  Btu  70 F − T0 = 300 =  0.032 32 ft ( )  ∆x hr  hr ⋅ ft ⋅0 F  1/ 24 ft T0 = 700 F − 12.20 F = 57.80 F i Q = −κ A 16 A cast iron frying pan ¼ inch thick is used to prepare some food on a stove top If the lower surface of the pan is at 6000F and the upper surface is 5000F, estimate the heat transfer through the pan per unit area Solution Using Fourier’s law of conduction and reading the value for thermal conductivity for cast iron from Table B.2E, i qA = κ ∆T  Btu   1000 F  Btu =  22.54  = 108,192  ∆x  hr ⋅ ft ⋅ F   1/ 48 ft  hr 17 Wind blows at 50 m/s around a cm diameter electric power line when the air temperature is -100C Estimate the heat loss of the power line per unit length if the surface temperature of the power line is 50C Solution Using Newton’s law of cooling and an approximate value from Table 1-4 for the convective heat transfer coefficient, i W  W  q l = hπ D∆T = 180  ( 3.14159 )( 0.05m ) (150 C ) = 424 m ⋅ C m  18 Water at 600F flows through a copper tube of inch inside diameter (ID) at aspproximately ft/s Determine the heat transfer to the water per foot of tube length if the inside surface temperature of the tube is 1800F Solution Using Newton’s law of cooling and an approximate value from Table 1-4 for the convective heat transfer coefficient, i   Btu Btu 1  q l = hπ D∆T =  616 ft  (1200 F ) = 19, 352.2  ( 3.14159 )  hr ⋅ ft ⋅ F  hr ⋅ ft  12   10 © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ 19 An automobile is moving on a highway at 100 km/hr Estimate the heat loss per unit area from the auto’s roof, which is at 300C, if the air temperature is 150C Solution Using Newton’s law of cooling and an approximate value from Table 1-4 for the convective heat transfer coefficient for 30 m/s, i W  W  q A = h∆T ≈  75  (150 C ) = 1125 m  m ⋅ C 20 Estimate the heat loss per unit area of a vertical south-facing wall of a large office building when the air temperature is -100F and the wall temperature is 50F Solution Using Newton’s law of cooling and an approximate value from Table 1-4 for the convective heat transfer coefficient for still air, i   Btu Btu q A = h∆T =  0.79  (15 F ) = 11.85 hr ⋅ ft ⋅ F  hr ⋅ ft  21 A well-clothed person walks into a large auditorium that is empty If the auditorium walls are at an average temperature of 550F and the average surface temperature of the person’s clothes is 850F, estimate the net radiation heat transfer between the person and the auditorium walls An average person can be assumed to have a surface area of 19.4 ft2 (1.8 m2) Solution Assuming black body radiation between the person and the auditorium walls, i   Btu Btu Q = A1σ (T14 − T24 ) = (19.4 ft )  0.174 x10−8 5454 R − 5154 R ) = 603.5 ( hr ⋅ ft ⋅ R  hr  22 A mercury-in-glass thermometer reads an outside temperature of 200C If the sky and surroundings of the thermometer have an average surface temperature of 50C, estimate the net radiation per unit area from or to the thermometer 11 © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Solution Assuming black body radiation i 4 − Tsky q A = σ (Ttherm ) ≈  4.67 x10−8 mW2 K  ( 2934 − 2784 ) = 79.2 mW2 The radiation is from the thermometer to the sky and surroundings 23 The surface of the sun seems to be about 10,0000F What would you guess the rate of heat emission from the sun to be for an area of the sun that measures ft2? Solution i   Btu Btu q A ≈ σ T =  0.174 x10−8 10, 4600 R ) = 20.9 x106 4 ( hr ⋅ ft ⋅ R  hr ⋅ ft  24 A radiation pyrometer is a device that uses radiant heat to measure the temperature of a surface Assume that a pyrometer has a surface area of cm2 and is at a temperature of 200C when directed towards a furnace opening having a temperature of 11000C Estimate the net rate of heat transfer towards the pyrometer if black body radiation is assumed Solution i W   Q = σ A (T14 − T24 ) =  5.67 x10 −8  ( 0.0005m )(13734 − 2934 K ) m ⋅K   i Q = 100.5W Section 1-5 25 Predict the concentration of ammonia in air mm from an interface of vapor ammonia and air if the interface area is 1500 mm2 and the evaporation rate is found to be 0.002 g/min Assume the ammonia and air are at 200C 12 â 2016 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Solution Assume the evaporation rate of ammonia occurs where the ammonia is a liquid and its concentration is 100% Using Table B-3 the density is 602 kg/m3 The diffusivity from Table 1-5 is about 0.236 cm2/s Then, using Fick’s law ݉ሶ = −ߩ‫ܣ‬ॊ஺஻ i m = 0.002 డట డ௫ g kg   cm   Ψ −   = − (1.5 x10−3 m2 )  602   0.236   m  s   5mm   Ψ = 0.99992 = 99.9992% 26 Liquid mercury is contained in a lead beaker as shown in Figure 1-14 Estimate the amount of mercury that migrates by diffusion into the beaker after 48 hours if the concentration of mercury is 2% at a distance 0.01 cm into the beaker wall from the inside surface Assume that the system is at 200C and neglect evaporation to air Solution Assume the mercury has 100% concentration at the surface of the beaker Then the density of diffusing mercury is 12,816 kg/m3 The approximate value for the diffusivity of mercury into lead is given in Table 1-5 Using Fick’s law ሶ Δ‫ܥ‬ = 13 â 2016 Cengage Learningđ May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ i  kg  m   1.0 − 0.02   m =  12,816  (π )( 0.1m )( 0.08m )  2.5 x10−19   m  s   0.01cm    i kg kg m = 7.89 x10−13 = 2.84 x10 −9 s hr In 48 hours, roughly the amount of mercury diffusing is 2.84 x 48 x 10-9 = 136 x 10-9 kg = 0.136 mg mercury 27 A ft radius spherical container holds helium gas at 20 psia and 950F Estimate the amount of helium lost through diffusion in 24 hours if the sphere is made of silicon dioxide, SiO2 (glass) and if it is assumed that at a point 1/8 inches into the container wall from the inside the concentration of helium is zero Solution The density of helium may be read from Table B-3E, 0.0101 lbm/ft3 at 800F The diffusivity is, from Table 1-5, 2.4 x 10-10 cm2/s = 0.372 x 10-10 in2/s = 0.258 x 10-12 ft2/s Assume the helium’s concentration is 1, 100% at the inside surface and zero at the outside Then, using Fick’s law ሶ Δ‫ܥ‬ ݉ = −ߩ‫ܣ‬ॊ஺஻ Δ‫ݔ‬ i  lbm  ft    2 m =  0.0101  ( 4π )( ft )  0.258 x10−12   ft  s   1/ 96 ft    i lbm lbm m = 50.297 x10−12 = 0.181x10−6 s hr 28 Water in a tightly closed flask evaporates at a rate of 0.00038 lbm/hr Estimate the specific humidity of the air inch above the liquid water-air surface if the dry bulb temperature is 700F The water-air surface is in2 Solution Assume the water has a density of 0.001579 lbm/ft3, using saturated vapor data from Table B-6E at 800F since Table B-6E does not list values for 700F From Table 1-5 the approximate diffusivity of water into air is 0.256 cm2/s = 0.000277 ft2/s Using Fick’s law ݉ሶ = −ߩ‫ܣ‬ॊ஺஻ Δ‫ܥ‬ Δ‫ݔ‬ 14 © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ i  lbm     ft   − Ψ  lbm m =  0.001579   ft   0.000277   = 0.00038 ft   144 s   1/ 12 ft  hr   lbmw mw mw Ψ = 0.031 = = lbmt mT mw + mda 0.031( mw + mda ) = mw 0.969mw = 0.031mda ω= mw 0.031 = = 0.032 mda 0.969 Section 1-6 29 Determine the number of boundary and initial conditions needed to solve the differential equation ∂ 2T (x, y, t) ∂ 2T ( x, y, t ) g   ∂T ( x, y, t ) + + =  κ α  ∂x ∂y ∂t Solution Four (4) boundary conditions need to be specified; for T=f(x) and for T=f(y) One initial or time condition for T=f(t) 30 Determine the number of boundary and initial conditions needed to solve the differential equation d 2T (r)   dT (r) +  =0 dr  r  dr Solution Two (2) boundary conditions needed 31 Determine the number of boundary and initial conditions needed to solve the differential equation  ∂T (x, y, t) ∂T ( x, y, t )  ∂ T(x, y, t) +v = κ  ∂x dy ∂y   ρ cP  u 15 © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Solution Three boundary conditions needed, for T(x) and for T(y) One initial or time condition needed for T(t) 32 Determine the number of boundary and initial conditions needed to solve the differential equation  ∂u (x, y) ∂v( x, y )  ∂p ∂ 2u (x, y) +v = − +  ∂x ∂y  ∂x ∂y  ρ u Solution Four (4) boundary conditions needed; two for u(x), one for v(y), and one for p(x) 16 © 2016 Cengage Learning® May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/ .. .Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ What is meant by reversible heat transfer? Reversible heat transfer is an... https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Solution Three boundary conditions needed, for T(x) and for T(y) One initial... https://TestbankDirect.eu/ Solution Manual for Heat and Mass Transfer 2nd Edition by Rolle Full file at https://TestbankDirect.eu/ Two kg of water are heated from 150C to 3000C at 101 kPa Determine the amount of heat