Fundamentals of heat and mass transfer frank p incropera david p dewitt solution manual ch10

PROBLEM 10.1 KNOWN: Water at atm with Ts – Tsat = 10°C FIND: Show that the Jakob number is much less than unity; what is the physical significance of the result; does result apply to other fluids? ASSUMPTIONS: (1) Boiling situation, Ts > Tsat PROPERTIES: Table A-5 and Table A-6, (1 atm): Water Ethylene glycol Mercury R-12 hfg (kJ/kg) cp,v (J/kg⋅K) Tsat (K) 2257 812 301 165 2029 2742* 135.5* 1015* 373 470 630 243 * Estimated based upon value at highest temperature cited in Table A-5 ANALYSIS: The Jakob number is the ratio of the maximum sensible energy absorbed by the vapor to the latent energy absorbed by the vapor during boiling That is, ( ) Ja = c p ∆T / h fg = cp,v ∆Te / h fg v For water with an excess temperature ∆Ts = Te - T∞ = 10°C, find Ja = ( 2029 J / k g ⋅ K ×10K ) /2257 ×103 J/kg Ja = 0.0090 Since Ja p o leads to Tl′ > Tsat,v and heat is transferred into the bubble causing evaporation with the formation of vapor Hence, the bubble begins to grow (c) Consider the specific conditions Tsat,v = 101°C and Tl = Tsat ( po ) = 100°C and calculate the radius of the bubble using the appropriate properties in Eq (2) rb = × 58.9 ×10 −3 rb = 0.032mm N  N  / (1.0502 −1.0133 ) bar × 10 /bar  m  m2  < Note the small bubble size This implies that nucleation sites of the same magnitude formed by pits and crevices are important in promoting the boiling process PROBLEM 10.4 KNOWN: Long wire, mm diameter, reaches a surface temperature of 126°C in water at atm while dissipating 3150 W/m FIND: (a) Boiling heat transfer coefficient and (b) Correlation coefficient, Cs,f, if nucleate boiling occurs SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate boiling PROPERTIES: Table A-6, Water (saturated, atm): Ts = 100°C, ρ l = 1/vf = 957.9 kg/m , ρ f = -6 1/vg = 0.5955 kg/m , cp,l = 4217 J/kgK, àl = 279 ì 10 Ns/m , Prl = 1.76, hfg = 2257 kJ/kg, σ = -3 58.9× 10 N/m ANALYSIS: (a) For the boiling process, the rate equation can be rewritten as h = q′′s / ( Ts − Tsat ) = h= q ′s / ( Ts − Tsat ) πD 3150W/m W / (126 − 100 ) °C = 1.00 × 106 /26°C = 38,600W/m2 ⋅ K π × 0.001m m < Note the heat flux is very close to q ′′max , and nucleate boiling does exist (b) For nucleate boiling, the Rohsenow correlation may be solved for Cs,f to give 1/3 1/6  g ( ρl − ρ v )   c p,l ∆Te   µ l h fg   Cs,f =      n ′′  q σ  s     h fg Prl   Assuming the liquid-surface combination is such that n = and substituting numerical values with ∆Te = Ts –Tsat , find 1/6 1/3  9.8 m ( 957.9 − 0.5955 ) kg    279 × 10−6 N ⋅ s / m × 2257 ×103 J/kg   s2 m3  Cs,f =   ×    1.00 × 106 W / m 58.9 × 10− N / m    4217J/kg ⋅ K × 26K     2257 ×103 J / k g ×1.76    Cs,f = 0.017   < COMMENTS: By comparison with the values of Cs,f for other water-surface combinations of Table 10.1, the Cs,f value for the wire is large, suggesting that its surface must be highly polished Note that the value of the boiling heat transfer coefficient is much larger than values common to single-phase convection PROBLEM 10.5 KNOWN: Nucleate pool boiling on a 10 mm-diameter tube maintained at ∆Te = 10°C in water at atm; tube is platinum-plated FIND: Heat transfer coefficient SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling PROPERTIES: Table A-6, Water (saturated, atm): Ts = 100°C, ρ l = 1/vf = 957.9 kg/m , ρ v = -6 1/vg = 0.5955 kg/m , cp,l = 4217 J/kg⋅K, µl = 279 × 10 N⋅s/m , Prl = 1.76, hfg = 2257 kJ/kg, σ = -3 58.9 × 10 N/m ANALYSIS: The heat transfer coefficient can be estimated using the Rohsenow nucleate-boiling correlation and the rate equation 1/2  µl hfg  g ( ρ l − ρ v )  q′′ h= s =   ∆ Te ∆ Te  σ      Cs,f h fg Pr n  l   cp,l ∆ Te From Table 10.1, find Cs,f = 0.013 and n = for the water-platinum surface combination Substituting numerical values, 1/2 279 ×10−6 N ⋅ s / m × 2257 ×103 J/kg  9.8m/s ( 957.9 − 0.5955 ) k g / m    h= − 10K   58.9 × 10 N / m   4217J/kg ⋅ K ×10K    0.013 × 2257 ×103 J / k g × 1.76    h = 13,690 W / m ⋅ K × < COMMENTS: For this liquid-surface combination, q ′′s = 0.137MW/m2 , which is in general agreement with the typical boiling curve of Fig 10.4 To a first approximation, the effect of the tube diameter is negligible PROBLEM 10.6 KNOWN: Water boiling on a mechanically polished stainless steel surface maintained at an excess temperature of 15°C; water is at atm FIND: Boiling heat transfer coefficient SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling occurs PROPERTIES: Table A-6, Saturated water (1 atm): Tsat = 100°C, ρ l = 957.9 kg/m , ρ v = 0.596 -6 -3 kg/m , cp,l = 4217 J/kgK, àl = 279 ì 10 Ns/m , Prl = 1.76, σ = 58.9 × 10 N/m, hfg = 2257 kJ/kg ANALYSIS: The heat transfer coefficient can be expressed as h = q ′′s / ∆ Te where the nucleate pool boiling heat flux can be estimated using the Rohsenow correlation 1/2  c ∆ Te  p,l  g ( ρl − ρ v )  q′′s = µ l hfg   σ      Cs,f h fg Pr n  l   From Table 10.1, find for this liquid-surface combination, Cs,f = 0.013 and n = 1, and substituting numerical values, 1/2  9.8m/s2 ( 957.9 − 0.596) k g / m3   q′′s = 279 ×10−6 N ⋅ s / m × 2257 ×103 J/kg    58.9 ×10−3 N / m  4217 J / k g ⋅ K× 15°C   0.013 × 2257kJ/kg ×1.76    × q′′s = 461.9kW/m Hence, the heat transfer coefficient is h = 461.9 ×103 W / m /15 °C = 30,790 W / m ⋅ K < COMMENTS: Note that this value of q ′′s for ∆Te = 15°C is consistent with the typical boiling curve, Fig 10.4 PROBLEM 10.7 KNOWN: Simple expression to account for the effect of pressure on the nucleate boiling convection coefficient in water FIND: Compare predictions of this expression with the Rohsenow correlation for specified ∆ Te and pressures (2 and bar) applied to a horizontal plate ASSUMPTIONS: (1) Steady-state conditions, (2) Nucleate pool boiling, (3) Cs,f = 0.013, n = PROPERTIES: Table A-6, Saturated water (2 bar): ρ l = 942.7 kg/m , cp,l = 4244.3 J/kg⋅K, µl = -6 -3 230.7 × 10 N⋅s/m , Prl = 1.43, hfg = 2203 kJ/kg, σ = 54.97 × 10 N/m, ρ v = 1.1082 kg/m ; Saturated -6 water (5 bar): ρ l = 914.7 kg/m , cp,l = 4316 J/kgK, àl = 179 ì 10 Ns/m , Prl = 1.13, hfg = -3 2107.8 kJ/kg, σ = 48.4 × 10 N/m, ρ v = 2.629 kg/m ANALYSIS: The simple expression by Jakob [51] accounting for pressure effects is n 0.4 h = C( ∆ Te ) ( p / pa ) (1) where p and pa are the system and standard atmospheric pressures For a horizontal plate, C = 5.56 and n = for the range 15 < q ′′s < k W / m2 For ∆Te = 10°C, p = bar h = 5.56 (10 ) ( 2bar/1.0133bar )0.4 = 7,298W/m ⋅ K, q ′′s = k W / m2 p = bar h = 5.56 (10 ) ( 5bar/1.0133bar )0.4 = 10,529W/m ⋅ K, q ′′s = 105kW/m2 < < where q ′′s = h∆T e The Rohsenow correlation, Eq 10.5, with Cs,f = 0.013 and n = 1, is of the form 1/2  cp, l ∆Te   g ( ρl − ρv )    q′′s = µ l h fg  (2)  n σ   C h Pr   s,f fg l   1/2 p = bar: q ′s = 230.7 × 10 −6 N⋅s m × 2203 ×10  9.8 m ( 942.7 − 1.1082) kg   J  s m   −3 kg  54.97 × 10 N / m       4244.3J/kg ⋅ K × 10K  ×  J  0.013 × 2203 × 10 × 1.431    kg q′′s = 232 k W / m p = bar: < < q′′s = 439 k W / m COMMENTS: For ease of comparison, the results with pa = 1.0133 bar are: ( Correlation/p (bar) Simple Rohsenow q′′s k W / m2 ) 56 73 135 232 105 439 Note that the range of q ′′s is within the limits of the Simple correlation The comparison is poor and therefore the correlation is not to be recommended By manipulation of the Rohsenow results, find that m the (p/po) dependence provides m ≈ 0.75, compared to the exponent of 0.4 in the Simple correlation PROBLEM 10.8 KNOWN: Diameter of copper pan Initial temperature of water and saturation temperature of boiling water Range of heat rates (1 ≤ q ≤ 100 kW) FIND: (a) Variation of pan temperature with heat rate for boiling water, (b) Pan temperature shortly after start of heating with q = kW SCHEMATIC: ASSUMPTIONS: (1) Conditions of part (a) correspond to steady nucleate boiling, (2) Surface of pan corresponds to polished copper, (3) Conditions of part (b) correspond to natural convection from a heated plate to an infinite quiescent medium, (4) Negligible heat loss to surroundings PROPERTIES: Table A-6, saturated water (Tsat = 100°C): ρ " = 957.9 kg / m3 , ρν = 0.60 kg / m3 , c p," = 4217 J / kg ⋅ K, µ" = 279 × 10 −6 N ⋅ s / m , Pr" = 1.76, h fg = 2.257 × 10 J / kg, σ = 0.0589 N / M -6 Table A-6, saturated water (assume Ts = 100°C, Tf = 60°C = 333 K): ρ = 983 kg/m , = 467 ì 10 -6 -1 -6 N⋅s/m , k = 0.654 W/m⋅K, Pr = 2.99, β = 523 × 10 K Hence, ν = 0.475 × 10 m /s, α = 0.159 × -6 10 m /s ANALYSIS: (a) From Eq (10.5), ∆ Te = Ts − Tsat = Cs,f h fg Pr n " cp," 1/  q /µ h A    s " fg s × 1/   g ( ρ" − ρν ) / σ   2 For n = 1.0, Cs,f = 0.013 and As = πD /4 = 0.0707 m , the following variation of Ts with qs is obtained S u tfa c e te m p e tu re (C ) 125 120 115 110 < 105 100 20000 40000 60000 80000 100000 H e a t te (W ) As indicated by the correlation, the surface temperature increases as the cube root of the heat rate, permitting large increases in q for modest changes in Ts For q = kW, Ts = 104.7°C, which is barely sufficient to sustain boiling 11 (b) Assuming 10 < RaL < 10 , the convection coefficient may be obtained from Eq (9.31) Hence, with L = As/P = D/4 = 0.075m, Continued … PROBLEM 10.8 (Cont.) 1/  9.8 m / s × 523 × 10−6 K −1 ( T − T )( 0.075m )3  k  0.654 W / m ⋅ K  s i   = h =   0.15 Ra1/ 0.15   L − 12 0.075m L     0.475 × 0.159 × 10 m / s2 ( = 1.308 2.86 × 107 ) 1/ (Ts − Ti )1/ = 400 (Ts − Ti )1/ With As = πD /4 = 0.0707 m , the heat rate is then ( ) q = hAs ( Ts − Ti ) = 400 W / m ⋅ K / 0.0707 m ( Ts − Ti ) 4/3 With q = 8000 W, < Ts = Ti + 69°C = 89°C COMMENTS: (1) With (Ts – Ti) = 69°C, RaL = 1.97 × 10 , which is within the assumed Rayleigh number range (2) The surface temperature increases as the temperature of the water increases, and bubbles may nucleate when it exceeds 100°C However, while the water temperature remains below the saturation temperature, the bubbles will collapse in the subcooled liquid PROBLEM 10.63 KNOWN: Copper sphere of 10 mm diameter, initially at 50°C, is placed in a large container filled with saturated steam at atm FIND: Time required for sphere to reach equilibrium and the condensate formed during this period SCHEMATIC: ASSUMPTIONS: (1) Laminar film condensation, (2) Negligible non-condensibles in vapor, (3) Sphere is spacewise isothermal, (4) Sphere experiences heat gain by condensation only PROPERTIES: Table A-6, Saturated water vapor (1 atm): Tsat = 100°C, ρ v = 0.596 kg/m , hfg = 2257 kJ/kg; Table A-6, Water, liquid (Tf ≈ (75 + 100)°C/2 = 360K): ρl = 967.1 kg/m , c p, l = 4203 -6 J/kgK, àl = 324 ì 10 N⋅s/m , kl = 0.674 W/m⋅K; Table A-1, Copper, pure ( T = 75° C) : ρ sp = 8933 kg/m , cp,sp = 389 J/kg⋅K ANALYSIS: Using the lumped capacitance approach, an energy balance on the sphere provides, E& in − E& out = E& st & h ′fg = h D A s ( Tsat − Ts ) = ρsp c p,sp Vs m dTs dt (1) Properties of the sphere, ρ sp and cp ,sp , Will be evaluated at Ts = ( 50 +100 ) ° C / = 75° C, while water (liquid) properties will be evaluated at Tf = ( Ts + Tsat ) / = 87.5 °C ≈ 360K From Eq 10.26 with Ja = c p, l ∆T / h fg where ∆T = Tsat - Ts , find 1 + 0.68 4203 J × 100 − 75 K /2 × 10 J / k g   = 2328 kJ (2) ( )     kg  kg ⋅ K kg kJ h ′fg = h fg (1 + 0.68Ja ) = 2257 To estimate the time required to reach equilibrium, we need to integrate Eq (1) with appropriate limits However, to perform the integration, an appropriate relation for the temperature dependence of hD needs to be found Using Eq 10.40 with C = 0.815, 1/4  g ρ ( ρ − ρ ) k3 h′  l l v l fg  hD = 0.815   µ l ( Tsat − Ts ) D    Substitute numerical values and find, 1/4  m / s × 967.1kg/m ( 967.1 − 0.596 ) k g / m ( W / m ⋅ K )3 × 2328 × 10 J / k g  h D = 0.815   −6 324 × 10 N ⋅ s / m ( Tsat − Ts ) × 0.010m   −1/4 hD = B ( Tsat − Ts ) where B = 30,707W/m ⋅ ( K ) 3/4 (3) Continued … PROBLEM 10.63 (Cont.) Substitute Eq (3) into Eq (1) for hD and recognize Vs / A s = π D / π D = D / , dT B ( Tsat − Ts )−1/4 ( Tsat − Ts ) = ρsp c p,sp ( D / ) s dt (4) Note that d(Ts) = - d(Tsat – Ts); letting ∆T ≡ Tsat – Ts and separating variables, the energy balance relation has the form ρsp c p,sp ( D / ) ∆T d ( ∆T ) t dt = − (5) ∆To ∆T3 / B where the limits of integration have been identified, with ∆ To = Tsat − Ti and Ti = Ts(0) Performing the integration, find ∫ t=− ∫ ρsp cp,sp ( D / 6) B ⋅  1/4  ∆T − ∆ T1/4 o    1− / Substituting numerical values with the limits, ∆T = and ∆To = 100-50 = 50°C, t=− 8933kg/m3 × 389J/kg ⋅ K ( 0.010m/6) 30,707 W / m ⋅ K / ×  01/4 − 501 /  K1/4   < t = 2.0s To determine the total amount of condensate formed during this period, perform an energy balance on a time interval basis, E in − E out = ∆E = E final − E initial Ein = ρsp cp,sp V ( Tfinal − Tinitial ] (6) where Tfinal = Tsat and Tinitial = Ti = Ts(0) Recognize that E in = M h ′fg (7) where M is the total mass of vapor that condenses Combining Eqs (6) and (7), M= M= ρsp cp,sp V h ′fg [Tsat − Ti ] 8933kg/m3 × 389J/kg ⋅ K (π / 6)( 0.010m )3 2328 × 103 J/kg [100 − 50 ] K M = 3.91 ×10 −5 kg < COMMENTS: The total amount of condensate could have been evaluated from the integral, t t q t h D A s ( Tsat − Ts ) dt dt = h ′fg h ′fg & =∫ M = ∫ mdt ∫ giving the same result, but with more effort PROBLEM 10.64 KNOWN: Saturated steam condensing on the inside of a horizontal pipe FIND: Heat transfer coefficient and the condensation rate per unit length of the pipe SCHEMATIC: ASSUMPTIONS: (1) Film condensation with low vapor velocities PROPERTIES: Table A-6, Saturated water vapor (1.5 bar): Tsat ≈ 385K, ρ v = 0.88 kg/m , hfg = 2225 kJ/kg; Table A-6, Saturated water (Tf = (Tsat + Ts)/2 ≈ 380K): ρl = 953.3 kg/m , c p, l = 4226 -6 J/kgK, àl = 260 ì 10 Ns/m , kl = 0.683 W/m⋅K ANALYSIS: The condensation rate per unit length follows from Eq 10.33 with A = π D L and has the form &′= m & m = h D (π D )( Tsat − Ts ) / h ′fg L where hD is estimated from the correlation of Eq 10.42 with Eq 10.43, 1/4 g ρ ( ρ − ρ )k h ′  l l v l fg  hD = 0.555   µ l ( Tsat − Ts ) D    where J J h ′fg = h fg + cp,l ( Tsat − Ts ) = 2225 ×103 + × 4226 ( 385 − 373 ) K kg kg ⋅ K h ′fg = 2244kJ/kg Hence, 1/4 kg kg    9.8m/s × 953.3 ( 953.3 − 0.88 ) ( 0.683W/m ⋅ K ) 2244 ×10 J/kg  m m hD = 0.555   −   260 × 10 N ⋅ s / m ( 385 − 373) K × 0.075m   hD = 7127W/m ⋅ K It follows that the condensate rate per unit length of the tube is & ′ = 7127W/m ⋅ K (π × 0.075m )(385 − 373)K / 2225 ×103 J / k g = 9.06 ×10−3 kg / s ⋅ m m < PROBLEM 10.65 KNOWN: Horizontal pipe passing through an air space with prescribed temperature and relative humidity FIND: Water condensation rate per unit length of pipe SCHEMATIC: ASSUMPTIONS: (1) Drop-wise condensation, (2) Copper tube approximates well promoted surface PROPERTIES: Table A-6, Water vapor (T∞ = 37°C = 310K): pA,sat = 0.06221 bar; Table A-6, Water vapor (pA = φ⋅pa,sat = 0.04666 bar): Tsat = 305K = 32°C, hfg = 2426 kJ/kg; Table A-6, Water, liquid (Tf = (Ts + Tsat )/2 = 297K): c p, l = 4180 J / kg ⋅ K ANALYSIS: From Eq 10.33, the condensate rate per unit length is &′= m q ′ h L (π D )(Tsat − Ts ) = h′fg h′fg where from Eq 10.26, with Ja = c p, l ( Tsat − T s ) / h fg , h ′fg = h fg [1 + 0.68Ja ] = 2426 kJ 1 + 0.68 × 4180J/kg ⋅ K ( 305 − 288 ) K/2426k J/kg  kg  h ′fg = 2474kJ/kg Note that Tsat is the saturation temperature of the water vapor in air at 37°C having a relative humidity, φ = 0.75 That is, Tsat = 305K and Ts = 15°C + 288K For drop-wise condensation, the correlation of Eq 10.44 yields hdc = 51,104 + 2044Tsat 22° C < Tsat < 100°C where the units of h dc and Tsat are W/m ⋅K and °C hdc = 51,104 + 2044 ( 32° C) = 116,510 W / m2 ⋅ K Hence, the condensation rate is & ′ = 116,510 W / m ⋅ K ( π × 0.025m )( 305 − 288 ) K/2474× 10 J/kg m & ′ = 6.288 ×10−2 kg/s ⋅ m m < COMMENTS: From the result of Problem 10.54 assuming laminar film condensation, the condensation rate was m& ′film = 4.28 ×10 −3 k g / s ⋅ m which is an order of magnitude less than for the rate assuming drop-wise condensation PROBLEM 10.66 KNOWN: Beverage can at 5°C is placed in a room with ambient air temperature of 32°C and relative humidity of 75% FIND: The condensate rate for (a) drop-wise and (b) film condensation SCHEMATIC: ASSUMPTIONS: (1) Condensation on top and bottom surface of can neglected, (2) Negligible noncondensibles in water vapor-air, and (b) For film condensation, film thickness is small compared to diameter of can PROPERTIES: Table A-6, Water vapor (T∞ = 32°C = 305 K): pA,sat = 0.04712 bar; Water vapor (pA = φ⋅pA,sat = 0.03534 bar): Tsat ≈ 300 K = 27°C, hfg = 2438 kJ/kg; Water, liquid (Tf = (Ts + Tsat)/2 = 289 K): c p," = 4185 J/kg⋅K ANALYSIS: From Eq 10.33, the condensate rate is  = m h (π DL )( Tsat − Ts ) q = h ′fg h ′fg where from Eq 10.26, with Ja = c p," (Tsat – Ts)/hfg, h ′fg = h fg [1 + 0.68 Ja ] h ′fg = 2438 kJ / kg 1 + 0.68 × 4185 J / kg ⋅ K (300 − 278 ) K / 2438 kJ / kg  h ′fg = 2501 kJ / kg Note that Tsat is the saturation temperature of the water vapor in air at 32°C having a relative humidity of φ∞ = 0.75 (a) For drop-wise condensation, the correlation of Eq 10.44 with Tsat = 300 K = 27°C yields h = h dc = 51,104 + 2044 Tsat 22°C < Tsat ≤ 100°C where the units of h dc are W/m ⋅K and Tsat are °C, h dc = 51,104 + 2044 × 27 = 106, 292 W / m ⋅ K Hence, the condensation rate is  = 1.063 × 105 W / m ⋅ K (π × 0.065 m × 0.125 m )( 27 − ) K / 2501 kJ / kg m <  = 0.0229 kg / s m Continued … PROBLEM 10.66 (Cont.) (b) For film condensation, we used the IHT tool Correlations, Film Condensation, which is based upon Eqs 10.37, 10.38 or 10.39 depending upon the flow regime The code is shown in the Comments section, and the results are Reδ = 24, flow is laminar <  = 0.00136 kg / s m Note that the film condensation rate estimate is nearly 20 times less than for drop-wise condensation COMMENTS: The IHT code identified in part (b) follows: /* Results, NuLbar 0.5093 Part (b) - input variables and rate parameters Redelta hLbar mdot D L Ts 24.05 6063 0.001362 0.065 0.125 278 Tsat 300 */ /* Thermophysical properties evaluated at Tf; hfg at Tsat Prl Tf cpl h'fg hfg kl mul nul 7.81 289 4185 2.501E6 2.438E6 0.5964 0.001109 1.11E-6*/ // Other input variables required in the correlation L = 0.125 b = pi * D D = 0.065 /* Correlation description: Film condensation (FCO) on a vertical plate (VP) If Redelta