CHE2163-Heat and Mass Transfer 2013 Heat and Mass Transfer
Heat Exchangers
1. (Incropera 11.17) A concentric tube heat exchanger of length L=2 m is used to thermally process a pharmaceutical product flowing at a mean velocity of um,c=0.1 m s−1 with an inlet temperature of Tc,i = 20◦C. The inner tube of diam- eter Di=10 mm is thin walled, and the exterior of the outer tube (Do = 20 mm) is well insulated. Water flows in the annular region between the tubes at a mean velocity um.h = 0.2 m s−1 with and inlet temperature Th,i = 60◦C. Properties of the pharmaceutical product are ν = 10 × 10−6 m2 s−1, κ = 0.25 W m−1K−1, ρ = 1100 kg m−3 and Cp = 2460 J kg−1K−1. Evaluate water properties at T ̄h = 50◦C.
(a) Determine the value of the overall heat transfer coefficient U.
Copyright By PowCoder代写 加微信 powcoder
(b) Determine the mean outlet temperature of the pharmaceutical product when
the exchanger operates in the counterflow mode.
(c) Determine the mean outlet temperature of the pharmaceutical produce when the exchanger operates in the parallel-flow mode.
2. (Incropera 11.27) In a diary operation, milk at a flow rate of 25o L h−1 and a cow-body temperature of 38.6◦C must be chilled to a safet-to-store temperature of 13◦C or less. Ground water at 10◦C is available at a flow rate of 0.72 m3h−1. The density and specific heat of the milk are 1030 kg m−3 and 3860 J kg−1K−1 respectively.
(a) Determine the US product of a counterflow hear exchanger required for the chillin process. Determine the length of the exchanger if the inner pipe ha a 50-mm diameter and the overall heat transfer coefficient U = 1000 W m−2 K−1.
(b) Determine the outlet temperature of the water.
3. (Incropera 11.64) A concentric tube heat exchanger uses water, which is available at 15◦C, to cool ethylene glycol from 100 to 60◦C. The water and glycol flow rates are each 0.5 kg s−1. What are the maximum possible heat transfer rate and effectiveness of the exchanger? Which is preferred, a parallel-flow or counterflow mode of operation based on required area?
4. (Extension problem:Incropera 11.56) Saturated process steam at 1 atm is con- densed in a shell-and-tue heat exchanger (one-shell, two tube passes). Cooling water enters the tubes at 15◦C with an average velocity of 3.5 m s−1. The tubes are thin walled and made of copper with a diameter of 14 mm and length of 0.5 m. The convective heat transfer coefficient for condensation on the outer surface of the tubes is 21800 W m−2K−1.
(a) Find the number of tubes/pass required to condense 2.3 kg s−1 of steam.
(b) Find the outlet water temperature.
(c) Find the maximum possible condensation rate that could be achieved with this heat exchanger using the same water flow rate and inlet temperature.
1. a) 135 W m−2 K−1 (NuDi=5.96, NuDh=22.22, check entrance effects for internal tube), b) Tc,o = 33◦C (ε = 0.323, Cr = 0.11), c) Tc,o = 31.2◦C (ε = 0.322).
2. a) UA = 785 W K−1 (ε = 0.895,CR = 0.330,NTU = 2.842) b) Tc,o = 18.4◦C.
3. a) 1.13×105W, ε = 0.47, b) Tc,o = 40.4◦C, c)ACF/APF = 0.79 (NTUPF = 0.95,NTUCF = 0.75).
4. a) ≈ 189 (ReD = 50933,NuD = 286.5,U = 7900 W m−2 K−1,ε = 0.144,NTU = 0.155) , b) Tc,o = 27.2◦C , c) 16 kg s−1.
PROBLEM 11.17
KNOWN: Inlet temperatures of pharmaceutical product and water in a concentric tube heat
exchanger. Tube diameters and fluid velocities.
FIND: (a) Value of the overall heat transfer coefficient, U, (b) Mean outlet temperature of the pharmaceutical product for counterflow operation, (c) Mean outlet temperature of the pharmaceutical product for parallel-flow operation.
SCHEMATIC: Water
Th,i = 60°C Tc,i = 20°C
Th,i = 60°C
um,c = 0.1 m/s
Parallel flow
Counterflow L=2m
Di = 10 mm
um,h = 0.2 m/s
um,c = 0.1 m/s
Pharmaceutical product
um,h = 0.2 m/s
Tc,i = 20°C Pharmaceutical product
ASSUMPTIONS: (1) Negligible heat transfer between the heat exchanger and the surroundings, (2) Constant properties, (3) Negligible conduction resistance posed by the thin-walled inner tube, (4) Smooth tube surfaces.
PROPERTIES: Pharmaceutical product (given): ν = 10 × 10-6 m2/s, k = 0.25 W/m⋅K, ρ = 1100 kg/m3 and cp = 2460 J/kg⋅K. Table A.6, water (Th = 50°C): ν = 5.54 × 10-7 m2/s, k = 0.643 W/m⋅K, ρ = 987.9 kg/m3, Pr = 3.56 and cp = 4181 J/kg⋅K.
ANALYSIS: (a) The Prandtl number of the pharmaceutical product is Prc = ρcpν/k = (1100 kg/m3 × 2460 J/kg⋅K × 10 × 10-6 m2/s)/0.25 W/m⋅K = 108. The overall heat transfer coefficient is U = (1/hi + 1/ho)-1. For the flow in the inner tube, ReDi = um,cDi/ν = (0.1 m/s × 0.01 m)/10 × 10-6 m2/s = 100. The thermal entrance length is xfd,t = 0.05ReDiPrDi = 0.05ReDi(ρcpν/k)Di = 0.05 × 100 × (1100 kg/m3 × 2460 J/kg⋅K × 10 × 10-6 m2/s/0.25 W/m⋅K) × 0.01m = 5.41 m. Therefore, entrance effects are important. The Hausen correlation is appropriate for determining the average Nusselt number associated with the inner tube flow.
NuDi =3.66+ 0.0668(D/L)ReDiPr =3.66+ 0.0668×(0.01/2)×100×108 =5.96 1+0.04⎡(D/L)Re Pr⎤2/3 1+0.04⎡(0.01/2)×100×108⎤2/3
Therefore,
hi = NuDik = 5.96×0.25W/m⋅K =149W/m2 ⋅K Di 0.01m
Continued…
Problem 11.17 (Cont.)
For the annular flow, ReDh = um,h(Do – Di)/ν = 0.2 m/s × 0.01 m/5.54 × 10-7 m2/s = 3610. Hence, the Gnielinski correlation is appropriate for use. The friction factor for the annular region is obtained from Eq. 8.21 and is
Therefore,
f =(0.790lnReDh −1.64) =(0.790ln(3610)−1.64) =0.0428
ho = NuDhk = 22.22×0.643W/m⋅K =1430W/m2 ⋅K Do − Di 0.01m
(f /8)(ReDh −1000)Pr 1/2 ( 2/3 )
(0.0428/8)(3610−1000)3.56 1/2 ( 2/3 )
1+12.7( f /8) Pr −1
1+12.7(0.0428/8) 3.56 −1
The overall heat transfer coefficient is
U = (1/hi + 1/ho)-1 = [1/(1430 W/m2⋅K) + 1/(149 W/m2⋅K)]-1 = 135 W/m2⋅K. (b) The heat capacity rate of the cold (pharmaceutical product) stream is
⎡0.1m/s×1100kg/m3 ×π ×(0.01m)2 ⎤
C =m& c =⎡u ρπD2/4⎤c =⎢ ⎥×2460J/kg⋅K=21.3W/K
c cp ⎣m,c i ⎦p ⎢ 4 ⎥ ⎣⎦
The heat capacity rate of the hot (water) stream is C = m& c = ⎣ u ρ π D 2 − D 2 / 4 ⎦ c
c cp ⎡m,c (o i)⎤p
⎡0.2m/s×987.9 kg/m3 ×π ×⎡(0.02m)2 −(0.01m)2⎤⎤
⎦⎥×4181 J/kg⋅K =195 W/K ⎣⎦
Therefore, Cr = Cmin/Cmax = (21.3 W/K)/(195 W/K) = 0.11 and the number of transfer units is NTU = UA/Cmin = UπDiL/Cmin = (135 W/m2⋅K × π × 0.01 m × 2m)/21.3 W/K = 0.398. The effectiveness of the counterflow heat exchanger is obtained from Eq. 11.29a and is
ε = 1−exp[−NTU(1−Cr)] = 1−exp[−0.398(1−0.11)] =0.323 1−Cr exp[−NTU(1−Cr)] 1−0.11exp[−0.398(1−0.11)]
Continued…
T =T + εCmin(Th,i −Tc,i)=20°C+
Problem 11.17 (Cont.)
The heat transfer rate is
q=m&ccp(Tc,o −Tc,i)=εCmin(Th,i −Tc,i) fromwhich
0.323×21.3W/K×(60°C−20°C)
ρu πD2 /4 c 1100kg/m3 ×0.1m/s× π(0.01m)2 /4 2460J/kg⋅K
m,c(i)p () (c) For parallel-flow operation,
1−exp⎡−NTU(1+C )⎤ 1−exp⎡−0.398(1+0.11)⎤
ε= ⎣ r ⎦= ⎣ ⎦=0.322
1+Cr 1+0.11 Therefore, the cold stream outlet temperature is
T =T + εCmin(Th,i −Tc,i)=20°C+ 0.322×21.3W/K×(60°C−20°C)
ρu πD2 /4 c 1100kg/m3 ×0.1m/s× π(0.01m)2 /4 2460J/kg⋅K m,c(i)p ()
COMMENTS: There is little difference in the outlet temperature of the pharmaceutical product. However, if the outlet temperature of the pharmaceutical product cannot exceed some critical value, the heat exchanger should be operated in parallel-flow since the ultimate outlet temperature associated with a very long concentric tube apparatus would be determined by the conservation of energy principle.
PROBLEM 11.27
KNOWN: Cooling milk from a dairy operation to a safe-to-store temperature, Th,o ≤ 13°C, using
ground water in a counterflow concentric tube heat exchanger with a 50-mm diameter inner pipe and overall heat transfer coefficient of 1000 W/m2⋅K.
FIND: (a) The UA product required for the chilling process and the length L of the exchanger, (b) The outlet temperature of the ground water, and (c) the milk outlet temperatures for the cases when the water flow rate is halved and doubled, using the UA product found in part (a)
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat loss to surroundings, and (3) Constant properties.
PROPERTIES: Table A-6, Water eT = 287 K, assume T = 18oCj: ρ = 1000 kg / m3, c c,o
cp= 4187J/kg⋅K;Milk(given):ρ=1030kg/m3, cp=3860J/kg⋅K.
ANALYSIS: (a) Using the effectiveness-NTU method, determine the capacity rates and the minimum
Hot fluid, milk:
m&h =ρh∀h =1030kg/m ×250liter/h×10 m /liter×1h/3600s=0.0715kg/s
Ch =m&hch =0.0715kg/s×3860J/kg⋅K=276W/K
Cold fluid, water:
C c = m& c c c = 1 0 0 0 k g / m 3 × e 0 . 7 2 / 3 6 0 0 m 3 / s j × 4 1 8 7 J / k g ⋅ K = 8 3 7 W / K
It follows that Cmin = Ch. The effectiveness of the exchanger from Eq. 11.20 is ε= q = ChdTh,i−Th,oi=b38.6−13gK=0.895
qmax Cmin dT −T i b38.6-10gK h,i c,i
The NTU can be calculated from Eq. 11.29b, where Cr = Cmin/Cmax = 0.330, l Fε−lI
N T U = C r − 1 l n GH ε C r − 1 JK
l F0.895−lI
N T U = 0 . 3 3 0 − 1 l n GH 0 . 8 9 5 × 0 . 3 3 0 − l JK = 2 . 8 4 2
Continued …
PROBLEM 11.27 (Cont.)
From Eq. 11.24, find UA
UA =NTU⋅Cmin=2.842×276W/K=785W/K < and the exchanger tube length with A = π DL is
L= UA /πDU = 785W/K/π0.050m×1000W/m2⋅K=5.0m < (b) The water outlet temperature, Tc,o, can be calculated from the heat rates,
C dT −T i=C dT −T i (3) h h,i h,o c c,o c,i
276W/Kb38.6−13gK=837W/KdT −10iK c,o
T =18.4oC < c,o
(c) Using the foregoing Eqs. (1 - 3) in the IHT workspace, the hot fluid (milk) outlet temperatures are evaluated with UA = 785 W/K for different water flow rates. The results, including the hot fluid outlet temperatures, are compared to the base case, part (a).
1, halved flow rate Base, part (a)
2, doubled flow rate
Tc,o (°C) 14.9
Th,o (°C) 25.6
COMMENTS: (1) From the results table in part (c), note that if the water flow rate is halved, the milk will not be properly chilled, since Tc,o = 14.9°C > 13°C. Doubling the water flow rate reduces the outlet milk temperature by less than 1°C.
(2) From the results table, note that the water outlet temperature changes are substantially larger than those of the milk with changes in the water flow rate. Why is this so? What operational advantage is achieved using the heat exchanger under the present conditions?
(3) The water thermophysical properties were evaluated at the average cold fluid temperature,
T = dT + T i/ 2. We assumed an outlet temperature of 18°C, which as the results show, was a c c,i c,o
good choice. Because the water properties are not highly temperature dependent, it was acceptable to use the same values for the calculations of part (c). You could, of course, use the properties function in IHT that will automatically use the appropriate values.
(4) The value of the overall heat transfer coefficient, U, will change as the mass flow rate is varied. The answer to part (c) would be different if this variation is accounted for.
PROBLEM 11.64
KNOWN: Inlet temperature and flow rates for a concentric tube heat exchanger. Hot fluid outlet
temperature.
FIND: (a) Maximum possible heat transfer rate and effectiveness, (b) Preferred mode of operation. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state operation, (2) Negligible heat loss to surroundings, (3) Fixed overall heat transfer coefficient.
PROPERTIES: TableA-5,Ethyleneglycol(Tm =80°C):cp=2650J/kg⋅K;TableA-6,Water(Tm ≈ 30°C): cp = 4178 J/kg⋅K.
ANALYSIS: (a) Using the ε-NTU method, find
C m i n = C h = m& h c p , h = ( 0 . 5 k g / s ) ( 2 6 5 0 J / k g ⋅ K ) = 1 3 2 5 W / K .
Hence from Eqs. 11.18 and 11.6,
q = C ( T − T ) = (1 3 2 5 W / K ) (1 0 0 − 1 5 ) ° C = 1 . 1 3 × 1 0 5 W . max min h,i c,i
q=m&hcp,h(Th,i−Th,o)=0.5kg/s(2650J/kg⋅K)(100−60)°C=0.53×105W. < Hence from Eq. 11.19,
Since Tc,o < Th,o, a parallel flow mode of operation is possible. However, with (Cmin/Cmax) = ( m& h
cp,h/ m& c cp,c) = 0.63,
Fig. 11.10 → (NTU)PF ≈ 0.95 Fig. 11.11 → (NTU)CF ≈ 0.75.
Hence from Eq. 11.24
(ACF/APF)=(NTU)CF/(NTU)PF ≈(0.75/0.95)=0.79.
Because of the reduced size requirement, and hence capital investment, the counterflow mode of operation is preferred.
ε = q / qmax = 0.53×105 /1.13×105 = 0.47. (b) From Eq. 11.7,
T = T + q = 15°C + 0.53×105
m& c c p , c 0 . 5 k g / s × 4 1 7 8 J / k g ⋅ K
PROBLEM 11.56
KNOWN: Shell(1)-and-tube (two passes, p = 2) heat exchanger for condensing saturated steam at 1 atm. Inlet cooling water temperature and mean velocity. Thin-walled tube diameter and length prescribed, as well as, convective heat transfer coefficient on outer tube surface, ho.
FIND: (a) Number of tubes/pass, N, required to condense 2.3 kg/s of steam, (b) Outlet water temperature, Tc,o, (c) Maximum condensation rate possible for same water flowrate and inlet temperature, and (d) Compute and plot Tc,o and the condensation rate, m& h , for water mean velocity, um, in the range 1 ≤ um ≤ 5 ms/, using the heat transfer surface area found in part (a) assuming the shell-side convection coefficient remains unchanged.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible thermal resistance due to the tube walls.
PROPERTIES: Table A.6, Saturated steam (1 atm): Tsat = 100°C, hfg = 2257 kJ/kg; Water (assume Tc,o ≈ 25°C, Tm = (Th + Tc)/2 ≈ 295 K): ρ = 1/vf = 998 kg/m3, cc = cp,f = 4181 J/kg⋅K, μ = μf = 959 × 10-6 N⋅s/m2, k = kf = 0.606 W/m⋅K, Pr = Prf = 6.62.
ANALYSIS: (a) The heat transfer rate for the heat exchanger is
q = m& hhfg = 2.3kg s×2257×103 J kg = 5.191×106W (1)
Using the ε-NTU method, evaluate the following parameters: Water-side heat transfer coefficient:
ReD = umD = 3.5m s×0.014m =50,993 (2) μ ρ 959×10−6 N⋅s m2 998kg m3
h =kNu =k0.023Re0.8Pr0.4=0.606Wm⋅K×0.023(50,993)0.8(6.62)0.4=12,400W m2⋅K(3) iDDD D 0.014m
using the Dittus-Boelter equation for fully developed turbulent conditions.
Overall coefficient:
U=(1hi+1ho)−1=(112,400+121,800)−1=7900W m2⋅K (4) Effectiveness relations: With Cmin = Cc and m& c = ρ(πD2/4)umN,
q=εq =εC (T −T ) (5) max min h,i c,i
C =m& c =998kg m π×0.014 m 4 ×3.5m s×N×4181J kg⋅K=2248N (6) min cc 3( 22)
Continued...
q min( h,i m& h , m a x = m a x = C T
PROBLEM 11.56 (Cont.)
5.191×106W =ε×2248N(100−15)K
εN = 27.17 (7)
From Eq. 11.35a with Cr = 0, the effectiveness is
ε =1−exp(−NTU)=1−exp(−0.155)= 0.144 (8)
where, using As = πDLNP, NTU is evaluated as,
NTU= UAs =7900W m2⋅K(π×0.014m×0.5m)N×2=0.155
Cmin 2248 N Hence, using Eq. (7), the required number of tubes is
N = 27.17/ε = 205.8 ≈ 189 < and the total surface area is
As =πDLNP=π×0.014m×0.5m×189×2=8.31m2.
(b) The water outlet temperature with Cmin = 2248 N = 424,900 W/K,
T =T +q C =15oC+5.191×106W 424,900W K=27.2oC < c,o c,i min
(c) The maximum condensation rate will occur when q = qmax. Hence
c,i) 424,900W K(100−15)K
− T = 3 = 1 6 . 0 k g s . <
2257×10 J kg
(d) Using the IHT Heat Exchanger Tool, All Exchangers, Cr = 0, along with the Properties Tool for Water, the foregoing analysis was performed to obtain Th,o and m& h using the heat transfer surface area As = 8.31 m2 (part a) as a function of um.
35 30 25 20 15 10
Note that the condensation rate increases nearly linearly with the water mean velocity. The cold water outlet temperature decreases nearly linearly with um. We should expect this behavior from energy balance
considerations. Since hh is nearly two times greater than hc, U is controlled by the water side coefficient. Hence U will increase with increasing um.
COMMENTS: Note that the assumed value for Tm to evaluate water properties in part (a) was a good choice.
12345 Water mean velocity, um (m/s)
Outlet temperature, Tco (C) Condensation rate, mdoth*10 (kg/s)
Tco (C), mdoth*10 (kg/s)
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com