As the humid air flows through a device, e. g. heat exchanger, the flow of the dry air remains ...... flux is q = 6700 W/m2 and condensation rate is h mâ²â²x ...
1 HUMID AIR HEAT EXCHANGERS Markku J. Lampinen 1986. Translated from Finnish by Voitto W. Kotiaho.
1 PROPERTIES OF HUMID AIR
1.1 Thermodynamic properties Humid air is a mixture of dry air (nitrogen, oxygen, carbon dioxide and noble gases) and water vapor. The molar mass of dry air is Mi = 0,0290 kg/mol and of water vapor (water) Mh = 0,0180 kg/mol. According to the ideal gas equation, the partial density of dry air is i
pi Mi RT
(1)
where pi is the partial pressure of the dry air and the gas constant R = 8,314 J/mol K. Correspondingly, the partial pressure of the water vapor is h
ph Mh RT
(2)
where ph is the partial pressure of the water vapor. The density of the humid air is the sum of the densities of the dry air and water vapor = i + h
(3)
and the total pressure is the sum of the pressures of the dry air and water vapor p = p i + ph .
(4)
The humidity of air, that is, ratio of water vapor and dry air, is defined as x
h , i
(5)
which, by using equations (1), (2) and (3) can be expressed as x
p p Mh ph 0,6220 h 0,6220 h . Mi pi pi p ph
(6)
By solving ph we obtain ph
x p. 0,6220 x
(7)
The enthalpy of the humid air is the sum of the enthalpies of the dry air and the water vapor h = i hi + h hh
(8)
2 and, correspondingly, the specific heat cp = i cpi + h cph .
(9)
As the humid air flows through a device, e. g. heat exchanger, the flow of the dry air remains constant but the humidity of air may change, for example, because of condensation. Because of this, quantity hk, which is the enthalpy of the humid air per mass of dry air, has been taken into usage. It is defined as hk
h , i
(10)
which, by using equations (8) and (5) can be expressed as hk = hi + x hh .
(11)
i x with the dry airflow m i. In a humid airflow, there is a flow of water vapor m ihk . Correspondingly, the enthalpy flow of the humid air is m
As liquid water at 0oC is chosen to be the reference point of the enthalpy, the enthalpies of the dry air and the water vapor can be expressed approximately as hi = cpi t 1006 t, J/kg hh = 0 + cph t 2,50 106 + 1850 t , J/kg and it follows that hk = 1006 t + x (2,50 106 + 1850 t) , J/kgdry air.
(12)
where t is the temperature of the humid air in Celsius (oC) degrees. Each temperature T, which is lower than the boiling point at total pressure p, is corresponded by some partial pressure of the water vapor, so called saturation pressure ph. The partial pressure of the water vapor in an air-water vapor mixture cannot be bigger than ph at temperature T. In case that the partial pressure of the water vapor is equal to the saturation pressure, condensation of water takes place. The saturation pressure depends, with very high accuracy, only on the temperature and very little on the total pressure, hence p'h = p'h(T). The saturation pressure can be taken most accurately from the steam tables or can be calculated approximately from the following formula: T 372,79 p h (T) p 0 exp11,78 , T 43,15
where p0 = 105 Pa and temperature T is in Kelvin (K) degrees.
(13)
3 1.2 Kinetic properties Heat- and mass transfer depend on the thermal conductivity and the Kinematic viscosity of the humid air, and on the Diffusion coefficient D between the water vapor and the dry air. These properties depend on temperature T, pressure p and humidity x in the following way (D does not depend on the humidity): = (T, p, x) (14) = (T, p, x) (15) D = D(T, p). (16) There are empirical formulas for (14) and (15) in the literature (e. g. Tekniikan käsikirja, part 1, pages 367-369 and Tekniikan käsikirja, part 2, pages 746-758). The Diffusion coefficient D can be obtained approximately from the following formula Da
T1,75 , p
(17)
where a = 1,190 10-4 m2/s, p is the total pressure in Pa and T is the temperature in K. Instead of formulas, also tables can be used. On the following pages there are thermodynamic properties of saturated air at different temperatures (Table 1). In practical calculations properties of saturated air at saturation temperature are used as properties of the humid air. Example 1
Air humidity x = 0,120. The pressure of the corresponding water vapor is obtained from equation (7) 0,120 ph 105 1,617 104 Pa 0,6220 0,120 The saturation temperature of the air is 55,6oC and the nearest value in the table is 56oC. Properties of the saturated air corresponding to this temperature from table 1: = 0,02526 W/mK = 18,59 10-6 m2/s D = 32.4 10-6 m2/s. Partial densities for the saturated air at 56o can be taken from the same table i = 0,900 kg/m3 h = 0,109 kg/m3.
4 Table 1. Thermodynamic properties of saturated air at a total pressure of 100 kPa. Tempera-
ture C 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100
Humidity
kgH2O / kgdry air 0.003821 0.004418 0.005100 0.005868 0.006749 0.007733 0.008849 0.010105 0.011513 0.013108 0.014895 0.016892 0.019131 0.021635 0.024435 0.027558 0.031050 0.034950 0.039289 0.044136 0.049532 0.055560 0.062278 0.069778 0.078146 0.087516 0.098018 0.10976 0.12297 0.13790 0.15472 0.17380 0.19541 0.22021 0.24866 0.28154 0.31966 0.36468 0.41790 0.48048 0.55931 0.65573 0.77781 0.93768 1.15244 1.45873 1.92718 2.73170 4.42670 10.30306
Partial pressure of water vapor kPa 0.6108 0.7054 0.8129 0.9346 1.0721 1.2271 1.4015 1.5974 1.8168 2.062 2.337 2.642 2.982 3.360 3.778 4.241 4.753 5.318 5.940 6.624 7.375 8.198 9.010 10.085 11.161 12.335 13.613 15.002 16.509 18.146 19.92 21.84 23.91 26.14 28.55 31.16 33.96 36.96 40.19 43.65 47.36 51.33 55.57 60.5 64.95 70.11 75.61 81.46 87.69 94.3 101.325
Partial density of water vapor
Partial density of dry air
Specific enthalpy of the mixture
Specific heat of the mixture
kgH2O / m3
kgdry air/m3
kJ/kgdry air
J/kgK
0.004846 0.005557 0.006358 0.007257 0.008267 0.009396 0.01066 0.01206 0.01363 0.01536 0.01729 0.01942 0.02177 0.02437 0.02723 0.03036 0.03380 0.03758 0.04171 0.04622 0.05114 0.05650 0.06233 0.06867 0.07553 0.08298 0.09103 0.09974 0.1091 0.1193 0.1302 0.1419 0.1545 0.1680 0.1826 0.1981 0.2146 0.2324 0.2514 0.2717 0.2933 0.3162 0.3406 0.3666 0.3942 0.4235 0.4545 0.4873 0.5221 0.5588 0.5977
1.285 1.275 1.264 1.254 1.243 1.232 1.221 1.211 1.199 1.188 1.177 1.165 1.154 1.141 1.129 1.116 1.103 1.090 1.076 1.061 1.046 1.030 1.014 0.997 0.9791 0.9606 0.9411 0.9207 0.8999 0.8768 0.8532 0.8283 0.8021 0.7746 0.7456 0.715 0.6829 0.6489 0.6132 0.5755 0.5358 0.4939 0.4497 0.4031 0.3542 0.3026 0.2482 0.1909 0.1305 0.06694 0.000
9.55 13.06 16.39 20.77 25.00 29.52 34.37 39.57 45.18 51.29 57.86 65.02 72.60 81.22 90.48 100.57 111.58 123.72 136.99 151.60 167.64 185.40 204.94 226.55 250.45 277.04 306.64 339.51 373.31 417.72 464.11 516.57 575.77 643.51 721.01 810.36 915.57 1035.6 1179.42 1348.40 1560.80 1820.46 2148.92 2578.73 3155.67 3978.42 5236.61 7395.49 11944.39 27711.34
1010.8 1012.0 1013.4 1014.9 1016.7 1018.6 1020.8 1023.3 1026.0 1029.1 1032.5 1036.4 1040.7 1045.5 1050.9 1056.9 1063.5 1071.0 1079.3 1088.5 1098.9 1110.3 1123.2 1137.5 1153.4 1171.3 1191.3 1213.7 1238.9 1267.3 1299.4 1335.7 1377.0 1424.1 1478.2 1541.8 1613.2 1698.6 1799.4 1919.9 2066.4 2247.7 2476.7 2773.9 3170.8 3730.4 4574 5987 8820 17338
5 Table 1 (cont.). Prandtl number (Pr) and Schmidt number (Sc) calculated from the values in this table. Temperature
Heat of evaporation for water
C
Kinematic viscosity
10-6 m2/s 2500.8 2495.9 2491.3 2486.6 2481.9 2477.2 2472.5 2467.8 2463.1 2458.4 2453.1 2449.0 2442.0 2439.5 2434.8 2430.0 2425.3 2420.5 2415.8 2411.0 2406.2 2401.4 2396.6 2391.8 2387.0 2382.1 2377.3 2372.4 2367.6 2362.7 2357.9 2353.0 2348.1 2343.1 2338.2 2333.3 2328.3 2323.3 2318.3 2313.3 2308.3 2303.2 2298.1 2293.0 2287.9 2282.8 2277.6 2272.4 2267.1 2261.9 2256.7
Thermal conductivity
Pr
Sc
0.726 0.721 0.722 0.723 0.724 0.726 0.728 0.730 0.732 0.734 0.738 0.741 0.746 0.750 0.755 0.760 0.766 0.773 0.781 0.788 0.798 0.808 0.819 0.832 0.846 0.862 0.879 0.898 0.920 0.944 0.970 1.000 1.033 1.070 1.113 1.161 1.213 1.276 1.347 1.430 1.527 1.642 1.783 1.958 2.184 2.490 2.933 3.650 5.057 9.219
0.597 0.600 0.602 0.605 0.605 0.607 0.608 0.608 0.608 0.608 0.608 0.606 0.606 0.604 0.602 0.603 0.601 0.599 0.597 0.594 0.592 0.591 0.588 0.586 0.583 0.582 0.579 0.576 0.574 0.571 0.569 0.564 0.562 0.558 0.554 0.549 0.545 0.539 0.534 0.528 0.520 0.511 0.504 0.493 0.481 0.470 0.455 0.440 0.422 0.403 0.380
W/mK 10-6 m2/s
kJ/kg 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100
Diffusion coefficient air-water
13.25 13.43 13.61 13.79 13.97 14.15 14.34 14.52 14.71 14.89 15.08 15.27 15.46 15.65 15.84 16.03 16.22 16.41 16.61 16.80 17.00 17.20 17.39 17.59 17.79 17.99 18.19 18.39 18.59 18.79 18.99 19.19 19.38 19.57 19.76 19.94 20.10 20.28 20.44 20.58 20.71 20.81 20.90 20.96 20.99 20.99 20.94 20.84 20.69 20.47 20.08
22.2 22.4 22.6 22.8 23.1 23.3 23.6 23.9 24.2 24.5 24.8 25.2 25.5 25.9 26.3 26.6 27.0 27.4 27.8 28.3 28.7 29.1 29.6 30.0 30.5 30.9 31.4 31.9 32.4 32.9 33.4 34.0 34.5 35.1 35.7 36.3 36.9 37.6 38.3 39.0 39.8 40.7 41.5 42.5 43.6 44.7 46.0 47.4 49.0 50.8 52.8
0.02380 0.02413 0.02427 0.02440 0.02454 0.02466 0.02478 0.02490 0.02500 0.02511 0.02520 0.02529 0.02537 0.02544 0.02548 0.02556 0.02561 0.02565 0.02567 0.02569 0.02569 0.02568 0.02566 0.02563 0.02558 0.02552 0.02545 0.02536 0.02526 0.02514 0.02501 0.02487 0.02471 0.02455 0.02437 0.02418 0.02399 0.02379 0.02360 0.02341 0.02323 0.02307 0.02294 0.02285 0.02281 0.02283 0.02295 0.02318 0.02355 0.02409 0.02486
6 2 PRINCIPLES OF HEAT EXCHANGER DESIGN
2.1 Analogy between heat transfer- and mass transfer coefficients By studying the differential equations of the laminar flow, the following correspondence between the heat transfer- and mass transfer coefficients can be derived: Nu = G(Re, Pr) Sh = G(Re, Sc),
(18) (19)
where the dimensionless numbers have been defined as follows: vL , Reynolds number cp , Prandtl number Pr Sc , Schmidt number D L , Nusselt number Nu kL , Sherwood number Sh D Re
The heat transfer coefficient has been denoted by , and corresponding mass transfer coefficient by k. The analogy between these quantities means that the function G( . , . ) is the same in equations (18) and (19). If there is a model for the heat transfer coefficient, of the form (18), a model for mass transfer coefficient, based on that model, can be derived, by replacing Prandtl number by Schmidt number and Nusselt number by Sherwood number. The analogy between equations (18) and (19) requires (on the boundary value) that there is no convective flow on the surface. This requirement is not, however, fulfilled as water vapor condensation takes place on the surface. In that case there is a convective flow perpendicular to the surface, called Stefan-flow. Let us have a closer look at this flow and the reasons for it. Let us denote the velocity component of the humid air normal to the surface by u. With the velocities of the water vapor and dry air, u can be expressed as u = huh + iui .
(20)
Velocity u is the velocity of the center of gravity and it is a macroscopic quantity which indicates the convective flow. u is the velocity which can be measured for example by the Pitot-tube and which is called as Stefan-flow in this context. The magnitude of term huh on the surface indicates the density of the water vapor-flow condensing on the surface (kg/m2s) and it is not zero. How about the term iui? The movement of the dry air is away from the surface and its magnitude corresponds the air displaced in the condensation phenomenon where the water vapor becomes liquid, hence ui
h u h , v
(21)
7 where is the density of the liquid water. The minus mark in equation (21) is because the direction of ui is opposite to uh. Example 2 Water vapor condensation density huh = 1,9 10-3 kg/m2s. Temperature 56oC. On the basis of example 1 we get h = 0,109 kg/m3 and i = 0,900 kg/m3. uh ui
1,9 103 kg/m 2 s 0,109 kg/m
1,743 102 m/s
3
1,9 103 kg/m 2 s 3
1,9 106 m/s
1000kg/m = 0,109 kg/m3 + 0,900 kg/m3 = 1,009 kg/m3 u
0,109 1,743 102 0,900 1,9 106 m/s 1,885 103 m/s 1,009
As can be seen in this example, it holds that ui 0, which means that the surface is a so called semi-transparent surface. This means that the Diffusion coefficient of the water vapor is not directly proportional to the pressure difference but the dependence is more complicated. One study gives the following formula: h u h M h
p p p h (Ts ) k ln RT p ph
(22)
where ph (Ts ) is the pressure on the surface of the condensate. The derivation of formula (22) has been shown in the Mass Transfer course as well as proof for formulas (18) and (19). Example 3 A heat transfer model is Nu = A Rem Prn, that is, function G( . , .) has been given as G(Re, Pr) = A Rem Prn. On the basis of formula (19) it is in this case so that Sh = A Rem Scn. This can be developed forward: n
n
Sc Sc Sh A Re m Pr n Nu , and it follows that Pr Pr n
k L L / D , and this can be reduced to D c p / k Le1n , cp
where Le
D cp
Pr is so called Lewis number. Sc
(23)
8 2.2 Calculating surface temperature Let us consider a certain spot in the exchanger, where the humid airflow is at temperature T and at humidity x and the liquid or gas on the other side of the heat transfer wall is at temperature Tv. Let us denote the surface temperature of the condensate at this particular spot by Ts. The heat flux through the inner surface As and the corresponding outer surface Au can be expressed as follows, as the heat resistance of the condensate layer is neglected h A u (Ts ) u A u (T Ts ) m
As (Ts Tv ) , s 1 s s
(24)
h ( = huh) is the density of water vapor flow condensing on the outer surface, (Ts ) where m is the heat of evaporation for water at Ts, s is wall thickness, s is the heat conductivity, u is the outer (humid air side) heat transfer coefficient and s is the inner heat transfer coefficient.
By using notation G
As 1 1 Au s s s
(25)
the equation (24) can be expressed as h (Ts ) G (Ts Tv ) . u (T Ts ) m
(26)
In equation (24) we have obviously assumed that so called fin efficiency is unity. If there were no condensation, the fin efficiency could be easily taken into account by using, as effective outer surface, the actual outer surface multiplied by the fin efficiency. This kind of concept, the fin efficiency, could of course be introduced in the case of condensation too, but the problem is that the expression for the fin efficiency is a lot more complicated and there are no complete formulas for it in the literature. Hence we can continue the consideration by assuming that the fin efficiency is unity or by assuming that Au denotes the effective outer surface, which has been obtained by multiplying the actual outer surface by the fin efficiency. By substituting equation (22) into equation (26) we get u (T Ts ) (Ts ) M h
p p p h (Ts ) k ln G (Ts Tv ) 0 , R Tk p ph
(27)
where Tk is the average boundary layer temperature at which the air properties are taken. In example 1 this corresponded to the saturation temperature. The surface temperature Ts can be solved from equation (27). If the condensation term weren't there, Ts could be solved directly Ts
u T G Tv . u G
(28)
9 But because of the condensation, the value of Ts is bigger than that. This can be seen in equation (24). Temperature difference Ts - Tv grows as grows. Let us denote the left side of eq (27) by F(Ts). Hence the task is to find solution for equation F(Ts) = 0.
(29)
One useful method for solving this is so called split-half method.
Fig. 1 Solving equation (29) by split-half method. The zero point is in the range Tv Ts T. Always the range that gives F different signs in the endpoints, is split in half. Each step gives a range, diminished to half, where the zero point is located. The split-half points generated (Ts1, Ts2, Ts3,...) approach the zero point, lim Tsi Ts . i
Example 4 x = 0,12 , T = 353 K, Tv = 323 K, u = 80 W/m2K and G'' = 3500 W/m2K. Air properties at mean temperature (Tk 329 K). cp = i cpi + h cph = 0,900 1006 + 0,109 1850 = 1107 J/m3K, Le = D cp / = 32,4 10-6 1107 / 0,02526 = 1,420. 80 From equation (23): k 1,42210,33 0,0913m/s . 1107 (Ts ) (Tk ) 2370 103 J/kg (in more accurate calculations the heat of evaporation as a function of Ts can be represented as a formula). 0,12 ph 105 Pa 1,617 104 Pa 0,12 0,6220 By substituting the values into equation (27) we get 80(353 Ts ) 2,37 106 0,0180
105 p h (Ts ) 105 0,0913 ln 5 3500(Ts 323) 0 , 8,314 329 10 1,617 104
where ph (Ts ) can be calculated from equation (13): T 372,79 p h (T) p 0 exp11,78 . T 43,15 The solution is obtained by using the split-half method: Ts = 324,9 K. 324,9 372,79 p h (324,9 K) 105 exp11,78 1,350 104 Pa . The same 324 , 9 43 , 15 saturation pressure can be also read from Table 1 (324,9 K = 324,9 - 273,15 = 51,75oC).
The result can be correct only physically, as the following condition is satisfied: ph (Ts ) p h . In this case it is (1,350 104Pa < 1,617 104Pa).
10 In case that the result is ph (Ts ) p h , the calculation has to be repeated by neglecting the term m h , that is, the surface temperature is obtained from equation (28). In this case the heat transfer is dry and no condensation takes place. Once we have solved the surface temperature Ts, we can calculate the heat fluxes. In case of example 4 these are: - dry heat flux u(T - Ts) = 80 (353 - 324,9) W/m2 - humid heat flux h M h m
p p p h (Ts ) 105 0,0913 105 1,350 104 kg k ln 0,0180 ln R Tk p ph 8,314 329 105 1,617 104 m 2 s
1,889 103 kg/m 2 s h (Ts ) = 1,889 10-32370106 W/m2 = 4450 W/m2 m
- total heat flux q = (2250 + 4450) W/m2 = 6700 W/m2 The total heat flux can be obtained also from the formula G'' (Ts - Tv) = 3500 (324,9 - 323) = 6650 W/m2. The little difference between the results is due to calculation inaccuracies and due to that the value for the surface temperature, Ts = 324,9 K , is not quite the accurate value.
2.3 Calculating state change of air flow As humid airflow meets a surface, which is at a temperature lower than the air saturation point, the condensation of the water vapor takes place. The air gets dryer and colder. The total energy flux leaving the air is the same as the change in the enthalpy flow i h k , m
(30)
where hk = hk(B) - hk(A), hk(B) is the enthalpy of the humid air after the heat transfer element and hk(A) is the enthalpy before the element. Correspondingly, the change of the humidity is obtained from equation h A u m i x , m
(31)
where x = x(B) - x(A) , x(B) is the humidity after the heat transfer element and x(A) is the humidity before the element. By dividing the equations (30) and (31) side by side and by notating q = /Au, we get
11 q h k . h x m
(32)
The equation characterizes the state change of humid air at a certain spot of the exchanger. It has to be emphazised, however, that ratio hk/x is usually different at different spots of the h ) depends strongly on the state of the air. It has to exchanger, because the condensation ( m be noticed, too, that the temperature of the liquid or gas receiving the heat is different at different spots of the exchanger, and also this temperature affects the heating power and condensation. Compared to the design of dry exchangers, the biggest difference is that there are no ready formulas for the total conductance but the surface temperature has to be calculated at many spots so that the amount of spots is adequate. This is the way the total heat flux can be calculated and the exchanger can be designed. v = 160 kg/s, leaves a heat exchanger at 323 K. Example 5 Water, flowing at the rate of m The total outer surface of the exchanger is 1400 m2 and it is thought to be divided into four elements for calculations. Calculate the state of air and water i = 87,5 kgdry air/s. temperature after the first element as the dry airflow is m Temperature and other values for the air entering the exchanger are the same as in example 4.
In example 4, we obtained that surface temperature is Ts = 324,9 K , total heat h = 1,889 10-3 kg/m2s. The flux is q = 6700 W/m2 and condensation rate is m enthalpy of the air entering the exchanger is obtained from equation (12): hk(A) = 1006 (353 - 273,15) + 0,12 (2,50 106 + 1850 (353 - 273,15)) = 3,98 106 J/kgdry air.
1400 2 m 6700W/m 2 2,35 106 W 4
The enthalpy change from equation (30) 2,35 106 W 2,69 104 J / kg dry air i m 87,5 kg dry air / s and enthalpy after the first element is thus hk(B) = hk(A) + hk = (39,8 104 - 2,69 104) J/kgdry air = 3,71 104 J/kgdry air. Change in humidity from equation (31) h k
1,889 103 (1450/ 4) 0,00756 87,5 and humidity after the first element is thus x(B) = x(A) + x = 0,1200 - 0,00756 = 0,11244. x
Air temperature from equation (12): h x 2,50 106 3,71 105 0,11244 2,50 106 t k C C 74,05C . 1006 x 1850 1006 0,11244 1850 Therefore the air temperature in state B is T = (74,05 + 273,15) K = 347,2 K. The air has thus cooled down (353 - 347,2) K = 5,8 K in the first element. Correspondingly, the water has been heated up in the first element
12 2,35 106 K 3,51K v c pv 160 4186 m and so the water temperature in spot B is Tv(B) = (323 - 3,51) K = 319,49 K. Tv
The calculation of the example 5 can be continued in the same way from spot B to spot C. At first, new surface temperature Ts is calculated by using the values of spot B, and after that the condensation and total heat flux can be calculated. After that, the changes in humidity, air temperature and water temperature can be calculated. By continuing this way, the performance of the whole exchanger can be calculated. The smaller elements the exchanger is divided into, the more accurate is the result of the calculation. The previous example can be understood from the point of view of design so that the calculation procedure is continued, that is, elements are added, as long as desired entering water temperature, or heating power, is achieved. The numerical values in the example are from a case considering heat recovery from exhaust air of a paper machine.
2.4 Air cooler design example A factory produces a tube-and-fin heat exchanger. The tubes are made of Cu and fins of Al. The fin thickness is 0,2 mm, fin distribution 2,5 mm and the ratio of the outer surface (fin surface) and inner surface (inner surface of the tubes) is Au/As = 27. The distribution of the tubes (15/13) in the direction of the airflow is 30 mm and perpendicular to the airflow 60 mm. The outer heat transfer area of one row is 22,2 m2 per one square meter of the face area. i = 2,4 kgdry air/s, is to be cooled from state (25oC, = 50%) down to state An airflow, m (8oC, = 90%). The liquid in the tubes is water-glycol mixture and it flows at the rate of 4,0 kg/s and enters the exchanger at -5oC. Design the heat exchanger.
The first task is to find out the surface temperature of the condensate in the intake (Ts1) and in the exit (Ts2). After the surface temperatures have been calculated, the outer area is obtained from the formula Au (33) G ln where (T Tv1 ) (Ts 2 Tv 2 ) ln s1 . (34) Ts1 Tv1 ln Ts 2 Tv 2 Temperature Tv1 is the temperature of the water glycol mixture at the first tube row and Tv2 is its temperature at the last tube row (Tv2 = 268,15 K = -5oC). The calculations step by step: 1) The properties of the air ph1 = 0,5 3,17 103Pa = 1,585 103 Pa 1,585 103 x 1 0,6220 5 0,01002 10 1,585 103 hk1 = 1006 25 + 0,01002 (2,50 106 + 1850 25) = 50660 J/kgdry air ph2 = 0,9 1072 Pa = 965 Pa
13
x 2 0,6220
965
6,06 103
10 965 hk2 = 1006 8 + 0,00606 (2,50 106 + 1850 8) = 23290 J/kgdry air 5
2) Heating power of the exchanger i (h k1 h k 2 ) 2,4 (50660 23290)W 65,7 kW . m 3) The outlet temperature of the water glycol mixture, Tv1 Let the mass fraction of the glycol be 30% in the mixture in which case the freezing temperature of the mixture is -15oC and specific heat cpg = 3,65 kJ/kgK.
g c pg (Tv1 Tv2 ) , hence m 65,7 K 4,5 K 4,0 3,65 and thus Tv1 = 268,15 + 4,5 = 272,65 K. Tv1 Tv 2
4) The heat transfer coefficient of the water glycol mixture For a 30% mixture, the Kinematic viscosity is g = 4,0 10-6 m2/s, thermal conductivity g = 0,465 W/mK and density g = 1045 kg/m3. g g c pg 1045 4,0 106 3650 Pr 32,8 g 0,465 Let the arrangement of the tubes be such that the velocity of the mixture in the tubes is 1,35 m/s. In that case v d 1,35 0,013 4390 . g 4,0 106 Nu = 0,037(Re0,75-180)Pr0,42 = 0,037(43900,75-180)32,80,42 = 57,6 g 0,465 Nu 57,6 2060 W/m 2 K . and thus s d 0,013 Let it be noted that for pure water in the same conditions, the heat transfer coefficient would have been approximately 3900 W/m2K, which is a lot better than for water glycol mixture. Re
5) Outer heat transfer coefficient The outer heat transfer coefficient for this particular exchanger calculated approximately from the following formula u 25,0 v W/m 2 K , where v is the face velocity in m/s. Let us choose that the face velocity is v = 2,0 m/s. Hence u 25,0 2,0 35,4 W/m 2 K .
6) The face area The mean temperature of the air is (25 + 8)/2 = 16,5oC = 289,65 K and the mean humidity x = (0,01002 + 0,00606)/2 = 0,00804. 0,00804 ph 105 Pa 1276Pa 0,6220 0,00804 pi = (105 - 1276) Pa = 98720 Pa pM 98720 0,0290 i i i 1,189 kg/m 3 . RT 8,314 289,65 The volume flow of the humid air
14 m i / i 2,40 / 1,189 2,02 m 3 /s V and the face area / v 2,02 m 3 /s / 2,0 m/s 1,01m 2 . A V op
7) The mass transfer coefficient The saturation temperature at the mean conditions of the airflow (part 6 above) is approximately 10oC (Tk = 283,25 K). Let us take the properties according to saturated air at this temperature. i = 1,232 kg/m3 h = 0,0094 kg/m3 cp = 1,2321006 + 0,00941850 = 1257 J/m3K. Le = Dcp/ = 23,310-61257/0,02466 = 1,188 35,4 From equation (23) k 1,18810,33 0,0316m/s . 1257 8) Conductance G'' From equation (25) G
1 1 75,9 W/m 2 K . 0 , 001 1 27 388 2060
9) Surface temperatures Heat of evaporation (Ts ) (Tk ) 2,48 106 J/kg. The surface temperatures are ruled by equation (27). Surface temperature Ts1 is obtained from equation 105 0,0316 105 p h (Ts1 ) 35,4(298,15 Ts1 ) 2,48 106 0,0180 ln 8,314 289,65 105 1585 75,9 (Ts1 272,65) 0 which gives Ts1 = 282,9 K. Surface temperature Ts2 is obtained from equation 105 0,0316 105 p h (Ts 2 ) 35,4(281,15 Ts 2 ) 2,48 106 0,0180 ln 8,314 289,65 105 965 75,9 (Ts 2 268,15) 0 which gives Ts2 = 273,8 K. 10) Outer heat transfer area The logarithmic mean temperature difference ln from equation (34) (282,9 272,65) (273,8 268,15) ln 7,72 K . 282,9 272,65 ln 273,8 268,15 The outer heat transfer area from equation (33) 65,7 103 Au 112 m 2 . 75,9 7,72 11) The number of the rows and the length of the exchanger in flow direction For the face area, 1,01 m2 was chosen (part 6). Thus there is outer heat transfer area per one row as much as 1,01 22,2 = 22,4 m2. The number of tube rows in the direction of the airflow z 112 m2/22,4 m2 = 5,0. Hence z = 5.
