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Example 1: A single effect evaporator is to be used to concentrate a food solution containing 15% (by mass)dissolved solids to 50% solids. The feed stream enters the evaporator at 291 K with a feed rate of 1.0 kg s−1. Steam is available at a pressure of 2.4 bar and an absolute pressure of 0.07 bar is maintained in the evaporator. Assuming that the properties of the solution are the same as those of water, and taking the overall heat transfer coefficient to be 2300 W m−2K−1, calculate the rate of steam consumption and the necessary heat transfer surface area. Working in units of kg s−1, the overall material balance becomes 1.0 = V + L Substituting into the component material balance for xF = 0.15 and xL = 0.50 gives 0.15 × 1.0 = 0.50 L => L = 0.3kg s−1 , V = 0.7kg s−1 from steam tables: (If the steam and condensate remain saturated at 2.40 bar) hS = 2715kJ kg−1 and hC = 530kJ kg−1 The feed enthalpy is determined by its temperature. Assuming the feed to be pure water, h F is equal to hf at 291 K and therefore hF = 75.5kJ kg−1. The enthalpies of the vapour and liquor streams are a function of the pressure within the evaporator: hV = 2572 kJ kg−1 (hg at 0.07 bar) and hL = 163kJ kg−1 (hf at 0.07 bar). The enthalpy balance: S(2715 − 530) = (0.70 × 2572) + (0.30 × 163) − (1.0 × 75.5) S = 0.812 kg s−1 and Q = 0.812(2715 − 530)kW = 1774kW The temperature of steam at 2.4 bar is TS = 126.1◦C and the temperature of saturated liquid water at the evaporator pressure of 0.07 bar is TE = 39.0◦C. Thus, from the rate equation, the heat transfer area is Example 2: An aqueous solution at 15.5◦C, and containing 4% solids, is concentrated to 20% solids. A single effect evaporator with a heat transfer surface area of 37.2 m2 and an overall heat transfer coefficient of 2000 W m−2 K−1 is to be used. The calandria contains dry saturated steam at a pressure of 200 kPa and the evaporator operates under a vacuum of 81.3 kPa. If the boiling point rise is 5 K, calculate the evaporator capacity. 1 At 200 kPa the steam and condensate enthalpies are hS = 2707 kJ kg−1, hC = 505 kJ kg−1, Ts=120.2◦C. The pressure within the evaporator is 101.3 − 81.3 = 20.0 kPa at which the boiling point of water is 60.1◦C. The evaporator temperature is now 60.1◦C plus the boiling point elevation and therefore TE = 65.1◦C. => S = 1.862 kg s−1 From steam tables, the feed enthalpy at 15.5◦C is hF = 65 kJ kg−1. Vapour enthalpy: hV = 2609 kJ kg−1 (hg at 0.20 bar) + (1.91 × 5) kJ kg−1 = 2618.6 kJ kg−1 Cp (water vapour at 60.1◦C)= 1.91 kJ kg−1 K−1 The enthalpy of the concentrated liquor stream at the evaporator temperature is h L = 272 kJ kg−1 (hf at 65.1◦C). The component balance becomes 0.04F = 0.20L => F = 5L , V = 4L S(hS − hC) = 4LhV + LhL − 5LhF 1.862(2707 − 505) = (4L × 2618.55) + 272L − 325L L = 0.393 kg s−1 and the evaporator capacity is F = 1.97 kg s−1 EXAMPLE 3. Single effect evaporator: steam usage and heat transfer surface A single effect evaporator is required to concentrate a solution from 10% solids to 30% solids at the rate of 250 kg of feed per hour. If the pressure in the evaporator is 77 kPa absolute, and if steam is available at 200 kPa gauge, calculate the quantity of steam required per hour and the area of heat transfer surface if the overall heat transfer coefficient is 1700 J m-2 s-1 °C-1. Assume that the temperature of the feed is 18°C and that the boiling point of the solution under the pressure of 77 kPa absolute is 91°C. Assume, also, that the specific heat of the solution is the same as for water, that is 4.186 x 103 J kg-1°C-1, and the latent heat of vaporization of the solution is the same as that for water under the same conditions. From steam tables, the condensing temperature of steam at 200 kPa (gauge)[300 kPa absolute] is 134°C and latent heat 2164 kJ kg-1; the condensing temperature at 77 kPa (abs.) is 91°C and latent heat is 2281 kJ kg-1. Mass balance (kg h-1) 2 Solids Liquids Total Feed 25 225 250 Product 25 58 83 Evaporation 167 Heat balance Heat available per kg of steam = latent heat + sensible heat in cooling to 91°C = 2.164 x 106 + 4.186 x 103(134 - 91) = 2.164 x 106 + 1.8 x 105 = 2.34 x 106 J Heat required by the solution = latent heat + sensible heat in heating from 18°C to 91°C = 2281 x 103 x 167 + 250 x 4.186 x 103 x (91 - 18) = 3.81 x 108 + 7.6 x 107 = 4.57 x 108 Kg J h-1 Now, heat from steam = heat required by the solution, Therefore quantity of steam required per hour = (4.57 x 108)/(2.34 x 106) = 195 kg h-1 Quantity of steam/kg of water evaporated = 195/167 = 1.17 kg steam/kg water. Heat-transfer area Temperature of condensing steam = 134°C. Temperature difference across the evaporator = (134 - 91) = 43°C. Writing the heat transfer equation for q in joules/sec, q = UA DT (4.57 x 108)/3600 = 1700 x A x 43 A = 1.74 m2 Area of heat transfer surface = 1.74 m2 (It has been assumed that the sensible heat in the condensed (cooling from 134°C to 91°C) steam is recovered, and this might in practice be done in a feed heater. If it is not recovered usefully, then the sensible heat component, about 8%, should be omitted from the heat available, and the remainder of the working adjusted accordingly). EXAMPLE 4. Concentration of tomato juice in a climbing film evaporator Tomato juice is to be concentrated from 12% solids to 28% solids in a climbing film evaporator, 3 m 3 high and 4 cm diameter. The maximum allowable temperature for tomato juice is 57°C. The juice is fed to the evaporator at 57°C and at this temperature the latent heat of vaporization is 2366 kJ kg-1. Steam is used in the jacket of the evaporator at a pressure of 170 kPa (abs). If the overall heattransfer coefficient is 6000 J m-2 s-1 °C-1, estimate the quantity of tomato juice feed per hour. Take heating surface as 3 m long x 0.04 m diameter. Mass balance: basis 100-kg feed Solids Liquids Total Feed 12 88 100 Product 12 31 43 Evaporation 57 Heat balance Area of evaporator tube pDHL = p x 0.04 x 3 = 0.38 m2 Condensing steam temperature at 170 kPa (abs) = 115°C from Steam Tables. Making a heat balance across the evaporator q = UA DT = 6000 x 0.38 x (115 - 57) = 1.32 x 105 J s-1 Heat required per kg of feed for evaporation = 0.57 x 2366 x 103 = 1.34 x 106 J Rate of evaporation = (1.32 x 105)/ 1.34 x 106) = 0.1 kg s-1 Rate of evaporation = 360 kg h-1 Quantity of tomato juice feed per hour = 360 kg Problem 1. A single-effect evaporator is used to concentrate 7 kg/s of a solution from 10 to 50 per cent of solids. Steam is available at 205 kN/m2 and evaporation takes place at 13.5 kN/m2. If the overall heat transfer coefficient is 3 kW/m2 K, calculate the heating surface required and the amount of steam used if the feed to the evaporator is at 294 K and the condensate leaves the heating space at 352.7 K. The specific heat capacity of a 10 per cent solution is 3.76 kJ/kgK, the specific heat capacity of a 50 per cent solution is 3.14 kJ/kg K. Problem 2. 1.9 kg/s of a liquid containing 10 per cent by mass of dissolved solids is fed at 338 K to a forward-feed double-effect evaporator. The product consists of 25 per cent by mass of solids and a mother liquor containing 25 per cent by mass of dissolved solids. The steam fed to the first 4 effect is dry and saturated at 240 kN/m2 and the pressure in the second effect is 20 kN/m2. The specific heat capacity of the solid may be taken as 2.5 kJ/kg K, both in solid form and in solution, and the heat of solution may be neglected. The mother liquor exhibits a boiling point rise of 6 deg K. If the two effects are identical, what area is required if the heat transfer coefficients in the first and second effects are 1.7 and 1.1 kW/m2 K respectively? Problem 3. An evaporator, working at atmospheric pressure, is used to concentrate a solution from 5 per cent to 20 per cent solids at the rate of 1.25 kg/s. The solution, which has a specific heat capacity of 4.18 kJ/kg K, is fed to the evaporator at 295 K and boils at 380 K. Dry saturated steam at 240 kN/m2 is fed to the calandria, and the condensate leaves at the temperature of the condensing stream. If the heat transfer coefficient is 2.3 kW/m2 K, what is the required area of heat transfer surface and how much steam is required? The latent heat of vaporisation of the solution may be taken as being the same as that of water. Problem 4. A liquid with no appreciable elevation of boiling-point is concentrated in a tripleeffect evaporator. If the temperature of the steam to the first effect is 395 K and vacuum is applied to the third effect so that the boiling-point is 325 K, what are the approximate boilingpoints in the three effects? The overall transfer coefficients may be taken as 3.1, 2.3, and 1.1 kW/m2 K in the three effects respectively. Problem 5. A single-effect evaporator with a heating surface area of 10 m2 is used to concentrate a NaOH solution flowing at 0.38 kg/s from 10 per cent to 33.3 per cent. The feed enters at 338 K and its specific heat capacity is 3.2 kJ/kg K. The pressure in the vapour space is 13.5 kN/m2 and 0.3 kg/s of steam is used from a supply at 375 K. Calculate: (a) The apparent overall heat transfer coefficient. (b) The coefficient corrected for boiling point rise of dissolved solids. (c) The corrected coefficient if the depth of liquid is 1.5 m. Problem 6. A triple-effect backward-feed evaporator concentrates 5 kg/s of liquor from 10 per cent to 50 per cent solids. Steam is available at 375 kN/m2 and the condenser operates at 13.5 kN/m2. What is the area required in each effect, assumed identical, and the economy of the unit? The specific heat capacity is 4.18 kJ/kgK at all concentrations and that there is no boiling-point rise. The overall heat transfer coefficients are 2.3, 2.0 and 1.7 kW/m2 K respectively in the three effects, and the feed enters the third effect at 300 K. 5 Problem 7. A single-effect evaporator is to produce a 35% solids tomato concentrate from a 6% solids raw juice entering at 18°C. The pressure in the evaporator is 20 kPa(absolute) and steam is available at 100 kPa gauge. The overall heat-transfer coefficient is 440 J m-2 s-1 °C-1, the boiling temperature of the tomato juice under the conditions in the evaporator is 60°C, and the area of the heat-transfer surface of the evaporator is 12 m2. Estimate the rate of raw juice feed that is required to supply the evaporator. [ 536 kgh-1 ] Problem 8. Estimate (a) the evaporating temperature in each effect, (b) the reirements of steam, and (c) the area of heat transfer surface for a two effect evaporator. Steam is available at 100 kPa gauge pressure and the pressure in the second effect is 20 kPa absolute. Assume an overall heattransfer coefficient of 600 and 450 J m-2 s-1 °C-1 in the first and second effects respectively. The evaporator is to concentrate 15,000 kg h-1 of raw milk from 9.5 % solids to 35% solids.Assume the sensible heat effects can be ignored, and that there is no boiling-point elevation. [ (a) 1st. effect 94°C, 2nd. effect 60°C, (b) 5,746 kgh-1, 0.53 kg steam/kg water (c) 450 m2 ] 6 STEAM TABLE - SATURATED STEAM Temperature Pressure(Absolute) (°C) (kPa) Enthalpy (sat. vap.) (kJ kg-1) Latent heat Specific volume (kJ kg-1) (m3 kg-1) 2501 2499 2497 2492 2487 2483 2478 2473 2468 2464 2459 2454 2449 2445 2440 2435 2431 2407 2383 2359 2334 2309 2283 2257 2244 2230 2217 2203 2189 2174 2160 2145 2114 2083 2015 1941 206 193 180 157 138 121 106 93.9 82.8 73.3 65.0 57.8 51.4 45.9 40.0 36.6 32.9 19.5 12.0 7.67 5.04 3.41 2.36 1.67 1.42 1.21 1.04 0.892 0.771 0.669 0.582 0.509 0.393 0.307 0.194 0.127 Temperature Table 0 1 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 40 50 60 70 80 90 100 105 110 115 120 125 130 135 140 150 160 180 200 0.611 0.66 0.71 0.81 0.93 1.07 1.23 1.40 1.60 1.82 2.06 2.34 2.65 2.99 3.36 3.78 4.25 7.38 12.3 19.9 31.2 47.4 70.1 101.4 120.8 143.3 169.1 198.5 232.1 270.1 313.0 361.3 475.8 617.8 1002 1554 2501 2503 2505 2509 2512 2516 2520 2523 2527 2531 2534 2538 2542 2545 2549 2553 2556 2574 2592 2610 2627 2644 2660 2676 2684 2692 2699 2706 2714 2721 2727 2734 2747 2758 2778 2793 7 Pressure Table 7.0 9.7 12.0 14.0 15.8 17.5 21.1 24.1 29.0 32.9 40.3 45.8 60.1 75.9 93.5 99.6 102.3 104.8 107.1 109.3 111.4 113.3 115.2 116.9 118.6 120.2 127.4 133.6 138.9 143.6 147.9 151.6 167.8 179.9 1.0 1.2 1.4 1.6 1.8 2.0 2.5 3.0 4.0 5.0 7.5 10.0 20.0 40.0 80.0 100 119 120 130 140 150 160 170 180 190 200 250 300 350 400 450 500 750 1000 2514 2519 2523 2527 2531 2534 2540 2546 2554 2562 2575 2585 2610 2637 2666 2676 2680 2684 2687 2690 2694 2696 2699 2702 2704 2707 2717 2725 2732 2739 2744 2749 2766 2778 8 2485 2479 2473 2468 2464 2460 2452 2445 2433 2424 2406 2393 2358 2319 2274 2258 2251 2244 2238 2232 2227 2221 2216 2211 2207 2202 2182 2164 2148 2134 2121 2109 2057 2015 129 109 93.9 82.8 74.0 67.0 54.3 45.7 34.8 28.2 19.2 14.7 7.65 3.99 2.09 1.69 1.55 1.43 1.33 1.24 1.16 1.09 1.03 0.978 0.929 0.886 0.719 0.606 0.524 0.463 0.414 0.375 0.256 0.194