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The oil system of a ceramic press requires cooling approximately 220 L/h of oil from 55°C (131°F) to 35°C (95°F). The heat exchanger needs to be sized and the specifications for the cooling water determined.
It will be convenient to make some unit changes in order to facilitate the estimation of the heat to be dissipated, namely:
oil flow = 2201 l/h = 2.201 m3/h = 2.215 * 35.3145 cu.ft./h = 77.728 cu.ft./h
If the respective densities of the oil at 55°C (131°F) and 35°C (95°F) are 57.277 lb/cu.ft and 57.997 lb/cu.ft, taking an average density value of 57.637 lb/cu.ft, we can estimate the mass of oil to be cooled will be:
mass of oil to be cooled = 77.728 * 57.637 = 4480 lb/hr
Like density, specific heat will vary with temperature. If the specific heats of oil at 55°C (131°F) and 35°C (95°F) are 0.4577 btu/lb°F and 0.4370 btu/lb°F respectively, taking an average specific heat value of 0.4474 btu/lb°F we can estimate the heat to be dissipated as follows:
Q = heat to be dissipated = 4480 lb/h * 0.4474 * (131-95°F) = 72156 btu/h = 18184 kcal/h
While not necessarily establishing a single alternative, it is indicated that cooling water is available at 28°C and a temperature difference of 5°C across the heat exchanger is also acceptable, albeit somewhat arbitrarily. We can then immediately determine the water flow rate required for cooling. Returning to English units, the cooling water will enter at 28°C (82.4°F) and exit at 33°C (91.4°F), therefore
water flow rate = Q / temperature difference = 72156 btu/hr / ( 500 * (91.4-82.4)°F) = 16 US GPM = 3641 l/h
The required exchange area/surface A (sq.ft.) comes directly from the Netwon equation :
Q (btu/h) = A (sq.ft.) * U (btu / hr * sq.ft. * °F) * LMTD
A (sq.ft.) = Q (btu/h) / ( U (btu / hr * sq.ft. * °F) * LMTD )
In our case, the logarithmic mean temperature is 23.6°F = LMTD (131, 95, 91.4, 82.4). Selecting a conservative value for U, e.g., 37-40 Btu/hr * sq.ft. * °F , the required heat exchange surface area is:
Required exchange surface A = 72156 btu/h / ( 37 * 23.6) = approximately 83 sq.ft. = 7.7 m2
A pulp and paper mill needs to cool 300 m³/h (1322 US GPM) from 71°C (159.8°F) to 65°C (149°F) using a closed cooling circuit. The water-to-water heat exchanger needs to be sized based on the availability of cooling water at 36.2°C (97.2°F).
As in the previous case, density and specific heat will vary with temperature. However, we can quickly estimate the heat to be dissipated, Q.
Q = 1322 USGPM * 500 * (159.8°F - 149°F) = 7138800 btu/h = 1799093 kcal/h
that is, 1800000 kcal/h = 300 m3/h * 10^3 * (71°C - 65°C)
Assuming a temperature difference of 46.4°F (25.78°C) across the heat exchanger, the cooling water flow rate to dissipate that heat load comes directly from the formula
Cooling GPMs = 7138800 btu/h / ( 500 * 46.4°F) = 307.7 US GPM = 69880 l/h
Note: Outlet temperature = 36.2° + 25.78° = 62°C
Q (verification) = 69880 l/h * (62 - 36.2) = 1802904 kcal/h
In this case, the logarithmic mean temperature is 27.43°F = LMTD (159.8, 149, 97.2, 143.6). Selecting a conservative value for U, e.g., 197-200 Btu/hr * sq.ft. * °F , the required heat exchange surface area is:
Required heat exchange surface A = 7138800 btu/h / ( 197.32 * 27.43) = approx. 1319 sq.ft. = approx. 123 m2
A yeast factory wishes to estimate the heat dissipated in a yeast production process using aerobic fermentation. The heat generated in the fermentation process is approximately 3.5 kcal per gram of yeast solids. The estimate is desired for a process using approximately 5 tons of molasses.
In order to obtain an order of magnitude estimate within the scope of this material, we will outline a reasoning with a small number of assumptions, applicable or not to each specific case.
Broadly speaking, the total heat generated during fermentation is composed of three sources:
Q1 = the heat generated by the fermentation process itself
Q2 = sensible heat
Q3 = the sensible heat of the air used (2500 CFM)
Q1 Fermentation process heat
Starting with 5 tons of molasses and assuming an 80% yield of fresh yeast and a respective content of 30% dry-dry yeast on compressed yeast, we obtain the kg of dry-dry yeast (solids) directly:
5000 kg molasses * 0.8 fresh yeast/molasses * 0.3 dry-dry yeast/fresh yeast = 1200 kg of dry-dry yeast or solids.
Since the heat generated in the fermentation process is on the order of 3.5 kcal per g of solids, it results for that shift
1200 kg * 10^3 * 3.5 kcal = 4,200,000 kcal
maximum load occurring (10% in one hour) =
Q1Max = 4200000 * .1 = 420,000 kcal/h
Q2 Sensible heat
From the mixture of 5000 kg of molasses with water we will obtain approximately 5500 liters of specific density 1.23 kg/l and specific heat 1 Kcal/g °C, which we will take from 80°C to 30°C, then Q2 comes out directly
Q2 = 5500 L * 1.23 kg/L * 1 * (80 - 30°C) = 338250 kcal/h
maximum load occurring (10% in one hour) =
Q2Max = 338250 * .1 = 33,825 kcal/h
Q3 Sensible heat of the air
Basically, it consists of following the evolution of the air mass from 80°C to 30°C and estimating the heat contribution by the difference in enthalpies.
The feed air is at 80°C and 34.5°C wet-bulb temperature, corresponding to an enthalpy of 61.77 Btu/lb. The outlet enthalpy corresponding to saturated air at 30°C is 50.67 Btu/lb, so the difference comes out directly:
enthalpy difference = 61.77 - 50.67 = 11.1 btu/lb
Air mass 2500 CFM * 0.071 llb/cu.ft * 60 = 10638 lb/h
Q3 = 10638 lb/h * 11.1 btu/lb = 118085 lb/h = 29760 kcal/h
Finally, the maximum hourly load Qmax
Qmax = Q1max + Q2max + Q3max = 420,000 kcal/h +33,825 kcal/h + 29,760 kcal/h = ca. 483600 kcal/h
An industrial facility needs to size a cooling circuit to serve four (4) air compressors, each rated at 150 HP and 565 CFM. An upper limit on the heat to be dissipated and an estimate of the required water flow rate are desired. Further data on the compressors are unavailable.
In the cooling of many machines, such as compressors and vacuum pumps, it is common to estimate the heat to be dissipated by assuming the entire horsepower is transferred as the thermal load. In our case, with four (4) units of 150 HP each, using a conversion factor of 2545 Btu/hr/HP, the following thermal load is obtained:
Q = 4 * 150 HP * 2545 Btu/hr/HP = 1,527,000 Btu/hr = 384,829 kcal/h.
Note that compressor specifications and type are typically unavailable, so the load is usually sized for the worst-case scenario. For example, in the case of single-stage reciprocating compressors, approximately 10% of the heat would be dissipated directly into the atmosphere; slightly less, 5%, in a two-stage system or screw compressor. It is not advisable to consider these small percentages as a positive factor.
Furthermore, depending on the type of compressor involved, the target chilled water temperature could vary, although only slightly, between 90°F and 95°F, or possibly lower. Intercoolers and aftercoolers typically require the lowest temperatures, usually associated with air drying. On the other hand, in a screw compressor, the primary function (80% of the heat load) might be oil cooling, which typically requires higher temperatures to maintain its properties. In another instance, the primary requirement, according to factory/head office information, might be the provision of chilled water for air drying, which could specify, say, 78°F (25.5°C).
In the absence of specifications or compressor type, a target chilled water temperature of 90°F (32.2°C) with a range of 10°F could be considered, so the inlet temperature to the tower would be around 100°F (37.7°C), with the following associated flow rate:
US GPM flow rate = Q / (500 * range) = 1:527,000 btu/hr / (500 * 10) = approximately 306 US GPM = 69360 l/h
In an oxygen generation process, an air/water heat exchanger must cool 10,594 CFM of air from approximately 167°F to 95°F before it enters a molecular sieve bank.
While the calculation should strictly be done using the enthalpy difference at each end of the process, a preliminary estimate/order of magnitude of the heat to be dissipated can be made, namely:
Q = 10,594 CFM * 60 * 0.0715 lb/cu.ft. * Cp = 10,594 CFM * 60 * 0.0715 lb/cu.ft. * 0.24 = 785,372 Btu/hr = 197,926 kcal/hr.
If the cooling water reaches 86°F (30°C) and returns at 104°F (40°C), the corresponding LMTD would be on the order of 27.75 IP units or 15.42 SI units. The flow rate/order of magnitude is easily derived, namely:
flow rate = Q / (500 * range) = 785,372 btu/hr / (500 * (104-86)) = 87.3 US GPM = c. 19800 l/h.
Without information from the heat exchanger manufacturer, a considerable heat exchange area would be anticipated . Based on an overall heat transfer coefficient of approximately 15 IP units, the order of magnitude required for capacity planning/performance troubleshooting could be determined. Insufficient heat exchange surface area in the original equipment will likely be detected.
Almost thirty years ago, we addressed the restoration of crossflow cooling towers in technical publications , demonstrating numerically/step-by-step what users had already experienced firsthand: thermal deficits exceeding 30% of the capacity values published in the catalog. Something similar can be seen in the technologies of all types of evaporative condensers and in almost all counterflow cooling towers, regardless of their origin. Typically, an industrial-grade cooling tower is manufactured to operate non-stop, 24/7, or on the order of 8,784 hours per year . Unfortunately, in the race for competitiveness, many unit manufacturers are promoting/supplying pseudo-industrial units with a clear bias (to the specialist!) towards air conditioning circuits, with target performance levels that may not even reach 1,600 hours per year . The first and fatal distinction is the aforementioned underperformance (30%), fills that don't last a year, questionable manufacturing materials, fans unsuitable for industrial service, among other issues. As an industrial manager, I would ensure that in every critical service/production contract award, I consider the resulting ratio, i.e., 5.49 x (8784 / 1600), in all proposal comparisons. Industrial supply requires cooling capacity and a service history of no less than twenty (20) years, with operating costs, consumption, sustainability, and other factors perfectly documented based on global standards that are, fortunately, publicly available. Basically, anyone who doesn't know this today simply doesn't want to know!
We have gradually been able to resolve these "setbacks" in a large number of industries and bring the unit capacities up to the levels originally requested by users. In the case of counterflow systems, we use packings of our own manufacture with considerably greater cooling capacity and revise the air section, including increasing the dimensions of the inlet windows—something that initially seems paradoxical; to make it work better, we essentially "drill holes," something that isn't part of what one would call "supply." Such is the importance that, without making these openings, in both counterflow and crossflow systems, the pressure drop can be greater than that experienced through the packings (horsepower of the motor is lost to parasitic pressure drops)!
In the case of evaporative condensers, it is quite possible to eliminate the tube bundles by packing them with structured plastic packings, fully recovering the rest of the unit—i.e., casing, liquid distribution, air section, and connections—without practically any significant modifications. This can be done by attaching a shell and tube or plate condenser.
In the case of cross-flow towers we have carried out similar conversions, e.g. 700 m3/h, successfully, eliminating the gravity feed system (classic imbalances and overflows of the troughs without entering the tower!), installing slightly pressurized liquid distribution systems and packing structured packings and countercurrent droplet eliminator modules, thus TOTALLY recovering the casing or enclosure, air section intact, connection with practically no significant modification beyond hydraulic calibration by third parties.
CASE STUDIES (CONDENSING APPLICATIONS)
APPLICATION # 1 COMPARISON WATER VS. AIR CONDENSATION 37.1 CHILLER TONS Te = 6°C, -4°C, -14°C AMMONIA CONDENSATION
For the purpose of carrying out a massive technological restructuring in a supermarket chain , a comparison of the performance coefficients of two competing condensing systems is desired, e.g., a water-cooled condenser versus an air-cooled condenser . The analysis of a small installation with approximately 37 tons of refrigeration (in the evaporator) will include estimates of the compression ratio, the amount of refrigerant required, displacement, as well as the aforementioned coefficients and specific power (for the refrigeration effect, i.e., kW per ton) for different suction temperatures.
Although the Mollier tables have been approximated by polynomials of varying degrees and/or straight lines to simplify the calculations, the results obtained show a very good correlation with values obtained using both specialized systems and direct manufacturer information. For primarily historical reasons, and to obtain the largest possible number of archival data sets, the results are presented using Freon 12 refrigerant. Identical tables were created for Freon 22 and ammonia, essentially confirming the same results.
Of the six contexts presented in the PDF insert, we will detail the step-by-step calculation for the following data set:
- load on the evaporator: 37.1 tons
- Suction temperature: 42.8°F (6°C)
- Condensation temperature 100.4°F (38°C)
- 11.5°C superheated and 7°C for subcooling, 20.7°F and 12.6°F respectively
Compression ratio
We can satisfactorily approximate the absolute pressure value for saturated freon 12 using, say, a polynomial of degree 4, namely:
Psia=23.8661104+T*0.50551626+T*T*4.1262/10^3+T*T*T*1.4967/10^5+T*T*T*T*9.5163/10^9
The direct evaluation of said polynomial for the suction temperature 42.8°F and condensation temperature 100.4°F results in 54.27 psia and 132.33 psia respectively, estimating then the compression ratio 2.44 = 132.33 psia/ 54.27 psia
Amount of refrigerant required
We can approximate the total enthalpy and the enthalpy of saturated liquid using simple straight lines.
H(T) =T*0.1017+78.258 (total)
H(T) =T*0.2291+8.2855 (liquid)
Direct evaluation of these formulas for the chosen operating temperatures, i.e., 42.8°F and 100.4°F, yields 82.6 Btu/lb and 31.3 Btu/lb, respectively. Therefore, by difference, the cooling effect per pound of refrigerant is:
Cooling effect per pound of refrigerant = 82.6 – 31.3 = 51.3 btu/lb
Based on the load on the evaporator , i.e., the refrigeration effect, we can then estimate the amount of refrigerant required:
amount of refrigerant required = 37.1 * 12,000 btu/h tr / 51.3 = 8674 lb/h = 144.6 lb/min
Displacement and volumetric efficiency of a generic compressor
Based on an approximation of the saturated refrigerant density given by
D (T)=0.611+0.0121*T+T^2*9/10^5+T^3*3/10^7+T^4*9/10^10-T^5*2/10^11+T^6*2/10^13
We can obtain an estimate of the saturated refrigerant density for 42.8°F, and immediately approximate the compressor displacement as follows:
compressor displacement = 144.6 lb/min / 1.3186 lb/cu.ft. = 109.6 CFM
Approximating volumetric efficiency as a function of the compression ratio as a straight line:
volumetric efficiency = 98.336 - 5.5395 * compression ratio
=98.336 - 5.5395* 2.44 = 84.8 %
In short, an estimate of the compressor displacement involved will be given by:
Compressor displacement = 109.6 CFM / 0.848 = 129.2 CFM
Estimation of the heat to be dissipated in the condenser
Based on the superheating and subcooling data, we must retrace our steps in three stages, namely:
cooling towards condensation, i.e., from 121.1°F and 132.33 psia, to identical pressure and 100.4°F
condensation at 100.4°F
subcooled 7°C, i.e. from 100.4°F to 87.8°F
From tables of superheated refrigerant at 123.33 psia for 121.1°F and 100.4°F, enthalpy values of 90.86 and 87.08 are obtained respectively, then the contribution of this section is approximately 3.78 btu/lb or 32789 btu/h.
Estimating the latent heat as the difference in the evaluation of the polynomials, already presented, at the two extremes, i.e., total enthalpy and saturated liquid, results in approximately 57.2 btu/lb or 496014 btu/h.
Finally, the subcooled section, directly evaluated using the saturated liquid polynomial, yields approximately 2.9 btu/lb or 25040 btu/h
The heat to be dissipated in the capacitor , i.e., the sum of the three sections, is then:
Q = 32789 btu/h.+ 496014 btu/h + 25040 btu/h = 553843 btu/h
resulting in a Q/refrigeration effect ratio = 1.24x, which is quite realistic.
COP and specific power calculations (per chiller ton)
Estimating the ideal work by difference results in:
W = Q – cooling effect = 553843 btu/h – 445200 btu/h = 108643 btu/h = ca. 42.7 HP or closer to the actual operating conditions, i.e. friction and efficiencies, on the order of 55 HP
Naturally, all this arithmetic is completely automated by each manufacturer and optimized according to the characteristics of each piece of equipment. The deviations observed in this simple model, for introductory purposes, affect the entire spectrum of suction and condensing temperatures fairly evenly, so the results, which can certainly be refined with specialized calculation systems, are consistent for the aforementioned purpose and summarized, for example, in the following figure, prepared for an even smaller installation (i.e., 3.75 chiller tons in the evaporator):
With this "scenario" (inherent energy surcharges of 30% or more) readers from critical industries will again be able to analyze whether remote condenser technologies ("air-cooled condensers") should really be included in industrial (versus commercial) equipment/installations.
The conclusions, i.e. energy consumption easily 30% higher or more for the air condenser alternative vs. water condenser, basically originate from two aspects, one related to the operating principle of each alternative and another of a "meteorological" nature, as we will discuss below.
The dry-bulb temperature , the basis for calculating the dimensions of air-cooled condensers, forces us to work with refrigerant condensation temperatures between 16 and 23°C above the inlet air temperature ; assuming an air temperature of 35°C, we are obliged to target condensation temperatures classically between:
tc = t + 16°C = 35°C + 16°C = 51°C
tc = t + 35°C= 35°C +23° C = 58°C
The wet-bulb temperature , the basis for calculating the dimensions of water-cooled or evaporative condensers, allows working with condensation temperatures between, say, 7 and 14°C above that temperature; assuming a wet-bulb temperature of 25.5°C, we can translate the target condensation temperature classically to between
tc = t + 7°C = 25.5°C + 7°C = 32.5°C
tc = t + 14°C = 25.5°C +14° C = 39.5°C
It turns out that dry-bulb temperature profiles and wet-bulb temperature profiles (typically more "constant" throughout the day) commonly differ between 5.5°C and 16.7°C, depending on the location. It is this difference that the water-based condensation alternative exploits spectacularly, allowing for significantly lower condensation temperatures.
Given that cooling capacity increases or decreases by 1% for every degree Celsius that the condensing temperature decreases or increases, the following corollary is clear: i.e., with the same suction temperature, the best-case scenario of air-cooled condensing (e.g., 51°C) versus a mediocre or pessimistic scenario of water-cooled condensing at 38°C will impact the installation by 12-13%. In other words, energy consumption, with the same evaporator tonnage, decreases by the same proportion as the condensing temperature decreases. The combined effect of the various optimizations is energy penalties of 30%, 50%, and more for remote/air-cooled condenser alternatives.
Beyond the "exact" value of kW per chiller ton , ideal or real, the conclusions are equivalent (30-50%), there are also a number of drawbacks from an installation point of view:
- For obvious reasons, air-cooled condenser equipment is noticeably more bulky , e.g., air flow rates two to three times higher per ton of refrigeration, e.g., 600-900 cfm/ton vs. 250 cfm/ton or even less ; the heat exchange area can be ten times larger than the water-cooled alternative.
- At a time when minimizing both the refrigerant charge and the circuit length is sought , the remote condenser and even the evaporative condenser do not offer a good alternative; the use of condensers (e.g., close to the central equipment) cooled by water allows, for example, the installation of the refrigeration machines in underground machine rooms, while the connection to the heat dissipation equipment, i.e., cooling tower, is made by means of a simple and harmless water line .
A small refrigeration facility wishes to synchronize available resources, i.e., three ammonia compressors, approximately 70 chiller tons each, three shell and tube condensers of 42.5 m² each, and two existing cooling cells, each of approximately 2.20m x 2.10m dimensions.
Although the person in charge of the refrigeration installation has detailed different operating contexts for the refrigeration equipment – see insert in PDF format – it has been estimated that in most cases, the daily operation of the plant will only require the operation of two of the compressors, each delivering 70 chiller tons (in the evaporator) and equipped with 100 HP motors. Therefore, the heat to be dissipated can be directly deduced:
Q = [70 tons * 12000 btu/h per ton + 100 HP * 2545 btu/h per HP] * 2 = 2189000 btu/h = 551664 kcal/h
It has been indicated that each existing shell and tube condenser offers approximately 42.5 m² of heat exchange surface area. Assuming all three are operational, the following heat exchange area will be available:
A = 3 * 42.5 m2 * 10.76 sq.ft. / m2 = 1371.9 sq.ft.
Since the installation is planned to operate with a condensing temperature of 95°F (35°C), it is proposed to receive cooling tower water at 82.5°F (28.06°C) and a condenser water outlet temperature of 90°F (32.22°C), resulting in the following LMTD:
LMTD = =((Tc-Ti)-(Tc-To))/LN((Tc-Ti)/(Tc-To)) = 8.185
Assuming the condensers operate at a conservative water velocity, we can consider values of U = 200 btu/h sq.ft. °F, so the heat they can exchange will be given, using Newton's equation, by:
Q = 1371.9 sq.ft. * 8.185 * 200 = 2245848 btu/h = 565990 kcal/h
Unless additional heat exchange surface is anticipated, in principle the cooling tower sector will simply need to be restored to handle the following flow rate:
Flow rate (US gpm) = Q / ( 500 * ( 90 – 82.5) = 2245848 / ( 500 * (90-82.5)) = ca. 600 US gpm
The survey of the existing structures, measuring 2.10m x 2.20m, indicates that a capacity of 1,372,884 BTU/h per cell , or 366.1 GPM at 90/82.5/twb°F (83,141 L/h at 32.22/28.06/twb°C), is easily achievable. Therefore, unless a capacity/plant expansion is planned in the short term, retrofitting the existing units will suffice.
With a target condensation temperature of 105°F (approx. 40.6°C), a 7000 kW (approx. 1585 tonnes of refrigeration) facility wishes to examine the evolution of the required heat exchange area under different cooling water supply conditions.
The reasoning is identical to the previous one and is illustrated for an arbitrary data set below.
Q = 7000 kW = 7000 * 3410 = 23,870,000 btu/h
Assuming a condensation temperature of 105°F, a water inlet temperature to the condenser of 90°F and an outlet temperature of 100°F, the following LMTD results =
LMTD = =((Tc-Ti)-(Tc-To))/LN((Tc-Ti)/(Tc-To)) = 9.102
Speculating a conservative value of U = 200, we can estimate the exchange surface as in the previous case, using the Netwon equation:
A = Q / ( U * LMTD) = 23,870,000 btu/h / ( 200 * 9,102) = 13113 ssq.ft. = 1219 m2
The attached PDF file shows the evolution of the required exchange area dictated by the available cooling tower water temperatures, from approximately 83.8°F (28.8°C) to 90.8°F (32.7°C).
While we have preserved the original dataset, we would anticipate a "radical/non-negotiable" reframe of context, i.e., a target condensation temperature of approximately 95°F. See the basis for the urgent argument/reframe in applications 7 and 8 below.
An estimate of the cooling system is desired, i.e., tentative operating framework, required exchange surface, flow rates involved, and other factors for three generating plants: one of 100 MW and two others, rural, of 10 MW and 25 MW respectively.
For introductory purposes, it will suffice to outline just one application, for example, the generation of 10 MW. Indeed, assuming a thermal efficiency of 35%, the heat to be dissipated in the condenser can be expressed as:
Q = 10 * 10^3 kW * 3410 btu/h per kW / 0.35 = ca. 95,000,000 btu/hr
The typical steam condenser environment ranges from 3"Hg to 4"Hg, corresponding to 115°F (46.1°C; 1,473 psia) and 125.4°F (51.9°C; 1,965 psia). Taking an arbitrary intermediate value, e.g., 122°F (50°C, corresponding to 3.64 psia and 1.79 psia), and assuming cooling water at 86°F (30°C) leaving the condenser at 106°F (41.1°C), the corresponding LMTD is obtained:
LMTD = =((Tc-Ti)-(Tc-To))/LN((Tc-Ti)/(Tc-To)) = 24.66
Speculating a value of U = 630, we can estimate the exchange surface as in the previous case, using the Netwon equation:
A = Q / ( U * LMTD) = 95,000,000 btu/h / ( 630 * 24.66) = 6115 sq.ft. = 569 m2
Cooling cells are then associated to handle 9500 US GPM 106/86/twb, verifying then, both for the condenser and for the cooling system:
Q = 500 * 9500 GPM * (106 – 86) = 95,000,000 btu/h
That is, the equivalent of approximately 100,000 lb/h = 45,360 kg/h of steam. The development of the applications for 25 MW and 100 MW is identical, at least for the purposes of a first estimate.
In a partially documented manufacturing process, a certain quantity of raw material is combined with 1700 kg of alcohol. The reaction process results in the condensation of 150 kg/h of steam and 400 kg of alcohol, which can be condensed over a period of 4 hours.
An estimate of the total heat to be dissipated to achieve this condensation is desired.
Mass of vapor (of water) to be condensed
150 kg/hr = 150 kg/hr * 2.2046 = 331 lb/hr
Assuming heat of vaporization on the order of 1,000 btu/hr, the heat to be dissipated Q1 is:
Q1 = 331 lb/hr * 1,000 btu/hr = 330,690 btu/hr = 330,690 btu/hr / 3.968 = 83,340 kcal/hr
It has been indicated that the alcohol used has a molecular weight of 130.26 and heat of vaporization of 11.4 kcal/mol. From this information, using a simple rule of three, we express the heat of vaporization in kcal/kg or btu/lb, namely:
heat of vaporization = 11.4 kcal/mol * 1000 gr / 130.26 = 87.5 kcal/kg = 87.5 kcal/kg * (4.186/2.326) = 157.5 btu/lb
The condensation of the alcohol fraction will contribute the following hourly load
Q2 = 100 kg/hr * 2.2046 lb * 157.5 btu/lb = 34,723 btu/hr = 34,723 btu/hr / 3.968 kcal = 8,750 kcal/hr
A small food processing plant needs to condense approximately 1,000 kg/h of steam from the evaporation stage using an existing small barometric condenser. An order of magnitude estimate of the heat to be dissipated by cooling water is desired.
Roughly, we can estimate the heat dissipation based on the mass of steam to be condensed and a vaporization rate of approximately 1,000 Btu/lb, as follows:
Q = 1,000 kg/hr * 2.2046 lb * 1,000 Btu/lb = 2,204,600 Btu/hr = 2,204,600 Btu/hr / 3.968 = c. 556,000 kcal/hr
The service flow rate corresponding to the cooling fluid is determined from the already calculated thermal load, the characteristics of the fluid (gravity and specific heat) and the temperature difference of the liquid through the barometric condenser, namely:
GPM = Q / ( ( Tout - Tin) * 500 * Cp * Sp)
Since the fluid used is water (Cp=1, Sp=1) and proposing a temperature difference of 10°F, the following flow rate results:
GPM = Q / ( ( Tout - Tin) * 500 * Cp * Sp) = 2,204,600 btu/hr / ( 10 * 500) = 441 US GPM
It could then be proposed to operate the system receiving cold water at 90°F (32.22°C) returning at 100°F (37.8°C) which would make it possible to aim for a target condensation temperature on the order of 105°F (40.6°C) which corresponds to an operation at 2.24 "Hg, in some way a measure of the efficiency transferred or recovered to the condenser.
During capacity planning for an industrial refrigeration (ammonia) system, a plate condenser was found to be receiving chilled water at 29.4°C, returning it at 33°C, with a target condensing temperature of 35.1°C. To improve system performance, the possibility of extracting as much heat as possible from the existing system is to be explored.
By calculating the logarithmic mean temperature (LMTD), it can be quickly seen that the plant design has a serious bias/orientation/dimensioning towards the "air conditioning sector". Indeed, the logarithmic mean temperature for this data set of 29.4/33/35.1°C corresponds to a value of 3.62.
Typically, in an industrial/ammonia refrigeration system, a lower value would be proposed for the cold water temperature. Even by only slightly lowering the target condensing temperature, the following data set, 28/32.2/35°C, yields a logarithmic mean temperature value of 4.58.
The implication/conclusion is that by modifying the operation with the indicated recommendation, it is possible to extract almost 30% more heat from the condenser (1.27x = 4.58 / 3.61).
Whether it is a plate or shell-and-tube condenser, the reasoning (Newton's Law) is the same. If many installations/projects kept this principle in mind, they would not need to "draw" capacities with fanciful overall heat exchange coefficients. There have been responsible plate capacitor suppliers who, in order to maintain science and competitiveness, have been forced to publish capacities with two sets of data: the correct one, which would put them at a great disadvantage (but honest), and another, called "oversized," which basically equates their offer to compete with fantasy numbers from other suppliers.
If central air conditioning and industrial refrigeration systems operated according to the best practice guidelines that gave rise to the various specializations—specifically, maintaining the lowest possible target and actual condensing temperatures—the energy saved would be greater than that potentially produced by any of the alternative energy sources currently being considered. We have not found a single person, whether in the industrial, HVAC, or institutional sectors, who denies our assertion!
Similar to the capacity planning illustrated for central air conditioning installations, it is possible to establish comparative operating charts, especially considering that the industrial refrigeration installation is typically in service 365/7/24, as is probably the case with most industrial installations.
Whether the installation is "large" or "small," the impact of penalties/excessive consumption, translated into annual costs in the tens or hundreds of thousands of dollars, is quite significant. The following table illustrates the financial losses incurred by the user.
INSTALLATION A
Condensation Temperature 105 °F 40.6 °C 1.58 BHP/ton (15 kg/cm2)
Cooling Effect 1200 tons (in evaporation) 1896 HPs 1415 kW (214 psig)
USD/kW h
365 24 12394746 kWh 0.2 2478949 US$/year
Condensation Temperature 95 °F 35 °C 1.37 BHP/ton (12.7 kg/cm2)
Cooling Effect 1200 tons 1644 HPs 1227 kW (181 psig)
USD/kW h
365 24 10747343 kWh 0.2 2149469 US$/year
Penalty 329481 US$/year
INSTALLATION B
Condensation Temperature 105 °F 40.6 °C 1.58 BHP/ton (15 kg/cm2)
Cooling Effect 220 tons 347.6 HPs 259 kW (214 psig)
USD/kW h
365 24 2272370 kWh 0.2 454474 US$/year
Condensation Temperature 95 °F 35 °C 1.37 BHP/ton (12.7 kg/cm2)
Cooling Effect 220 tons 301.4 HPs 225 kW (181 psig)
USD/kW h
365 24 1970346 kWh 0.2 394069 US$/year
Penalty 60405 US$/year
The harm passed on to the user is several times any amount of honest supply/investment!
Fortunately, today it's possible to carry out capacity planning for any industrial refrigeration system , eliminating or at least reducing these daunting hidden costs. To the classic industry recommendation—that is, to operate with the lowest possible condensing temperatures/pressures—we can add the unequivocal recommendation of the extremely simple conversion/migration of existing evaporative condensers to condenser configurations (e.g., case and tube) and cooling towers.
Besides numerous other drawbacks, the CAPACITY PLANNING OF AN EVAPORATIVE CONDENSER reveals the terrible THERMAL COLLAPSE that occurs shortly after installation due to the increased thickness of the scale film ( direct evaporation on the tubes, similar to a low-pressure boiler). As grim as it is graphic:
With this "scenario" (a collapse of nearly 50% capacity), readers in critical industries will once again be able to analyze whether evaporative condenser technologies should really be included in industrial (versus commercial) equipment/installations.
Here is the detailed calculation, which is conclusive (Technical note: it is Newton with the design coefficients of the original manufacturers ).
EC Sizer B&B/CTI SA
Q 756049 kcal/h 3000002 btu/h 880 kW
Twb 25 °C 77.0 °F C-1766
Tcondensing 35.72 °C 96.3 °F 1/10/13
chiller tons 200.0 tons (15k) 250 tons (12k) IF at 2.1 US GPM per chiller ton
water flow 600.0 US GPM at 3 GPM per chiller ton 486.0 US GPM at 2.43
37.9 lps 30.7 lps
face area sq.ft. 100.0 sq.ft. at 6 gpm/sq.ft. 81.0 sq.ft. area (alt.)
tentative air flow 60000.0 CFM at typ. 600 fpm 101941 m3/h 48600.0 CFM (alt.)
l/g = 1.2
delta enthalpy 11.7 btu/lb Evaporative Condenser Calculation (R-717)
CAPACITY COLLAPSE
Hin 45.0 btu/lb
Hout = Hin + delta 56.7 btu/lb
Note the conservative value in the choice
saturated air out at 88.7 °F of the U design, classic within the
manufacturers of good schools.
LTD 12.6 °F 7.0 °C
OR 100 cu.ft. 3/4" coil
required area 2388.4 sq.ft. 222.0 m2 2388.4 sq.ft. 37.3
+20% safety factor 2866.0 sq.ft. 266.4 m2 2866.0 sq.ft. 44.8
Q capacitor thickness mm Cooling Effect % ability
Original Capacity 756049 kcal/h 0 200 chiller tons 100
Capacity 0.8mm 551916 kcal/h 0.8 146 chiller tons 73
Capacity 1mm 415827 kcal/h 1 110 chiller tons 55
Capacity 2mm 287299 kcal/h 2 76 chiller tons 38
Capacity 2mm & re. 244204 kcal/h 65 chiller tons 32
The harm passed on to the user is several times any amount of honest supply/investment!
Sample keywords: hydrogen sulfide EGSB tannery (no commas!)
