Exercise 2. (session: )

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1 1 EEN-E3001, FUNDAMENTALS IN INDUSTRIAL ENERGY ENGINEERING Exercise 2 (session: ) Problem 3 will be graded. The deadline for the return is on 7.2. at 12:00 am (before the exercise session). You can hand in your answer either into the brown mail box in K1 building next to the room 148, or straight to the course assistant before the next session ( return also possible). Problem 1 The exit temperature and relative humidity of the air used for drying cellulose are 76 C and 60 %. What is the required amount of energy per produced air dry ton of cellulose [MJ/ADt], and per kg of removed water [MJ/kg H2O] in the dryer? The temperature of the inlet air is 20 C and relative humidity 80 %. The temperature of the incoming and outgoing cellulose is about 60 C, and the dry-matter content before drying is 50 % and after drying 90 %. The properties of saturated air in temperatures 20 C and 76 C are given in the table below. Tehtävä 1 Sellun kuivauksessa käytetyn kuivatusilman poistolämpötila on 76 C ja suhteellinen kosteus 60 %. Mikä on kuivauksen energiantarve tuotettua ilmakuivaa sellutonnia kohti [MJ/ADt] ja poistettua vesikiloa kohti [MJ/kg H2O]? Ilman alkulämpötila on 20 C ja suhteellinen kosteus 80 %. Sellun lämpötila radanvientiaukoissa on noin 60 C ja kuiva-ainepitoisuus ennen kuivausosaa on 50 % ja kuivausosan jälkeen 90 %. Alla olevassa taulukossa on esitetty kylläisen kostean ilman ominaisuudet 20 C ja 76 C lämpötiloissa. Temperature Humidity Vapour pressure Lämpötila Kosteus Vesihöyryn osapaine C kg H2o/kg da kpa 20 0, , , ,19 Solution:

2 2 The calculations are performed for 1 ton of cellulose: 1 ADt of cellulose contains after drying 900 kg dry matter and 100 kg water. The dry matter content before drying is 50 % (900 kg dry matter and 900 kg water). Incoming cellulose: moist. cont. 50 % dry matter 900 kg water 900 kg Dried cellulose ton: dry mat. cont. 90 % dry matter 900 kg water 100 kg The amount of water to be removed is thus m H2O = = 800 kg H2O/ADt. We form the energy and mass balance of the drying: m dmc p,dm t in + m w,inc p,w t in + m ah in + Φ = m dmc p,dm t out + m w,outc p,w t out + m ah out m dm + m w,in + m a(1 + x in ) = m dm + m w,out + m a(1 + x out ) where the subindex dm is the dry matter of cellulose, w the water in the cellulose, a dry air, and in is the incoming stream and out the outgoing stream. The specific enthalpy h is in the equation calculated for 1 kg of dry air [kj/kg da]. The unit of the absolute humidity of moist air x is kg H2O/kg da. Because the dry-matter flow of the cellulose is constant and the temperature of the cellulose is t in = t out, the energy balance is contracted into: m w,inc p,w t in + m ah in + Φ = m w,outc p,w t out + m ah out Φ = m a(h out h in ) m evapc p,w t from which the required thermal power per produced ton of cellulose can be calculated. The required mass flows and enthalpies have to be calculated first (m evap = evaporated water). The relative humidity of air φ is defined by the following equation: φ = p h(t) p h (T), (1) where p h(t) is the actual partial pressure of the water and p h (T) is the partial pressure of the water in saturated air in the same temperature. The connection between the partial pressure and the absolute humidity in the air x [kg H2O/kg da], is received from: p h = x 0,622+x p, (2) where p is the total pressure of the air. The enthalpy of moist air per kg of dry air [kj/kg da] is calculated using: h = c p,a t + x(l + c p,h ) 1,006 t + x( ,85 t), (3)

3 3 where c p,a is the average specific heat capacity of dry air in the range 0 t C [kj/kg da C], t is the temperature of air in Celsius-degrees, l the heat of evaporation of water in 0 C [kj/kg H2O], and c p,h the average specific heat capacity of water vapor in the range 0 t C [kj/kg H2O C]. From equation (1) we get the actual partial pressures of incoming and outgoing water vapor in the air: p h,in = φ in p h (t in ) = 0,8 2,337 = 1,87 kpa p h,out = φ out p h (t out ) = 0,6 40,19 = 24,11 kpa From equation (2) we get the absolute humidities: x in = 0,622 p h,in 0,622 1,87 kpa = p p h,in (100 1,87) kpa = 0,012 kg H2O kg da x out = 0,622 p h,out 0,622 24,11 kpa = p p h,out (100 24,11) kpa = 0,198 kg H2O kg da From equation (3) we get the enthalpies per kg of dry air: h in = 1, ,012 ( ,85 20) = 50,6 kj/kg da h out = 1, ,198 ( ,85 76) = 599,5 kj/kg da The mass flow of dry air used in the drying can be solved by using the amount of evaporated water and the change in the air humidity: m evap = m a(x out x in ) m a = m evap 800 kg H2O /ADt = = 4301 kg x out x in 0,198 0,012 kg H2O /kg da /ADt da Finally we calculate the required heat per produced ton of cellulose: Φ = m a(h out h in ) m evapc p,w t Φ = 4301 (599,5 50,6) 800 4,19 60 = kj MJ = ADt ADt The energy per evaporated kg of water is 2160 / 800 = 2,7 MJ/kg H2O. The heat of evaporation of water in 60 C is about 2,4 MJ/ kg H2O. So, the most of the energy goes for evaporating water and the rest for heating up the humid air. The I,x-diagram for humid air can also be used for solving the enthalpies: Air in: t = 20 C, x = 0,012 kg H2O/kg da h 50 kj/kg da Air out: t = 76 C, x = 0,198 kg H2O/kg da h 600 kj/kg da

4 The dotted straight lines represent the wet bulb temperatures. The wet bulb temperature is the equilibrium temperature found between the boundary surface of a liquid and the air with a certain temperature and absolute humidity. 4

5 5 Problem 2 A certain chemical is vaporized with recovery heat. The heat is recovered from moist exit air, which temperature is 70 C and relative humidity 30 %. The surface temperature on the exit-air-side of the evaporator tubes is measured to 40 C. Is water from the exiting air condensed on the evaporator tubes? Determine the amount of water condensing from the exiting air on the evaporator tubes [g H2O/kg da], when the temperature of the air after the evaporator is measured to 50 C. The ratio between the mass flows of the vaporizing chemical and the exiting air is 1:28, and the heat of evaporation of the chemical is 900 kj/kg. Tehtävä 2 Eräs kemikaali höyrystetään talteenottolämmöllä. Lämpö otetaan talteen kosteasta poistoilmasta, jonka lämpötila on 70 C ja suhteellinen kosteus on 30 %. Höyrystinputkien poistoilmanpuoleiseksi pintalämpötilaksi mitataan 40 C. Tiivistyykö poistoilmasta vettä höyrystinputkien pinnalle? Määritä poistoilmasta höyrystinputkien pintaan tiivistyvän veden määrä [g H2O/kg da], kun poistoilman lämpötilaksi höyrystimen jälkeen mitataan 50 C, höyrystyvän kemikaalin ja poistoilman välinen massavirtojen suhde on 1:28 ja kemikaalin höyrystymislämpö on 900 kj/kg. Solution: The values for the condition of the humid exhaust air is looked from a Mollier-i,x-diagram, when the temperature of the air is 70 C and relative humidity 30 %. The received value for the moisture is 0,064 kg H2O/kg da and enthalpy 238 kj/kg da. The dew point of the exiting air is 44 C (can be found from the saturation curve at the current absolute humidity of the exiting air). At the dew point and temperatures below it water is condensed. The surface temperature of 40 C is below the dew point of the exiting air, so water is condensed on the evaporator surfaces at the boundary layer between the air and the tubes. The amount of condensing water is calculated by solving the enthalpy of the exiting air from the energy balance: Φ = m chem r = m air(h in h out ) h out = h in m chem m air r = kj 900 = kg The air humidity in this enthalpy and 50 C temperature is found from the Mollier-diagram. The moisture is 0,060 kg H2O/kg da, and the amount of condensing water is thus: m cond = x = 0,064 0,060 = 4 g/kg da

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7 7 Problem 3 * A wood-fuel is dried with hot air in a dryer. The rate of evaporating water is 0,28 kg/s. Ambient air, with temperature 15 C and relative humidity 40 %, is used for the drying. The air is heated to 90 C before feeding it into the drying chamber. The mass flow of the air is 15 kg/s. a) What is the absolute humidity of the exiting air from the drying chamber [kg H2O/kg da]? b) What is the specific heat consumption of the dryer [kj/kg H2O(evaporated)]? c) What is the temperature of the outgoing air from the dryer, if drying process is adiabatic i.e. air moisturizes adiabatically? Hint: use mollier i,x diagram. Tehtävä 3 * Puupolttoaineen kuivaamossa kuivatetaan puuta kuumalla ilmalla. Puusta haihtuu vettä 0,28 kg/s. Kuivaamiseen käytetään ulkoilmaa, jonka lämpötila on 15 C ja suhteellinen kosteus 40 %. Ilmaa lämmitetään 90 C:een ennen kuivuriin syöttämistä. Kuivan ilman massavirta on 15 kg da/s. a) Mikä on kuivurista ulos tulevan ilman absoluuttinen kosteus [kg H2O/kg da]? b) Mikä on kuivurin ominaislämmönkulutus [kj/kg H2O(haihdutettu)]? c) Mikä on kuivurista poistuvan ilman lämpötila, jos kuivausprosessin oletetaan olevan adiabaattinen t.s. ilman kostutus on adiabaattinen? Vinkki: käytä mollier i,x diagrammia. Problem 4 Live steam is generated with a boiler in a certain power plant by combusting moist wood chips with the boiler efficiency of 0,80. The price of the purchased chips is 40 /ton of moist chips, and their water content is 50 % (on wet basis, w.b, i.e the mass of water of the mass of the moist wood). The heating value of the wood chips is decided to be raised by drying the chips to 20 % (w.b) moisture content. The drying will be performed with a dryer using natural gas as fuel. The specific heat consumption of the dryer is 3,6 MJ/ kg H2O. Calculate the production costs of the live steam in both cases [ / of produced live steam], when the investment costs are ignored. Is the use of this kind of dryer profitable here? The boiler efficiency in the power plant can be assumed as constant in both cases. The heating value of natural gas is 50 MJ/kg and price 0,65 /kg. The lower heating value of the dry matter in wood is 19 MJ/kg. The lower heating value of moist wood q GHV [kj/kg] can be calculated by using the following equation:

8 8 q GHV = q i (100 w) 24,40 w, 100 where the water content w is in mass-% of the moist fuel, and the unit of the lower heating value q i of the dry-matter is kj/kg. Tehtävä 4 Erään voimalaitoksen kattilalla tuotetaan tuorehöyryä polttamalla kosteaa puuhaketta kattilahyötysuhteella 0,80. Ostetun hakkeen hinta on 40 /tonni kosteaa haketta, ja vesipitoisuus on 50 % (w.b, eli veden massan osuus kostean aineen kokonaismassasta). Hakkeen lämpöarvoa päätetään kasvattaa kuivaamalla haketta 20 %:n kosteuteen (w.b) maakaasua käyttävällä kuivurilla. Kuivurin ominaislämmönkulutus on 3,6 MJ/kg H2O. Laske tuotetun tuorehöyryn kustannukset kummassakin tapauksessa [ / tuotettua tuorehöyryä], kun investointikuluja ei huomioida. Onko kuivurin käyttö kannattavaa tässä? Voimalaitoksen kattilahyötysuhde voidaan olettaa pysyvän muuttumattomana molemmissa tapauksissa. Maakaasun lämpöarvo on 50 MJ/kg ja hinta 0,65 /kg. Puuhakkeen kuiva-aineen tehollinen lämpöarvo on 19 MJ/kg. Kostean puupolttoaineen tehollinen lämpöarvo q GHV [kj/kg] lasketaan kaavalla: (100 w) q GHV = q i 24,40 w, 100 jossa vesipitoisuus w on paino-%:ina kosteasta polttoaineesta ja kuiva-aineen tehollisen lämpöarvon q i:n laatuna on kj/kg.

9 9 Solution: Let s compare the fuel costs with and without the dryer. Without the dryer: Calculate the heat value of wet fuel, as the wood chips are wet (50%, w.b): q GHV = kj kg (100 50) 24,40 50 = kj 100 kg Calculate the energy cost included in the wet fuel: C wf = 40 /t wf 8280 MJ/t wf 3600 MJ = 17,39 Cost of live steam production is thus: C s1 = C wf η b = 17,39 0,80 = 21,74 With the dryer: Since the mass of the dry mattter does not change during the drying, all mass flows can be calculated in dry matter basis. The following descriptions are used: w = water, dm = dry matter, wf = wet fuel (50%), and df = dried fuel (20%). Calculate first the fuel moisture content (moisture ratio, kg H2O/kg dm) before and after drying: w = m w m w + m dm, where w denotes moisture, m w denotes mass of water and m dm denotes mass of dry matter. m w m dm = w = u = moisture ratio 1 w moisture ratio before drying: u wf = 0,50 1 0,50 = 1,0 kg H2O/kg dm moisture ratio after drying: u df = 0,20 1 0,20 = 0,25 kg H2O/kg dm Water is evaporated during drying: u = 1,0 0,25 = 0,75 kg H2O /kg dm

10 10 Thus, the energy consumption of the dryer in wood chips dry-matter basis is: 3600 kj 0,75 kg H2O = 2700 kj kg H2O kg dm kg dm Calculate the heat value of the fuel after the drying, that is in moisture content of 20% (w,b): q GHV = kj kg (100 20) 24,40 20 = kj 100 kg Calculate the required fuel flow when 1 live steam is produced. 1 m df = 3600 MJ q GHV η b m df = MJ t df 0,80 t df 3600 MJ = 0,3059 Required wood chip dry matter per one live steam produced is thus: m dm = m df (1 w) m dm = 305,9 kg df (1 0,20) = 244,7 kg dm Costs of natural gas and wood chips: Requirement of natural gas used in the dyer per wood chips dry matter: m ng = 2,7 MJ/kg dm 50 MJ/kg = 0,054 kg ng kg dm Thus, the requirement of natural gas per produced live steam: 0,054 kg ng kg dm 244,7 kg dm = 13,214 kg ng Cost of natural gas is then; C ng = 0,65 kg ng 13,214 kg ng = 8,589 Calculate the cost of wet wood chips in dry matter basis: m dm = m wf (1 w) = 0,5 m wf

11 11 = 1 m dm 0,5 m wf 2 40 t wf = 80 t dm Cost of wood chips in the basis of produced live steam: C wf = 0,080 kg dm 244,7 kg dm = 19,576 Total costs with the dryer: C s2 = C ng + C wf = 8, ,576 = 28,17 C s2 > C s1, This, it is not profitable to use the dryer. Alternative way to calculate the costs with the dryer: Calculate the costs and the mass of the produced live steam in the basis of one ton of wet wood chips: Total costs are then: C s2 = 40 /t wf+17,55 /t wf 2,043 /t wf = 28,17 /