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Doğal gazlı kat kaloriferi dizaynı

The Design of natural gas domestic heater

  1. Tez No: 21713
  2. Yazar: TAHSİN ÖZDEN
  3. Danışmanlar: PROF. DR. OSMAN F. GENCELİ
  4. Tez Türü: Yüksek Lisans
  5. Konular: Makine Mühendisliği, Mechanical Engineering
  6. Anahtar Kelimeler: Doğal gaz, Isıtma sistemleri, Isıtıcılar, Tasarım, Natural gas, Heating systems, Calorifics, Design
  7. Yıl: 1992
  8. Dil: Türkçe
  9. Üniversite: İstanbul Teknik Üniversitesi
  10. Enstitü: Fen Bilimleri Enstitüsü
  11. Ana Bilim Dalı: Belirtilmemiş.
  12. Bilim Dalı: Belirtilmemiş.
  13. Sayfa Sayısı: Belirtilmemiş.

Özet

ÖZET Katkaloriferi bireysel ısıtmada en yaygın kullanılan ısıtma araçlarından biridir. Bunda gerek ekonomikliğin ve gerekse kullanıcının özgürlüğünün katkısı büyüktür. Bu tez çalışması ile doğalgazlı katkaloriferi dizaynına çalışmış, teorik hesaplamaların yanında bir deney düzeneğinde deney sonuçları da alınarak mukayese edilmiştir.

Özet (Çeviri)

SUMMARY THE DESIGN OF NATURAL GAS DOMESTIC HEATER INTRODUCTION. The designed domestic heater is planned for a heat capacity of 19.5 kW an<^ with an atmospheric burner.Also the boiler of the domestic heater is planned as steel construction. PART 1. THE BURNING OF THE NATURAL GAS. Heat values of the natural gas are calculated according to heat value of each one of the element of it. The values of the elements of the natural gas are given below. Element The ^-S*1 heat value (q) The low heat value (qO Methane-CH 37706-33043 Ethane-C n 66060-60434 Propane-c\fL 94042-86515 Butane-C.H * 121874-112448 Pentane C5H^2 149781-138492 The natural gas is burnt as shown below CxHy+ (x+ -^-) 02 y xC02+ -*- H20 (1.1) 1.1. Thermal Calculations 1_The high heat value (HQ) : Ho=CH4^CH4+C2H6^C2H6+C3H8^C3H8+ C4H10 ^C^^2 ' qc H (1.2) U5tt12 Ho=0, 85. 3 770 6+0, 07. 6 6 060+0, 03. 9 4042+0, 02. 121874+ 0,01. 149781 Hq= 43431 kJ/Nrn32- Low heat Valve (Hu) : VCH4-W+C2H6'q,C9H,+C3H8^C,Hfi+C4H10-C3cAH-n+C5H12' 4 2 6 3 8 4 10 q'C5H12 (I-3) H =0,85.33943+0,07.60434+0,03.86515 +0,02.112448+0,01. 138492 H = 39311 kJ/Nm3 u 3- The theoretical amount of air (Vho) - EU + (x+ -^ICH-O | (1.4 v loo i _jl_c0 + ti ho 21 2 2 2 4XY2 There is not CO, H_ and 0 in. natural gas, so. v = ^00|(x+ _Y_)C H | ho 21 4 x y = ^ Id- -^)CH4+ (2+ -f-)C2H6+(3+ -f-)C3H8 + (4+-^-)C4H10+(5+-^ )C5H12| V. = ^^- 12.0,85 + 3,5.0,07 + 5.0,03 + 6,5.0,02+ 8.0,01 ho 21 VhQ=10,98 Nm3 (air)/Nm3 (fuel) 4- The excess- air coefficient (n) The excess- air coefficient can be found by Oswald Diagram which was prepared for n.Also n can be calculated from (1.5) n = - ± ± - = o,21 (1 - ) (1.5) n 2 copyO for natural gas so 0,21-02 VIfor 4%0-; 0,21 n= 0,21-0,04 = 1,2 5- The actual amount of air (Vh) V n“Vho tl-6) =1,2.10,98 =13,18 Nm3(air)/Nm3 (fuel) 6- The theoretical amount of smoke (Vgo) Vgo= (x+ İ-)CxV0'79- Vho+N2 Î1.7) = (1+ - )CH.+ (2 + - )C_H+.(3 + - ) C,Hfi +(4+ - ) 2 4 2 26v 2 38 4 C4H10+ (5+ -f)C5H12 + 0'79-Vho+N2 V = 3.0,85+5.0,07+7.0,03+6,5.0,02 +11.0,01+0,79. 10,98+0,0458 =12,12 Nm3 ( smoke) /Nm3 (fuel) 7- The actual amount of smoke (Vg) VVgo+(n-1) Vho (1-8) =12,12+ (1,2-1).10,98 =14,32 Nm3 (smoke) /Nm3 (fuel) 8-Determining Thermal Efficiency (-y) Thermal efficiency can be calculated as ”^=l-£ 2 Here -»*j is thermal efficiency and £z is thermal losses vixThermal loss must be calculated 8.1. Loss in unburned fuel (Z ) y For natural gas z =o 8.2. Loss in Ash (z,) For natural. gas zd=o 8.3. Loss in incomplete Burning (Z ) For natural gas Z =0 4. Loss in hot Wall (Z ) zc is assumed (0,5-12)%. zc is calculated as 6,3 % for this project. 8.5. Loss in Chimney (Z.) Zk=V. (İb“İa) (1.9) b g H lx-*> u Here, i = Enthalpy of chimney smoke | kJ/Nm3 | i = Enthalpy of air | kJ/Nm3 | T, = Temperature of chimney smoke T - Temperature of environment air Fox Tb = 216°C ib= 280KJ/Nm3 from table EK 2 For T = 20°C io= 4.18 KJ/Nm3 from EK 5 Hu = 39311 kJ/Nm3 V = 14,32 Nm3/Nm3 : o Vlll7 _ 14,32 (280-4,18) Zb 39311 = 0,1005 zb= 10 % 8-6 Loss in Uncontinuous activity (Zo) 3. The domectic heater is assumed active continuously S. Z = 0 o' a Thermal efficiency ('h*) i* = i- £ z = 1-(Z + y = l-(0+0+0+0, 063+0, 10+0) 1-(Z +Z-.+Z +Z +Z, + Z ) v y d e c b a' ”7 = 0,837 9- Determining Of Fuel Amount (B) B= Q'"7 (1.10) H u Q= 70.000 kJ/h ^ = 0,837 Hu= 39311 kJ/Nm3 B= 70.000. 0,837 39311 B= 1,49 Nm3/h 10- Total Air And Smoke Amount 10-1, Total Air Amount (Vfa) vh = Vh.B (1.11) IX= 13,18. 1,49 = 19,64 Nm3/h 10-2- Total Smoke Amount (V ) g V = V.B (1.12) g ? g = 14,32.1,49 = 21,34 Nm3/h 1-3 DETERMINING OF CONVECTION SURFACES First of all the temperatures of boiler are assumed. The assumed temperatures are shown at Figure 1. The formulas are given below about determining of con vection surfaces. K.A.ATm= Vg. (igo-igl) (1.13) AT - AT-, In C±T± K= hk + hR (1.15) Tm Tm ir°>75 - -0,0088 ( ~^-)2\. - - 100 100 d0,25 hfc =0,95 |4,19.0,3. ^ -0,0088 ( -S-) 2 |. -^oc(l.lS) = qC02 +*H20 (1#17) Here, K : Total heat-transfer coefficient A : Heat-transfer surface &T : Average temperature V : Total smoke amount g i : Enthalpy at: To igl: Enthalpy at ^ xFigure: 1.Determined Temperatures Of Boiler XIh, = heat-transfer coefficient by convection h = heat-transfer coefficient by radiation V : Velocity of smoke d : Diameter of canal inwhich smoke passes qCQ2: Transferee} heat per m2 by C02 qH p.: Transfered heat per m2 by H20 Under these formulas, Aol = °>882 m2 A12= 0,575 m2 A23= 0,688 m2 1-4. Calculation Of Load Losses (AP) General formula of calculation of loqd loss in boiler is, Ap=^Px+ APq + APk +Apb+4Py (1.18) Here, A : Total load loss Ap : Load loss at grid Apo: Load loss at fire place Ap,,: Load loss at convection surfaces ^p,: Load loss at chimney Ap : Load loss for ascending( + ) or descending(-)smoke Apl = ° for natural gased system^ A^p =yip gH Fo= Density of smoke at temperature of fire place (T0) g is taken as (9,81 m/ sec2) HQ = High.t of the heat transferred surface /**=/* - ~, j& is taken 1,3 kg/m3 (1.19) m Apk = Aps+ Apc (1.20) APS = R-! (1.21) R : Specific friction loss R is found from EK- ^ p : Load loss in. friction *s xix^Pc : Load loss of discharge APC : T./>' - (1-22) ~f^ = Rate of surface areas. (1.23) (1.24) h : Hight of heat transferred surface T : Average temperature of air T : Average temperature of smoke Under these formulas, Ap=APl+Ap0+Apk+4pB+ A py = 0+2,188+4,9+0,0378-2,3494 = 4,78 Pa 1-5. Chimney Calculation /\ p is calculated 3&^,1S pa- Also 5 Pa for wind, 5 Pa for fireplace and extra 15% of up is taken for chimney calcu lation. So; P = 1,15 Ap + 5 + 5 = 1,15. 4,78+10 = 15,5 Pa Now, we must decide that natural or forced circulation is used for this system. h- £* (1.25) 3455 _ 3535 273+th 273+tb h: Required hight of chimney t, : Air temperature t. : Chimney temperature xmh is found as 3,39m This length is suitable for natural circulation. PART II. BURNING AT ATMOSPHERIC BURNER As shown in figure 2. primary air is entered into natural gas when natural gas is being sprayed from nozzle. Then the secondary air is taken at fire place. Because of the burner mades of stainless steel, corrosion or thermal and mechanical tension do. not occur. Primary Air JSo_zzi£_ Natural gas ft \ Secondary Air <n m? Figure: 2. Burning At Atmospheric Burner PART III THE EXPERIMENT Because of absence of the naturgas pipe line near the experiment place, a gas fuel which has same lower thermal value with nature gas is needed. It is obtained by weakening the L.P.G. 3-1, The weakening of L.P.G The average lower thermal value of LPG is about 105000 kJ/Nm3. If air is mixed wrth LPG, agas fuel which has same lower thermal value with naturgas, is obtained L.P.G consists of 70% butane and 30% prophane. Lower thermal values of Butane and prophane are 112.448 kJ/Nm3 and 86515. ckJ/Nm3 XXVqLPG = °'7' ^Butane + °> Mprophane (3'1} qLPG = °'7' 112448 + 0,3.86515 qLPG = 104668 kJ/Nm3 If 2,5m 3air is mixed to 1.5 m3 LPG, 2,5.0 + 1,5.104668,,,. qsystem 4 ^.^>> qo,ro,om = 39251 kJ/Nm3 found. ^system 3.2. Result Of Experiment The thermal efficiency is rate of taken heat to given heat. Given heat. - o\ q =V.q s g ^system = 1,77.39251 = 69521 kJ/h Taken Heat Qt= m.C. £T (3.4) Here, m : Flow rate of the water (kg/min) C : Specific heat |C is taken 1 Kcal/kg°C) AT : Temperature difference (°C) The calculation is repeated with 5 min-. intervals for one hour at end of one hour the taken heat is found 13898 Kcal/h, (= 58177 kJ/h) So, nn = - taken (3.5) ^given 58177 / 69521 nj = 0,837 ->-? is found 83,7 % XV

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