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ae VALUE (I) Definition of Calorific Value (CV) It is defined as the total q volume of the fuel at STP. uantity of heat liberated after the complete combustion of a unit mass or a Calorific Value (cv) = Amount of heat liberated Amount of fuel used Ad op (II) Units of calorific value oe System Unit Solid/ liquid fuel | Gaseous fuel CGS cal/gm cal/em* SI joule/kg joule/m? MKS kcal/kg kcal/m? (III) Significance of calorific value 3 Calorific value indicates the fuel efficiency. Highly eacnte “= efficient fuel has higher calorific value as it is capable of producing large amount of heat energy by utilizing small a] E t : 1. ae (2) quantity of fue igh oe Valu Low Calorific Value % 1.2.1 Types or Classification of the 2 ey Cy ifi ‘Gross Calorific Value | Net Calorific Value Calorific Value ae (@cv) (NCV) State the two values of the calorific values , ith help of a suitable chart. 2 (2F3)Fig. 1.2.1 : Calorific values New Syllabus w.e.f Academic Year 24-25) (M1-01) (Fuels and Combustion)....Page no, (1-6) —— —, MuU-Sem 1) Applied Chemistry (MU= w 1.2.2 pion, etal Nome iges Calorific Value ee = of High Calorific Value High calorific value (HCV) also called as gross calorific value (GCV) is defined as ‘the total amount of the heat produced when a unit mass (in the case of a solid fuel) or a unit volume (in the case of a liquid or gaseous fuel) is burnt completely and the products of combustion are cooled to the room temperature.’ (11) Significance of High Calorific Value (i) The total heat produced by the complete combustion of a fuel is directly proportional to its calorific value. (ii) Therefore a good fuel should have a high value of HCV. wa 1.2.3 Low Calorific Value/ Net calorific Value : 3 ‘ (2 Marke), f (1) Definition of low calorific value Low calorific value (LCV) also called as net calorific value (NCV) is defined as ‘the net quantity of the heat produced when a unit mass (in the case of a solid fuel) or a unit volume (in the case of a liquid or gaseous fuel) is burnt completely and the products of combustion are permitted to escape into the atmosphere NCV or LCV = [GCV- heat released for condensation of steam] (II) Derivation of mathematical relation between GCV and NCV Mathematical relation: NCV = GCV — (mass of the hydrogen/unit weight of the fuel burnt x g x latent heat of steam) Note: Justification of relation Mol. We. i : ea If the % hydrogen content in a fuel is H, the net calorific values can be calc a as oviewa: 2 gmof hydrogen will produce 18 gm steam : 1 gm of hydrogen will produce 18/2 = 9 gm steam Then, H% or H/100 gm of hydrogen will produce = 9 x (H/100) gm steam = 0.09H gm steam Heat released for the condensation of 0.09H gm steam = 0.09H x L, [Lr= Latent heat of water] Hence, NCV = [GCV — heat released for condensation of steam] = [Gcv - 0.09xHxLj] *, NCV = GCV - 0.09 x H x 587 cal/gm or kcal/kg (New Syllabus w.e.f Academic Year 24-25) (M1- -01) ee Given, latent heat of steam (water) = 587 cal/gm oe (Fuels and Combustion)....Page no, (1g ta) — Applied Chemisty (Mu-Sem 1) M7] soln. : Given ‘ C= 82%, H = 6% O=8 %, S = 0.05%, N = 3%, Ash = 0.5% Required : GCV. NCV @ GCV = 700 =Hy[ 8080 C + 34500 (a eS) +2240 s| = wl 80 x 82 + 34500 (6 - 8). 2240 x05 | = $361.8 KCal/kg aD NCV = [HCV-(0.09 Hx 587)] = $044.82 KCal/kg sae ey MU - Q. 6(b), May 14, 5 Marks 1.5 g of a coal sample was analysed for nitrogen content by Kjeldahi’s method. The liberated ammonia required 14 ml of 0.1 N H,SO, solution for neutralization. In a separate expetiment using Bomb calorimeter, 1.5 g of the same sample gave 0.3 g of BaSO,. Calculate percentage nitrogen and sulphur in the sample. M Soin. : Given C= 80%, H=4%,0 = 6%, S= Required : GCV, NCV @ GCV = inl C+34500 (4 =3 +2240 s |kcalme 3,N=2%, Ash =5% ash = 0, = 8356.05 ~ (0.09 x 5.5 x 587) Required : GCV, NCV = 8065.48 Keal/kg Ans. () GCV ISSR E-AMU - Q. 6(b), May 10, 5 Marks = Tas 880 C+ 34500 (i H = +2240 s| A sample of coal has the following composition by r mass, C = 85%, H = 7%, O = 3.5%, S = 3.5%. ‘~ = 700 | 8080 x 66434500 (4 3), pee N = 2.1%andAsh = 4.4%. Calculate HCV and LCV using Dulong formula, = 5538.9 KCal/kg M sotn. : HCV ~ (0.09 x Hx 587) Given : C= 85%, H=7 %, 0 =3.5% 5538.9 - (0.09 x 4587) S =3.5%,N=2.1%, ash = 4.4% Required : HCV, LCV 5327.58 KCal/kg +. Ans. UEx. 1.2.8 J 1(c), May 18, 3 Marks A sample of coal has the followi Z composition : = T5 | 8080 x 85 + 34500(7 dD NCV iT u 5 L dQ ICY = Too [ s080 C + 34500 (a a + 2240 s| =) + 2240 x 35 C = 10%, O = 23%, H = 5%, S = 1.5%, N = 0.4%, Ash = 0.1%. Calculate the G.C.V. of this fuel. = 9232.025 KCal/kg M1 Soin. : Given: C= 70%, H = 5%, O = (1) LCV = [HCV -0.09 x H x 587] S = 1.5%, N =0.4%, ash = 0.1% = [9232.025 — (0.09 x 7 x 587) Required : GCV = 8062.215 Keal/kg () GCV Homwork Problems fe) Q.1 Calculate HCV and LCV of the following two = Too Ta5| 8080 C + 34500 (H = 2 + 2240 s| coal samples with : (A) C= 85%, H=7%,0=3%,S = = =| 8080 x 70 + 34500 (s a + 2240 x 15] 3.5%, N = 2,1 % and Ash = 4.4 % 100 (B) C = 80 %, H = 7%, 0 = 2.8%, S = = 6422.725 KCal/kg . Ans. 3.4%, N = 9% and Ash = 3.5 % (Ans. : HCV = 9232.03 kCal/kg, LCV = 8862.02 kCal/kg) [ian eescIMU - Q. 1(c), Dec. 17, 3 Marks Calculate the higher and lower calorific values of coal | g,2 A sample of coal has the following sample containing 84% carbon, 1.5% sulphur, composition by mass : 0.6 Nitrogen, 5.5% oan and 8.4% oxygen. C= 70%, H = 9%, O = 4%, S = 2%, N= 1% and Ash = 14% M soln. : Calculate gross calorific values of the fuel Seen C= y = %,O =8.4% using Dulong’s formula. Given : C = 84%, H=5.5 %, : 8.4% g Sonar Rare S = 1.5%, N = 0.6%, ash = 0.1% (Ans. : HCV = 8808 kCal/kg, Required : HCV, LCV LCV = 8438.2 kCal/kg) 2) + 2240 s| A coal sample contains C = Lots O= ae Tox | 8080 C+ 34500 ee H = 5%, N = 0.4% Ash = 0.1 %. Calculate GCV and NCV of the fuel. () HCV = (MU - Q. 1(g), May 17, 3 Marks) 8.4 | gogo x 84 + 34500 (5.5 - 8 84) , 2240 x 13 = 100L (New wy Byllabus wi ef ReEooIs Year 24-25) (M1-01) 
