UNIT 1 UNITS AND DIMENSIONS

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2 UNIT 1 UNITS AND DIMENSIONS Unit is any measure or amount used as a standard for measurement. It is a means of the measurable extent of a physical quantity. The derived unit is a combination of primary units. The derived quantity is a combination of two or more primary dimensions. Their units are a combination of primary units. TABLE 1 PRIMARY DIMENSIONS AND UNITS IN DIFFERENT SYSTEM OF UNITS Quantity Units Symbol Dimension FPS MKS CGS FPS CGS CGS Length Foot metre centimeter ft m cm L Mass Pound kilogram Gram lb kg g M Time Second second Second S S S T Temperature Fahrenheit Celsius Celsius 0 F 0 C 0 C K

3 TABLE 2 DERIVED UNITS IN SI Quantity Units (symbolic abbreviation in bracket) Dimension SI MKS/CGS FPS (based on SI) Area square metre (m 2 ) square centimetre (cm 2 ) square feet (ft 2 ) L 2 Volume cubic metre (m 3 ) cubic centimetre (cm 3 ), litre (L) cubic feet (ft 3 ) L 3 Density kilograms per cubic metre grams per cubic centimetre (g cm -3 ) pounds per cubic foot (lb ft -3 ) ML -3 (kg m -3 ) Mass flow rate kilogram per second (kg s -1 ) grams per second (g s -1 ) pounds per hour (lb h -1 ) MT -1 Quantity Units (symbolic abbreviation in bracket) Dimension SI MKS/CGS FPS (based on SI) Molar flow rate moles per second (mol s -1 ) moles per second (mol s -1 ) pounds per hour (lb h -1 ) nt -1 Volumetric flow rate cubic metres per second (m 3 s -1 ) cubic centimetres per second (cm 3 s -1 ), litre per second (L s -1 ) cubic feet per hour (ft 3 h -1 ) L 3 T -1

4 Mass flux moles per square metre per second (mol s -1 m -2 ) moles per square metre per second (mol s -1 m -2 ) pound moles per hour per square foot (lb-mol h -1 ft -2 ) nt -1 L -2 Force newton (N) kilogram-force (kgf ) pound force (lbf) MLT -2 Pressure newton per square metre or pascal (N m -2 or Pa) kilogram force per square centimetre (kgf cm -2 ) pound force per square foot (lb ft -2 ) ML -1 T -2 Quantity Units (symbolic abbreviation in bracket) Dimension SI MKS/CGS FPS (based on SI) Energy joule (j) metre kilogram force (m kgf) foot pound force (ft lbf) ML 2 T -2 Power watts or joules per second (W or J s -1 ) horse power (HP) foot pound force per second (ft lb s -1 ) ML 2 T -3 Heat, Enthalpy Joules (J ) British kilocalories (Kcal ) Thermal Unit (Btu) ML 2 T -2 Heat capacity joules per kilogram per degree kelvin (J kg -1 K -1 ) kilocalories per kilogram per degree celsius (kcal kg -1 0 C -1 ) Btu per pound per degree Fahrenheit (Btu lb -1 0 F -1 ) L 2 T -2 K -1 Molar heat capacity Joules per mole per degree kelvin (J mol -1 K -1 ) Kilocalories per mole per degree celsius (kcal mol -1 0 C -1 ) Btu per pound per degree Farenheit (Btu lb mol -1 0 F -1 ) ML 2 T -2 K -1 N -1 Heat flux joules per square metre per second (J s -1 m -2 W m -2 ) kilocalories per hour per square metre (kcal h -1 m -2 ) Btu per hour per square foot (Btu h -1 ft- 2 ) MT -3

5 Conversion factors 1) Length: 1ft = 12 inches 1 inch = 2.54 cm, 1ft = cm, 1 yard= 36 inch = 3 feet, 1 mile=1.609 km 2) Mass: 1 kg = 1000 g, 1 lb = g = kg 1 ton = 1000 kg 3) Temperature: K= C+ 273 C= 5/9 ( F-32) R = F ) Pressure: 1 bar = 1x10 5 N/m 2 = 750 mmhg 1 N/m 2 = 1 Pa 1atm = x 10 5 N/m 2 = 760 mmhg = 33.91ft water = psi = inches Hg 1 mm Hg = 1 torr 5) Force: 1 kgf = N 6) Energy: 1 calorie =4.184 J 1 Btu = 252 calorie 1 J = 1 kgm 2 /s 2 = 10 7 erg 1 Btu =1055 J 7) Volume: 1cc = 1 ml 1 L = 1000 cc 8) Power: 1 Hp = W. 1W = 1J/s. 1 W = 1 N.m 9) Density: 1 g/cc = 1000 kg/m 3 = 1 kg/l Value of Universal gas constant R= J/(gmol.K) = (cal/gmol.k) = atm.l/(gmol.k) = m 3 Pa/(gmol.K) = m 3 KPa/(kgmol.K) Temperature scale: 1) C= R

6 2) C= K 3) C =1.8 F 4) K =1.8 R Pressure scale: 1) Absolute pressure = gauge pressure + atmospheric pressure 2) Absolute pressure = gauge pressure + barometric pressure 3) Absolute pressure = atmospheric pressure vacuum pressure

7 UNIT 2 GRAM ATOM Used to specify the amounts of chemical elements. It is defined as the mass in grams of an element which is equal numerically to its atomic weight. gram atoms of an element = wt.in grams atomic weight (1) Similarly, the mass in kilograms of a given element that is numerically equal to its atomic weight is called a kilogram-atom. Similarly, kilogram atoms of element kg atoms of an element = wt.in kg atomic weight (2) For chemical compounds, a mole is defined as the amount of substance equal to its molecular weight / formula weight. GRAM MOLE Used to specify the amount of chemical compounds.it is defined as the mass in grams of substance that is equal numerically to its molecular weight. gmoles of compound = wt.in molecular g weight (3) Gram mole of a substance is the mass in grams of the substance that is numerically equal to its molecular weight. Similarly, kgmoles of compound = wt.in molecular kg weight (4) Mole is defined as the amount of substance equal to its molecular weight. (1) Calculate the kilogram atoms of carbon which weighs 36 kg Solution: 36 kg carbon Atomic weight of carbon = 12

8 wt.in kg 36 katom of carbon = = = 3 atomic weight of carbon 12 (2) Calculate the kilograms of Na of which the amount is specified as 3 katom. Solution: 3 katom Na Atomic weight of Na = 23 katom of Na = wt.in kg of atomic weight Na of Na Kg of Na = katom of Na x Atomic weight of Na = 3 x 23 = 69 (3) How many kilograms of ethane are there in 210 kmol? Solution: Basis: 210 kmol ethane. Atomic weights: C=12, H=1, Chemical formula of ethane = C 2 H 6 Molecular weight of ethane = 2x12+1x6 = 30 kmol of C 2 H 6 = wt.in kg of C 2 H 6 mol. weight of C 2 H 6 kg of ethane (C 2 H 6 ) = kmol of ethane x Molecular weight of ethane = 210 x 30 = 6300 kg 210 kmol of ethane = 6300 kg ethane (4) Convert 88 kg of carbon dioxide into its amount in molar units. Solution: Basis: 88 kg of carbon dioxide Molecular formula of carbon dioxide = CO 2 Atomic weights: C=12, O=16 Molecular weight of CO 2 = (1x12) + (2x16) = 44 wt.in kg of C O 88 kmol of CO 2 2 = = = mol. weight of CO g of CO 2 = 2 kmol CO 2

9 (5) Find the moles of oxygen present in 500 grams Solution: Basis: 500 g of oxygen Molecular weight of O 2 = 2 x 16 = kmoles of O 2 = = (6) Convert 499 g of CuSO 4.5H 2 O into moles. Solution: Basis: 499 g of CuSO 4.5H 2 O Atomic weights: Cu=63.5, S=32, O= 16 and H = 1 Molecular weight of CuSO 4.5H 2 O = (1x63.5)+(1x32)+(4x16)+5(2x1+1x16) = Moles of CuSO 4.5H 2 O = 499 = 2 mol kmoles of CuSO 4.5H 2 O = = The relationship of a compound and its constituents is given for some compounds as follows: Each mole of NaOH contains one atom of Na 1 mol of NaOH 1 atom of Na 1 atom of H Each mole of NaOH contains 1 atom of Na. The sign refers to equivalent to and not equal to. Similarly for H 2 SO 4 and S 1 mol of H 2 SO 4 = 1 atom of S (atom is written for gram-atom). 1 mol of H 2 SO 4 = 1 atom of S (atom is written for gram-atom) 1 kmol of H 2 SO 4 = 1 katom of S (katom is written for kilogram-atom) i.e., each mole of H 2 SO 4 contains 1 atom of S. For CuSO 4.5H 2 O and CuSO 4 1 mol of CuSO 4.5H 2 O 1 mol CuSO 4 1 kmol of CuSO 4. 5H 2 O 1 kmol of CuSO 4 (7) How many kmoles of H 2 SO 4 will contain 64 kg S

10 Solution: Basis: 64 kg of S Atomic weight of S = 32 kg of S 64 atoms of S = = = katom 32 2 Each moles of H 2 SO 4 contains one atom of S. 1 kmol of H 2 SO 4 1 katom of S 1 moles of H 2 SO 4 = x 2 = 2 kmol 1 (8) Find kmoles of K 2 CO 3 that will contain 117 kg of K? Solution: Basis: 117 kg of K Atomic weight of K = atoms of K = = 3 katom 39 Each mole of K 2 CO 3 contains 2 atom of K 2 atom of K 1 mole of K 2 CO 3 2 katom of K a kmol of K 2 CO 3 1 moles of K 2 CO 3 = x 3 = kmol (9) How many kilograms of carbon are present in 64 kg of methane? Solution: Basis: 64 kg of methane Atomic weight of C = 12 Molecular weight of CH 4 = 16 1 katom of carbon 1 kmol of CH 4 12 kg of carbon 16 kg of CH 4 i.e., in 16 kg of CH 4, 12 kg of carbon are present. So amount of carbon present in 64 kg of methane = 12 x 64 = (10) Find equivalent moles of Na 2 SO 4 in 644 kg of Na 2 SO 4.10H 2 O crystals Solution: Basis: 644 kg of Na 2 SO 4.10H 2 O crystals Molecular weight of Na 2 SO 4 = 2x23+1x32+4x16 = 142 Molecular weight of Na 2 SO 4.10H 2 O = (2x23)+1x32+4x16+10 (2x1+1x16) = moles of Na 2 SO 4.10H 2 O = = 2 mol 322

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Materials and Energy Balance in Metallurgical Processes. Prof. S. C. Koria. Department of Materials Science and Engineering

Materials and Energy Balance in Metallurgical Processes Prof. S. C. Koria Department of Materials Science and Engineering Indian Institute of Technology, Kanpur Module No. # 01 Lecture No. # 02 Measurement