Conduction: Conduction is process of transfer of energy from more energetic molecule to those with low energetic molecule of same object or objects in physical contact with one another until the equilibrium is reached. Conduction of Heat is occurred due to molecular vibration without appreciable displacement of particles of the object. Examples of conduction: Heat flow through the brick wall of a furnace, the metal sheet of a boiler and the metal wall of a heat exchanger tube. Basic Law governing the heat transfer by conduction may be given as Rate of Heat Transfer = Driving Force Thermal Resistance ............................................(1) Driving force is temperature drop across solid surface Thermal Resistance is conductivity of material Theramal Resistance = Thickness of the body Proportionality Constant×Surface area .............................(2) Theramal Resistance = L KmA The thermal resistance can be obtained from Fourier’s Law Fourier’s Law “The rate of heat transfer through a uniform material is directly proportional to the area and negative gradient in the temperature and inversely proportional to the length of the path of heat flow” Rate of heat flow ∝ Area×Temparature Difference Thickness of the body q ∝ A∆t L q = KmA∆t L Where q is Heat flow through the body per unit time A is area Surface area Δt is Temperature difference between faces of body L is thickness of the body Conduction Mr. Sandesh N. Somnache Assistant Professor Department of Pharmaceutics PES’s Rajaram and Tarabai Bandekar College of Pharmacy, Farmagudi, Pionda, Goa-403401 Email:
[email protected] Pharmaceutical Engineering Sandesh N Somnache 1 Km is Proportionality constant ∆t = t1 − t2 If Fourier’s Law is applied to the thin section of wall having thickness dL, dQ dθ = − kAdt dL .................................(3) Where, Q is Heat Transferred θ is time k is proportionality constant (thermal Conductivity Constant) t is temperature dt/dL is temperature gradient At steady state of heat transfer, dt/dL is constant (q) ⸫ q = − kAdt dL ..........................(4) Here negative sign indicates decrease in temperature in direction of flow of heat. Where q is rate of heat transfer ⸫ qdL A = − kdt Integrating above equation between limit, L=0 when t= t1 and L= L when t= t2, q ∫ dL L 0 = − ∫ kdt t2 t1 or q ∫ dL L 0 = ∫ kdt t1 t2 qL A = km(t1 − t2 ) qL A = km∆t ...........................(5)
Pharmaceutical Engineering Sandesh N Somnache 2 q = ∆t L/kmA But, L kmA = Resistance q = ∆t Resistance Rate of Heat Transfer = Driving Forse Thermal Resistance Compound Resistance in series Driving Force Δt = Δt1+ Δt2+ Δt3 Overall Resistance R = R1 + R2 + R3 According to Fourier’s Law, R = L1 k1A + L2 k2A + L3 k3A If q is entire heat pass through the resistances in the series, q = q1 + q2 + q3 Using the principles conduction, the rate of heat transfer q may be expressed as q = ∆t R1+R2+R3 The contribution of temperature drops to the total temperature and individual resistance to the total resistance can be expressed mathematically as ∆t: ∆t1: ∆t2: ∆t3 ∷ R: R1: R2: R3
Pharmaceutical Engineering Sandesh N Somnache 3 Heat Flow Through a Cylinder Rate of Heat transfer (q) through a thin layer of cylinder can be written as, q = coefficient temparature difference length of heat flow i.e. thickness (area) q = −k dt dr (2πrN) Where 2πrN is area of heating surface i.e. the interior of the cylinder and N is length of cylinder Rearranging the above equation, dr r = − 2πNk q dt Integrating above equation ∫ dr r r 2 r 1 = − 2πN q ∫ kdt t 2 t 1 We will get q = 2πNkm(t1−t2) In ( r2 r1 ) But for flat wall q = kmAm(t1−t2) L ⸫ kmAm(t1−t2) L = 2πNkm(t1−t2) In ( r2 r1 ) Or Am = 2πN(r2−r1) In( r2 r1 ) But Am = 2 prmN ⸫ 2πrmN = 2πN(r2−r1) In( r2 r1 ) ⸫ rm = (r2−r1) In( r2 r1 ) ⸫ rm = (r2−r1) 2.303 log( r2 r1 )
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