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2300_les05 - II. Energy and the First Law of Thermodynamics...

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Unformatted text preview: II. Energy and the First Law of Thermodynamics A. Energy Transfer as Heat Transfer. Energy is Conserved. 1. Energy can cross the boundary of a closed system by only two mechanisms: heat transfer and work transfer. 2. The change in energy of a closed system is equal to the net heat transferred to the system minus the net work performed by the system. Net work out, Wout,net = Wout - Win ∆Etotal = ∆Esystem + ∆Esurroundings = 0 ∆Esystem = −∆Esurroundings Closed system ∆Esys = Qin,net − Wout ,net (2-34); (4-17) Net heat in, Qin,net = Qin - Qout Lesson 5, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics B. Heat transfer 1. Heat is energy transferred across the boundary of the system due to a difference in temperature between the system and the surroundings. a. The heat flux, q, is the rate of heat transfer per unit area of surface. The units are W/m2 or kW/m2 or Btu/(ft2 h) b. The heat transfer rate (kW) is calculated from the heat flux as Q = ∫ qdA A A is the area over which heat transfer is occurring. Lesson 5, Geof Silcox, Chemical Engineering, University of Utah 1 II. Energy and the First Law of Thermodynamics c. The energy transferred (kJ) between times t1 and t2 is t Q = ∫ 2 Qdt t1 (2-15) d. The energy transferred (kJ) by heat during a process is Q = ∫ δQ 1 2 Lesson 5, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 2. Conduction heat transfer (transfer of energy by putting a hotter object in contact with a cooler system) Fourier’s law Wall (2-52) Qcond dT = −kt A dx T1 Qcond Steady state, kt = constant Qcond ⎛T −T ⎞ = −kt A ⎜ 2 1 ⎟ ⎝L⎠ L x T2 kt = thermal conductivity, W/(m K) kt(air) = 0.026, kt(water) = 0.613, kt(copper) = 401 Steady state kt = constant Lesson 5, Geof Silcox, Chemical Engineering, University of Utah 2 II. Energy and the First Law of Thermodynamics 3. Convection heat transfer (transfer of energy between a solid surface and a moving fluid) Newton’s law of cooling, h = heat transfer coefficient. Wall L T1 Qconv Qconv = hA(T2 − Tf ) (2-53) Flow & fluid free conv, air forced conv, air h, W/m2K 5-12 10-300 T2 Tf Lesson 5, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 4. Radiation heat transfer (transfer of energy due to emission and absorption of electromagnetic radiation) Wall 4 Qrad = σε 2 A T24 − Tsur L T1 ( ) (2-57) T2 Tf Tsurface σ = Stefan-Boltzmann constant = 5.67x10-8 W/(m2 K4). ε2 = emissivity of surface 2, 0 < ε ≤ 1. The emissivities of skin, concrete, and stainless steel are 0.95, 0.91, 0.3. Lesson 5, Geof Silcox, Chemical Engineering, University of Utah 3 II. Energy and the First Law of Thermodynamics 5. Heat transfer through a simple wall inside T = Ti Wall T1 outside T = To Qconv,o Qconv,i Qrad,i Qcond L x T2 Qrad,o It is common practice in the building industry to represent all three modes of heat transfer by a single, overall heat transfer equation. R = “R value” or overall thermal resistance, hrft2-°F/Btu or m2-°C/W. Q= A( Ti − To ) R Lesson 5, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 6. Application A poorly insulated home with 1300 ft2 of ceiling is located in a region with an 8-month heating season. The average outdoor temperature during the heating season is 40°F and the inside temperature is fixed at 70°F. The owner will spend $850 to increase the Rvalue of the insulation in the ceiling from 11 to 40 (hr-ft2°F/Btu). The house is heated with electricity that costs 10 cents / kWh. How much energy will the owner save each year and how long will it take for the saved energy to pay for the new insulation? Lesson 5, Geof Silcox, Chemical Engineering, University of Utah 4 II. Energy and the First Law of Thermodynamics 6. Application Heat loss rate with existing insulation: Q= A(Ti − To ) 1300 ft 2 (70 − 40 )o F = = 3545 Btu / h R 11 ( ft 2 o F h / Btu ) Heat loss rate with new insulation: A(Ti − To ) 1300 ft 2 (70 − 40 )o F Q= = = 975 Btu / h R 40 ( ft 2 o F h / Btu ) Energy saved in one year: Btu h 24 h 30 day 8 mo = 4339 kWh Energy saved = Btu day mo yr yr 3414 kWh Lesson 5, Geof Silcox, Chemical Engineering, University of Utah ( 3545 − 975 ) II. Energy and the First Law of Thermodynamics 6. Application $ saved in one year: $ saved = 4339 kWh $0.10 / kWh = $434 / yr yr Time to recover cost of insulation: $850 = 1.96 heating seasons $434 / y Lesson 5, Geof Silcox, Chemical Engineering, University of Utah 5 ...
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This note was uploaded on 09/02/2010 for the course PHYS 2300 taught by Professor Silcox during the Fall '09 term at Utah.

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