Consider a volume Ω enclosing a mass M and bounded by a surface δω. d dt. q n ds. The Work done by the body on the surroundings is

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1 The Energy Balance Consider a volume enclosing a mass M and bounded by a surface δ. δ At a point x, the density is ρ, the local velocity is v, and the local Energy density is U. U v The rate of change total energy in is: ρ v s.n d dt ρudv The heat flow from the body is q n ds The Work done by the body on the surroundings is v T n ds + ρg vdv T is the stress tensor and S in the surface (normal) Since for the body U = Q W An equivalent form is d dt ρudv + q n ds v T n ds = + ρg vdv If our control volume is a differential cube, the differential equation describing the Energy Equation is: ρu + ρuv + q = v T + ρg v ChE 333 1

2 The first term is the local rate of energy change The second is the convective energy flow The third in the gradient of the heat The fourth is the sum of reversible work and dissipation The last is the work done by the gravitational acceleration. Other Conservation Laws Mass ρ + ρv = 0 Momentum ρv + ρvv T ρg = 0 Mechanical Energy This is obtained by taking the inner product of the momentum equation and the velocity to yield v ρv + ρvv T ρg = 0 The real Energy Equation The real Energy equation is obtained by subtracting the Mechanical Energy Balance from the complete Energy Equation, using the mass balance and recognizing that H = U + PV. ρ H + v H + q = τ v W s + S Σα =1 R α H α This is simplified recalling that H T p = C p ρc p T + v T + q = τ v W s + R α H α Σα =1 S ChE 333 2

3 Applications of the Energy Equation Solids with a Uniform Temperature Suppose a metal sphere of uniform temperature, T. Heat is transferred by convection with a heat transfer coefficient, h. The temperature of the surroundings is T a Therefore the energy balance is ρc p V dt = T T a small Biot number If the sphere is initially at T 0, how does one describe the cooling of the sphere? The equation is separable so that T T a = ρc p V dt It follows that the solution is ln T T a T 0 T a = Or more explicitly Adimensionalization If I had defined the following: θ = T T a and τ = ρc p V T 0 T a T T a T 0 T a = e ρc p V t ρc p V t The solution has a simple expression... Θ = e -t/τ ChE 333 3

4 Measurement of a Convective Heat Transfer Coefficient Suppose a sphere of radius R in a stagnant gas of infinite extent. The heat flux from the sphere through the gas is given by Fourier s law q r = k g dr The heat flow through a spherical shell is constant so r 2 q r = r 2 k g dr = C The boundary conditions are T = T s at r = R and T -> T a at r -> The solution becomes T T a T R T a = R r We can calculate the heat flux at the surface as q r r =R = k g dr r = R = k g T R T a R We can define a heat transfer coefficient as h q r T R T a = k g R The corresponding Nusselt Number for heat transfer in a stagnant gas is Nu = hd k g = 2 This represents a lower bound for convective heat transfer, given that there is no gas flow. We expect the heat transfer coefficient to be larger. Note the difference between the Nusselt and Biot numbers ChE 333 4

5 The experiment The thermal conductivity of air is Btu/ft- F. If the sphere was 1 cm. in diameter, then h = 0.014(1/2.54(12)) = 4.2 Btu/hr-ft 2 - F Note that h = 4.2 Btu/hr-ft 2 - F = Btu/sec-ft 2 - F If ρ = 557 lb/ft 3 and C p = Btu/lb.- F, then τ = τ = ρ Cp V = ρ Cp D for sphere h lb/ft Btu/lb.- F 1/( ) ft Btu/sec-ft2- F 6 = sec If the heat transfer coefficient were 5 times larger (Nu = 10) then τ = 23.4 sec, If the heat transfer coefficient were 10 times larger (Nu = 100) then τ = 2.3 sec, Note (page 827) for Copper: ρ: 8933kg/m3* = 557 lb/ft 3 and Cp= 385J/kgK* x10-4 = Btu/lb m o F and k= 401W/mK * = Btu/h ft o F What is wrong here? It should be clear that we cannot make a reasonable verification of a uniform temperature until we solve for the temperature field in the sphere. Bi = hl/k s where L= V/A = r/3 = d/6 (pg.216) for this problem Bi = hl k s = 4.2 Btu/hr-ft2- F 1/( ) Btu/hftF 6 = 1x10-4 <<1 ChE 333 5

6 The Total Thermal Energy Balance (neglecting Mechanical, Gravitational): E in - E out + E gen = E st = Rate of accumulated This is expressed as the rates of thermal energy changes. The first two terms (in - out) are the Heat Transfer Terms which would include conduction, convection and radiation. These are the surface terms, depending on the direction normal to the surface. Can be expressed in different coordinate systems, depending on symmetry. The other two are volume based. We will simplify the Generation (specific forms to be defined, as needed ) as q This may be a function of position (but not usually); however, these are not vectors. The rate of energy Accumulation in time (=0 at steady state) is: ρcp T/ This is not a vector. So far we have dealt with solutions excluding these last two terms (Generation and Accumulation) ChE 333 6

Consider a volume Ω enclosing a mass M and bounded by a surface δω. d dt. q n ds. The Work done by the body on the surroundings is.

The Energy Balance Consider a volume Ω enclosing a mass M and bounded by a surface δω. δω At a point x, the density is ρ, the local velocity is v, and the local Energy density is U. U v The rate of change