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.

Transcription

1 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 total energy in Ω is: ρ v s.n d dt Ω ρudv The heat flow from the body is q n d Ω The Work done by the body on the surroundings is Ω v T n d ince for the body U = Q W An equivalent form is + ρg vdv Ω d dt Ω ρudv Ω + q n d Ω v T n d = + ρg vdv Ω If our control volume is a differential cube, the differential equation describing the Energy Equation is: ρu + ρuv + q = v T + ρg v t ChE 333

2 The first term is the local rate of energy change The second is the convective energy flow The third 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 t Momentum ρv + ρvv T ρg = 0 t Mechanical Energy This is obtained by taking the inner product of the momentum equation and the momentum equation to yield v t 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 t H + v H + q = τ v W s + Σα = R α H α This is simplified recalling that H T p = C p ρc p t T + v T + q = τ v W s + R α H α Σα = ChE 333 2

3 Applications of the Energy Equation to teady tate Conduction The Energy Equation was ρc p t T + v T + q = τ v W s + R α H α Σα = But for systems at steady state where there is no motion, no shaft work done, and no chemical reaction time derivatives vanish the velocity, v, is zero the shaft work iz zero, and the reaction rate is zero.. This means that the energy equation has a very simple form Recall that Fourier's Law is a relation for the heat flux, q, q = k T so that q = 0 k T = 0 and it follows for constantk, that 2 T = 0 In rectangular Cartesian coordinates, the resulting equation becomes the steady state head conduction equation. 2 T x T y T z 2 = 0 ChE 333 3

4 Boundary Conditions Types of boundary conditions in heat transfer problems. Constant surface temperature On a surface, the temperature is constant if T(x, t) = T s 2. Constant heat flux a) At a surface, the flux is continuous, finite, and constant so that : q i = k T? x i b) At an adiabatic surface, the flux vanishes: q i = k T x i? = 0 3. Convective urface condition At any surface, the flux leaving one body is equal to the flux leaving the other, so that k T' x i = k T" x i 2 ChE 333 4

5 A imple teady tate Conduction Problem Consider a rectangular slab of infinite extent in the z-direction T2 T T T side is length L, the vetrical sides are of length W. The differential equation for steady heat conduction in 2 dimensions is: 2 T x T = 0 y 2 The boundary conditions are: T = T at y = W T = T 0 at y = 0 T = T 0 at x = 0 T = T 0 at x = L ChE 333 5

6 If the Temperature. T, and the independent variables, x and y, are made dimensionless, as Θ = T T 0 T T 0 and η = y W and ζ = x L The conduction equation becomes 2 Θ ζ 2 + L W 2 2 Θ η 2 = 0 With the boundary conditions transformed to Θ = at η = Θ = 0 at ζ = 0 Θ = 0 at η = 0 Θ = 0 at ζ = The method we use to solve this partial differential equation is "the method of separation of variables". Assume that the solution is of the form Θ = F ζ G η We obtain an equation of the form 2 F ζ G η + L ζ 2 W 2 2 F ζ G η η 2 = 0 ChE 333 6

7 We group the terms that depend on each individual independent variable so that G η d2 F ζ dζ 2 + L W 2 F ζ d2 G η dη 2 = 0 If we divide by FG and separate variables, we obtain F ζ d 2 F ζ dζ 2 = L W 2 G η d 2 G η dη 2 = constant = λ 2 For simplicity, let α 2 = L W ChE 333 7

8 The result is that we have two ordinary differential equations to solve: so that d 2 F ζ dζ 2 + λ 2 F ζ = 0 d 2 G η dη 2 λ α 2 G η = 0 The solutions to the pair are : F ζ = A sin λζ + B cos λζ G η = C sinh λ α η + D cosh λ α η The entire solution is of the form Θ = F ζ G η = A sin λζ + B cos λζ C sinh λ α η + D cosh λ α η If we recognize that since at ζ = 0, θ = 0, then B = 0 and the solution simplifies considerably. Θ = A' sinh λ α η sin λζ + B' cosh λ η sin λζ α ChE 333 8

9 The boundary condition at η = 0 gives Θ = 0 at η = 0 leads to 0 = A' sinh 0 sin λζ + B' cosh 0 sin λζ o that B' must be zero and the simplified solution is Θ = A' sinh λ η sin λζ α There are two constants left, λ and A', and two boundary conditions. The condition at ζ = leads to 0 = A' sinh λ α η sin λ And we must note that Either A' must vanish and the solution is trivial or 0 = sin λ This is true if and only if λ = nπ where n = 0,,2,3,... That means that there are a countable infinite number of solutions. To find the solution we need to add all the possible solutions and determine the coefficients (constants). Σ n= Θ = a n sinh nπ α η sin nπζ ChE 333 9

10 The coefficients may be determined by the last boundary condition. Θ = at η = Σ n= = a n sinh nπ α sin nπζ To determine the coefficient a n, we have to recognize the orthogonality property of sin functions, that is, To determine the coefficients, we can use the orthogonality properties of the sine and cosine functions. 0 sin(nπζ)sin(mπζ)dξ = 0 for m n π 2 for m = n We integrate Σ n= a n sinh nπ α 0 sin(nπζ)sin(mπζ)dζ = ()sin(mπζ)dζ 0 ChE 333 0

11 Remember that the first sine integral is non-zero if and only if n = m. Now the equation for a n is π a n 2 sinh nπ α = sin(nπζ)dζ 0 = cos (nπζ) 0 n or a n = 2 π n n sinh nπ α Finally the solution is Θ = Σ n= 2 n n π sinh nπ α η sinh nπ α sin nπζ ChE 333