# CHAPTER 8 APPROXIMATE SOLUTIONS THE INTEGRAL METHOD

## Presentation on theme: "CHAPTER 8 APPROXIMATE SOLUTIONS THE INTEGRAL METHOD"— Presentation transcript:

CHAPTER 8 APPROXIMATE SOLUTIONS THE INTEGRAL METHOD
• Obtain approximate solutions when (a) Exact solution is not available (b) Form of the exact solution is not convenient to use • Integral method is used in (a) fluid flow (b) heat transfer (c) mass transfer

8.1 Integral Method Applications: Mathematical Simplification
• Differential formulation: Basic laws are satisfied at every point • Integral formulation: Basic laws are satisfied in an average sense • Mathematical simplification: (a) Reduction of the number of independent variables and/or (b) Reduction of the order of the governing differential equation

(b) Integration of the governing differential equation
8.2 Procedure (1) Integral formulation: Construct the heat-balance integral using one of two methods: (a) Control volume formulation • Select a finite control volume • Apply conservation of energy (b) Integration of the governing differential equation • Write the heat equation • Multiply through by an infinitesimal area dA of the control volume

Each problem has its own heat-balance integral
• Integrate each term over the entire control volume • Make use of Leibnitz’s rule to integrate the term (8.1) Each problem has its own heat-balance integral

(2) Assumed temperature profile
• Not unique. Various possibilities: Example: polynomial for Cartesian coordinates • Must satisfy boundary conditions • Profile is expressed in terms of unknown parameter or variable (3) Determination of the unknown parameter or variable • Use the heat balance integral for the problem

8.3 Accuracy of the Integral Method
• Errors are acceptable • Different assumed profile give different solutions and different errors • Accuracy is not very sensitive to the form of the assumed profile • Not possible to identify the most accurate integral solution

8.4 Application to Cartesian Systems
• Formulation of the heat-balance integral for each problem by two methods: (a) Control volume formulation (b) Integration of the governing differential equation Example 8.1: Constant Area Fin Base temperature = To Tip is insulated Ambient Temperature = T

(2) Origin and Coordinates (3) Formulation and Solution
(1) Observations • Constant area fin • Insulated tip • Convection at surface • Specified base temperature (2) Origin and Coordinates (3) Formulation and Solution (i) Assumptions • Steady state • Fin approximations are valid (Bi  0.1) • No energy generation (4) Uniform h and T

(ii) Integral Formulation. Use two methods:
(a) Control volume formulation. Conservation of energy applied to control volume (a) (b) (c)

(b) and (c) into (a) (8.2) (d) Equation (d) is the heat-balance integral for the fin. (b) Integration of the governing differential equation (2.9)

Multiplying eq. (2. 9) by dx and integrating over the
Multiplying eq. (2.9) by dx and integrating over the length of the fin from x = 0 to x = L (e) (f) (8.2)

(iii) Assumed Temperature Profile
(g) (h) Applying (f) and (h) to (g) gives (g) becomes (i)

(iv) Determination of the unknown coefficient
a1 is determined by satisfying the heat-balance integral. (i) into (8.2) Performing the integration and solving for a1 (j)

(j) into (i) (8.3) Tip temperature is T * (L) (8.4) Fin heat transfer rate qf : Apply Fourier’s law at x = 0 and use eq. (8.3) (8.5)

(5) Comments. Exact solution:
(4) Checking Dimensional check: Each term in eq. (8.3) and eq. (8.5) is dimensionless Limiting check: If h = 0 no heat leaves the fin. Fin should be at the base temperature. Setting h = m = 0 in eqs. (8.3) and (8.5) gives T (x) = To and qf = 0 (5) Comments. Exact solution: (8.6)

(8.7) Solutions depend on the parameter mL. Table 8.1 gives the percent error as a function of this parameter. Table 8.1 Percent Error mL 0.5 1.0 1.5 2.0 3.0 4.0 5.0 10 T * 0.248 3.56 16.0 48.2 225.8 819 2619 5027 q * 0.013 1.52 5.27 11.09 24.63 36.8 46.4 70.9 100

• Heat transfer error is smaller than tip temperature
NOTE • Heat transfer error is smaller than tip temperature error • The error increases with increasing mL large mL = large L (long fin) • As L  , T* = -1/2 (exact solution: T* = 0),   error   • As L  , q* = 0 (exact solution: q* = 1),   error  100

Example 8.2: Semi-infinite Region with Time-Dependent Surface Flux
Initially at Ti Surface flux = (1) Observations • Semi-infinite region • Time dependent flux at the surface (2) Origin and Coordinates

(3) Formulation and Solution
(i) Assumptions • One-dimensional conduction • Uniform initial temperature • Constant properties (ii) Integral Formulation (a) Control volume formulation •  (t) = penetration depth or the thermal layer • At edge of this layer the temperature is Ti • Control volume: Extends from x = 0 to x = 

Conservation of energy for control volume
Evaluating each term (b) (c) (d) (b)-(d) into (a) (8.8)

Equation (8.8) is the heat-balance integral for this
problem. (b) Integration of the governing differential equation. The heat equation is (e) Multiplying both sides of (e) by dx and integrating from x = 0 to x =  (t) (f)

Use Leibnitz’s rule (g) Simplify using B.C. (1) (2) (3)

Substituting into (g) (h) However Substituting into (h) (8.8)

(iii) Assumed Temperature Profile
(8.9) B.C. give a0, a1 and a2 Equation (8.9) becomes (8.10)

(iv) Determination of the unknown variable  (t)
Heat-balance integral gives  (t). Substituting (8.10) into (8.8) or (i) The initial condition on  (t) is (j)

Integrating (i) and using (j)
Solving for  (t) (8.11)

(5) Checking Dimensional check: Equations (8.10), (i) and (8.11) are dimensionally consistent. Limiting check: (i) If the temperature remains at Ti. Setting in eq. (8.10) gives T (x,t) = Ti . (ii) If   , the penetration depth   . Setting  =  in eq. (8.11) gives  = .

(6) Comments Special case: Constant heat flux: = constant,
eq. (8.11) becomes (k) (k) into (8.10) (l) Surface temperature: set x = 0 in (l)

or Exact solution: Error = 8.6%.

8.5 Application to Cylindrical Coordinates
Example 8.3: Cylindrical Fin Base temperature = To Inner radius = ri Outer radius = ro Thickness = w (1) Observations • Constant fin thickness • Specified temperature at base • Insulated tip

(2) Origin and Coordinates (3) Formulation and Solution
(i) Assumptions • Steady state • Fin approximations are valid (Bi < 0.1) • Uniform h and T (ii) Integral Formulation (a) Control volume formulation Control volume: Entire fin.

Conservation of energy:
Fourier’s law: (b) Newton’s law: (c) (b) and (c) into (a) (8.12)

(b) Integration of the governing differential
equation: Fin eq. (2.24) (2.24) Rewrite (d) Multiplying (d) by 2r dr and integrating from r = ro to r = ri

or (e) or Insulated tip (f) (e) becomes (8.12)

(iii) Assumed Temperature Profile
(g) additional B.C. (h) (f) and (h) into (g) (i) Unknown is a1

(iv) Determination of the unknown coefficient
Use the heat-balance integral. Substituting (i) into (8.12). Performing the integration (j) (k)

where (l) (m) Substituting into (i) (8.13)

Heat transfer rate: Apply Fourier’s law at r = ri
Substituting (8.13) into (n) and introducing the dimensionless fin heat transfer rate q* (8.14)

(4) Checking (5) Comments
Dimensional check: Equations (8.13) and (8.14) are dimensionally correct. Limiting check: For ro = , the dimensionless heat transfer rate q* should vanish. Setting ro =  in eq. (8.14) gives q* = 0. (5) Comments • The fin equation (2.24) can be solved exactly. q* depends on two parameters: R and m ro. Table 8.2 compares the integral solution with the exact solution for R = 0.2

Percent error in q* for ri / ro = 0.2
• Integral solution becomes increasingly less accurate as m ro is increased Table 8.2 Percent error in q* for ri / ro = 0.2 m ro 0.2 0.5 1 2 3 4 % Error 1.7 3.5 16.8 33.0 43.4 51.2

Example 8.4: Semi-infinite Region with Temperature Dependent
8.6 Non-linear Problems Example 8.4: Semi-infinite Region with Temperature Dependent Properties A semi-infinite region Initial temperature = Ti Surface flux = qo Conductivity, density and specific heat depend on temperature

• Semi-infinite region
(a) (b) (c) (1) Observations • Semi-infinite region • Temperature dependent properties • Transient one-dimensional conduction

(2) Origin and Coordinates (3) Formulation and Solution
(i) Assumptions • One-dimensional transient conduction • Uniform initial temperature (ii) Integral Formulation Integrating the differential equation: Variable properties heat equation: (8.15)

 (t) = penetration depth
B.C. (1) (2) (3)  (t) = penetration depth Simplify (8.15), use Kirchhoff transformation (8.16)

Differentiating (8.16) (d) Thus or or Substituting into (8.15) (8.17)

where (8.18) Left term of eq. (8.17) is non-linear Transformation of B.C. B.C. (1) becomes or (e) S = diffusivity at surface temperature  (0,t)

BC (2) becomes (f) BC (3) becomes (g) Solve eq. (8.17) for  (x,t) using the integral method. Once  (x,t) is determined, the temperature distribution T(x,t) can be obtained from eq. (8.16) using (b) and (c)

Carrying out the integration gives
(8.19) Heat-balance integral: Integrating eq. (8.17). Multiply eq. (8.17) by dx and integrate from x = 0 to x=  (t) (h) Use Leibnitz’s rule (i)

where i = diffusivity at initial temperature i
Simplify (i) using 3 B.C. (j) However Substituting into (j) (8.20)

(iii) Assumed Temperature Profile
(k) B.C. (1), (2) and (3) give a0 , a1 and a2 Equation (k) becomes (8.21)

(v) Determination of the unknown variable  (t)
Use the heat-balance integral, (8.20). Eq. (8.21) into (8.20) or (l) Initial condition on  (t) (m) Integrate (l) and use (m) (8.22)

S depends on  (0,t). Setting x = 0 in (8.21) gives  (0,t).
Eliminating  (t) by using eq. (8.22) (8.23) Eq.(8.18) and (8.19) give S in terms of  (0,t). Using this result with eq. (8.23) gives S as a function of time. Equation (8.22) is then used to obtain  (t) which when substituted into eq. (8.21) gives  (x,t). The transformation eq. (8.19) gives the temperature distribution T (x,t).

(4) Checking (5) Comments
Dimensional check: Equations (8.21) and (8.22) are dimensionally correct. Limiting check: For the special case of constant properties the solution agrees with the result of Example 8.2. (5) Comments Improving accuracy: Using a cubic polynomial of the form (n)

Corresponding solution is
Need a fourth B.C. (o) (n) becomes (8.24) Corresponding solution is (8.25)

Example 8.5: Conduction with Phase Change
Semi-infinite region Initially solid at the fusion temperature Tf Surface at x = 0 is TO Tf (1) Observations • Semi-infinite region • Solid phase remains at Tf (2) Origin and Coordinates

(3) Formulation and Solution (i) Assumptions
• One-dimensional transient conduction • Constant properties • Uniform initial temperature (ii) Integral Formulation The heat equation is (a) Multiplying by dx and integrating from x = 0 to x = xi (t) (b)

Integrate and use Leibnitz’s rule
(c) Simplify using B.C. (1) (2) (3)

(iii) Assumed Temperature Profile
Use B.C. (2) and (3) (d) Equation (d) is the heat-balance integral for this problem. (iii) Assumed Temperature Profile (e) • BC (3) is not suitable. It leads to 2nd order DE for xi

• Alternate approach: Combine (2) and (3). Total
• Alternate approach: Combine (2) and (3). Total derivative of T (xi,t) in condition (2) is at x = xi (f) Setting dx = dxi in (f) (g) B.C. (3) into (g) (h)

Using B.C. (1), (2) and (h) give the coefficients a1 , a2
and a3 (i) (j) (k) where (l)

s (t) is unknown. Use the heat-balance integral.
(e) into (d) (m) Solve for xi (t) and use interface initial condition, xi (0) = 0 (n) where (o)

(4) Checking (5) Comments
Dimensional check: Eqs. (h), (m) and (n) are dimensionally consistent (5) Comments Problem is identical to Stefan’s problem. Exact solution is (p)

Table 8.3 compares the two solutions
xi / xie 1.000 0.4 1.026 0.8 1.042 1.2 1.052 1.6 1.059 2.0 1.064 2.4 1.068 2.8 1.070 3.2 1.072 3.4 1.073 4.0

Example 8.6: Semi-infinite Region with Energy Generation
Initial Temperature Ti = 0 At t > 0 apply Surface at x = 0 is at T (0,t) = 0 (1) Observations • Semi-infinite region • Transient conduction • Time dependent energy generation

(2) Origin and Coordinates
(3) Formulation and Solution (i) Assumptions • One-dimensional transient conduction • Constant properties • Uniform initial temperature (ii) Integral Formulation Integrating the differential equation: (a)

Multiplying by dx and integrating from x = 0 to
Use Leibnitz’s rule (c)

Simplify using B.C. (1) (2) Substituting into (c) (d) Equation (d) is the heat-balance integral for this problem.

(iii) Assumed Temperature Profile
Two additional B.C. (3) A fourth condition is obtained by evaluating differential equation (a) at x = . Using boundary condition (3), equation (a) becomes Integrate (f)

Define (g) (g) into (f) gives the fourth B.C. (4) The assumed profile becomes (h) Unknown is  (t). Use the heat-integral equation.

Substitute B.C. (4) and the assumed profile (h)
into (d) (i) Differentiate (g) (j) Substitute into (i) or (k)

Integration of (k) (l) The initial condition on  (t) is Substitute into (l) (m) (m) into (h) gives the transient temperature distribution

(4) Checking (5) Comments
Dimensional check: Equations (h) and (m) are dimensionally consistent. Limiting check: If the temperature remains at the initial value. Setting in (g) gives Q = 0. When this is substituted into (h) gives T (x,t) = 0 (5) Comments Special case: Constant energy generation rate, (g) gives

Substitute into (k) Equation (h) becomes

Download ppt "CHAPTER 8 APPROXIMATE SOLUTIONS THE INTEGRAL METHOD"

Similar presentations