Conductive oss through a Window Pane q T T 1 Examine the simple one-dimensional conduction problem as heat flow through a windowpane. The window glass thickness,, is 1/8 in. If this is the only window in a room 9x12x8 or 864 ft 3, the area of the window is 2 ft x 3 ft or 6 ft 2. Recall that q x is the heat flux and that k is the thermal conductivity T 2 q x = k dt dx x The energy at steady state yielded q x = k T 1 T 2 The room is well heated and the temperature is uniform, so the heat flow through the windowpane is Q = k T 1 T 2 A If the room temperature is 60 F and the exterior temperature is 20 F, and k is 0.41 Btu/hr-ft2- F then Q = 0.41 60 20 0.125 / 12 6 = 9444 Btu hr Questions Is this a large rate? How can you tell whether it is large or not? ChE 333 1
Energy balance on the Room How long does it take for the room temperature to change from 60 F to 45 F? To make this estimate, we need to solve an energy balance on the room. A simple analysis yields d dt VρC p T 1 T ref = Q Recognizing that heat capacity density are essentially constant, the equation becomes dt 1 dt = ka VρC p T 1 T 2 Note that τ = VρC p ka Data T 2 = 20 F T 10 = 60 F T 1 = 45 F k = 0.41 BTU/ft-h- F A = 6ft^2 V = 864 ft^3 = 0.125 in. ρ air = 0.07 lb/ft^3 C pair = 0.24 BTU/lb.- F and that τ has units of time. At the outset, T 1 = T 10 = 60 F The solution of the differential equation representing the energy balance is T 1 T 2 T 10 T 2 = e t τ To solve for the time required to get to 46 F, we need all the data in the table. t τ = ln T 1 T 2 T 10 T 2 = ln 45 20 60 20 = 0.47 It follows that t = 0.47τ = 1.75 minutes. ChE 333 2
Heat Conduction in a Composite Solid q T T 0 T 2 Examine the simple one-dimensional conduction problem as heat flow through a thermally insulated windowpane. Each layer of window glass thickness,, is 1/16 in. The insulation layer of air between the two panes is also 1/16 in. Recall that q x is the heat flux and that k is the thermal conductivity T 1 1 2 3 T3 The heat flow through the glass is given by x q x = k dt dx The energy at steady state yielded q x = k T 1 T 2 In layer 1 Q = k 1 T 0 T 1 δ 1 A In layer 2 Q = k 2 T 1 T 2 δ 2 A and in layer 3 by Q = k 3 T 2 T 3 δ 3 Then we can rewrite the equations in this form A Q δ 3 = T k 2 T 3 A ; Q δ 2 = T 3 k 1 T 2 A ; Q δ 1 = T 2 k 0 T 1 A 1 If we add the three equations, we obtain Q δ 1 + δ 2 + δ 3 = T A k 1 k 2 k 0 T 3 3 ChE 333 3
We can consider the thickness/conductivity as a resistance so that et Ri = δ i Ak i then δ 1 Ak 1 + δ 2 Ak 2 + δ 3 Ak 3 = R 1 + R 2 + R 3 = R t The heat flow is then of the following form : Q = q x = q x '' A = T 0 - T 3 Rt This is like a problem of current flow in a series circuit. In the single pane problem discussed in ecture 1, we noted that the resistance, δ/k, was 1/(192(0.41) = 0.254 hr-ft 2 - F/Btu. Recall that for the problem of cooling the room, τ was 1.75 minutes. The thermal conductivity of air is 0.014 Btu/hr-ft- F. so that δ 1 = δ 3 = k 1 k 0.0254 3 2 ; δ 2 k 2 = 1 96 0.014 = 0.744 as a consequence the reciprocal of the overall resistance is 0.744 + (0.0254) = 0.746. Then we see that τ = (1.min) (0.746/0.0254) = 29.37 min ChE 333 4
Tr T i Ta The Convective Boundary Condition Again consider a windowpane, but now there is a heat transfer limitation at one boundary described by a boundary condition. q x '' = - h T r -T i Conduction through the glass is described by The flux is constant at any cross-section so that we can write Solving for the temperatures we get q x '' = - k T i-t a δ T i - T a = δ k q x '' ; T r -T i = q x '' h T r - T a = δ k q x '' + q x '' Solving for q x, the relation becomes h = q x '' δ k + 1 h = q x A δ k + 1 h q x A = q x '' = T r- T a δ k + 1 h ChE 333 5
q x A = q x '' = h T r- T a hδ k + 1 Which modified shows a correction to the heat transfer coefficient modulated by the conduction problem Bi = hδ k The dimensionless number in the denominator is the Biot number, a ratio of the convective heat transfer coefficient to the equivalent heat transfer coefficient due to conduction.? What happens when we take into account the convection on both sides of the window? What do we do if we wish to transfer the heat transfer through the window sash (wooden frame around the window). ChE 333 6
Heat Transfer across a Composite Cylindrical Solid. In the case of heat transfer in a cylinder, there is radial symmetry do that heat conduction is important only in the radial direction. R 2 R 3 T 0 R 1 T 1 T2 T3 The heat flux in the radial direction is given by Fourier s law. q r '' = - k dt dr The total heat flow through any circular surface is constant Q = k 2πr dt = constant = C dr Rearranging we obtain a relation for the temperature gradient dt = C dr k 2πr which upon separation of variables is dt = C dr k 2π r An indefinite integration yields the temperature profile. T = The boundary conditions are C k 2π ln r + a 1 at r = R 1, T = T 1 ; at r = R 2, T = T 2 q" r2 = q" r2 at r = R 3, T = T 3 q" r3 = h(t 3 - T 0 ) ChE 333 7
so that T 1 T 2 = C k 1 2π ln R 1 ; T R 2 T 3 = C 2 k 2 2π ln R 2 R 1 It follows that T 1 T 0 = C 2π 1 k 1 ln R 2 R 1 + 1 k 2 ln R 3 R 2 + 1 h This can be expressed as Q = 2π 1 k 1 ln R 2 R 1 + 1 k 2 ln R 3 R 2 + 1 hr 3 T 1 T 0 Optimal Insulation on a Pipe Is there an optimal thickness for the exterior insulation? In the context of the problem just formulated, is there a best value for R 3? Note that Q = f(r 3 ). To find an extremum, dq = 0 and d2 Q 2 < 0 Some algebra yields: hr 3 = 1 dr 3 dr 3 k 2 It offers a critical radius for R 3 = k 2 /h beyond which the heat loss increases. ChE 333 8