Chapter 3 Part 2 One-Dimensional, Steady-State Conduction

Plane Wall with Energy Generation x Cold air TsTs L h k TsTs T0T0 h

The heat equation for a plane wall with constant uniform energy generation per unit volume and symmetrical boundary conditions can be developed by performing an energy balance around an element of volume ΔV as illustrated below x ΔxΔx Acc= E In - E Out + E Generated E In = E Out = E Generated =

integrate by separation of variables at x =0 we know that due to the symmetry so c = 0 integrate again

To determine the constant c’ we can use the boundary condition at x=L, which tells us that the amount of energy generated in half the plane wall as to be transfer to the air since there is no accumulation of energy

If T L is explicitly known than

Asymmetrical Boundary Conditions x Cold air T s,2 L h2h2 k Warmer air T s,1 h1h1

The same differential equation applies but the boundary conditions have changed. The general solution of the differential equation is where c 1 and c 2 are the constants of integration if the temperature T s,1 and T s,2 are known than T(-L)= T s,1 and T(L)= T s,2 the constant may be evaluated and are of the form

In which case the temperature distribution is and the heat flux at any position x can be determined using If the surface temperatures are not explicitly known we can use the boundary conditions and to determine the integration constants

Heat Generation in a Wall With One Face Insulated x Cold air TsTs L h k T0T0 Insulated Surface Solution is as per symmetrical boundary conditions

Example x Water T s,1 LALA h=1000W m -2 K -1 T0T0 Insulated Surface T s,2 A B LBLB A plane wall is a composite of two materials, A and B. The wall of material A has uniform heat generation, k A =75 W m -1 K -1 and thickness L A = 500 mm. The inner surface of the material A is well insulated, while the outer surface of material B is cooled by a water stream with and h=1000W m -2 K -1. Sketch the temperature distributuion that exists in the composite under steady-state condition

Assumptions: One dimensional wall system Constant conductivity for A and B Steady-State Inner surface A adiabatic Negligible contact resistance T(x)T(x) x LALA LA+LBLA+LB T s,1 T s,2

Determine the temperature T 0 of the insulated surface and the temperature T s,2 of the cooled surface. Steady State implies that the energy generated per unit area in section A is equal to the heat flux in section B and the heat flux to the water so In section A Hence

Radial System with Energy Generation Cold Fluid h r r0r0

The heat equation for a cylinder with constant uniform energy generation per unit volume can be developed by performing an energy balance around an element of volume ΔV as illustrated below Acc= E In - E Out + E Generated E In = E Out = E Generated =

Boundaries conditions : at r = 0 symmetry T = T se at r = r 1

Extended Surfaces Aim:Increase heat transfer rate Mode:Increase surface area General Conduction Analysis x xx S q3q3 A q1q1 q2q2

Assumptions:Steady State Temperature gradient in fin only in x direction No energy production No radiation Energy Balance Accumulation = Energy In – Energy Out + Energy Produced Energy in = q 1 Energy Out = q 2 + q 3

Divide by and take the limit to 0 If A and S are independent of x (uniform cross-section)

Solution Boundary ConditionsTemperature Distribution

Heat Transfer Rate Boundary Conditionsq

Graphical Solution Fin Efficiency Ratio of the actual heat transfer rate of the fin over the heat transfer rate if the entire fin temperature was equal to the base temperature. The fin efficiency is given as graphical function of (figure 3.18 and 3.19) the fin geometry a characteristic length a characteristic area the conduction the convective coefficient

When can we assume that the temperature gradient in fin only in x direction: if we consider a rectangular fin only if y x TcTc TsTs TT E/2

Energy balance on a slice of the fin TsTs TT assumption is valid if Biot Number