Unit 42: Heat Transfer and Combustion Lesson 6: Conduction-Convection Systems

Aim LO1: Understanding Heat Transfer Rates for Composite Systems.

Conduction-Convection Systems Heat that is conducted through a body must frequently be removed (or delivered) by some convection process. For example, the heat lost by conduction through a furnace wall must be dissipated to the surroundings through convection. In heat exchanger applications a finned-tube arrangement might be used to remove heat from a hot liquid – the heat is conducted through the material and finally dissipated to the surroundings by convection.

Conduction-Convection Systems Consider a one-dimensional fin exposed to a surrounding fluid at a temperature T ∞. A qxqx q x + dx dxdx L Z X Base t dq conv = hPdx(T-T ∞ )

Conduction-Convection Systems The temperature at the base of the fin is T 0. Energy on left face = energy out right face + energy lost in convection As previous… q = hA(T w – T ∞ ) Where the Area A in this equation is the surface area for convection. Let the cross sectional area be A and the perimeter be P.

Conduction-Convection Systems Energy on left face = q x = - kAdT/dx Energy out right face = q x+dx = - kAdT/dx| x+dx = - kA dT + d 2 T dx dx dx 2 Energy lost in convection = hPdx(T – T ∞ )

Conduction-Convection Systems Energy on left face = energy out right face + energy lost in convection - kAdT = - kA dT + d 2 T dx + hPdx(T – T ∞ ) dx dx dx 2 Thus… d 2 T dx - hP(T – T ∞ ) = 0 now let (T – T ∞ ) = θ dx 2 kA

Conduction-Convection Systems Thus… d 2 θ dx - hPθ = 0 dx 2 kA One boundary condition occurs when θ = θ 0 = (T 0 – T ∞ ) at x = 0 The other boundary conditions depend on the physical situation… CASE 1: the fin is very long and the temperature at the end of the fin is essentially that of the surrounding fluid CASE 2: the fin is of finite length and loses heat by convection from its end CASE 3: the end of the fin is insulated so that dT/dx = 0 at x = L

Conduction-Convection Systems If we let m 2 = hP/kA then we can derive a general solution for the equation… θ = C 1 e -mx + C 2 e mx For case 1 the boundary conditions are θ = θ 0 at x = 0 and θ = 0 at x = ∞ Thus…θ = T – T ∞ = e -mx i.e.θ = θ 0 e -mx θ 0 T 0 - T ∞

Conduction-Convection Systems For case 3 the boundary conditions are θ = θ 0 at x = 0 and dθ/dx = 0 at x = L Thus… θ 0 = C 1 + C 2 i.e.0 = m(- C 1 e -mL + C 2 e mL ) Solving for constant C 1 and C 2 gives… θ = e -mx + e -mx = cosh[m(L-x)]/cosh(mL) θ e -2mL 1 + e 2mL

Conduction-Convection Systems For case 2 the result is… T – T ∞ = cosh[m(L-x)] + (h/mK)sinh[m(L-x)] T 0 – T ∞ cosh(mL) + (h/mK)sinh(mL) Note all of the heat lost by the fin must be conducted into the base at x = 0. Using the equations for temperature distribution we can compute the heat loss…

Conduction-Convection Systems Thus… q = - kA dT dx x=0 An alternative method of integrating convection heat loss could be used… q = hP(T - T ∞ )dx = hPθdx 0 L L 0

Conduction-Convection Systems Thus for CASE 1… q = -kA(-mθe -mθ ) = √(hPkAθ 0 ) And for CASE 3… q = -kAθ 0 m = √(hPkA)θ 0 tanh(mL) 1 + e -2mL 1+ e 2mL For CASE 2… q = √(hPkA)(T 0 - T ∞ ) sinh(mL) + (h/mk)cosh(mL) cosh(mL) + (h/mk)sinh(mL)

Conduction-Convection Systems For this is has been assumed that… 1.that there is a substantial temperature gradient only in the x-direction which is only true if the fin is sufficiently thin 2.The convection coefficient h is seldom uniform over the entire surface of the fin. To deal with this in practical terms finite difference techniques must be employed.