## Echdip 2 years ago Consider a long, thin, uniform rod of constant cross-section whose temperature distribution is θ(x,t), surrounded by an atmosphere of constant uniform temperature θ0. The sides of the rod are not insulated so that heat is lost from the longitudinal surface at a rate h(θ(x,t)- θ0) per unit length, for some constant h. Derive the equation which describes the temperature distribution in the rod and show that it can be written ∂φ/∂t=k/ρc (∂^2 φ)/(∂x^2 ) - (h/ρAc) φ where φ(x,t) = (θ(x,t)- θ0), k is the thermal conductivity, ρ the density , c the thermal capacity and A the cross-sectional a

1. Echdip

Consider a long, thin, uniform rod of constant cross-section whose temperature distribution is θ(x,t), surrounded by an atmosphere of constant uniform temperature θ0. The sides of the rod are not insulated so that heat is lost from the longitudinal surface at a rate h(θ(x,t)- θ0) per unit length, for some constant h. Derive the equation which describes the temperature distribution in the rod and show that it can be written ∂φ/∂t=k/ρc (∂^2 φ)/(∂x^2 ) - (h/ρAc) φ where φ(x,t) = (θ(x,t)- θ0), k is the thermal conductivity, ρ the density , c the thermal capacity and A the cross-sectional area of the rod. (You may assume one-dimensional heat flow in the rod.) Such a rod of length l is initially maintained at a constant temperature θ1. Both ends of the rod are then brought to and maintained at temperature θ0 and the rod is surrounded by an atmosphere at temperature θ0. Show that the temperature distribution in the rod at any subsequent time t has the form θ(x,t)= θ_0+∑_(n=1)^∞▒〖B_n e^(-k/ρc ((n^2 π^2)/l^2 +h/KA)t ) Sin (nπx/l)〗 and determine Bn. (When attempting to solve the equation above you may find it easier to swap the φ term over to the other side of the equation.)

2. abb0t

1. Rate of heat loss by this element to the surroundings = h.φ.dx ; so that Heat loss to surroundings in time dt = h.φ.dx.dt Change in temperature due to heat loss to surroundings in dt = h.φ.dx.dt/(ρ.A.c.dx) 2. Rate of heat loss by conductivity along the rod = [dφ(x+dx)/dx – dφ(x)/dx].k.A Change in temperature of the element in time dt = [dφ(x+dx)/dx – dφ(x)/dx].dt.k.A/(ρ.A.c.dx)] 3. Total change in temperature dφ = [dφ(x+dx)/dx – dφ(x)/dx].dt.k.A/(ρ.A.c.dx)]+h.φ.dt/(ρ.A… giving: ∂φ/∂t = [dφ(x+dx)/dx – dφ(x)/dx].k/(ρ.c.dx)]+h.φ/(ρ.A.c) Now [dφ(x+dx)/dx – dφ(x)/dx]/dx = ∂²φ/∂x² giving finally: ∂φ/∂t = (k/ρc).∂²φ/∂x² + φ.h./(ρ.A.c)

3. Echdip

last part i did not get it

4. Echdip

∂φ/∂t = (k/ρc).∂²φ/∂x² + φ.h./(ρ.A.c isn't that suppose to be ∂φ/∂t = (k/ρc).∂²φ/∂x² - φ.h./(ρ.A.c