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题目

GEOL0012_25-26 **********GEOL0012 Moodle Test (unassessed) 2025/26**********

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Question text1. The one-dimensional heat conduction equation for a cooling dyke with no internal heating can be written: [math: ∂T∂t=κ∂2T∂x2]Where [math: T] is temperature, [math: t] is time, [math: x] is horizontal distance, and [math: κ] is the thermal diffusivity, and has a solution of the form: [math: T(x,t)=T02[erf(w−x2(κt))+erf(w+x2(κt))]] At t=0, T=T0 for –[math: w] < x < [math: w], and at t=0, T=0 for |x| > [math: w]. If the half-width of the dyke is [math: w=2.7m], centred on x = 0, and if T0 = 1500 oC and [math: κ] = 10-6 m2s-1 a) calculate the temperature at the centre of the dyke after one week and after one year (365 days) in degrees Celsius to three significant figures. HINT: Use a calculator, an online tool or MATLAB to calculate the error function (in MATLAB simply use erf(your value) ). After one week: [math: T=] Answer 1 Question 1[input] degrees Celsius [3]After one year: [math: T=] Answer 2 Question 1[input] degrees Celsius [3]b) Calculate the temperature of the dyke at the edges after 1 year in degrees Celsius to three significant figures.[math: T=] Answer 3 Question 1[input] degrees Celsius [4]

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We are given the 1D heat equation with no internal heating and a solution of the form: T(x,t) = (T0/2) [ erf((w − x) / (2 sqrt(κ t))) + erf((w + x) / (2 sqrt(κ t))) ] with w = 2.7 m, T0 = 1500 °C, κ = 1e-6 m^2/s. We need temperatures at the centre (x = 0) after one week and after one year, and the temperature at the edges (x = w) after one year. We also note that erf is the error function. Step 1: Temperature at the centre (x = 0) Plug x = 0 into the general solution: T(0,t) = (T0/2) [ erf(w / (2 sqrt(κ t))) + erf(w / (2 sqrt(κ t)......Login to view full explanation

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