python assignment

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timer Asked: Apr 5th, 2016

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i have a python assignment i need in 7 hours.


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130 Chapter 10. Numerical Differentiation and Integration Homepage fa) f(b) ) a 이용 NI 2 Figure 10.4: Iluustration of Simpson's Rule for integration of f(x) from a to b. Image from commons.wikimedia.org. where T should be in °C. The heat capacity of carbon solid in kJ/(mol·K) is given by Cp(T) = 0.1118+1.095 x 10-ST+489.1/T2 where T should be in K. Your contract with the refinery asks you to calculate the enthalpy change for each material (nitrogen gas and carbon solid) starting at a temperature of 20°C and raising the temperature to 100°C. Report the enthalpy change in kJ/mol. You should numerically approximate the integral using the Gaussian Quadrature algorithm included in Scipy (scipy. integrate . quadrature ()). Additionally, use the Trapazoid rule with 10 intervals to approximate the enthalpy change of nitrogen gas and report the difference between the Trapazoid rule and Gauss Quadrature. Problem 10.2 The calculation of the amount of energy required to change the temperature of a material is performed frequently in process engineering. For example, if we have a cubic meter of nitrogen or a kg of carbon, how much energy is required to raise the temperature by 1°C or 100°C? The calculation frequently requires integrating a polynomial between the starting and ending temperatures. A refinery has hired you to create a Python program that both automates these calculations and compares two different numerical methods for the approximation of integrals. Assuming a constant pressure process, the enthalpy change associated with a temperature change from Ti to T2 is: AA = C(T)dr where C (T) is the heat capacity as a function of temperature at constant pressure. The heat capacity of a material as a function of temperature is frequently given through a polynomial relationship. The refinery is particularily interested in two materials: (1) nitrogen gas and (2) carbon solid. The heat capcity of nitrogen gas in kJ/mol.°C) is given by C(T) = 0.0290 +0.2199 x 10-ST+0.5723 x 10-872 - 2.871 x 10-1273
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