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Ameya W.

Tutor for 2 years and currently a research assistant working on Mechanical Engineering

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Mechanical Engineering

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Question:

Derive the steady flow energy equation.

Ameya W.

Answer:

We first define variables involved in this. $$P_1$$= Pressure at the entrance of the system $$V_1$$= Specific volume of the working substance at the entrance $$U_1$$= Specific Internal Energy at the entrance $$C_1$$= Velocity at the entrance $$Z_1$$ = height above the datum at the entrance $$P_2,V_2,U_2,C_2,Z_2$$= Corresponding values at the exit $$Q$$= Heat supplied to the system $$W$$= Work done by the system We first evaluate total energy at the entrance which will be sum of all the energies Potential Energy=$$gZ_1$$ Kinetic Energy= $$0.5C_1^2$$ Internal Energy=$$U_1$$ Pressure Energy=$$P_1V_1$$ Heat Supplied=$$Q$$ Therefore total energy at entrance= $$gZ_1 +0.5C_1^2+U_1+P_1V_1+Q$$ Similarly, total energy at the exit=$$gZ_2 +0.5C_2^2+U_1+P_2V_2+W$$ We see that the Q is replaced by W here, that is because work is assumed to be extracted at the end of the process. Using first law of thermodynamics Total Energy at the entrance=Total energy at the exit $$gZ_1 +0.5C_1^2+U_1+P_1V_1+Q=gZ_2 +0.5C_2^2+U_1+P_2V_2+W$$ Thus this is the energy equation for steady-state flow. Note that all the energies here are defined as energies per unit mass.

Calculus

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Question:

Evaluate $$ \int^{+\infty}_{-\infty} \exp(-x^2) \, dx $$.

Ameya W.

Answer:

This integral is known as gaussian integral and it is rather tricky to evaluate this particular integral. We need to use polar coordinates to solve this particular problem. Here goes the solution Assume $$ I=\int^{+\infty}_{-\infty} \exp(-x^2) \, dx $$. Changing the variable to $$y$$. We also have $$ I=\int^{+\infty}_{-\infty} \exp(-y^2) \, dy $$. We multiply these to expressions to get $$I^2=\int^{+\infty}_{-\infty} \int^{+\infty}_{-\infty} \exp(-(x^2+y^2)) \, dxdy $$ Converting this to polar coordinates $$x^2+y^2=r^2$$ and $$dxdy=rdrd\theta$$ And $$r$$ varies from $$0\to \infty$$ and $$\theta$$ varies fro $$0 \to 2\pi$$ Substituting in the previous equation we have $$I^2=\int^{2\pi}_{0} \int^{\infty}_{0} \exp(-(r^2)) \, rdrd\theta$$ So $$I^2=2\pi*\int^{\infty}_{0} \exp(-(r^2)) \, rdr$$ Substitute $$s=-r^2$$, $$ds=-2rdr$$ We have $$\int^{\infty}_{0} \exp(-(r^2)) \, rdr=\int^{0}_{-\infty} 0.5*\exp(s) \, ds$$ So $$I^2=2\pi*\int^{0}_{-\infty} 0.5*\exp(s) , ds$$ Therefore $$I^2=\pi*(\exp(0)-\exp(-\infty))$$ $$I^2=\pi*(1-0)$$ Implying $$I^2=\pi$$ And $$I=\sqrt{\pi}$$

Differential Equations

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Question:

Explain how one can use Laplace transform to solve a differential equation? Solve the following equation using Laplace transform. y''-3y'+2y=0, with initial conditions y(0)=0,y'(0)=1

Ameya W.

Answer:

Laplace transform is an excellent mathematical tool which finds it's application in variety of fields of mathematics, solving Ordinary and partial differential equations is one of them. Depending on the type of differential equation it converts the differential equations into either algebraic equation or differential equations of lower order and thus making it a tad bit easier to solve the equation. We find out the solution to the transformed equation and take the Laplace Inverse of that solution to obtain the actual solution. Although it is difficult to evaluate the Laplace inverses, with certain amount of practice you can master the art of calculating inverses Laplace transform is generally denoted by L(y). This method is particularly helpful in solving second order ordinary differential equation We use the following formula to $$L(y'')= s^2*L(y)-s*y(0)-y'(0)$$ $$L(y')=s*L(y)-y(0)$$ Now, these are fairly standard formulae found in any differential equation textbooks. Now we use laplace transform to solve the given differential equation $$L(y''-3y'+2y)=L(0)$$ $$L(0)=0, L(y''-3y'+2)=L(y'')-3L(y')+2*L(y)$$ This is true since laplace transform is a linear transformation. We use the formula previously stated. Using the the initial conditions $$ L(y")=s^2*L(y)-1$$ and $$L(y')=s*L(y)$$. Substituting this in the previous equations we have $$s^2*L(y)-1 -3*s*L(y)+2L(y)=0$$ $$ L(y)*(s^2-3s=2)-1=0$$ $$L(y)=1/(s^2-3s+2)$$ $$L(y)=1/((s-2)(s-1))$$ Using partial fractions $$L(y)=(2/(s-2) +1/(s-1))$$ So $$y=L^1(2/(s-2) +1/(s-1))$$, where $$L^1$$ denotes laplace inverse. Therefore $$y=L^1(2/(s-2)) + L^1(1/(s-1))$$ By formula $$L^1(exp(ax))=1/(s-a)$$ So we get $$y=2*exp(2x)+exp(x)$$. Now I know there is a much simpler method to arrive at the same solution for this problem. But our goal here was to learn the usage of the helpful tool Laplace equation and we see that this method is very valuable when we move into more complex 2nd order ordinary differential equations.

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