jasoncurious

Thanks. So, the "head" concept is something like a "conservation of height" thing?

Hi guys, I am currently studying fluid mechanics. This is my first time doing the subject. I've been reading some books. Then I realised something: 1. Do we always have to use an annulus as a fluid element? What's so special about the ring? Was it because it's easier to analyse? 2. Regarding "head" (dimension L), is it easier to understand by expressing it in L? Thank you.

Hi there! As long as I can remember, an addition reaction consists of both an electrophile and nucleophile. Everytime, the nucleophile is "the one attacking" and the electrophile is "the one attacked". If nucleophilic addition reaction is named in such a way that the reaction is initiated by a nucleophile (it attacks), then electrophilic addition reaction must be occurring in such a way that the electrophile is the one that initiates the reaction. My question is, how does an electrophile initiate an addition reaction when it's always "attacked"? Thank you.

I saw my friend solving it using Bernoulli equation, guess everything is possible in mathematics

I thought the d.e has to be in the form of dy/dx+P(x)y=Q(x) for integrating factor to be implemented?

y=vx dy/dx=v+x dv/dx vx(v+x dv/dx)=x^3+(v^2 x^2)/x v^2*x+vx^2 dv/dx=x^3+v^2*x Eliminate the v^2*x: vx^2 dv/dx=x^3 Divide both sides with x^2: v dv/dx=x vdv=xdx Continue the integration: y^2=x^2(x^2+c), where c is a constant

So, even if the differential equation is not homogeneous, we can use y=u*v as well? This is howthe solution starts: yy'=x^3+(y^2)/x Let y=vx dy/dx=v+x dv/dx Then the rest is substitution.

Thank you, I think I was caught in the thought that "only homogeneous differential equations can be solved".

Convert the following differential equation into separable form by using suitable substitution: yy’=x3+(y2/x) When I try to prove it’s being homogeneous or not, I found that yy’=x3+ y’= (lx)4+(ly)2/ (lx)(ly) Factorise it: (l^2x2)+(y^2) / xy Is this considered homogeneous? the l stands for lambda

Sorry, but I can't help but wonder, isn't that every thermodynamic situation involves entropy?

Hi guys, happy holidays. I need some help in choosing the usage of H (enthalphy) and U (internal energy) when dealing with thermodynamics. Any guidelines?

So, the black one is the piston. Since it is weighted, the pressure produced is lost in pushing the piston upwards?

100g of CO are contained in a weighted piston-cylinder device. Initially, the CO is at 1000kPa and 200oC. It is then heated until 500oC. Determine the final volume of the CO treating it as an ideal gas. This is what my lecturer gave in his solution: V2=(mRT2)/P=(0.1kg*0.2968kPa.m3/kg.K*(500+273)K)/1000kPA My question is: why is the original pressure (1000kPa) been used instead of the final pressure? Thanks for your help.

Thanks, so I assume this type of to be contextual?

Hi all, there's this function: y=(x^2-6*x+5)/(x-1) which can be reduced to y=(x-5)(x-1)/(x-1) and then to y=x-5 The question asked me to determine the domain and range of this function. Here's the problem, when the function is y=(x^2-6*x+5)/(x-1), the domain is every real number except 1. But when it's been reduced to y=x-5, the domain should be every real number. However, the answer stated that the domain is still every real number except 1. How should I solve this problem?

Hi all, this is how my lecturer derive exergy: 1st law efficiency=W/Q Carnot efficiency=(T1-T2)/T1 W/Q=(T1-T2)/T1 W=(Q*(T1-T2))/T1 The W above refers to useful work. My question is, when we formulate the 1st law efficiency, do we have in mind that it is a reversible device (that we can equalize it with the Carnot efficiency)?

Hi all, I am currently learning Jacobian and I wonder where can we apply this thing in the engineering field?

Hi guys, it's me again. for i=1:n; fprintf('Enter the coefficient of b%d:',i) b(i)=input(''); end Just wondering what's the (' ') doing over there? I've already run it several times, and it seems that it is accumulating your input. Just want to know perspectives of you guys.

http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_06961.jpg.html?sort=3&o=3 http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_06982.jpg.html?sort=3&o=2 http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_06991.jpg.html?sort=3&o=1 http://s1301.beta.photobucket.com/user/jasoncurious/media/IMG_07002.jpg.html?sort=3&o=0 Sorry for the poor upload.

Hi guys, I am currently doing a Matlab program for scaled partial pivoting. I looked at my friend's example: % Partial Pivoting for i=1:n-1 for j = i+1:n if (a(j,i)) > (a(i,i)) u=a(i,; a(i,=a(j,; a(j,=u; v=b(i,1); b(i,1)=b(j,1); b(j,1)=v; end end end I was wondering what is the meaning of the :? Previously I have come across a few times this thing, but I don't know what it stands for. Is it some sort of Matlab keyword?

In order for me to comprehend this question, first I look at the polyhedron 2-dimensionally (x-y plane, z-y plane, x-z plane), then I combine the three ways of viewing the polyhedron and get an octahedron. Only then I realize why my friend chose to find the vertices at the three axes. Is this ok?

What my friend did is, he determined where each of the 8 planes intersect the x,y and z-axis. Then he used the coordinates as the vertices for the polyhedron, which is a solid with 8 triangular surfaces. For example, the plane x+y+z=1 intersects x-axis at (1,0,0), y-axis at (0,1,0), z-axis at (0,0,1) I was wondering what is the reason of doing so. Does this mean that (1,0,0),(0,1,0) and (0,0,1) is the furthest that the plane x+y+z=1 can reach? If so, what about (3,-3,1)? If I plug in (3,-3,1), I will be getting 1 as well.

I have this assignment asking us to find the polyhedron described by the equation |x|+|y|+|z|=1. The question also asked us to justify the answer. After some calculation, I got the result that the polyhedron is a cube. I have the feeling that I am wrong with the answer as the top student in our class has a different answer. Below are my attempt to solve the question: From http://en.wikipedia.org/wiki/Octant_(solid_geometry), I got 8 different plane equations: x+y+z=1 -x+y+z=1 -x-y+z=1 -x-y-z=1 x-y+z=1 x+y-z=1 x-y-z=1 -x+y-z=1 So, with the knowledge of plane algebra, the normal for each plane is: (1,1,1) (-1,1,1) (-1,-1,1) (-1,-1,-1) (1,-1,1) (1,1,-1) (1,-1,-1) (-1,1,-1) From this 8 coordinates, I reached my answer indicating that the polyhedron is a cube, but plugging each of this coordinate back into its respective plane equation does not produce 1. I am stuck here. Help from you guys are greatly appreciated. God bless you all.

Thanks guys. At least I've sth to fight my case.

I was thinking if there will be no change in energy at all, i.e deltaE=0, since the exapnsion is done against vacuum?