Recent content by fred2028
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Power factor of a transformer at 80%
PF is P / S in general. If the transformer is running at 80.1% of full load, the PF should stay the same right since PF = 0.801P / 0.801S = P / S since if the transformer is running at 80.1% full load, the output S and output P would both be 80.1% right? So basically, no matter what load...- fred2028
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- Power Power factor Transformer
- Replies: 2
- Forum: Engineering and Comp Sci Homework Help
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Calculate the transformer parameters
Question basically asks to find the parameters of a transformer via an open circuit and closed circuit test. We were taught that for OCT, V = (I * cos(theta)) * R V = (I * sin(theta)) * X To find the core losses. This makes sense, since we are only using the component of the current that...- fred2028
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- Parameters Transformer
- Replies: 1
- Forum: Engineering and Comp Sci Homework Help
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Why does the Rankine cycle require the condensation of water before reheating?
Thanks, so in this case, the pumping would require more power than the amount of power the condenser and boiler would save? Thanks for the reply!- fred2028
- Post #4
- Forum: Mechanical Engineering
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Why does the Rankine cycle require the condensation of water before reheating?
In a Rankine cycle such as a steam power plant, why does the liquid (water) have be condensed before being reheated first? Isn't the Q just extracted during the condensation stage, wasted into a nearby reservoir, and then re-added in the boiler? Could this water not be fed straight to the...- fred2028
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- Cycle Efficiency Rankine cycle
- Replies: 3
- Forum: Mechanical Engineering
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Understanding Basic Transformer Concepts
Oh, so what exactly does closed ckt test give you? R, or X? And open ckt? Thanks!- fred2028
- Post #5
- Forum: Engineering and Comp Sci Homework Help
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Understanding Basic Transformer Concepts
Does impedance not include resistance (Z = R + X)? I'm logically thinking that the closed circuit test gets the impedance because all of the I flows back to the source, and any resulting V drop would be due to the impedance (real and imaginary) in the transformer right? How about the open...- fred2028
- Post #3
- Forum: Engineering and Comp Sci Homework Help
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Understanding Basic Transformer Concepts
The questions in this pre-lab are pretty basic and conceptual, and have not yet been taught in class. However, I'd like to complete this prelab early since I will be busy later on. Because it is pretty conceptual, I don't have any attempt at them except Googling around. Quick explanations or...- fred2028
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- Transformer
- Replies: 7
- Forum: Engineering and Comp Sci Homework Help
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Substituting Functions: Simplifying Multivariate Expressions
OK thanks!- fred2028
- Post #3
- Forum: Calculus and Beyond Homework Help
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Substituting Functions: Simplifying Multivariate Expressions
This is more of a concept question, so the template is not followed. Say you're given T(x,y,z) = xy-z And z = x+y Basically, T is a function of x, y, and z while z is a function of x and y. If we substitute z into T, would T become T(x, y) or stay T(x, y, z)?- fred2028
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- Functions
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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Complex repeated roots for ODEs
I know that a 2nd order homo ordinary differential equation's solution is in the form of \[f(x) = {C_1}{e^{{a}t}} + {C_2}t{e^{{a}t}}\] for repeated real roots of the characteristic equation, and that the solution for a single complex root (and its conjugate) involves a cosine. I'm curious...- fred2028
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- Complex Odes Roots
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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Integral of a function multiplied by a Heaviside step function
But the Heaviside function merely states that it equals a constant (1) when greater than t = a and another constant (0) when less, so can't you treat it like a constant?- fred2028
- Post #11
- Forum: Calculus and Beyond Homework Help
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Integral of a function multiplied by a Heaviside step function
Oh thanks! My mistake in LaTeX for the 3rd question. I meant \[f(t)u( - t + 1)u(t - 1)\] so it only turns on when t < -1 and t > 1 at the same time, which is never. Then the graph would just be a constant 0 right?- fred2028
- Post #10
- Forum: Calculus and Beyond Homework Help
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Integral of a function multiplied by a Heaviside step function
that depends on the value of a. This isn't the part I am confused about, it's been explained to me already. I need help with my last 2 questions.- fred2028
- Post #7
- Forum: Calculus and Beyond Homework Help
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Integral of a function multiplied by a Heaviside step function
This just means that when integrating a function times a Heaviside the Heaviside can be pulled out.- fred2028
- Post #5
- Forum: Calculus and Beyond Homework Help
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Integral of a function multiplied by a Heaviside step function
Oh OK thanks, that makes my life a lot easier in many situations. Could you shed some light on my last 2 questions? Thanks!- fred2028
- Post #3
- Forum: Calculus and Beyond Homework Help