Recent content by Pete_L

Please, my Jotul 602N (wood stove) is located in my living room! I would prefer that it not look like my basement.

That's what would be a correct assessment, I think, and what I was afraid of.

Thank you. That's a good point that surrounding the stove with bricks would increase the temperature of the stove for a given size of fire in the stove. Wood stoves are most efficient running in the range of 400 deg. F to 600 deg. F. Wouldn't it be possible to make smaller fires in the stove...

Yes, thanks, I did see in my chart of thermal properties that water has a specific heat that is five times that of brick. The problem with using water is that the containers for it would have to be metal, and finding metal containers of the proper size to fit to the dimensions of my stove would...

Please see if you agree with how I am calculating the number of bricks required to reduce cooling off of my house overnight in the winter when I'm not there (sleeping) to feed logs into the stove. The estimated weight of each log that I would normally be burning in my stove is 8.6 pounds...

What I need to solve is dB of loss, or reflected power relative to incident power in dB, without regard to the amount of acoustic power involved. So regardless of any particular incident power, the reflected power is reduced by some dB amount.

Thank you, I'm not always as rigorous as I should be in solving a problem. -P

My approach is to separately calculate loss due to absorption and transmission. I came across an equation for approximating transmission loss only. Then I also can approximate absorption knowing what the material of the wall is and knowing the absorption coefficient for that material. But my...

To make this easier to respond to, here is how I derived the equation for Lt. Consider the fraction below, [(W1/W2) -1] / (W1/W2) If for example W1/W2 equals 10, this means 10 parts incident sound power to 1 part transmitted power. In the numerator, 10 minus 1 equals 9. The denominator is the...

In considering a sound wave (in air) incident on a wall, the transmission loss (TL) is calculated by the equation TL (dB) = 10*Log (W1/W2) where W1 = sound power incident on wall W2 = sound power transmitted through wall Based on this understanding of TL, I wish to determine the extent to...

Yes that is absolutely true. My initial thought was, okay, if the entire voltage of the voltage source drops across the resistor, then what is the motive force driving the electrons towards and away from the resistor? Following E=IR, there must be a very small voltage drop across each "feed" and...

The OP says, "Been away from this for awhile, but, if I recall correctly, you'd get the same voltage for each reading with just different signs (one negative and the other positive). Am I close?" No, connecting a 1st lead of the voltmeter to the negative terminal of the voltage source, then the...

Connect one lead of a voltmeter to the negative terminal of your voltage source. Now connect the other lead of the voltmeter to first one circled section of wire, and then to the other circled section. Do you get the same voltage reading for each of the two different connections to the circuit?

Yes, that is a correct description. The newly discovered sub-atomic particles have been confirmed to exist by being observed in a time interval, albeit a very short interval of time. Time is the fourth dimension of the physical world.

The voltage rating of a capacitor is the voltage that the capacitor can withstand before being destroyed. At this moment I forget whether that is peak or RMS voltage, but it is general practice to overrate capacitors in a circuit. So if one capacitor in a network of paralleled capacitors is...