Recent content by Sorbik

thanks for the help. got it to work

Ok. I have LdL = 3.19x10^-8*dt ∫LdL ==> L^2 / 2 = 3.19x10^-8 *t + C at time = 0, L = 1.7617m 1.7617^2 / 2 = 0 + C ===> 1.55179 = C at time = t, L = 2.0604m 2.0604^2 / 2 = 3.19x10^-8t + 1.55179 t = 17,894,377s ==> 4970.66 hours <-- time it takes to get another foot of ice ...

My calc understing is pretty poor. Lots of bad turns. I don't think this problem should have a ln(L) in it either after integrating or have the need to use indefinite integrals and solving for constants.

I'll rework this and try to keep symbols to the end.

great. It worked out using moles calculated from the temp K during morning time.

Ok. n=PV/RT n= [(101,325)(332.889)] / [(8.3144)(294.65)] = 13,768.25 mol both of these answers are wrong too. Not sure where I'm botching this

Homework Statement ice needs to be at least 6.76 feet thick. = 2.0604m Currently the ice is 5.78 feet thick = 1.7617m Later, there is a cold spell If the average temperature during the cold spell is 23.23°F =-4.8722C = 268.277K and the water temperature is 32.0°F = 0C = 273.15K...

I liked this problem because it's something you can use to talk about money savings w/ other people. Just making sure everything was done right though. Homework Statement the interior temperature of the room is typically 34.6°F ( the thermostat is typically set for 70.7°F. you propose to...

awesome!

So, 31,851.651MW * 0.373 = 11880.66582MW is the power output. That's still like... 20 times the amount the problem is expecting with the 535MW desired power output.

Got the change part. 3.44 change in C is also 3.44 change in K I'm still getting some massive number for Qh though in comparison to the initial power range in the problem. Also not sure if that initial 222ft width specification means anything. QH is max plant output before the fishes die.

delta T is max increase in temp the river smelt can survive which is 3.44C. This is converted to Kelvin to cancel with the Kelvin in the specific heat value of 4186 J/kg K For the efficiency, I was going to use 0.373 = 1 - Qc/Qh where Qc is the value I get from Q=mcΔT since it is the heat...

Thanks for the welcome. Help is scarce with online classes. Here's my work so far. the power I got at the end is pretty big... the original problem was only thinking about boosting power from 188MW to 535MW whereas I got 1.6MMW I'm still not sure how to apply the 222 ft portion of the...

Homework Statement My simplified version of the equation: Power plant proposes output power increase from 188MW to 535 MW Excess heat is dumped into river with smelt. Smelt can withstand a 3.44C temp increase River = 222ft wide at dump site. Do not know depth or cross section here...