Recent content by Jay_

I guess the main thing is like you mentioned they don't bother to fully analyze it. I will look at the other links. An interesting conversation on this would be understanding the conservation of energy in this scenario.

Derek Muller is the Veritasium YouTube channel host and he had a wager against UCLA professor Alexander Kusenko on whether a unpowered car going downwind can go faster than the wind. Derek Muller said it can go faster than the wind, while Alexander Kusenko said it only seems that way because...

So, I had a discussion with a friend of mine, neither of us are in controls but I was curious about an answer here. In a PID controller, we essentially take in an error value, do a mathematical operation on it and determine the input (controller output signal B) needed to the actuator to produce...

My professor said I needn't assume any difference in the outlet temperatures, so I went with that. I also assumed the coolant flow to be based on the bhp, which depends on the rpm. So depending on that I don't get much of a difference.

I can't. But I have that graph and I am getting it from literature. I showed it to my professor, he said its okay since we are also out of time. Now, what I would like to know is based on automotive engineering: 1. Does that graph in the previous link make sense? Specifically, is the...

As of now, in my city drive cycle the calculations seem correct. I get average coolant power = 12.3115 kW, fuel_power = 39.3417 kW, so its 31.29% However, for my highway driving, I get coolant power = 11.78 kW, fuel_power = 64.70 kW, so its just 18.61% Does this make sense? I had to estimate...

However, he posted saying this is for a 90 degF (or 32 degC) rise. Should I change the value based on delT?

Okay, I got a reference for the flow rate and it works. I get 31.2938% of the total fuel as the coolant heat energy, which I think is consistent with literature. From this link : http://talk.newagtalk.com/forums/thread-view.asp?tid=349481&DisplayType=flat&setCookie=1 The post by user Gerald J...

Okay. So that means its the coolant energy that is higher than the actual number. And since we have everything that is makes sense, we are down to changing the coolant flow rate?

How many miles were covered in this 26 minutes, and what was the gas mileage? Given that the average velocity was 12.2159 m/s, in 1597 seconds, we went 12.2159*1597 = 19508.7923 meters or 12.12 miles. Fuel calculation method 1 (based on mileage) Assuming city mileage of 19mpg, I burnt...

Okay sir, the drive cycle is not 12 minutes. Its 1597 seconds, or 26 mins. and 37 secs. Under these circumstances, would you say 18 kWh of fuel burnt is too less? Because my coolant calculation average power comes to 49 kW, which is close to your 50.8 kW. Total energy in 1597 seconds (0.443611...

What I have done as of now is calculate the instantaneous values of heat, added them all and compared it with the fuel burnt. I will get the average of the samples and compute the power, multiply it by time and get energy and see what that gives me. But on what basis are we choosing any of...

I don't have the car anymore to do the experiment. Presently, I have two (sensible) things to go with for the coolant flow rate : 1. flow rate in gpm = 0.015 * rpm 2. flow rate in gpm = 0.2*bhp I did some scouting calculations. I found the flow rate at rpm values of 1500, 2000, 3000, 3500 and...

I used the rule of thumb in this thread (post 2). http://www.eng-tips.com/viewthread.cfm?qid=161237 Using that, my flow rate shows from 1.1 L/sec to 3.1 L/sec which is better? I also made my cold side temperature greater (to 80 deg C). But I still get the coolant heat energy to be greater...

Quoting myself Is that the specific heat value at 300 K? I interpolated these : http://homepage.usask.ca/~llr130/physics/HeatCapcityOfAntiFreeze.html. At 27 deg C (300 K) the value is close to what I have in standard units. Regarding the flow rate, could someone give me a rule of thumb on...