Electrochemistry: cell vs battery

Tutoring chemistry, the distinction between cell and battery is noted.

In electrochemistry, a cell is a single unit of electrical energy production. A cell comprises an anode and cathode, plus the ingredients and the environment needed for the chemical reaction that outputs electrical energy.

A battery comprises more than one cell connected so that they work together to deliver energy to a circuit.

People have come to refer to single cells as batteries. I’d say that the button-style power sources found in calculators, watches, etc are cells. If a calculator contains two of them, those two cells constitute a battery.

The typical car battery really is one, since it contains six cells connected.

Source:

Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

Giancoli, Douglas C. Physics, fifth ed. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Auto batteries: reserve capacity vs Ah (Amp*hrs)

The tutor defines two terms relating to auto batteries.

The reserve capacity (RC) of a car battery is given in minutes. It’s the duration the battery can discharge 25A at ≥ 10.5V.

Ampere hours (Ah), on the other hand, is the product of steady discharge in Amps that is possible for a duration of hours before the battery voltage slips below 10.5V. Ah is often given in a format like 50Ah@20h, meaning that, over a 20 hour duration, the battery delivered a steady 2.5A at or above 10.5V.

Mathematically, reserve capacity can be converted to Ah as follows:

Ah ≈ RC*(5/12)

However, this equation might be more a “rule-of-thumb”, since the total amount of energy available from a battery decreases with speed of discharge. In other words, a battery that delivers 2.5A for 20h might likely not manage 50A for 1h.

Source:

www.pacificpowerbatteries.com

all-about-lead-acid-batteries.capnfatz.com

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physical chemistry: efficiency of internal combustion engine in hot vs cold weather

The tutor examines the idea that internal combustion engines are more efficient in cold weather.

An upper limit for efficiency of an internal combustion engine is

eff = (Tcombust – Tsurrounding)/Tcombust

where

Tcombust is the temp of the combustion cylinder

Tcombust, Tsurrounding both in degrees Kelvin (Celsius + 273).

Let’s imagine a diesel engine, whose average internal cylinder temperature might be around 1600°C. Then at outdoor temp of 25°C (293K) we have

eff = (1873 – 298)/1873 = 84.1%

Likewise, at outdoor temp -25°C we have

eff = (1873 – 248)/1873 = 86.8%

The 2.7% increase in efficiency at -25°C vs 25°C may be noticeable to an operator.

Source:

www.ncert.nic.in

White, J. Edmund. Physical Chemistry: College Outline Series. New York: Harcourt Brace Jovanovich, 1987.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Lifestyle: the Hydro bill

The tutor investigates higher household usage over last year.

For Nov11-Jan11, our daily household usage this year was 6kWh above last year. My wife points to the new electric fireplace downstairs as the culprit. More specifically, she means that it’s been left on while people weren’t in the room. (The people who left it on would be the culprits.)

To be sure, late December through early January was cold: the first week back to school, snow still lay around from December 23. Around here, cold temperatures sustaining that long are rare.

The heat from the electric fireplace was needed anyway. The fan, however, doesn’t produce heat, but just circulates it. I’ve looked around and suspect, from hearth.com, that the fan might use 120W (0.12kW). Left on 14 hours per day, it would use 0.12kWx14h=1.68kWh. That’s more than 25% of the 6kWh per day over what we used last year.

Where the other three quarters went, I’ll continue to investigate:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Household electricity: partial outage

The tutor discusses a surprising event last night.

Lately, our electricity has been blinking sometimes. My wife said we should call someone; I said I didn’t expect it was from our own wiring. Therefore, I reasoned, other people would likely be experiencing the same issue; eventually it would be detected and fixed. We never called anyone.

Last night the electricity blinked again. When it returned, not all of my household circuits worked. I could, for instance, turn on some lights, but not others. I checked the electrical panels; all the relevant breaker switches were “ON”. What, I wondered, was transpiring?

Around that time, I heard a heavy truck pull up across the street. Looking out, I saw it was a utility truck parking alongside a power pole. Next I heard the bucket rising workmen to the wires. I had a couple of little chores which I returned to, wondering if the workmen would fix my “partial” electrical outage.

About two minutes later I was plunged into total darkness; the workmen had obviously turned the electricity off so they could fix the wiring. I grabbed a flashlight (my wife had told me where to find it a couple of days back); setting it between my teeth, I continued my chores. The last one was assembling and putting out the garbage.

In the dark garage I loaded the bag and tied it. The main door didn’t work, of course, so I pulled the can through the side door, into the entrance of the house, and out the front.

Outside, I could hear the workmen. A swath of houses around them, including mine, lay dark. I rolled the garbage can up the driveway to the curb, then returned inside.

What seemed like fifteen minutes later, the lights came back on. Absurdly, every light in the house was suddenly on, since I’d tried so many earlier. The electricity was completely restored. I happily walked around the house, turning lights off.

I was still confused how, earlier, with all the circuit breakers “ON”, some circuits had been live but not all. Freshly equipped with electricity, I heated up a coffee in the microwave and went on the ‘net to find an answer.

While I wasn’t optimistic I’d find anything decisive, I should have been. The top option Google suggested – firstenergycorp.com – contains a likely explanation. It points out that, often, a house receives electricity from three wires; if one is damaged, the circuits that depend on that wire may be affected, while the other circuits continue to function.

To my mind, this was an interesting event and explanation:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Driving: Gas consumption and cost

The tutor explores a common topic among families.

Lately, I’ve been driving the kids to their activities and my wife to and from work. While there’s rarely been a dull moment, I’ve arrived at the topic of fuel consumption and cost.

In a typical week, one or both my kids have activities most days. We live in a small town; the drive is never far. Between work and activities, I’d say the van covers 100km a week. We only have one car.

Our van, according to fuelly.com, uses 14.2L/100km. So, we use about 14.2L per week. With gasoline about $1.20/L, the fuel cost should be around 14.2*$1.20=$17.04 per week. Just for round figures, that leads to a monthly expenditure of about $70 – yearly, $890.

I’ll be talking more about fuel economy and driving:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Residential electricity: poss. cost to go solar

The tutor offers an estimate about the price of going solar.

In an earlier post , I reported my research findings of possibly 22W per $100 for a residential solar application.

Here in BC, according to BC Hydro, the typical household use is 11000kWh per year, or about 917kWh per month. Use fluctuates from minute to minute; taking the average power, we get

    \[\frac{917kWh}{month}\times \frac{month}{(30*24h)}=1.274kW=1274W\]

So, typical household average power usage is 1274W.

At 22W per $100, we can estimate what solar power might cost, installed, to replace the Hydro bill:

    \[1274W\times \frac{\$100}{22W} = \$5790.91\]

So, maybe around $6 grand to replace Hydro with solar in your house?

There are several key variables to be considered surrounding this very rough estimate. The first is hours of sun: in Osyoos, which is desert, you can probably expect much more output from solar panels than on the rainy coast. Installation costs no doubt vary widely as well. Next, there’s usage: this example is for a typical house; it likely doesn’t fit many cases.

That being said, the $6 grand estimate is a great starting point – and quite a value, from a certain point of view.

HTH:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Residential electricity: what is a kiloWatt*hour (kWh)?

The tutor takes a detour from his solar energy series to explain kiloWatt*hour.

A kiloWatt*hour means 1 kiloWatt multiplied by 1 hour. (The asterisk means “times.”)

A Watt is energy used per second – aka, power. The unit is named for James Watt, developer of the steam engine.

To calculate the energy used, you must multiply the rate (aka, the power) by the time:

    \[Energy\ used=power \times time=Watts*seconds\]

The electric company doesn’t charge for power, but rather for energy. A simple conception is that each second, they multiply the rate you are using it (in Watts) by one second. They do this every second, then add up all those values to get your monthly (electrical) energy usage.

If you average out the rate of electrical use for a typical household, you might find it’s around 1275W. At the same time, there are 60*60*24*30=2592000s in a month. Therefore, the energy used might be

    \[Energy=Watts*seconds=1275W*2592000s=3304800000Ws\]

To make the numbers easier to process, the company divides the power by 1000, changing the Watts to kiloWatts (in Metric, kilo means 1000):

    \[3304800000Ws=3304800kWs\]

Next, they divide the seconds by 3600, converting the time to hours:

    \[3304800kWs=918kWh\]

That’s the meaning of kWh and why the unit is used:)

Source:

BC Hydro

Giancoli, Douglas C. Physics, fifth ed. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Solar energy: price per Watt

The tutor continues from his last post about the value of solar energy dollars.

In my last post I discussed the Wattage one might expect from solar energy for $100. Based on a graph from Scientific American, the equation for Watts per $100, W, at time t (with 1980 as year 0 and 2008 as year 28) turns out to be

    \[W=4.2e^{0.0787t}\]

Extrapolating, one can plug in 35 for t to get the predicted Watts per $100 today:

    \[W=4.2e^{0.0787(35)}=66Watts for \  \$100\]

In fact, just talking about the hardware itself, one can get more than 100W for $100 in 2015 (see sunelec.com, for example.) However, for a typical residential application the price per Watt might be more like $4.50, installed. Thus, the Watts available for use might be more like 22 per $100.

Interestingly, taking the average of the “theoretical” 100W for $100, along with the “practical” 22W per $100, one arrives at 61W per $100, which is fairly close to the 66W predicted by the graph.

I’ll be talking more about solar energy for households in coming posts:)

Source not already mentioned:

pv-magazine.com

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Solar energy: yield trend

The tutor begins an exploration of the changing feasibility of solar energy.

There is a graph provided by Scientific American that shows the Watts one might get for $100 from solar cells by year. As they point out, it’s a logarithmic graph; here I’ll show some analysis.

The two points I’ve fetched from the graph, in the format (year, watts) are (1980, 4.2) and (2008,38). The linear shape is accomplished by the relationship

    \[lnw=mt+b\]

where t=time, while w=Watts.

I’m imagining 1980 as year 0; the two points mentioned earlier become (0,4.2) and (28,38). Then b, the lnw intercept, is given by

    \[ln4.2=m(0) +b\]

giving

    \[ln4.2=1.4351=b\]

m is solved by the slope formula:

    \[m=\frac{ln38-ln4.2}{28-0}=0.0787\]

We arrive at

    \[lnw=0.0787t + 1.4351\]

Taking the exponential of both sides, we get

    \[w=e^{0.0787t + 1.4351}\]

This can be rewritten, by an exponent law, as

    \[w=e^{1.4351}e^{0.0787t}\]

which becomes

    \[Watts=4.2e^{0.0787t}\]

The exponential argument 0.0787t suggests growth of 7.9%. Rounding it to 8%, and applying the rule of 72, the Wattage per $100 should double every 9 years.

I’ll follow up some implications of this analysis in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.