Electrical heating elements are often used for teapots, irons, electric ovens, soldering irons etc. When projecting designs with electric heating elements you need to do some calculations that may seem difficult at first glance. But when looking more deeply this becomes simple task.

We know that electric heating is a result of current flow in wire with some resistance. Resulting heat is work done by electric current. Work A(J) can be calculated by formula:

A=U·I·t

Where U – Voltage(V), I – current(A), t – time(s). Then amount of heat produced in wire where electric current flows is calculated as follows:

Q=I²·R·t

Lets calculate simple problem. How much time we need to boil 2 liters of water?

Lets take voltage U=220V;

Heating element requires current I=4A;

Efficiency coefficient is 80%;

Starting temperature of water is 20°C;

Water specific heat – 4200(kJ/kgK).

First we have to calculate the amount of heat required to boil a water:

Q_n=C·m·(t_boil-t₀)=4200·2·(100-20)=672000J;

With efficiency of 80% we get that wee neet to produce

Q=Q_n/η=672000/0.8=840000J.

So we get that

Q=A=U·I·t

where time needed to boil:

t=Q/(U·I)=840000/(220·4)=954s=15min and 54s.

When we know work produced by current during time period – we can find the power:

P=A/t = UI = 880W

I bet this is really simple, isn’t it?

Blogsphere: TechnoratiFeedsterBloglines
Bookmark: Del.icio.usSpurlFurlSimpyBlinkDigg
RSS feed for comments on this post  |  TrackBack URI for this post

This entry was posted on Monday, February 5th, 2007 at 5:30 pm and is filed under Electronics Tutorial. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.