00:01
In this problem, we have to estimate the percentage savings in electricity.
00:04
So let's write the formula for the coefficient of performance as beta is equals to qi dot divided by wi dot.
00:17
So this can be written as th, th divided by th minus tl.
00:25
So from here we can write a relation for this wi dot as w i dot is equal to into t h minus t l divided by t h into q i dot now we know that this uh q a dot is equals to k into t h minus t l so by inserting this value into this equation we can write this w i dot divided by k is equal to into th minus tl all in power 2 divided by th.
01:10
So let's call it equation number 1.
01:13
Now let's write the values for two different values of tl.
01:19
That is 1 is a tl1 and it is equals to 20 plus 20 and 773 .15 into cal 1 and this is equals to 293 .15 kelvin.
01:39
The second one is a tl2 and this is equals to 25 plus a two hundred and seventy three point one cal one.
01:51
So this is equals to two hundred and ninety eight point one five kelvin.
01:59
Now we have to calculate this term in equation one for different values of th.
02:05
So let's see for th is equal to 45 degrees celsius.
02:12
Sorry 40 degrees searches.
02:14
Let's start from 40 degrees searches.
02:17
The equation one can be written as wi dot wi1.
02:23
Divided by k is equals to into there is 40, 40 minus 20 into cal 1 hall -rage power 2 divided by we have the value for th as a 2001093 .15 kelvin so this will give by the value for this term as a 1 .346 kelvin now for th is equal to 45 degrees celsius this quadrant can be written as a wi 1 .d.
03:11
Divided by k is equals to into 45 minus 20 into cal 1, all the power 2 divided by 293 .15 kelvin...