As I promised last time, I will give you the detail solution to our brain teaser on my previous post here in Learn Electrical Engineering for Beginners. So for those who are not yet through, you can still catch up.
Let's proceed to solved our first problem here with this topic.
Last time, this is the worded problem I had given to you.
Problem No.1 : A process equipment contains 100 gal of water at 25 degree celcius. It is required to bring it to boiling in 10 minutes. The heat loss is estimated to be 5%. What is the KW rating of the heater?
Here's the solution:
Since, we have given the mass of water subject for boiling ( 100 degree celcius) = 100 gal. I would like to convert it to gram so that our units are compatible when we substitute it with our general formula.
Given that, 1 gal = 3.786 liters and 1 liter of water is equivalent to 1kg. We can therefore convert it:
100 gal x 3.786 liters /1 gal x 1 liter/ 1 kg = 378.6 kg of water. But since I would like to convert it to gram. It is therefore equivalent to 378,600 grams. The specific heat of water is 1cal/g-C.
In our Heat Formula, we all know that it is,
Qo = mc delta T
Qo = 378, 600 ( 1 cal/g-C) ( 100-25 ) = 28395000 cal or 28,395 kcal
Let : Wo = Qo; where Wo = energy out
but 4, 186 joules = 1 kcal
Wo = 28, 395 kcal ( 4, 186 joules/ 1 kcal) = 118,861,470 joules
In order to get the energy in of the process equipment, we will be needing the formula below to get it:
Win = Wo/ %n, which comes from the efficiency formula. This formula will be touched in various topic here in Electrical Engineering course.
Substituting the values that we have will lead to:
Win = 118,861,470 joules / (1- .05) = 125,117,336.8 joules or 125.117 x 10 ^6 joules.
We are not yet through with our answer. The problem is asking for the KW rating of the equipment. In our previous formula of Power = Energy ( joules) / Time ( seconds). Since, the water will be boiled for 10 minutes as what stated in the above problem. It is therefore,
Rating of Equipment (KW) = 125, 117, 336.8 joules / 600 seconds ( in 10 minutes)
Rating of Equipment (KW) = 208, 529 watts or 208. 529 KW - ANSWER
Notice that I did not used rounded values in my substitution. I only rounded it off in my final answer because in board exams, multiple choice answers are very near with each other. Rounding off numbers while you are still in the process of computing will take a little difference in final answer. For you to get the accurate and exact number, I advice you to round off your numbers in your final answer only.
I hope you had learned something here today. Please catch me more here in Learn Electrical Engineering for Beginners. I'll be back shortly.
Cheers!
