LACTNET Archives

Lactation Information and Discussion

LACTNET@COMMUNITY.LSOFT.COM

Options: Use Forum View

Use Monospaced Font
Show Text Part by Default
Show All Mail Headers

Message: [<< First] [< Prev] [Next >] [Last >>]
Topic: [<< First] [< Prev] [Next >] [Last >>]
Author: [<< First] [< Prev] [Next >] [Last >>]

Print Reply
Subject:
From:
Reply To:
Lactation Information and Discussion <[log in to unmask]>
Date:
Sun, 20 Feb 2011 11:09:52 +0100
Content-Type:
text/plain
Parts/Attachments:
text/plain (45 lines)
Dear Karleen,
you see, the huge uncertainty on the last number (90% loss, i.e., 10% 
efficiency) is exactly why I took a different approach to circumvent 
that difficulty.
The efficiency estimate  is a  difficult one to give, as it has to 
include a lot of unknowns (e.g., size matching between the burner and 
the pot; pot with/without the lid, burner maintenance record, just to 
name a few which have a dramatic influence on the efficiency).
However 10% efficiency seems to me quite extreme even for emergency 
condition in Haiti...

So I made a little experiment with my (not so clean :-) 2.5 kW burner 
and in 28 min I managed to bring to boiling point 6 liters of water 
(starting from temperature 15 C) in a steel pot with a lid .
That means about 112 min to boil 24 liters.
So according to my previous calculations (4,5 g of LPG/min for my 
burner)  you would need to burn 112*4.5 = 504 g of LPG to boil 24 l of 
water.
So the efficiency should be 40%, which I find more reasonable.

All the best
Micaela


Il 20/02/2011 5.35, Karleen Gribble ha scritto:
> Hello Micaela
> Below are the calculations very helpfully worked out for me by Lactnetter Veronica Garea.
> As you can see, a considerable amount of fuel is required to formula feed.
> Karleen Gribble
> Australia
>
> Energy required to heat 24L of water from 15oC to boiling temperature at sea level, that is 100oC, is given by the equation Q=m cp (100-15)K (where Q is the amount of energy required, m is the mass of water, cp is the heat capacity of water, that is cp=4.817 kJ/(kg K) and the numbers in parenthesis represents the temperature difference). Then, Q= 23.5 kg x 4.817 kJ/(kg K) x 85 K = 9621.96 kJ per day of energy required. LPG has a specific caloric value of 46.1MJ/kg. However, this energy cannot be perfectly used to heat water, it is estimated that 90% of the energy will be lost. Thus the estimate of the amount of LPG per day required is 2kg per day or 14kg for one week.
>

             ***********************************************

Archives: http://community.lsoft.com/archives/LACTNET.html
To reach list owners: [log in to unmask]
Mail all list management commands to: [log in to unmask]
COMMANDS:
1. To temporarily stop your subscription write in the body of an email: set lactnet nomail
2. To start it again: set lactnet mail
3. To unsubscribe: unsubscribe lactnet
4. To get a comprehensive list of rules and directions: get lactnet welcome

ATOM RSS1 RSS2