# Thermodynamic (Solar Hot Water System) Problems?

1. Dec 21, 2013

1. The problem statement, all variables and given/known data

A solar hot water system is designed to supply 100% of the hot water demand of 0.6 m3/day when the isolation on the tilted collector is 22,000 kJ/m2 with a collector efficiency of 50%. The tap water temperature is 8ÂºC and the desired hot water temperature is 58ÂºC. Determine the required collector area.

2. Relevant equations

P0 = W0/t
W0 = C m Î”T
P0 = Îµ P

3. The attempt at a solution
I tried to solve this problem, but how can I use the 0.6 m3/day, somehow I was used it by replacing m3 by kg because density of water is 1 cal/gm*0C, but some illogical value raise as a result like Ac = 1.32*10-7 m2

Last edited: Dec 21, 2013
2. Dec 21, 2013

### SteamKing

Staff Emeritus
Since when is density measured in units of cal/gm-C?

3. Dec 21, 2013

Thermodynamic (Solar Hot Water System) Problems?miss print

I'm sorry its miss print.
I mean that the density of water is 1 kg/m3.
So according to the equation Ï= m/v when Ï=1 m=v
Hence we can replace 0.6 m3 by 0.6 kg

4. Dec 21, 2013

### SteamKing

Staff Emeritus
Is the density of water equal to 1 kg/m^3? That seems to be an awfully small number. Take a moment to visualize how large a volume a cube measuring 1 m on each side is.

5. Dec 22, 2013

Tring to solve the problem

Any mistake or faults you see in the following solve?
P0=W0/t
W0=C*m*Î”T
P0=Îµ*P
P=1/Îµ *(C.m.Î”T) ------ (1)
Îµ is 50% = 0.50
C is 1 cal/gm *0C
since Ï= m/V and Ïwater= 1 âˆ´ m=V , m=600 gm
Î”T=Tm-Ta = 58-8=50 0C
by replacing the values above in equation (1)
P=1/0.50 * (1 * 600 * 50)
P= 60,000 cal = 60,000 * 4.186 = 251160 J = 251,16 kJ
Ac=P/Q
Ac=251.16 / 22,000 = 0.0114 m2 Collector area
.
.
.
is this result is illogical or a step mistake was happened.

6. Dec 22, 2013

### SteamKing

Staff Emeritus
Hello! You're not listening! Water does not have a density of 1 kg per cubic meter! That is an absurdly low density! You can barely make a cup of coffee with 600 g of hot water!

7. Dec 22, 2013

### voko

They have interesting sizes of coffee cups where you live. Starbucks' largest is Trenta, at 920 ml, but it is for iced drinks only, I hear. Their largest, which is actually used to serve hot beverages, is Venti, 590 ml.

The customary kitchen "cup" unit is 250 ml (except in the US, where it is a tad less).

8. Dec 22, 2013

### SteamKing

Staff Emeritus
Yeah, but when the water only has a density of 1 kg/m^3 like the poster thinks, it's hard to get that full-bodied flavor you want in a premium cuppa Joe.

9. Dec 22, 2013

This forum make me bored meanwhile two of the big members have fighting without giving any respect to the new members, nevertheless I not get any interest.

10. Dec 22, 2013

### SteamKing

Staff Emeritus
You're still not paying attention to what is being said about major errors in your calculation.

WATER DOES NOT HAVE A DENSITY OF 1 KG / M$^{3}$!

11. Dec 24, 2013