Thermodynamic (Solar Hot Water System) Problems?

[gfxroad]

Homework Statement

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 8ºC and the desired hot water temperature is 58ºC. Determine the required collector area.

Homework Equations

P₀ = W₀/t
W₀ = C m ΔT
P₀ = ε P

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 A_c = 1.32*10^-7 m²

 

[SteamKing]

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

 

[gfxroad]

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

 

[SteamKing]

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

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.

 

[gfxroad]

Tring to solve the problem

Any mistake or faults you see in the following solve?
P₀=W₀/t
W₀=C*m*ΔT
P₀=ε*P
P=1/ε *(C.m.ΔT) ------ (1)
ε is 50% = 0.50
C is 1 cal/gm *⁰C
since Ï= m/V and Ï_water= 1 ∴ m=V , m=600 gm
ΔT=T_m-T_a = 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
A_c=P/Q
A_c=251.16 / 22,000 = 0.0114 m² Collector area
.
.
.
is this result is illogical or a step mistake was happened.
Thanks in advance.

 

[SteamKing]

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!

 

[voko]

You can barely make a cup of coffee with 600 g of hot water!

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).

 

[SteamKing]

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).

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.

 

[gfxroad]

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.

 

[SteamKing]

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$^{3}$!

 

[gfxroad]

It's gotten bro, you know it's passed, our questions not on this point, and no need to this red bold things, I'm not haha. Anyway very thanks for your helps. Bye Bye this forum.