Need Help creating formula: COOLING CONCRETE USING ICE

[Mykro]

Hello,

If anyone is up to the challenge.. I have this formula below that I need to modify but don't know how... And I really need the boards help..

What I need is instead of the formula returning: "T = final temperature of concrete mixture (deg F)"
I need it to return: "Wi, Weight of the Ice (Lb)"

I need to know how much ice to add to the mix to reach a target cool temperature...

I have a spreadsheet that I use to calculate concrete mix designs and I want to incorporate this new formula into it..


T = [0.22(TsWs + TaWa + TcWc) + TwWw ¡V 112Wi]
[0.22 (Ws + Wa + Wc) + Ww + Wi]

T = final temperature of concrete mixture (deg F)

Tc = temperature of cement (deg F)

Ts = temperature of fine aggregate (deg F)

Ta = temperature of coarse aggregate (deg F)

Tw = temperature of added mixing water (deg F)

Wc = weight of cement (lb)

Ws = weight of fine aggregate (lb)

Wa = weight of coarse aggregate (lb)

Ww = weight of mixing water (lb)

Wi = weight of ice (lb)

 

[thisoldman]

There is something screwy with the formula you found and posted—what's with the iV?

I used the formula from
https://www.cemstone.com/userfiles/file/PCA-Hot Weather Concreting_2.pdf

T = (0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa - 112*Mi) / (0.22*(Ma + Mc) + Mw + Mwa + Mi)

T is the target temperature in °F.
Ma, Mc, Mw, Mwa. and Mi are the weight in pounds of the aggregate, the cement, the mixing water, the free water (moisture content of aggregate), and the ice.
Ta, Tc, Tw, and Twa are the temperatures in °F of the aggregate, cement, mixing water, and the free water (moisture content of aggregate).

I came up with this formula (the final line below) to determine the weight of the ice:

T*(0.22*(Ma + Mc) + Mw + Mwa) + T*Mi + 112*Mi = 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa
T*Mi + 112*Mi = T*(0.22*(Ma + Mc) + Mw + Mwa) + 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa
Mi*(T + 112) = T*(0.22*(Ma + Mc) + Mw + Mwa) + 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa

Mi = (T*(0.22*(Ma + Mc) + Mw + Mwa) + 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa) / (T + 112)

If the temperatures are given in °C, replace 112 with 80.

 

[Mykro]

There is something screwy with the formula you found and posted—what's with the iV?

I used the formula from
https://www.cemstone.com/userfiles/file/PCA-Hot Weather Concreting_2.pdf

T = (0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa - 112*Mi) / (0.22*(Ma + Mc) + Mw + Mwa + Mi)

T is the target temperature in °F.
Ma, Mc, Mw, Mwa. and Mi are the weight in pounds of the aggregate, the cement, the mixing water, the free water (moisture content of aggregate), and the ice.
Ta, Tc, Tw, and Twa are the temperatures in °F of the aggregate, cement, mixing water, and the free water (moisture content of aggregate).

I came up with this formula (the final line below) to determine the weight of the ice:

T*(0.22*(Ma + Mc) + Mw + Mwa) + T*Mi + 112*Mi = 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa
T*Mi + 112*Mi = T*(0.22*(Ma + Mc) + Mw + Mwa) + 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa
Mi*(T + 112) = T*(0.22*(Ma + Mc) + Mw + Mwa) + 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa

Mi = (T*(0.22*(Ma + Mc) + Mw + Mwa) + 0.22*(Ta*Ma + Tc*Mc) + Tw*Mw + Twa*Mwa) / (T + 112)

If the temperatures are given in °C, replace 112 with 80.

Thanks Thisoldman,

Actually I don't need to determine the weight of the ice...

To reach the final temperature of the mix the user inputs the weights and temperatures of the cement and aggregates, and then inputs the weight of the ice until he reaches temperature he wants for the final temperature of the mix...

The formula is even simpler..

All that is really needed is to input the weights and temperatures for the "Dry" ingredients, and input for the weights and temperatures for the "Wet" ingredients along with the weight and temperature for the Water... Thats it..

Michael

 

[thisoldman]

Try this:

<tbody>
[TD="class: xl66"][/TD]
[TD="class: xl68"]A[/TD]
[TD="class: xl68, width: 64"]B[/TD]
[TD="class: xl68, width: 64"]C[/TD]

[TD="class: xl68"]1[/TD]
[TD="class: xl64"] Material
[/TD]
[TD="class: xl65"] lbs [/TD]
[TD="class: xl65"] °F [/TD]

[TD="class: xl68"]2[/TD]
[TD="class: xl66"]Cement[/TD]
[TD="class: xl66, align: right"]590[/TD]
[TD="class: xl66, align: right"]72[/TD]

[TD="class: xl68"]3[/TD]
[TD="class: xl66"]Fine Aggregate[/TD]
[TD="class: xl66, align: right"]1265[/TD]
[TD="class: xl66, align: right"]95[/TD]

[TD="class: xl68"]4[/TD]
[TD="class: xl66"]Coarse Aggregate[/TD]
[TD="class: xl66, align: right"]1865[/TD]
[TD="class: xl66, align: right"]93[/TD]

[TD="class: xl68"]5[/TD]
[TD="class: xl66"]Mixing Water[/TD]
[TD="class: xl66, align: right"]250[/TD]
[TD="class: xl66, align: right"]68
[/TD]

[TD="class: xl68"]6[/TD]
[TD="class: xl66"]Ice[/TD]
[TD="class: xl66, align: right"]40[/TD]
[TD="class: xl66, align: center"]#N/A[/TD]

[TD="class: xl68"]7[/TD]
[TD="class: xl66"][/TD]
[TD="class: xl66"][/TD]
[TD="class: xl66"][/TD]

[TD="class: xl68"]8
[/TD]
[TD="class: xl66"] Predicted °F
[/TD]
[TD="class: xl67, align: right"] 78.0
[/TD]
[TD="class: xl66"][/TD]

</tbody>

Formula in B8:
=(0.22*(SUMPRODUCT(B2:B4,C2:C4))+B5*C5-112*B6)/(0.22*SUM(B2:B4)+SUM(B5:B6))

I wonder how accurate this will be: there's no correction for the moisture content of the sand and gravel. If you could somehow keep track of the actual versus predicted...

 

[Mykro]

Thisoldman,

Genius..! Looks like you solved it.! I really appreciate your help.. Thank You very much..

I'm going to experiment with this formula and see how accurate is over time...

I'll post back with my results..


Thank You..
Michael

 

[thisoldman]

I'm happy to have helped.

The results do seem reasonable when I plug in different values.