Measuring P.P.G.

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hector

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Hi there !

I have bought a few pounds of light DME , but I don't know its p.p.g .

Therefore , I did a Test in two different ways to measure that .

1 - I dissolved 30 grams of the Extract in 250 mL of Water . After dissolving

the total Volume was more than 250 mL . I reduced its Temperature to 20C .

My Hydrometer showed then 1.042 .

2 - I dissolved 30 grams of the Extract in 150 mL of Water , then diluted it with Water to 250 mL and

reduced its Temp. to 20C .

My Hydrometer showed this time 1.046 .


Which Result should I accept as the p.p.g of this DME ?


Hector
 
The latter one would be more accurate in terms of the process you used. Your volumes and weights seem in the right ratio. Is your hydrometer the kind that uses 60F (16C) as reference or is it really for 20C? If it's the former, and you haven't adjusted, then you're real measurement is 1.047. Also make sure you calibrate your hydrometer. What you're doing right now is how I learned my hydro is off by 4 pts. I measured some DME at 1.048. Then I calibrated in water at 60F and it read 1.004. All DME that I am aware of has a listed PPG of 1.044 or 1.045.
 
You put 30 grams (30/453.59237 = 0.0661387 pound) of DME in 250 mL of water obtaining 280 grams of solution. You measured the apparent specific gravity as 1.042. This means the density is 1.042*.998203 = 1.04013 grams/cc and the volume of 280 grams of solution is 280/1.04013 = 269.197 cc. This is .269197/3.78541178 = 0.0711143 gal so your solution is 0.0661387/0.0711143= 0.930034 and, as you got 42 points the ppppg =42/0.93 = 45.1613

For comparison, if you made a 10°P solution with pure sucrose (10 grams of sucrose per 100 grams of solution) the SG would be 1.04003, the density, 1.03816, the volume of the solution 96.3245 cc ~ 0.0254462 gal, the pounds of sucrose 0.02205 and the ppppg = 46.16

In the second experiment you dissolved 30 grams DME (0.0661387 pound) in 250 mL (0.066043 gal) made up to 250 mL (0.066043 gal for 1.00145 ppg and measured SG =1.046. Thus the ppppg is 46/1.00145 = 45.9334


So you got 45.2 and 45.9 - very close to one another and to the 46.16 value for pure sucrose. You are to be commended for your skill at reading a hydrometer!

In summary, both methods gave very close to the same answer. As volumes are more difficult to read accurately than a good analytical balance your first method would be preferred if you had an accurate way to measure density. Given the accuracy with which a hydrometer can be read one method is, as you have demonstrated, as good as the other.
 
Also make sure you calibrate your hydrometer. What you're doing right now is how I learned my hydro is off by 4 pts. I measured some DME at 1.048. Then I calibrated in water at 60F and it read 1.004. All DME that I am aware of has a listed PPG of 1.044 or 1.045.

I calibrated my Hydrometer in Water at 20C and it showed 1.005 !

So , I think the p.p.g of this Extract should be 41 .

Hector
 
No. It's actual extract and it should be about 46. It is interesting that you got two readings right about the right answer with an instrument that is off by 0.005. Something does not compute here.
 
No. It's actual extract and it should be about 46. It is interesting that you got two readings right about the right answer with an instrument that is off by 0.005. Something does not compute here.

Agree - I've never heard of DME being any anything other than 44 +/-1 PPG.

OP - Did your hydrometer come with instructions that say what temp standard is used, or is it printed on the paper/scale inside the glass? It would actually be worse if it were 60F. If it is on the 60F standard, which is 15.6 C, not 20 C, then it would really be reading a point higher (1.006). I'd check to make sure you are using the right standard anyway. I think most hydrometers use the 60F standard.

You can always just apply a temp correction if you don't want to have to control the sample to the reference temp.
 
OP - Did your hydrometer come with instructions that say what temp standard is used, or is it printed on the paper/scale inside the glass?

The calibration Temperature of my Hydrometer is 20C and it is printed on

the Scale inside the glass .


Hector
 
By the way, do you know the brand and type of DME you bought?

The Seller gave me a sheet of Paper as " Certificate of Analysis " .

It's written on this Paper :

Dried Malt Extract Powder

Colour : Light

pH ( 10% Solution ) : 5.31

Acidity as Lactic Acid : 0.95%

Moisture : 2.44

Reducing Sugar as Maltose : 64.23%

Protein ( Dry Basis ) : 4.2%

Ash : 1.5%

Produced in INDIA

Hector
 
The number I gave is for completely dry malt extract. If a sample contains moisture then its ppppg will be reduced by the moisture content. And lo we see:
Moisture : 2.44

That will reduce the ppppg

Reducing Sugar as Maltose : 64.23%
Not terribly helpful as clearly there are other (we hope) sugars in there which are not reducing sugars.

Protein ( Dry Basis ) : 4.2%
That's quite a bit of protein but it does not detract from extract if it is soluble.

Ash : 1.5%

Minerals which were presumably dissolved. They still count as extract.

Produced in INDIA
No comment.
[/QUOTE]

I'd suggest checking the hydrometer at close to the point where you want to make the measurement. Dissolve 30 grams of sucrose (table sugar is pure enough) in 270 grams of water (tare a beaker, add the sugar, add some water, dissolve, make up to 300 grams). This is a 10 °P solution with apparent SG 1.04003. If the hydrometer reads 0.005 low at that level then the paper has slipped and you can correct your readings based on the now known bias use the methodology I outlined earlier and calculate the actual extract. I expect that the error at 10 °P will match approximately the error you got with distilled water but brewing hydrometers (good ones) are calibrated for the surface tension of wort - not distilled water.

As an aside - sucrose was chosen for the basis of the Plato (and Brix and Balling) scale because it does not pick up water of hydration as enthusiastically as maltose.
 
To ajdelange :

I dissolved 30 grams of table sugar in 150 mL of Water and

diluted it to 300 mL .

After reducing its Temperature to 20C , the Hydrometer

read 1.045 .

So , the correct S.G. of this solution is 1.040 , as you said .

I still believe that the p.p.g. of the Extract is 42 .

Hector
 
A solution of 30 grams of sucrose in 300 mL of DI water solution is a 9.6457 °P solution with SG = 1.03856. If your hydrometer reads 1.045 in this it is reading high by 1.045 - 1.03856 = 0.00644 (throw it away). Thus the original reading of 1.046 should be corrected to 1.0386 and you got 38.6 points from 1.00415 ppg (measurement 2) for 38.4 ppppg. That seems low even given the moisture content. I suspect errors in reading volume are responsible. With class B glassware you can expect ± 0.6% error if the liquids are attemperated in a water bath. With "student" glassware, who knows. That's why I suggested doing it by weight.

I just wish people would forget about this ppppg stuff and stick with Plato.
 
A solution of 30 grams of sucrose in 300 mL of DI water solution is a 9.6457 °P solution with SG = 1.03856. If your hydrometer reads 1.045 in this it is reading high by 1.045 - 1.03856 = 0.00644 (throw it away). Thus the original reading of 1.046 should be corrected to 1.0386 and you got 38.6 points from 1.00415 ppg (measurement 2) for 38.4 ppppg. That seems low even given the moisture content. I suspect errors in reading volume are responsible. With class B glassware you can expect ± 0.6% error if the liquids are attemperated in a water bath. With "student" glassware, who knows. That's why I suggested doing it by weight.

I just wish people would forget about this ppppg stuff and stick with Plato.

1 - I will NEVER throw my Hydrometer away . It's VERY expensive ( 20-25 $ )

2 - What do you mean by this Sentence : " Thus the original reading of 1.046 should be corrected to 1.0386 and you got 38.6 points " ?!

As I know now that this Hydrometer shows 0.006 high , then I can correct my Measurments

and 1.046 would be 1.040 ( NOT 1.0386 ) .

3 - On the Glassware I use it's written ( DIN A ) ± 0.15 ml in 20C .

Hector
 
1 - I will NEVER throw my Hydrometer away . It's VERY expensive ( 20-25 $ )

It may have cost you $25 but if it is off by 1.5 °P it isn't worth $25 IMO.

2 - What do you mean by this Sentence : " Thus the original reading of 1.046 should be corrected to 1.0386 and you got 38.6 points " ?!

What it means is that if your hydrometer reads 1.045 when it is in the sucrose solution whose SG is 1.03856 it is reading 0.00644 SG high and that you should subtract 0.00644 (0.006) from any reading in that specific gravity range. Thus 1.046 must be corrected to 1.0396. That's 39.6 points.

As I know now that this Hydrometer shows 0.006 high , then I can correct my Measurments

Yes, I suppose so but you really should get your money back. That's pretty bad performance. The error is 0.004 SG (1 °P) at 0 °P and 0.006 (1.5 °P) at 10 °P. What is it at 5°P? At 17°P. The hydrometers I use are good to better than 0.1°P at any level I've checked them against and I don't think I paid more that $25 apiece for them.

and 1.046 would be 1.040 ( NOT 1.0386 ) .
.
Roger that. I shouldn't try to do even simple math in my head. So you got 39.6 points, not 38.6. Thus the extract yield of the DME is 39.6/1.00415 = 39.4 ppppg. That still seems low but given that volume rather than weight measure was used and the possibility of water pickup during handling of the malt is suppose it's possible.

3 - On the Glassware I use it's written ( DIN A ) ± 0.15 ml in 20C .

Assuming that this is a 300 mL volumetric flask that's 0.05% and consistent with Class A which makes the liquid measure a smaller part of the error budget but it's still best to do these measurements based on weight. A decent balance ought to read 100 grams to a precision of 0.0001 gram (0.0001%) or so and properly calibrated should give an accuracy close to that. The Plato system (from which our SG measurements are derived) is a w/w system because of this.

Let's wrap this up with the formula

°P = -616.868 + 1111.14*SG -630.272*SG*SG + 135.997*SG*SG*SG

which you can use to compute grams extract per 100 grams solution from SG (the apparent 20/20 specific gravity) to do as many or as few additional calculations of the type I've outlined here as you wish. °P is the number of grams of extract in 100 grams of solution i.e. the sucrose concentration (all extract is modeled as sucrose) on a w/w basis. This is the "official" ASBC polynomial and is thus accepted throughout the north American brewing industry.
 
To ajdelange :

I'd like to make a 300 mL Sugar Solution which has the S.G. = 1.090 .

How much Sugar do I need to dissolve ?

Hector
 
To ajdelange :

I'd like to make a 300 mL Sugar Solution which has the S.G. = 1.090 .

How much Sugar do I need to dissolve ?

Hector

Program that does the calculations is in use (collecting data) so I can't give you the actual number at the moment but I can tell you how to do it.

Step 1: Paste
= ((135.997*A1 -630.272)*A1 + 1111.14)*A1 -616.868

into cell A2 of a spreadsheet.

Step 2: Type 1.090 into cell A1. A2 now shows the Plato strength of the solution. This is the number of grams of sugar you will need for each 100 grams of solution.

Step 3: Mutilply the density of water at 20 °C (0.998203) times the desired S.G. This is the density of the solution you will have.

Step 4: Multiply this times the 300 mL desired volume. That's how much 300 mL will weigh.

Step 5: Divide by 100 (that's how many units of 100 grams you have)

Step 6: Multiply by the Plato value. That's the number of grams sucrose required.

As I keep noting you'll get a result closer to what you intend if you multiply the Plato value times 3 and put that much sugar in a beaker, add water, stir and make up to 300 grams with more DI water.
 
ajdelange:

First of all, thanks for the lesson on the Plato scale and extract chemistry.

Your following point made me wonder something:

The error is 0.004 SG (1 °P) at 0 °P and 0.006 (1.5 °P) at 10 °P. What is it at 5°P? At 17°P.

I would think that because a hydrometer is based on density, there would be a linear relationship between the float height and SG reading. I realize that SG and density are not the same, but I don't think that would affect the linearity (much). As far as I can tell the markers on a hydrometer are evenly spaced. And thus the paper-position calibration would be a constant offset. On the other hand your comment about being calibrated for wort surface tension effects would seem to indicate the possibility of a non linearity since I assume the surface tension would vary with sugar concentration. But I doubt the effect is that significant.

It seems the error he reported with water was 5 points and the error with sucrose solution was pretty close to 5 as well.

So, I wonder if it is linear or not? Thoughts?
 
Model the hydrometer as having total weight W and consisting of a bulb with a stem attached. Define some arbitrary point along the length of the stem close to the bulb but definitely past the transition, i.e. where the rest of the stem is of uniform cross sectional area, A, as where the stem starts and let the stem extend L units beyond this. The bulb itself, the transition from bulb to stem and the part of the stem up to the arbitrary stem origin point (beyond which there are L linear units of stem) has volume V. When the instrument is immersed in a liquid of density p the stem sticks out of the sample to a height h. The higher the density the higher than h. The calibration is the mapping between h and p.

At equilibrium the downward force on the hydrometer is W, its weight. The upward forces are the buoyancy of the bulb and L-h units of stem plus the buoyancy of h units of stem in air. The liquid displaced is V + (L-h)*A and so the buoyancy from displacement of liquid is p*(V + (L-h)*A). The volume of air displaced is h*A and so the buoyancy from air displacement is a*h*A where a is the density of air. At equilibrium the buoyant forces equal the gravitational force so
W= p*(V + (L-h)*A) + a*h*A

Solving this for (and you'd better check me on this)

h = (L*A + V - W/p)/(A*(1- p/a))

Since p >> a we can approximate p/a = 0 and simplify

h = (L*A + V - W/p)/A = L + V/A - W/p*A

Rather than concern ourselves with the small inaccuracy we simply call readings obtained without the correction for air density "apparent" and proceed from there. The ASBC tables and polynomial are based on apparent specific gravity.

The most important thing in terms of your question here is the h is the sum of 2 constants minus a term that is inversely proportional to the density p. Thus h is not a linear function of density. But the deviation from linearity should be small depending on the design of the hydrometer. If we want h0 to be how far the stem sticks out when water is measured (p0 ~ 1) and h1 to be the length of the scale at the maximum calibrated density p1 then we can wrie

h0 = L + V/A - W/p0*A
and
h1 = L + V/A - W/p1*A

and subtracting these gives

h1-h0 = W/p0*A - W/p1*A = (W/A)*(1/p0 - 1/p1)

so it is clear that the main design parameter is W/A. To see how much non linearity there is let's suppose that you want a hydrometer that reads from 1.000 (p0 = 0.008203) to 1.080 (p1=1.07806) and that you want the scale to be 10" (25.4 cm) long. Then W/A = 25.4/(1/.998203 - 1/1.07806) = 342.283. If you want the stem to stick out of water 2 cm then

L + V/A = 2 + 342.283/0.998203 = 344.900


and

h = 344.9 - 342.283/p = 344.9-342.283/(0.998293*SG) = 344.9 - 342.9/SG

describes the hydrometer.

If you put that formula in a spreadsheet and type in 1.00 for the SG it tells you the stem will be 2 cm out of the liquid - that's what we designed for. If you enter 1.010 it will compute 5.396 i.e. 3.395 cm above the 0 mark. If you enter 1.020 it will compute 8.723 or 6.723 beyond the reference mark. Now if the change were linear doubling the points would double the displacement to 6.790. But it didn't. It went to 6.723 cm i.e. 0.067 cm less than linearity would predict. Going to 1.04 gives h = 15.188 cm or 13.188 past the 0 mark. That's 4 times 10 points so a linear instrument would show 4*3.395 = 13.580. At midscale the non linearity amounts to almost 4 mm. But as the scale is 80/254 = 0.314 point ( i.e the difference between 1.040 and 1.0403) you can see that the non linearity doesn't amount to much.

All this ignores surface tension effects, of course. Slippage of the paper inside the stem or misplacement of it at manufacture are doubtless the major causes of error and are probably pretty uniform across the scale. If I were making the hydrometer we just designed I'd position the scale so that the error (assuming linear markings) was 0.15 point at 1.040 and -0.15 point at 1.000 and 1.080)
 
Just as Info :

I used the Formula which " ajdelange " wrote for measuring the amount of Sugar .

I put 1.092 as the S.G. and the Formula told me 72 grams .

So , I dissolved 72 grams of Sugar in 200 mL of Water and

diluted it to 300 mL .

After reducing its Temperature to 20C , the Hydrometer read 1.097 .

As you see , the Error is always + 0.005 .

Hector
 
Very nice analysis. Looks sound to me, even if the results are counter-intuitive (to me anyway). I can't seem to think of where the non-linearity originates.

It seems like saying the height is non-linear with gravity points is different than saying it's non-linear with density. E.g. the density difference between 1.010 and 1.020 is not 2x.

Solving this for (and you'd better check me on this)

h = (L*A + V - W/p)/(A*(1- p/a))

Since p >> a we can approximate p/a = 0 and simplify

h = (L*A + V - W/p)/A = L + V/A - W/p*A

I solved differently, but still got the same answer by assuming a = 0.
 
Very nice analysis. Looks sound to me, even if the results are counter-intuitive (to me anyway). I can't seem to think of where the non-linearity originates.

It originates from the fact that density, or SG appear in the denominator. A linear function has the form y(p) = m*p + b. This is in the form y = m*p^-1 + b. That's not linear.

It seems like saying the height is non-linear with gravity points is different than saying it's non-linear with density. E.g. the density difference between 1.010 and 1.020 is not 2x.

True but points are still a linear function of density

pts = SG*1000 - 1000 = p*998.203 - 1000

In terms of specific gravity

h = L + V/A - (W/A)/(S*0.998203) = L + V/A - (1/.998203)*(W/A)/(1 + q)
where q = pts/1000 (e.g. 1.040 = 1 + 40/1000)


If we also define b = L + V/A and m = (1/.998203)*(W/A) then

h = b - m/(1 + q)

This can be written (Taylor series expansion about 1) as

h = b - m + m*q - m*q^2 + m*q^3...

= b - m + m*(pts/1000) - m*(pts/1000)^2 + m*(pts/1000)^3 -...

This makes it plain that h is a non linear function of points. For example, for SG = 1.040 we'd have

h = b - m + m*(0.040 + 0.0016 - 0.000064 + 0.0000026...)

This makes it very easy to see what the non linearity is in points. Most of it is from the squared term - 1.6 points. The other terms are insignificant.
 
It originates from the fact that density, or SG appear in the denominator. A linear function has the form y(p) = m*p + b. This is in the form y = m*p^-1 + b. That's not linear.

Sorry - I should be more specific. The math was clear and the non-linearity was noted. I was talking about the connection to the physical world. It seemed like a linear change in density (weight over volume) would have resulted in a linear change in buoyancy which it seems would have resulted in a linear change in height. I'll have think about that some more.
 
If the density changes by 10% then the buoyant force must be reduced by 10% to restore equilibrium. The buoyant force is V*p where V is the immersed volume and p the density, and must equal the weight, W, of the instrument. Thus W=V*p and the immersed volume is V=W/p. That's non linear in density as is easily verified. For water (p=1) V=W. For wort of gravity 1.1 (10% change), V= W/1.1 = 0.90909*W, a 9.09% change. As linearity is a mathematical concept I can't think of any other way to explain it than through the math.
 
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