Using powers in calculations?
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Good day,
Som insight why there are two different evaluations for “basically” the same calculus…
u=symunit;
V1 = .15*u.m^3;
p1 = 2*u.bar;
V2 = .02*u.m^3;
gma=1.4;
p2=p1*V1^gma/V2^gma
Output for this coud p2 =((2*(0.1500*[m]^3)^1.4000)/(0.0200*[m]^3)^1.4000)*[bar].
And for…
u=symunit;
V1 = .15*u.m^3;
p1 = 2*u.bar;
V2 = .02*u.m^3;
gma=1.4;
p2=p1*(V1/V2)^gma
Is this one - p2 =33.5827*[bar]
How the program is calculating powers? What's the difference between these two cases?
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Antworten (1)
Steven Lord
am 2 Sep. 2020
They give the same results, just in different forms.
u=symunit;
V1 = .15*u.m^3;
p1 = 2*u.bar;
t1 = (12+273)*u.K;
V2 = .02*u.m^3;
gma=1.4;
p2=p1*V1^gma/V2^gma
p3=p1*(V1/V2)^gma
Now check:
isAlways(p2 == p3) % true
simplify(p2)
simplify(p3) % these two simplify calls give the same result
The bottle of soda sitting next to me says it contains 2 liters, 2 quarts 3.6 fluid ounces, or 67.6 fluid ounces. That's three different ways to describe the same amount of soda.
3 Kommentare
Steven Lord
am 2 Sep. 2020
My guess is that the symunit functionality in Symbolic Math Toolbox is unwilling to work with fractional dimensions unless you explicitly tell it to simplify the result. (cubic meters)^(1.4) is meters^(4.2) -- what does that represent?
When you divide cubic meters by cubic meters before raising the result to the 1.4 power, you're raising a unitless quantity to that power.
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