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I'm trying to get the R+ Solutions to a school problem with solve. Everything works fine with Dom -> Reals but when I try Dom -> PositiveReals it throws no solutions even if they exists.

(* Data *)

epsilon = 0.3;
Tinf = Quantity[295.15, "Kelvins"];
Q = Quantity[5000, "Watts"/"Meters"^2];
h = Quantity[10, "Watts"/("Meters"^2*"Kelvins")];
sigma = Quantity[5*10^-8, "Watts"/("Meters"^2*"Kelvins"^4)];

(* Equations *)

Qconv[T_] := h (T - Tinf)
Qrad[T_] := sigma*epsilon*(T^4 - Tinf^4)
Solve[Q == Qconv[T] + Qrad[T], T, Reals]

Executing this code throws

{{T -> Quantity[-1055.54, "Kelvins"]}, {T -> Quantity[605.245, "Kelvins"]}}

But when I limit the solution to R+

Input: Solve[Q == Qconv[T] + Qrad[T], T, PositiveReals]
Ouput: {}

Why is this happening ? How do I fix this "error" ?

Edit

As explained in comments Element[Quantity[#, "Kelvins"], PositiveReals] never returns True.

Is there a way to Take the QuantityMagnitude[] for Domain comparation ?

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    $\begingroup$ Element[Quantity[605.2454186793566`, "Kelvins"], PositiveReals] does not return True. It's somewhat ambiguous to look at signs of quantities: if Quantity[20, "DegreesFahrenheit"] is positive, is it's corresponding value in Celsius positive too? $\endgroup$
    – Greg Hurst
    Commented Oct 11, 2021 at 23:17
  • $\begingroup$ Thanks. I thought because Kelvin is an absolute scale it doesn't have those problems but it seems actually it has them. But can we take the QuantityMagnitud at certain Unit (ex. Celsius, Kelvin, Fahrenheit)? $\endgroup$
    – Eduardo Jiménez
    Commented Oct 11, 2021 at 23:24
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    – bbgodfrey
    Commented Oct 11, 2021 at 23:25

1 Answer 1

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I can recommend you to add the restriction $T>0$ into the Solve parameters, for instance:

epsilon = 0.3;
Tinf = Quantity[295.15, "Kelvins"];
Q = Quantity[5000, "Watts"/"Meters"^2];
h = Quantity[10, "Watts"/("Meters"^2*"Kelvins")];
sigma = Quantity[5*10^-8, "Watts"/("Meters"^2*"Kelvins"^4)];

(*Equations*)

Qconv[T_] := h (T - Tinf)
Qrad[T_] := sigma*epsilon*(T^4 - Tinf^4)
Solve[{Q == Qconv[T] + Qrad[T], T > 0}, T, Reals]

In this way the solver finds this particular positive real solution only.

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  • $\begingroup$ +1 The inequality should contain a Quantity, i.e., Solve[{Q == Qconv[T] + Qrad[T], T > Quantity[0, "Kelvins"]}, T]. It works with just 0 but if you used any other threshold, it wouldn't work without the Quantity $\endgroup$
    – Bob Hanlon
    Commented Oct 12, 2021 at 5:24

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