Let's say that you are carrying a ton of stabilized metahelium (theoretical exhaust velocity of 30 km/s, but nobody has figured out how to keep it stable for more than a few hours), and a ton of water. Adding the water to the metahelium WILL gain you more delta-v than if you just expended the metahelium by itself, BUT you would get even MORE delta-v by just using two tons of metahelium instead. Thus, it is only if your primary fuel is ultra expensive/rare enough that you can not simply pack in as much as you can carry (e.g. tritium/helium 3, or antimatter) that you are better off diluting it.
Again, I must say that minimizing energy expenditure is rarely the primary goal in engineering. But your extreme case of an exhaust velocity in range of the total mission delta-v for an interplanetary mission, I must say that for launching earth satellites efficiently, it may be desirable to slow down the exhaust.
Let us see what we can do with one ton of metahelium (assuming of course that we can produce it in ton batches, and that it really has the stated energy density). We will also subtract 10% of P from M to correct for propellant tank mass.
The rocket equation is:
(M+P)/M = e^ (dv/c)
Solving for M yields:
M= P/ (e^ (dv/c) – 1)
P1 = 1,000Kg
dv = 10,000m/s
c1 = 30,000m/s
10%P1 = 100Kg
M1 = P1/ (e^ (dv/c1) – 1) = 1,000Kg/ (e^ (10,000m/s/30,000m/s) – 1) = 2,528Kg
Subtract the 100Kg mass of the propellant tanks.
Total useful mass to orbit is 2,528Kg – 100Kg = 2,428Kg
Remember 2,428Kg
Now let us water down the methahelium and see what we can get out of it.
First we must find the energy density of methahelium.Â
KE = .5mv^2
KE/m = .5v^2
KE/Kg = .5*(30,000m/s) ^2 = 450MJ/Kg
Now let us calculate the energy density of a propellant that produces an exhaust velocity of 7,000m/s.
KE/Kg = .5*(7,000m/s) ^2 = 24.5MJ/Kg
Now we can find the ratio of water to methahelium, by dividing methahelium’s energy density by our desired energy density and subtracting one.
(450MJ/Kg)/ (24.5MJ/Kg)-1 = 17.4
Let us check this calculation
450MJ/(1Kg methahelium + 17.4Kg water)= 450MJ/18.4Kg
450MJ/18.4Kg = 24.
Now we can recalculate payload to orbit:
M= P/ (e^ (dv/c) – 1)
P2 = 18.4*P1 = 18,400Kg
dv = 10,000m/s
c2 = 7,000m/s
10%*P2 = 1,840Kg
M2 = P2/ (e^ (dv/c2) – 1) = 18,400Kg/ (e^ (10,000m/s/7,000m/s) -1) = 5,799Kg
Subtract the 1,840Kg mass of the propellant tanks.
5,799Kg – 1,840Kg = 3,959
So our final payload mass to orbit is 63% greater using the same amount of methahelium.
From a cost point of view however, the benefit may not be so apparent. For example, if our propellant tanks are disposable they must cost less than 38% the cost per Kg of methahelium if we are going to brake even(for this specific propellant ratio). And, if the designers are building a missile, storage and transport issues may rule in favor of the pure methahelium since the weight of the propellant and its tanks will be over an order of magnitude lower than our chosen ratio.
Which only goes to show that energy expenditure is not the only factor to conceder.