Good day people
I am modelling a water bottle rocket. Using the conservation of mass :
$$-{\rho}vA + \frac{d}{dt}∫dM = 0 \tag{1}$$
Since the mass, O2 pressure, O2 volume and velocity change over time :
$$ \rho = \frac{M}{V} \tag{2} $$
and $$PV = MRT $$ R in J/kg.K : Assume constant T $$ M = PV/RT \tag{3} $$
Plugging (2) and (3) into (1) we find :
$$ \frac{d}{dt} ∫d(\frac{PV}{RT}) = \frac{PVvA}{RTV} $$
Since $RT$ is constant :
$$ \frac{1}{RT}\frac{d}{dt}∫d(PV) = \frac{PvA}{RT} $$
Simplifying ( sort of )
$$ \frac{1}{P}\frac{d}{dt}∫d(PV) = vA$$
This is where is start to doubt my math :
$$\frac{d}{dt}∫\frac{1}{P}d(PV) = vA $$ : Can I put P inside the integral?
Assuming I can. $$d(PV) = PdV + VdP $$using chain rule
thus
$$ \frac{d}{dt}∫(\frac{PdV}{P} + \frac{VdP}{P}) = vA$$
after integrating :
$$\frac{d}{dt}(V_1-V_0 + Vln(\frac{P_1}{P_0}) = vA $$: Where I have values for V0 and P0. but I do not know what the value for V is.( coefficient of the ln term)
multiplying by dt on both sides :
$$ d(V_1-V_0 + Vln(\frac{P_1}{P_0}) = vA dt $$: And this is where I am stuck. Substituting my values : $V_1 = 0.057 ; V_0 = 0.848 ; P_1 = 101.3 ; P_0 = 1500 $
$$ d(0.791 - 2.69V) = vA dt $$
integrating :
$$ ∫d(0.791 - 2.69V) = ∫vA dt $$
$$ 0.791 -2.69(V-V_0) = A∫vdt $$
simplifying
$$ V = \frac{A}{-2.69}∫vdt - \frac{0.791}{-2.69}+V_0 $$
$$ V = -0.105∫vdt-0.294+0.057 $$
$$ V = -0.105∫vdt-0.237 $$
Does this seem correct?