The high pressure Nitrogen (N2) fed to the Liquid Oxygen (LOX) and Isopropyl Alcohol(IPA) enters the tanks at high speed on pressurization. Gas at high velocity may impoinge the surface of the liquid and form sub-surface bubbles of nitrogen. A second issue with LOX is the diffusion rate of nitrogen into LOX. This could result in a lower concentration of oxygen and subsequently a lower combustion efficiency.
lends itself to the notion that where $d\forall = 0$ and $\frac{dB}{dt} = 0$,
$$\sum_{i}^{n}\bigg[\frac{\partial}{\partial t}\int_{cs}\rho b \: \big(\boldsymbol{V} \cdot \hat{n}\big)_i \:dA \bigg] = 0$$For this analysis, the extant property will be mass which will result in a return dimension of velocity.
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import numpy
"""
Conversion Factors
"""
# flowrate_mass_LOX = 1.13 # kg/s
# flowrate_volume_LOX = flowrate_mass_LOX / density_LOX
# flowrate_volume_LOX / area_entrance
constant_gas_nitrogen = 2.968e2 # J/kg*K
constant_k_nitrogen = 1.4
constant_viscosity_nitrogen = 1.76e-5
temperature = 300 # K
# Metrics
diameter_entrance = (4/16) * (2.54/100) # meters
diameter_exit = 3/32 * 2.54 / 100 # meters
diameter_inside_thread = 0.5 * 0.0254 # meters
area_entrance = 1/4 * numpy.pi * diameter_entrance ** 2
area_exit = 1/4 * numpy.pi * diameter_exit**2
length_exit_max = (diameter_inside_thread - diameter_entrance) / 2
velocity_exit = 0 # m/s
velocity_sound_nitrogen = numpy.sqrt(constant_gas_nitrogen * constant_k_nitrogen * temperature)
reynolds_desired = length_exit_max / (0.06 * diameter_exit)# laminar values are <=2100 , will not be developed. Munson 416
pressure_std = 14.7 # psi
pressure_feed_LOX = 500 # LOX side feed pressure psi
pressure_feed_IPA = 800 # IPA side feed pressure psi
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print(length_exit_max * 100 / 2.54)
print(str(round(reynolds_desired,2)))
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density_nitrogen = 1.14e0 * pressure_feed_LOX / pressure_std
velocity_exit = 2*(pressure_feed_LOX - pressure_std)/density_nitrogen
round(velocity_exit,2)
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The model used in this notebook relys on three distinct models. The first model, the inlet, uses an orifice to choke the maximum flow and has two control surfaces. The second type of section models the intermediate steps with a control surface on either end of the cylinder and radially symmetric control surfaces as a row inbetween. The last section is the end of the diffuser and only considers one end control surface and the radial exits at the end. The general stacking is as such:
inlet -> intermediate -> intermediate -> intermediate -> end
Where: $\hat{V} = \hat{n}$
$A_{in} = \frac{\pi}{4}D_{in}^{2}$
$A_{out} = \frac{\pi}{4}D_{out}^{2}$
$V_{in} = f\big(P_{feed}, P_{tank}\big)$
$\rho = $ density of nitrogen
$B = $ mass, $b = 1$.
This function reduces to:
$$\frac{\partial}{\partial t}\big(\rho b A_{in} \: \boldsymbol{V}_{in}\big)=\frac{\partial}{\partial t}\big(\rho b A_{out}\boldsymbol{V}_{out}\big)$$Since $\dot{m}_{in} = \dot{m}_{out}$, we can rely on compressible flow, $\frac{P_{in}}{\rho_{in}^k} = \frac{P_{out}}{\rho_{out}^k}$
$$$$Where: $\hat{V} = \hat{n}$
$A_{in} = \frac{\pi}{4}D_{out}^{2}$
$A_{out} = A_{in}$
$V_{in} = V_{out,\;n-1}$
$\rho = $ density of nitrogen
$B = $ mass, $b = 1$.
Since the radial holes will be designed to achieve steady flow, $\hat{V}_{rad} = \hat{n}_{rad}$
Where: $\hat{V} = \hat{n}$
$A_{in} = \frac{\pi}{4}D_{out}^{2}$
$V_{in} = V_{out,\;n-1}$
$\rho = $ density of nitrogen
$B = $ mass, $b = 1$.
Since the radial holes will be designed to achieve steady flow, $\hat{V}_{rad} = \hat{n}_{rad}$
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