Determine a rough estimate of the time it takes for a 3" diameter 5.15" long cylinder of liquid oxygen to boil off on a hot summer day in Brothers, OR (T~38C).
The temperature of the outermost carbon fiber layer is equal to the ambient air temperature ($T_s = T_{inf}$).
The temperature of the innermost (PTFE) layer is equal to the liquid oxygen (LOX) temperature ($T_1 = T_{lox}$).
The Nomex Honecomb layer is considered to be composed of mostly air.
Convection and radiation effects are neglected.
Material properties are constant.
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# Import packages here:
import math as m
import numpy as np
from IPython.display import Image
import matplotlib.pyplot as plt
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# Properties of Materials (engineeringtoolbox.com, Cengel, Tian)
# Conductivity
Kair = 0.026 # w/mk
Kptfe = 0.25 # w/mk
Kcf = 0.8 # transverse conductivity 0.5 -0.8 w/mk
# Fluid Properties
rhoLox = 1141 # kg/m^3
TLox = -183 # *C
# Latent Heat of Evaporation
heOxy = 214000 # j/kg
# Layer Dimensions:
r1 = 0.0381 # meters (1.5")
r2 = 0.0396 # m
r3 = 0.0399 # m
r4 = 0.0446 # m
r5 = 0.0449 # m
L = 0.13081 # meters (5.15")
# Environmental Properties:
Ts = 38 # *C
T1 = -183 #*C
The following analysis was performed using steady heat conduction analysis of multilayered cylinders (Cengel, pg 156). The heat transfer rate is given by:
$$\dot{Q} = \frac{T_s -T_1}{R_{total}}$$where $R_{total}$ is the total thermal resistance expressed as:
$$R_{total} = R_{PTFE} + R_{CF1} + R_{Air}+ R_{CF2}$$where:
$$R_{material} = (\frac{ln(r_{outer} / r_{inner})}{2*\pi*L*K_{material}})$$
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Rptfe = m.log(r2/r1)/(2*m.pi*L*Kptfe)
Rcf1 = m.log(r3/r2)/(2*m.pi*L*Kcf)
Rair = m.log(r4/r3)/(2*m.pi*L*Kair)
Rcf2 = m.log(r5/r4)/(2*m.pi*L*Kcf)
Rtot = Rptfe + Rcf1 + Rair + Rcf2
print('Total Thermal Resistance equals: ', "%.2f" % Rtot, 'K/W')
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#Heat transfer rate:
Qrate = (Ts - T1)/Rtot
print('Calculated Heat Transfer rate equals: ',"%.2f" % Qrate, 'W')
The energy balance on a thin layer of liquid at the surface is expressed by: (Cengel, pg. 841)
$$\dot{Q}_{transferred} = \dot{Q}_{latent, absorbed}$$or $$\dot{Q} = \dot{m}_v*h_{e, oxygen}$$
where $\dot{m}_v$ is the rate of evaportation. Solving for $\dot{m}_v$ gives us:
$$\dot{m}_v = \frac{\dot{Q}}{h_{e, oxygen}}$$
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EvapRate = Qrate/heOxy
print ('The rate of evaporation is', "%.6f" % EvapRate, 'kg/s')
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VLox = m.pi*(r1)**2*L
mLox = rhoLox*VLox
print('The mass of the liquid oxygen in tank is: ', "%.2f" % mLox, 'kg')
Finally, the amount of time that it takes for all 0.68kg of LOX to boil off is:
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Tboiloff = mLox/EvapRate/60
print('%.2f' % Tboiloff, 'minutes' )
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