Homework#2
1/24/25, 9:08 PM
Homework #2
Synopsis
In expansion and compression problems, we currently have two strategies:
1. Solving problems using ideal gas properties
2. Solving problems using real gas properties
Regardless of the cases above, the task is the same we need to identify the number of
degrees of freedom of the initial and final states of the fluid and we need to identify the
pathway that the fluid takes during the compression.
Reversible, adiabatic expansion will take 1 of 2 paths:
1. isoenthalpic pathway - when the fluid is expanded across a valve with no work
performed.
2. isoentropic pathway - when the fluid is expanded across a turbine where work is
extracted.
Reversible, adiabatic compression can only take 1 path:
1. isoentropic pathway - when the fluid is compressed when either boundard or shaft
work.
Functions used:
1. np.interp
2. pd.read_csv
3. plt.plot
Objectives
1. More practice with unit conversion
2. Working with real data and using interpolation as a means to extract quantities
3. Working with analytical functions
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Tutorial
Let's start by importing our standard libraries
In [ ]: import matplotlib.pyplot as plt
import numpy as np
from matplotlib import rc
import os
import pandas as pd
from IPython.display import Markdown as md
path2files = os.getcwd() + '/Homework2 - data'
Loading Property Data:
As a quick refresher from HW#1 that property databank data can be loaded using the
pandas.read_csv function by telling the function that the text file is tab ('\t') delimited.
This tells the functions that the columns in the text file are separated by tabs. The
resulting data will be stored in a dataframe which acts like an excel worksheet where the
columns of data can be operated on and plotted. Below I load in the 1 bar isobar data
from the nist property databank
In [ ]: meth6bar = pd.read_csv(os.path.join(path2files,'Methane.txt'), delimiter='\t')
nitrogen_isobar_1bar = pd.read_csv(os.path.join(path2files,'Nitrogen1isobar.txt'),
nitrogen_isobar_4bar = pd.read_csv(os.path.join(path2files,'Nitrogen4isobar.txt'),
#nitrogen_isobar_2000bar = pd.read_csv(os.path.join(path2files,'nitrogen_isobar_20
Pandas Keys:
To access a column in a pandas dataframe, we need to use its keys. Keys are usually
strings in dataframes. To know the formatting of the keys, we can use the .keys()
property of the dataframe which returns a list of the strings corresponding to the keys.
In [ ]: print(nitrogen_isobar_1bar.keys())
Index(['Temperature (C)', 'Pressure (bar)', 'Density (mol/l)',
'Volume (l/mol)', 'Internal Energy (kJ/mol)', 'Enthalpy (kJ/mol)',
'Entropy (J/mol*K)', 'Cv (J/mol*K)', 'Cp (J/mol*K)', 'Sound Spd. (m/
s)',
'Joule-Thomson (K/bar)', 'Viscosity (uPa*s)', 'Therm. Cond. (W/m*K)',
'Phase'],
dtype='object')
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Extracting data and plotting:
Now that we know the keys we can extract properties of interest and plot them or
perform calculations with them. Copying and pasting the strings within the list above
makes this task simple. Note that the column associated with each key is accessed by
the correspond key string passed to the entry in the key dictionary using the
dataframe[key] format. In the example below, we plot the enthalpy vs. the temperature.
In [ ]: key1 = 'Temperature (C)'
key2 = 'Enthalpy (kJ/mol)'
key3 = 'Entropy (J/mol*K)'
key4 = 'Density (kg/m3)'
plt.figure(dpi=300,figsize=(1.5,1.5))
Temperature = nitrogen_isobar_1bar[key1]
Enthalpy = nitrogen_isobar_1bar[key2]
plt.plot(Temperature, Enthalpy)
plt.xlabel(key1)
plt.ylabel(key2)
Out[ ]:
Text(0, 0.5, 'Enthalpy (kJ/mol)')
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Interpolation:
Often when working with real gas properties, we need to interpolate because the data is
only available at finite increments and the value that we seeks is not available in our
table. For this we can use the np.interpolate command.
Let's say that I want to find the enthalpy at 25 C using the isobar above, I do that using
the code below and print the result in a formatted f-string (this makes things nice and
tidy). Note that the formatting string works {python_variable}:.1f which is a floating
point number that will print off values up to only one decimal place.
In [ ]: H = np.interp(25, Temperature, Enthalpy)
print(f'The enthalpy at 25 C is {H:.1f} kJ/kg')
The enthalpy at 25 C is 8.7 kJ/kg
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We can confirm that the interpolation was successful by plotting the interpolated value.
The red star lands on the line so the interpolation is successful.
In [ ]: plt.figure(dpi=300,figsize=(1.5,1.5))
plt.plot(Temperature,Enthalpy, '-r')
plt.plot([25],[H],'r*')
plt.xlim(0,100)
plt.ylim(0,500)
plt.xlabel(key1)
plt.ylabel(key2)
Out[ ]:
Text(0, 0.5, 'Enthalpy (kJ/mol)')
Interpolation works for most things, but sometimes when the data is non-monotonic the
interpolation can fail (where there are multiple y-values for the same x-value). In these
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cases, it is best to use masking for the region of the data that you anticipate finding the
answer.
Problem #1: An adiabatic compressor operates at
steady state to compress nitrogen from 1 to 4 bar.
Using the NIST property database, compute the work
required for this compression. Note that the nitrogen
enters the compressor at −80 o C .
1. State your assumptions and write in the space provided
below the simplified energy and entropy balance expression for
the compression.
dU
dt
= 0 = Σ(Ṅ H̄) + Ws because Q̇ = 0 and dV = 0
Ws = Ṅ(H̄2 − H̄1 )
dS
˙ =0
= 0 = Σ(Ṅ S̄) because Sgen
dt
S¯1 = S¯2
2. Calculate the enthalpy change associated with the reversible
compression and compare it to the enthalpy change for the
equivalent ideal gas. Compute the temperature of the outlet
gas in both cases
In [ ]: print(nitrogen_isobar_1bar.keys())
Index(['Temperature (C)', 'Pressure (bar)', 'Density (mol/l)',
'Volume (l/mol)', 'Internal Energy (kJ/mol)', 'Enthalpy (kJ/mol)',
'Entropy (J/mol*K)', 'Cv (J/mol*K)', 'Cp (J/mol*K)', 'Sound Spd. (m/
s)',
'Joule-Thomson (K/bar)', 'Viscosity (uPa*s)', 'Therm. Cond. (W/m*K)',
'Phase'],
dtype='object')
In [ ]: t1 = nitrogen_isobar_1bar[key1]
entro1 = nitrogen_isobar_1bar[key3]
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entha1 = nitrogen_isobar_1bar[key2]
s1 = np.interp(-80,t1,entro1)
s2=s1
h1 = np.interp(-80,t1,entha1)
entha4 = nitrogen_isobar_4bar[key2]
entro4 = nitrogen_isobar_4bar[key3]
t4 = nitrogen_isobar_4bar[key1]
h2 = np.interp(s2,entro4,entha4)
t2 = np.interp(s2,entro4,t4)
print(f'Enthalpy change: {h2-h1} kJ/mol')
print(f'Enthalpy change: {(h2-h1)/28.02*1000} kJ/kg')
print(f'Real Gas temp out: {t2+273.15}K or {t2}C')
Enthalpy change: 2.728166440702675 kJ/mol
Enthalpy change: 97.36496933271503 kJ/kg
Real Gas temp out: 287.29680370289384K or 14.146803702893845C
In [ ]: R = 8.314
cp = 28.318
ti1 = -80+273.15
ti2 = 4**(R/C_p)*ti1
print(f' ideal gas out temp = {ti2} K or {ti2-273.15}C')
print(f'Ideal enthalpy change = {cp*(ti2-ti1)/1000} kJ/mol')
ideal gas out temp = 290.1723468190625 K or 17.022346819062534C
Ideal enthalpy change = 2.747478817222213 kJ/mol
3. Calculate the corresponding amount of work for the
compression if the feed is 5000 mol/s
In [ ]: w = (h2-h1)*5000
print(f'Work: {w}kW')
Work: 13640.832203513375kW
Problem #2: A cylinder, initially charged with nitrogen
at 2000 bar and 25 C , is vented through a valve. The
result of this adiabatic expansion is that both the
temperature and pressure of the gas within the
cylinder decrease. The gas exits the cylinder
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expanding to atmospheric pressure. The volume of
the cylinder is 1 m3 .
1. If there is negligable heat transfer during the expansion, what
is the final temperature within the cylinder? What mass of
nitrogen remains in the cylinder? Assume nitrogen is an ideal
gas with a heat capacity Cp = 28.318 J/mol ∗ K )
In [ ]: tn2 = ((1.0135/2000)**(8.315/28.318))*(25+273.15)
print(f'Final temp: {tn2}K')
n = (1.01*10**5)/(8.314*tn2)
print(f'Mass in the cylinder: {n*28.02/1000}kg')
Final temp: 32.12662273156453K
Mass in the cylinder: 10.595328134508662kg
2. Repeat (1) using real gas properties of nitrogen.
In [ ]: Nitrogen2000isobar = pd.read_csv(os.path.join(path2files,'Nitrogen2000isobar.txt')
Nitrogen101isobar = pd.read_csv(os.path.join(path2files,'Nitrogen101isobar.txt'),
In [ ]: ty1 = Nitrogen2000isobar[key1]
entroy1 = Nitrogen2000isobar[key3]
enthay1 = Nitrogen2000isobar[key2]
s1 = np.interp(25,ty1,entroy1)
print(s1)
s2=s1
h1 = np.interp(25,ty1,enthay1)
entha101 = Nitrogen101isobar[key2]
entro101 = Nitrogen101isobar[key3]
t101 = Nitrogen101isobar[key1]
h2 = np.interp(s2,entro101,entha101)
t2 = np.interp(s2,entro101,t101)
print(f'Enthalpy change: {h2-h1} kJ/mol')
print(f'Gas temp out: {t2+273.15}K')
dens = Nitrogen101isobar['Density (mol/l)']
dfinal = np.interp(s1,entro101,dens)
print(f'Mass in the cylinder is {dfinal*28.02}kg')
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121.53529739776953
Enthalpy change: -10.844793054021956 kJ/mol
Gas temp out: 77.32999999999998K
Mass in the cylinder is 337.966259144853kg
Problem #3: In order to fuel booster, SpaceX must fill
the fuel tanks with ∼ 800 tons of liquid methane and
∼ 3000 tons of liquid oxygen. The tanks hold the
saturated liquids at 6 bar while the ship awaits
launch. A launch at sea level results in the firing of all
33 raptor engines with the exhaust gas velocity of 3.2
km/s. Using the NIST Property database for liquid
methane and liquid oxygen and assuming the fuel to
oxidizer ratio during combustion is given by the ratio
of volumes of liquid methane and oxygen in the tanks,
answer the following questions.
1. What are the temperatures and densities of the liquid
methane and liquid oxygen in the tanks? Plot the isobars of
liquid methane as density (ρ) in kg/m3 vs. temperature (T) o C
and identify the saturated liquids.
In [ ]: mmmeth = 16.04
mmox = 32
In [ ]: keydense = 'Density (mol/l)'
keytemp = 'Temperature (C)'
methtemp6bar = meth6bar[keytemp]
Dens = meth6bar[keydense]*mmmeth
for i in range (methtemp6bar.size):
checkTemp = methtemp6bar[i]
if checkTemp == methtemp6bar[i+1]:
print(f'Tcrit for methane: {checkTemp}C')
print(f'Density at Tcrit: {Dens[i]}kg/m^3')
Tcritmeth6 = checkTemp
denscritmeth = Dens[i]
break
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ox6bar = pd.read_csv(os.path.join(path2files,'Oxygen6bar.txt'), delimiter='\t')
keydenseox6 = 'Density (mol/l)'
keytempox6 = 'Temperature (C)'
Tcox6 = ox6bar[keytemp]
Densox6 = ox6bar[keydense]*mmox
densityox6 = Densox6*16.01*0.001
for i in range (Tcox6.size):
checkTempox6 = Tcox6[i]
if checkTempox6 == Tcox6[i+1]:
print(f'Tcrit for oxygen: {checkTempox6}C')
print(f'Density at Tcrit: {Densox6[i]}kg/m^3')
Tcritox6 = checkTempox6
denscritox = Densox6[i]
break
mask2 = Tcox6 <-80
mask1 = methtemp6bar <-80
plt.figure(dpi=300,figsize=(1.5,1.5))
plt.xlabel(keytempox6)
plt.ylabel('Density (kg/m3)')
plt.plot([Tcritox6],[denscritox],'r*')
plt.plot([Tcritmeth6],[denscritmeth],'r*')
plt.plot(Tcox6[mask2],Densox6[mask2],label = 'Oxygen')
plt.plot(methtemp6bar[mask1],Dens[mask1],label = 'Methane')
plt.legend(fontsize=4,frameon=False)
Tcrit for methane: -134.42C
Density at Tcrit: 379.07331999999997kg/m^3
Tcrit for oxygen: -161.69C
Density at Tcrit: 1026.976kg/m^3
Out[ ]:
<matplotlib.legend.Legend at 0x15ad7f970>
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2. During launch, the methane and oxygen are pressurized by
separate turbopumps to 860 bar and 430 bar respectively
before being pumped into the combustion chamber. Construct
a Temperature (T) vs. Entropy (S) diagram for this compression
and indicate the temperature and entropy of the gas exiting the
turbopump.
In [ ]: meth860bar = pd.read_csv(os.path.join(path2files,'Methane860bar.txt'), delimiter='
ox430bar = pd.read_csv(os.path.join(path2files,'Oxygen430bar.txt'), delimiter='\t'
In [ ]: #do the equation balances, end result is constant entropy
entropymeth860 = meth860bar['Entropy (J/mol*K)']
tempmeth860 = meth860bar['Temperature (C)']
entropymeth6 = meth6bar['Entropy (J/mol*K)']
entropyox430 = ox430bar['Entropy (J/mol*K)']
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tempox430 = ox430bar['Temperature (C)']
entropyox6 = ox6bar['Entropy (J/mol*K)']
In [ ]: satmeth6 = meth6bar [Dens == denscritmeth]
satox6 = ox6bar [Densox6 == denscritox]
In [ ]: #smeth = np.interp(Tcritmeth6,methtemp6bar,entropymeth6)
smeth = float(satmeth6['Entropy (J/mol*K)'].iloc[0])
sox = float(satox6['Entropy (J/mol*K)'].iloc[0])
print(f'Entropy meth: {smeth}J/molK')
print(f'entropy ox: {sox}J/molK')
Tmeth860 = np.interp(smeth,entropymeth860,tempmeth860)
Tox430 = np.interp(sox,entropyox430,tempox430)
print(f'Temp meth: {Tmeth860}C')
print(f'Temp ox: {Tox430}C')
Entropy meth: 12.472J/molK
entropy ox: 105.89J/molK
Temp meth: -107.10666666666667C
Temp ox: -149.3116030534351C
In [ ]: print('Different units for easier grading:')
print(f'Entropy meth: {smeth/mmmeth} J/gK')
print(f'entropy ox: {sox/mmox} J/gK')
print(f'Temp meth: {Tmeth860+273.15} K')
print(f'Temp ox: {Tox430+273.15} K')
Different units for easier grading:
Entropy meth: 0.7775561097256858 J/gK
entropy ox: 3.3090625 J/gK
Temp meth: 166.0433333333333 K
Temp ox: 123.83839694656487 K
In [ ]: plt.figure(dpi=300,figsize=(1.5,1.5))
plt.plot(entropymeth860, tempmeth860, label = 'Methane 860 bar')
plt.ylabel(keytempox6)
plt.xlabel('Entropy (J/mol*K)')
plt.plot([smeth],[Tmeth860],'r*')
plt.plot(entropymeth6, methtemp6bar, label = 'Methane 6bar' )
plt.plot([smeth],[Tcritmeth6],'r*')
plt.plot(entropyox430, tempox430, label = 'Oxygen 430bar')
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plt.plot([sox],[Tox430],'r*')
plt.plot(entropyox6, Tcox6, label = 'Oxygen 6bar' )
plt.plot([sox],[Tcritox6],'r*')
plt.legend(fontsize=4,frameon=False)
Out[ ]:
<matplotlib.legend.Legend at 0x15ae13310>
3. Assuming a liquid flow rate of 930 kg/s, what is the work
required for the compression in both turbopumps? Assume
negligible heat transfer in the turbopumps. Calculate the total
work of compression required for all engines.
˙
^
^
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Ẇ = ṁ(H^2 − H^1 )
Find mass flows and enthalpies for each species.
In [ ]: massmeth = 800*907.185
massox = 3000*907.185
mtot = massox+massmeth
rmassox = massox/mtot
rmassmeth = 1-rmassox
print(rmassmeth)
0.21052631578947367
In [ ]: mflowmeth = 930*rmassmeth
mflowox = 930*rmassox
print(mflowmeth)
195.78947368421052
In [ ]: print(sox)
print(smeth)
105.89
12.472
In [ ]: enthalpyoxin = ox6bar['Enthalpy (kJ/mol)']
enthalpyoxout = ox430bar['Enthalpy (kJ/mol)']
enthalpymethin = meth6bar['Enthalpy (kJ/mol)']
enthalpymethout = meth860bar['Enthalpy (kJ/mol)']
h2ox = np.interp(Tox430,tempox430,enthalpyoxout)
h1ox = float(satox6['Enthalpy (kJ/mol)'].iloc[0])
h2ox = h2ox/mmox*1000
h1ox = h1ox/mmox*1000
print(h2ox)
print(h1ox)
print(h2ox-h1ox)
Wox = mflowox*(h2ox-h1ox)
print(f'Power from oxygen: {Wox} kJ/s')
h2meth = np.interp(Tmeth860,tempmeth860,enthalpymethout)/mmmeth*1000
h1meth = float(satmeth6['Enthalpy (kJ/mol)'].iloc[0])/mmmeth*1000
Wmeth = mflowmeth*(h2meth-h1meth)
print(h2meth-h1meth)
print(f'Power from meth: {Wmeth} kJ/s')
Wtot = Wox+Wmeth
print(Wtot)
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print('This section is just for me, the answers are in the next set of print state
-55.986283396946554
-95.97812499999999
39.99184160305344
Power from oxygen: 29362.43107171555 kJ/s
210.51390674452594
Power from meth: 41216.40700471771 kJ/s
70578.83807643325
This section is just for me, the answers are in the next set of print statem
ents!
In [ ]: #above is just working it out, here it is simplified:
print(f'Power from the methane: {Wmeth/1000} MW ')
print(f'Power from the oxygen: {Wox/1000} MW ')
print(f'Total power: {Wmeth/1000+Wox/1000} MW ')
Power from the methane: 41.21640700471771 MW
Power from the oxygen: 29.362431071715548 MW
Total power: 70.57883807643326 MW
In [ ]: # multiply by time to get total energy
print(f'Energy from the methane: {Wmeth/1000*112.32658846529814} MJ ')
print(f'Energy from the oxygen: {Wox/1000*112.32658846529814} MJ ')
print(f'Total Energy: {(Wmeth/1000+Wox/1000)*112.32658846529814} MJ ')
Energy from the methane: 4629.698387637158 MJ
Energy from the oxygen: 3298.1817113332754 MJ
Total Energy: 7927.880098970433 MJ
In [ ]: #h1ox = np.interp(sox,entropyox6,enthalpyoxin)
#h1meth = np.interp(smeth,entropymeth6,enthalpymethin)/mmmeth*1000
In [ ]: #byhand
dhmethhand = 712.984
dhoxhand = 234.031
In [ ]: ###
#Wmethhand = mflowmeth*dhmethhand
#print(Wmethhand)
#Woxhand = mflowox*dhoxhand
#print(Woxhand)
#Wtothand = Wmethhand+Woxhand
#print(f'total wattage: {Wtothand}kJ/s')
#print(Wtothand*1000)
#print((Wtothand*1000)*10**(-6))
#print(Wtot*1000*10**(-6))
4. How long does it take for the tanks to empty?
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In [ ]: t = (mtot)/(mflowmeth+mflowox)/33
print(f'it takes {t}s or {t/60}mins to empty the tank')
it takes 112.32658846529814s or 1.8721098077549692mins to empty the tank
5. The combustion chamber is optimized so that the
combustion gases expand to atmospheric pressure prior to
exiting the nozzle. This ensures that all of the enthalpy of the
pressurized liquids entering the combustion chamber is
converted to kinetic energy. Based on the exit velocity of the
combustion gases, estimate the amount of heat generated as a
product of combustion in J/mol of methane.
^ = KE
Due to assumptions, we can assume the following: ΔH^vap + ΔH comb
^
We can find ΔH^vap and KE, so we can find ΔHcomb
In [ ]: print('The following has a lot of space where I work out enthalpy change in differ
The following has a lot of space where I work out enthalpy change in differe
nt ways, look at the end for my final answer
In [ ]: methaneatm = pd.read_csv(os.path.join(path2files,'Methane101bar.txt'), delimiter=
methaneatmenth = methaneatm['Enthalpy (kJ/mol)']
methaneatmtemp = methaneatm['Temperature (C)']
In [ ]: oxyatm = pd.read_csv(os.path.join(path2files,'Oxygen101bar.txt'), delimiter='\t')
oxyatmenth = oxyatm['Enthalpy (kJ/mol)']
oxyatmtemp = oxyatm['Temperature (C)']
In [ ]: plt.plot(methaneatmtemp,methaneatmenth)
Out[ ]:
[<matplotlib.lines.Line2D at 0x15aea0d60>]
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In [ ]: plt.plot(oxyatmtemp,oxyatmenth)
Out[ ]:
[<matplotlib.lines.Line2D at 0x15af187f0>]
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In [ ]: # Initialize variables
topenthox = None
bottomenthox = None
# Loop through the temperature array
for i in range(oxyatmenth.size - 1): # Avoid out-of-bounds index
check = oxyatmtemp[i]
print()
# Check for constant temperature at the "bottom point"
if check == oxyatmtemp[i + 1]: # Temperature constant
bottomenthox = oxyatmenth[i] # Interpolate enthalpy at the bottom point
# Search for when temperature stops being constant
for j in range(i, oxyatmenth.size - 1): # Inner loop avoids out-of-bounds
check2 = oxyatmtemp[j]
print(check2)
if check2 != oxyatmtemp[j + 1]: # Temperature changes
topenthox = oxyatmenth[j] # Interpolate at the top point
break
break # Exit the outer loop once top and bottom points are found
print(topenthox)
print(bottomenthox)
# Calculate enthalpy difference
if topenthox is not None and bottomenthox is not None:
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denthox = topenthox - bottomenthox
print(f"Enthalpy Difference (Hvap): {denthox}")
else:
print("Could not find both top and bottom points.")
h430ox = np.interp(Tox430,tempox430,enthalpyoxout)
h860meth = np.interp(Tmeth860,tempmeth860,enthalpymethout)
print(h430ox)
-182.96
-182.96
2.55
-4.2675
Enthalpy Difference (Hvap): 6.8175
-1.7915610687022898
In [ ]: # Initialize variables
topenthmeth = None
bottomenthmeth = None
# Loop through the temperature array
for i in range(methaneatmenth.size - 1): # Avoid out-of-bounds index
check = methaneatmtemp[i]
print()
# Check for constant temperature at the "bottom point"
if check == methaneatmtemp[i + 1]: # Temperature constant
bottomenthmeth = methaneatmenth[i] # Interpolate enthalpy at the bottom p
# Search for when temperature stops being constant
for j in range(i, methaneatmenth.size - 1): # Inner loop avoids out-of-bo
check2 = methaneatmtemp[j]
print(check2)
if check2 != methaneatmtemp[j + 1]: # Temperature changes
topenthmeth = methaneatmenth[j] # Interpolate at the top point
break
break # Exit the outer loop once top and bottom points are found
print(topenthmeth)
print(bottomenthmeth)
# Calculate enthalpy difference
if topenthmeth is not None and bottomenthmeth is not None:
denthmeth = topenthmeth - bottomenthmeth
print(f"Enthalpy Difference (Hvap): {denthmeth}")
else:
print("Could not find both top and bottom points.")
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Homework#2
1/24/25, 9:08 PM
#not done, need to midterm study
-161.48
-161.48
8.1952
0.00016787
Enthalpy Difference (Hvap): 8.19503213
In [ ]: moldenthmeth = denthmeth
moldenthox = denthox
moldenthtot = moldenthox+moldenthmeth
In [ ]: denthmeth = denthmeth/mmmeth*1000
denthox = denthox/mmox*1000
denthtot = denthox+denthmeth
print(denthtot)
723.9591025561098
In [ ]: #total moles
moltot = 1000*massox/(mmox)+1000*massmeth/(mmmeth)
print(moltot)
130294728.41334164
In [ ]: v = 3.2*1000
KE = 0.5*v**2
molKE = KE/moltot
print(molKE)
0.039295526859363966
In [ ]: enthcomb = KE-denthtot
print(f'Enthalpy of combustion is {enthcomb}kJ/kg')
Enthalpy of combustion is 5119276.040897444kJ/kg
In [ ]: h430ox = h430ox/mmox*1000
h860meth = h860meth/mmmeth*1000
Hliquid = h430ox*rmassox + h860meth*rmassmeth
vapsatmeth6_dens = np.max(meth6bar['Density (mol/l)'][meth6bar['Phase'] == 'vapor'
vapsatox6_dens = np.max(ox6bar['Density (mol/l)'][ox6bar['Phase'] == 'vapor'])
#from NIST
SatTempmeth6 = -134.42
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Homework#2
1/24/25, 9:08 PM
SatTempox6 = -161.69
satvapormeth6 = meth6bar[meth6bar['Density (mol/l)']==vapsatmeth6_dens]
satvaporox6 = ox6bar[ox6bar['Density (mol/l)']==vapsatox6_dens]
Svapmeth6 = float(satvapormeth6['Entropy (J/mol*K)'].iloc[0])
Svapox6 = float(satvaporox6['Entropy (J/mol*K)'].iloc[0])
h860meth_vap = np.interp(Svapmeth6,entropymeth860,enthalpymethout)
h430ox_vap = np.interp(Svapox6,entropyox430,enthalpyoxout)
h860meth_vap = h860meth_vap/mmmeth*1000
h430ox_vap = h430ox_vap/mmox*1000
Hvap = h430ox_vap*rmassox + h860meth_vap*rmassmeth
dHvap = Hvap-Hliquid
dHcomb = KE-dHvap
In [ ]: print(f'Enthalpy of combustion is {dHcomb} kJ/kg')
print('FINAL ANSWER')
Enthalpy of combustion is 5119497.60997892 kJ/kg
FINAL ANSWER
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