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Saya Rawhul Ihsan Setyoko dengan NPM 2106639371, mahasiswa Teknik Perkapalan 2021, saya lahir di jakarta 13 Agustus 2003 | Saya Rawhul Ihsan Setyoko dengan NPM 2106639371, mahasiswa Teknik Perkapalan 2021, saya lahir di jakarta 13 Agustus 2003 | ||
+ | == My Concious Effort on Numerical Method Learning and Its Application in Hydrogen Storage Design == | ||
+ | = | ||
− | = | + | = |
− | |||
− | + | == Design & Optimization of Pressurized Hydrogen Storage == | |
+ | |||
+ | |||
+ | '''Study Case''' | ||
+ | |||
+ | Design and Optimization | ||
+ | of Prezzurized Hydrogen Storage | ||
+ | |||
+ | |||
+ | Capacity : 1 L | ||
− | + | Pressure : 8 bar | |
+ | Cost should not exceed Rp. 500000 | ||
− | |||
Line 58: | Line 68: | ||
4. Result Interpretation | 4. Result Interpretation | ||
+ | |||
+ | |||
+ | |||
+ | == Thickness == | ||
+ | |||
+ | using two different layers material based on research, Alumunium and GFRP, we can generate the codes | ||
+ | |||
+ | import math | ||
+ | def calculate_optimal_thickness(volume, pressure, tensile_strength_aluminum, tensile_strength_gfrp): | ||
+ | # Constants | ||
+ | safety_factor = 4 # Safety factor for design | ||
+ | |||
+ | # Calculation | ||
+ | total_thickness = (volume * pressure) / (2 * math.pi * (tensile_strength_aluminum / safety_factor + tensile_strength_gfrp / safety_factor)) | ||
+ | thickness_aluminum = total_thickness * (tensile_strength_aluminum / safety_factor) / (tensile_strength_aluminum / safety_factor + tensile_strength_gfrp / safety_factor) | ||
+ | thickness_gfrp = total_thickness - thickness_aluminum | ||
+ | return thickness_aluminum, thickness_gfrp | ||
+ | |||
+ | # Input values | ||
+ | volume = 1 # Volume in liters | ||
+ | pressure = 8 * math.pow(10, 5) # Pressure in Pascals (1 bar = 10^5 Pascals) | ||
+ | |||
+ | # Convert volume to cubic meters | ||
+ | volume = volume / 1000 | ||
+ | |||
+ | # Convert pressure to Pascals | ||
+ | pressure = pressure * math.pow(10, 5) | ||
+ | |||
+ | # Material properties | ||
+ | tensile_strength_aluminum = 276 * math.pow(10, 6) # Tensile strength of aluminum in Pascals | ||
+ | tensile_strength_gfrp = 500 * math.pow(10, 6) # Tensile strength of glass fiber reinforced plastic in Pascals | ||
+ | |||
+ | |||
+ | # Calculate optimal thickness | ||
+ | aluminum_thickness, gfrp_thickness = calculate_optimal_thickness(volume, pressure, tensile_strength_aluminum, tensile_strength_gfrp) | ||
+ | |||
+ | # Output | ||
+ | print(f"The optimal thickness for a hydrogen tube with a {volume} m^3 capacity and {pressure} Pa pressure is split between aluminum thickness: {aluminum_thickness} | ||
+ | meters and GFRP thickness: {gfrp_thickness} meters.") | ||
+ | |||
+ | |||
+ | The optimal thickness for a hydrogen tube with a 1 L capacity and 0.8 mPa pressure is split between '''aluminum thickness: 3 mm GFRP thickness: 4 mm''' | ||
+ | |||
+ | == Height and Diameter == | ||
+ | |||
+ | next we determine the height and diameter of the tube | ||
+ | |||
+ | import math | ||
+ | |||
+ | def calculate_optimal_dimensions(volume, pressure): | ||
+ | # Constants | ||
+ | safety_factor = 4 # Safety factor for design | ||
+ | |||
+ | # Calculation | ||
+ | radius = math.sqrt(volume / (math.pi * math.sqrt(2))) # Optimal radius | ||
+ | height = (2 * pressure * safety_factor) / (radius * math.sqrt(2)) # Optimal height | ||
+ | diameter = 2 * radius # Optimal diameter | ||
+ | return height, diameter | ||
+ | |||
+ | # Input values | ||
+ | volume = 1 # Volume in liters | ||
+ | pressure = 0.8 * math.pow(10, 6) # Pressure in Pascals (1 MPa = 10^6 Pascals) | ||
+ | |||
+ | # Convert volume to cubic meters | ||
+ | volume = volume / 1000 | ||
+ | |||
+ | # Calculate optimal height and diameter | ||
+ | optimal_height, optimal_diameter = calculate_optimal_dimensions(volume, pressure) | ||
+ | |||
+ | # Output | ||
+ | print(f"The optimal dimensions for a hydrogen tube with a {volume} m^3 capacity and {pressure} Pa pressure are height: {optimal_height} meters and diameter: | ||
+ | {optimal_diameter} meters.") | ||
+ | |||
+ | '''The optimal dimensions for a hydrogen tube with a 1 L capacity and 0.8 mPa pressure are height: 0.305 meters and diameter: 0.12 meters.''' | ||
+ | |||
+ | |||
+ | == Surface Area == | ||
+ | import math | ||
+ | |||
+ | def calculate_surface_area(height, diameter): | ||
+ | radius = diameter / 2 | ||
+ | base_area = math.pi * radius**2 | ||
+ | lateral_area = 2 * math.pi * radius * height | ||
+ | surface_area = 2 * base_area + lateral_area | ||
+ | return surface_area | ||
+ | |||
+ | def calculate_cross_sectional_area(diameter): | ||
+ | radius = diameter / 2 | ||
+ | cross_sectional_area = math.pi * radius**2 | ||
+ | return cross_sectional_area | ||
+ | |||
+ | # Example usage | ||
+ | height = 30 # in cm | ||
+ | diameter = 12 # in cm | ||
+ | |||
+ | surface_area = calculate_surface_area(height, diameter) | ||
+ | cross_sectional_area = calculate_cross_sectional_area(diameter) | ||
+ | |||
+ | print("Surface Area:", surface_area, "cm^2") | ||
+ | print("Cross-sectional Area:", cross_sectional_area, "cm^2") | ||
+ | |||
+ | '''Surface Area: 1357.1680263507906 cm^2 | ||
+ | |||
+ | Cross-sectional Area: 113.09733552923255 cm^2''' | ||
+ | |||
+ | |||
+ | == Cost == | ||
+ | |||
+ | [[File:Gfrp Prices.jpg]] | ||
+ | |||
+ | after determining the Surface Area, we can determine the cost for all material that we used, Glass Fiber Reinforced Plastic and Aluminium, when the calculation is done, we get '''$7.12 or Rp105.848 for GFRP and Rp278.000 for Aluminium, so the total cost is Rp383.484''' |
Latest revision as of 18:10, 19 April 2024
INTRODUCTION Saya Rawhul Ihsan Setyoko dengan NPM 2106639371, mahasiswa Teknik Perkapalan 2021, saya lahir di jakarta 13 Agustus 2003
Contents
My Concious Effort on Numerical Method Learning and Its Application in Hydrogen Storage Design
=
=
Design & Optimization of Pressurized Hydrogen Storage
Study Case
Design and Optimization of Prezzurized Hydrogen Storage
Capacity : 1 L
Pressure : 8 bar
Cost should not exceed Rp. 500000
Design Optimization and Material Choosing
in some research, the best material for optimazing the performance of the storage is to use layers, layers are the combine system of two or more material to maximise the effect of both material, in this case we use 2 material :
1. GFRP (Glass Fiber Reinforced Plastic)
2. Alumunium
To maximize the performance of the storage, we can use 2 layer of material, those material are metallic and composite polymeric layers thicknesses, number of composite layers and winding angle of filament-wound composite, were studied. Using Numerical Method Finite Element Analysis on a parametric study including 64 cases, shooting minimum tank weight and cost was performed. Using different thickness of composite polymeric layers is significant the same over all composite layers thickness Optimal design was found to be 8 mm thickness of Aluminum liner and one composite layer of high toughness GFRP material with thickness of 6 mm at angle 65°
Using Sorption - Based Storage
The timing between supply and energy demand cause calls for reliable storage system. In this current years, energy storage system become further significant with the widespread implementation of renewable energy. Hydrogen is being a renewable energy with the most demands past the current years, that makes hydrogen storage keep being upgraded to provide the most efficient storage at minimum price, with the research I did from reading many sources, that one journal says that hydrogen storage with sorption based storage system exibit huge potential due to a high energy density. An average temperature lift of 25 °C, energy density of 40 kWh/m3 , and energy efficiency of 73% were obtained.
Do the Life Cycle Assessment method
Life cycle assessment method for determine the tube life cycle from the storage material until it’s not being used anymore, the main factor from this method is knowing whether this storage are toxic to the nature around. the step of doing this assessment are :
1. Objective and scope analysis
Ensure that when the product has passed the limit of use / old age, then the product will be treated in such a way that it does not damage and disturb nature.
2. Inventory Analysis
environmental influences in each stage of the analysis
3. Environmental Impact Analysis
material impact on the environment
4. Result Interpretation
Thickness
using two different layers material based on research, Alumunium and GFRP, we can generate the codes
import math def calculate_optimal_thickness(volume, pressure, tensile_strength_aluminum, tensile_strength_gfrp): # Constants safety_factor = 4 # Safety factor for design
# Calculation total_thickness = (volume * pressure) / (2 * math.pi * (tensile_strength_aluminum / safety_factor + tensile_strength_gfrp / safety_factor)) thickness_aluminum = total_thickness * (tensile_strength_aluminum / safety_factor) / (tensile_strength_aluminum / safety_factor + tensile_strength_gfrp / safety_factor) thickness_gfrp = total_thickness - thickness_aluminum return thickness_aluminum, thickness_gfrp
# Input values volume = 1 # Volume in liters pressure = 8 * math.pow(10, 5) # Pressure in Pascals (1 bar = 10^5 Pascals)
# Convert volume to cubic meters volume = volume / 1000
# Convert pressure to Pascals pressure = pressure * math.pow(10, 5)
# Material properties tensile_strength_aluminum = 276 * math.pow(10, 6) # Tensile strength of aluminum in Pascals tensile_strength_gfrp = 500 * math.pow(10, 6) # Tensile strength of glass fiber reinforced plastic in Pascals
# Calculate optimal thickness aluminum_thickness, gfrp_thickness = calculate_optimal_thickness(volume, pressure, tensile_strength_aluminum, tensile_strength_gfrp)
# Output print(f"The optimal thickness for a hydrogen tube with a {volume} m^3 capacity and {pressure} Pa pressure is split between aluminum thickness: {aluminum_thickness} meters and GFRP thickness: {gfrp_thickness} meters.")
The optimal thickness for a hydrogen tube with a 1 L capacity and 0.8 mPa pressure is split between aluminum thickness: 3 mm GFRP thickness: 4 mm
Height and Diameter
next we determine the height and diameter of the tube
import math
def calculate_optimal_dimensions(volume, pressure): # Constants safety_factor = 4 # Safety factor for design
# Calculation radius = math.sqrt(volume / (math.pi * math.sqrt(2))) # Optimal radius height = (2 * pressure * safety_factor) / (radius * math.sqrt(2)) # Optimal height diameter = 2 * radius # Optimal diameter return height, diameter
# Input values volume = 1 # Volume in liters pressure = 0.8 * math.pow(10, 6) # Pressure in Pascals (1 MPa = 10^6 Pascals)
# Convert volume to cubic meters volume = volume / 1000
# Calculate optimal height and diameter optimal_height, optimal_diameter = calculate_optimal_dimensions(volume, pressure)
# Output print(f"The optimal dimensions for a hydrogen tube with a {volume} m^3 capacity and {pressure} Pa pressure are height: {optimal_height} meters and diameter: {optimal_diameter} meters.")
The optimal dimensions for a hydrogen tube with a 1 L capacity and 0.8 mPa pressure are height: 0.305 meters and diameter: 0.12 meters.
Surface Area
import math
def calculate_surface_area(height, diameter): radius = diameter / 2 base_area = math.pi * radius**2 lateral_area = 2 * math.pi * radius * height surface_area = 2 * base_area + lateral_area return surface_area
def calculate_cross_sectional_area(diameter): radius = diameter / 2 cross_sectional_area = math.pi * radius**2 return cross_sectional_area
# Example usage height = 30 # in cm diameter = 12 # in cm
surface_area = calculate_surface_area(height, diameter) cross_sectional_area = calculate_cross_sectional_area(diameter)
print("Surface Area:", surface_area, "cm^2") print("Cross-sectional Area:", cross_sectional_area, "cm^2")
Surface Area: 1357.1680263507906 cm^2
Cross-sectional Area: 113.09733552923255 cm^2
Cost
after determining the Surface Area, we can determine the cost for all material that we used, Glass Fiber Reinforced Plastic and Aluminium, when the calculation is done, we get $7.12 or Rp105.848 for GFRP and Rp278.000 for Aluminium, so the total cost is Rp383.484