Difference between revisions of "Ayudya Arindari Murahardjo"
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return pressure - 8 # Tekanan harus sama dengan atau kurang dari 8 bar | return pressure - 8 # Tekanan harus sama dengan atau kurang dari 8 bar | ||
− | # Batasan volume | + | # Batasan volume |
def volume_constraint(x): | def volume_constraint(x): | ||
volume = x[0] | volume = x[0] | ||
Line 81: | Line 81: | ||
return volume - 1 # Volume harus sama dengan atau kurang dari 1 liter | return volume - 1 # Volume harus sama dengan atau kurang dari 1 liter | ||
− | # Batasan batas harga | + | # Batasan batas harga |
def cost_constraint(x): | def cost_constraint(x): | ||
volume = x[0] | volume = x[0] | ||
Line 88: | Line 88: | ||
return 500000 - cost # Total biaya harus kurang dari atau sama dengan Rp 500.000,- | return 500000 - cost # Total biaya harus kurang dari atau sama dengan Rp 500.000,- | ||
− | # Initial guess | + | # Initial guess |
− | x0 = [0.5, 6] # [Volume, Tekanan] | + | x0 = [0.5, 6] # [Volume, Tekanan] |
− | # Batasan | + | # Batasan |
− | constraints = [{'type': 'ineq', 'fun': pressure_constraint}, | + | constraints = [{'type': 'ineq', 'fun': pressure_constraint}, |
{'type': 'ineq', 'fun': volume_constraint}, | {'type': 'ineq', 'fun': volume_constraint}, | ||
{'type': 'ineq', 'fun': cost_constraint}] | {'type': 'ineq', 'fun': cost_constraint}] | ||
− | # Optimisasi | + | # Optimisasi |
− | result = minimize(objective, x0, constraints=constraints) | + | result = minimize(objective, x0, constraints=constraints) |
− | # Print hasil optimisasi | + | # Print hasil optimisasi |
− | print("Status:", result.success) | + | print("Status:", result.success) |
− | print("Volume Optimal (liter):", result.x[0]) | + | print("Volume Optimal (liter):", result.x[0]) |
− | print("Tekanan Optimal (bar):", result.x[1]) | + | print("Tekanan Optimal (bar):", result.x[1]) |
Revision as of 08:53, 9 June 2023
Contents
Introduction
Halo!
Perkenalkan, nama saya Ayudya Arindari Murahardjo, akrab disapa Arin. Saya merupakan mahasiswa semester 4 Program Studi Teknik Perkapalan Universitas Indonesia.
Resume Pertemuan 1 (26/5/2023)
Pada pertemuan pertama mata kuliah Metode Numerik, saya belajar mengenai pemahaman tentang "cosciousness", yakni semua orang harus memiliki kesadaran dalam melakukan segala sesuatu termasuk mempelajari Metode Numerik. Terdapat study case pada pertemuan pertama, yaitu mahasiswa diminta untuk menyelesaikan persamaan (x-1)^2/x-1 jika x=1. Pada hal ini, tidak terdapat jawaban yang mutlak atau eksak (1 solusi) karena pada hakikatnya di dalam dunia ini tidak terdapat suatu hal yang pasti.
Semakin kita dewasa, kita semakin kian mengerti akan arti hidup ini, begitu juga dengan kepercayaan yang selama ini kita anut. Mungkin sebagian besar orang memiliki pemahaman yang mereka yakini itu benar dan tidak ada salahnya memilih jalan hidup masing-masing selagi kita tetap "conscious"
Design & Optimization of Pressurized Hydrogen Storage
Design & optimization of pressurized hydrogen storage with maximum cost Rp 500.000,-
Capacity
Volume : 1 liter
Pressure : 8 bar
WEEK 1 PROGRESS
Designing and optimizing a pressurized hydrogen storage system with a 1-liter capacity and 8-bar pressure within a budget of Rp 500.000,- involves careful consideration of materials, dimensions, and cost optimization. Here's a design and optimization approach:
Material Selection
To meet the budget constraint, consider using high-density polyethylene (HDPE) as the material for the storage system. HDPE is cost-effective and offers good chemical resistance.
Container Design
Shape: Design a cylindrical container, as it is a common and practical shape for pressurized storage. Dimensions: Determine the container dimensions based on the desired volume and pressure. The container's volume is fixed at 1 liter, and the pressure is 8 bar.
Wall Thickness: Calculate the required wall thickness using the Barlow's formula: t = (P * D) / (2 * S), where P is the pressure (8 bar), D is the diameter of the container, and S is the allowable stress for HDPE. Ensure the calculated wall thickness is within the manufacturing capabilities and budget constraints.
Optimization Strategies
Material Cost: Compare prices from different HDPE suppliers to select the most cost-effective option. Manufacturing Process: Consider extrusion or injection molding processes for HDPE container fabrication, as they can be cost-effective for producing cylindrical shapes.
Size Optimization: Optimize the dimensions of the container to minimize material usage and manufacturing costs while still meeting the required volume and pressure specifications. This can be achieved by adjusting the diameter and height of the container.
Safety Considerations: Incorporate safety features into the design, such as pressure relief devices and adherence to safety standards and regulations for hydrogen storage.
Final Result Design & Optimization of Pressurized Hydrogen Storage
Fundamental Steps
To calculate the design of an optimal hydrogen storage tube with a 1-liter volume and 8-bar pressure specification, we can follow these steps:
1. Determine the desired dimensions: Since the volume and pressure specifications are given, the next step is to calculate the dimensions of the storage tube.
2. Convert the volume to cubic meters: 1 liter is equal to 0.001 cubic meters.
3. Convert the pressure to pascals: 1 bar is equal to 100,000 pascals.
4. Apply the ideal gas law: The ideal gas law equation, PV = nRT, can be used to calculate the volume of the storage tube. However, we need additional information such as the number of moles of hydrogen (n) and the temperature (T) to proceed with the calculation. Without this information, we cannot determine the exact dimensions of the storage tube.
5. Consider the material and safety factors: Once you have the necessary dimensions, you will need to select a suitable material that can withstand the pressure and store hydrogen safely. Materials such as high-strength steel or composite materials may be considered.
The Calculation
from scipy.optimize import minimize
# Fungsi tujuan yang ingin kita maksimalkan def objective(x): return -x[0] # Maksimalkan volume hydrogen
# Batasan tekanan def pressure_constraint(x): volume = x[0] pressure = x[1] return pressure - 8 # Tekanan harus sama dengan atau kurang dari 8 bar
# Batasan volume def volume_constraint(x): volume = x[0] pressure = x[1] return volume - 1 # Volume harus sama dengan atau kurang dari 1 liter
# Batasan batas harga def cost_constraint(x): volume = x[0] pressure = x[1] cost = 200000 + 500000 * (volume - 1) + 300000 * (pressure - 8) return 500000 - cost # Total biaya harus kurang dari atau sama dengan Rp 500.000,-
# Initial guess x0 = [0.5, 6] # [Volume, Tekanan]
# Batasan constraints = [{'type': 'ineq', 'fun': pressure_constraint}, {'type': 'ineq', 'fun': volume_constraint}, {'type': 'ineq', 'fun': cost_constraint}]
# Optimisasi result = minimize(objective, x0, constraints=constraints)
# Print hasil optimisasi print("Status:", result.success) print("Volume Optimal (liter):", result.x[0]) print("Tekanan Optimal (bar):", result.x[1])