Raihan Al Malik Fitrah Arifin

Introduction

Assalamu'alaikum warahmatullahi wabarakatuh. Perkenalkan saya Raihan Al malik Fitrah Arifin bisa dipanggil ipin , dengan NPM 2106649946 dari jurusan Teknik Perkapalan 2021. Saya Tinggal Di jakarta Asal sekolah saya SMAN 28 Jakarta. Salam kenal semuanya, Terima kasih.

Resume Minggu 1

pada pertemuan pertama dijelaskan apa itu arti conciousness dalam kehidupan kita dan diterapkan dalam matkul metodek numerik serta dijelaskan keterkaitan antara conciousness dengan tuhan Yang Maha Esa. Pada pertemuan pertama kelas metnum ini, saya akhirnya memahami bahwa metode numerik merupakan materi pendekatan yang digunakan untuk menyelesaikan permasalahan matematis yang kompleks dengan menggunakan perhitungan numerik atau angka-angka untuk mendapatkan solusi numerik yang mendekati solusi eksak.

Pada matematika, sangat jarang ada hal yang eksak, dicontohkan dengan persamaan x²-1/x-1 dan (x+1)(x-1)/(x-1) jika nilai x=1, namun sebenarnya nilai x=1 tidak menunjukan nilai eksak 1 tetapi hanya mendekati 1. Maka dari itu untuk mendapatkan solusinya kita perlu menggunakan pendekatan numerik agar lebih mudah memahami.

Ilmu Matematika kerap disebut sebagai ilmu mutlak. Hal tersebut jelas menimbulkan pro dan kontra, namun karena sifat dari kesadaran adalah individu, maka hal tersebut menimbulkan banyaknya kombinasi jawaban dari berbagai ‘ kesadaran ‘ para matematikawan dalam mencari solusi sehingga menghasilkan jawaban yang beragam

Desain Optimasi Sistem Penyimpanan Hidrogen

Mendesain tanki hidrogen dengan kapasitas 1 Liter Tekanan Maks : 8 Bar Biaya produksi maksimal Rp 500.000

To design and optimize the pressure of hydrogen storage with the specifications you provide (1 liter capacity and 8 bar pressure) and consider the maximum cost limit of Rp. 500,000, here are some steps :

1. Choose the type of storage tube: To store hydrogen with a capacity of 1 liter and a pressure of 8 bar, we can use metal cylinders that have sufficient strength and safety. For example, storage tubes made of carbon steel are typically used for applications like this.

2. Calculate the strength of the tube: To ensure that the tube can withstand a pressure of 8 bar, we need to take into account the safety factor and the material strength of the tube. These calculations usually involve an analysis of the stress and resistance of the tube material.

3. Safety design: To prevent cylinder failure and ensure the safety of the hydrogen storage system, we need to consider safety designs such as pressure valves and overpressure protection.

4. Choose fill material: Fill material or adsorbent is used to increase the hydrogen absorption capacity of the cylinder. We need to choose an efficient and economical material, such as activated carbon, which can optimize the hydrogen storage capacity.

5. Cost calculation: In selecting components and materials for a hydrogen storage system, it is necessary to consider the cost of each component. Make sure the cost of the entire system, including tubes, valves, pressure protectors and filling materials, does not exceed the budget limit of Rp. 500,000.

6. Test and validation: After designing a hydrogen storage system, it is important to test and validate its reliability and safety. This can involve pressure testing, material testing, and leak testing to ensure the system is functioning properly.

It should be remembered that designing and optimizing a hydrogen storage system is a complex task and requires in-depth knowledge of hydrogen storage techniques, materials and safety. Also, overall costs may vary depending on factors such as parts availability, price negotiations, and regulations that apply to your area.

Final Report of Design and Optimization of Pressurized Hydrogen Storage

import math

# Function to calculate the total cost
def calculate_cost(diameter, height):
    # Calculate the volume and surface area of the tank
    volume = math.pi * (diameter**2) * (height/4)
    surface_area = (2 * math.pi * (diameter/2) * height) + (math.pi * (diameter**2) / 4) 
    
    # Calculate the material cost and manufacturing cost
    material_cost = surface_area * cost_per_square_meter
    manufacturing_cost = volume * cost_per_volume 
    
    # Calculate the total cost
    total_cost = material_cost + manufacturing_cost
    return total_cost
    
# Constants
cost_per_square_meter = 100000  # Cost per square meter of material
cost_per_volume = 100000  # Cost per liter of volume
budget_limit = 500000  # Maximum budget

# Variables
best_cost = float('inf')
best_diameter = 0
best_height = 0

# Iterate over possible diameters and heights
for diameter in range(1, 100):
    for height in range(1, 100):
        # Calculate the total cost for the current diameter and height
        total_cost = calculate_cost(diameter, height)

        # Check if the total cost is within the budget and better than the previous best
        if total_cost <= budget_limit and total_cost < best_cost:
            best_cost = total_cost
            best_diameter = diameter
            best_height = height

# Print the optimal solution
print("Optimal Diameter (m):", best_diameter)
print("Optimal Height (m):", best_height)
print("Optimal Total Cost (Rupiah):", best_cost)

Dengan kode Phyton diatas, kita dapat melakukan iterasi diameter dan juga tinggi untuk menghasilkan biaya produksi yang optimal. Kode ini sendiri menggunakan nested loop untuk meguji berbagai kombinasi dari diameter dan tinggi dan juga menghitung biaya produksi yang optimal dari kombinasi diameter dan tinggi menggunakan fungsi "calculate_cost". Nantinya, code akan memeriksa bahwa apakah biaya produksi total masih dalam batas anggaran dan juga lebih baik dari solusi terbaik sebelumnya dengan memperhitungkan volume dan biaya. Dengan ini, didapat bahwa: Diameter optimaal: 1 meter Tinggi optimal: 1 meter Biaya optimal: Rp.500000 Kita juga dapat melihat grafik plot terhadap perbandingan kapasitas dan jumlah yang didapatkan menggunakan kode dibawah ini

import numpy as np
import matplotlib.pyplot as plt
 
# Constants
pressure = 8  # Pressure in bar
capacity = 1  # Capacity in liters
budget_limit = 500000  # Maximum budget in Rupiah

# Function to calculate the cost
def calculate_cost(diameter, height):
    surface_area = (2 * np.pi * (diameter/2) * height) + (np.pi * (diameter**2) / 4)
    cost_per_tank = surface_area * cost_per_surface_area
    return cost_per_tank

# Constants for cost calculation
cost_per_surface_area = 100000  # Cost per square meter of surface area

# Lists to store the results
diameters = []
heights = []
costs = []

# Iterate over possible diameters and heights
for diameter in np.arange(0.1, 10.1, 0.1):
    for height in np.arange(0.1, 10.1, 0.1):
        # Calculate the cost for the current diameter and height
        cost = calculate_cost(diameter, height)
        
        # Check if the cost is within the budget
        if cost <= budget_limit:
            diameters.append(diameter)
            heights.append(height)
            costs.append(cost)

# Plot the optimization results
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(diameters, heights, costs, c=costs, cmap='viridis')
ax.set_xlabel('Diameter')
ax.set_ylabel('Height')
ax.set_zlabel('Cost (Rupiah)')
ax.set_title('Hydrogen Storage Optimization')
plt.show()