Muhammad Fajri Shiddiq
Biodata
Nama :
Muhammad Fajri Shiddiq
NPM :
2106727941
TTL :
Palembang, 12 Februari 2004
Program Studi :
Teknik Mesin
Review Pertemuan pembelajaran
Pekan 1 Dalam menjadi manusia dibutuhkan suatu kesadaran akan apapun yang diperbuat. Pada mata kuliah mekanika fluida, proses pembelajaran akan berdasar pada konsep dengan komponen penilaian yang akan dijelaskan lebih lanjut pada Review Pekan 1
Pekan 2 Manometer merupakan alat yang menggunakan kolom cairan untuk mengukur tekanan. Penjelasan lebih lanjut terdapat pada Review pekan 2
Pekan 3 Hasil dari pengamatan penggunaan manometer, terdapat perbedaan tekanan pada pipa ketika salah satu valve dibuka. Kemudian di akhir pekan kelas diadakan kunjungan ke lab agar dapat mengamati secara langsung macam alat pengukuran tekanan, yaitu manometer analog, manometer U, pressure tranducerr. Penjelasan lebih lanjut mengenai pengamatan serta jawaban dari tugas manometer terdapat pada Review pekan 3
Pekan 4 Pembahasan mengenai geometri gerak fluida tanpa adanya perhitungan haya eksternal merupakan pengertian dasar dari kinematika fluida. Aliran sendiri memiliki tipe dan karakteristik yang bermacam. Penjelasan lebih lanjut dan hasil analisis langsung salah satu fenomena aliran pada Rumah Kabeda terdapat pada Review Pekan 4
Pekan 5 Pembahasan mengenai transport reynold, mulai dari pengertian dasar, dan kenapa pada aplikasi kinematika fluida hal ini penting. Penjelasan lebih lanjut terdapat pada Review Pekan 5
Python Code
Storage system design and optimization occurs in several stages, from existing systems to meeting required limits. First, based on the required volume (1 liter), determine the plate size to use, taking into account the minimum plate area using the following calculation:
1. Import the math library and the minimize module from scipy.optimize.
2. Define the objective(x) function that takes x as an argument, which is an array containing the radius and height. This function calculates the surface area of a cylinder using the given formula.
3. Define the constraint(x) function that takes x as an argument, which is an array containing the radius and height. This function calculates the volume of the cylinder and ensures that it is equal to 1000.
4. Set the initial_guess for the radius and height.
5. Define the bounds for the radius and height. In this case, the radius and height must be non-negative.
6. Define the volume constraint (volume_constraint) as an equality (eq) constraint using the constraint function defined earlier.
7. Minimize the surface area of the cylinder by calling the minimize function with the objective function, initial_guess, 'SLSQP' method, bounds, and volume_constraint. Save the result in the result variable.
8. Retrieve the optimal values for the radius, height, and minimum surface area from the result object.
9. Print the results using the print statement, including the optimal radius, optimal height, and minimum surface area.
import math
from scipy.optimize import minimize
def objective(x):
radius, height = x
return 2 * math.pi * radius**2 + 2 * math.pi * radius * height
def constraint(x):
radius, height = x
return math.pi * radius**2 * height - 1000
initial_guess = [1, 10] # Initial guess for the radius and height
# Define the bounds for the radius and height
bounds = [(0, None), (0, None)]
# Define the volume constraint
volume_constraint = {'type': 'eq', 'fun': constraint}
# Minimize the surface area subject to the volume constraint
result = minimize(objective, initial_guess, method='SLSQP', bounds=bounds,
constraints=volume_constraint)
optimal_radius = result.x[0]
optimal_height = result.x[1]
min_surface_area = result.fun
print(f"\n\nOptimal Radius: {optimal_radius} cm")
print(f"Optimal Height: {optimal_height} cm")
print(f"Minimum Surface Area: {min_surface_area} cm²\n\n")
These calculations yield the following values:
● Optimal Radius: 5.419261255088046 cm
● Optimal Height: 10.838519182022262 cm
● Minimum Surface Area: 553.5810443894838 cm²
Furthermore, the calculation is performed with a given budget variable, ie Rp. 500,000. Use the following calculations for the calculations:
1. Import the `minimize` function from the `scipy.optimize` module.
2. Set the required `price_per_unit` and `capacity_per_unit` variables.
3. Set the maximum budget (`max_budget`).
4. Define the objective function (`objective_function`) that takes a variable `x` and returns the negative value of `x` multiplied by `capacity_per_unit`.
5. Define the budget constraint function (`budget_constraint`) that takes a variable `budget` and returns the difference between `max_budget` and the product of `price_per_unit` and `budget`.
6. Create the budget constraint equation (`budget_constraint_eq`) as an inequality constraint using the `ineq` type and the `budget_constraint` function.
7. Set the initial value (`x0`) to 0.
8. Define the bounds for the optimization problem as a list containing a tuple `(0, None)`.
9. Solve the optimization problem using the `minimize` function with the objective function, initial value, 'SLSQP' method, bounds, and constraints.
10. Store the solution in the `solution` variable.
11. Display the results using the `print` function, including the status of the optimization (whether an optimal cost is found or not), the number of units that can be purchased (`solution.x[0]`), the total storage capacity obtained, and the total budget required.
from scipy.optimize import minimize
# Required price and capacity price_per_unit = 500000 # Rupiah capacity_per_unit = 1 # Liter
# Maximum budget max_budget = 500000
# Objective function def objective_function(x): return -x * capacity_per_unit
# Budget constraint def budget_constraint(budget): return max_budget - (price_per_unit * budget)
budget_constraint_eq = {'type': 'ineq', 'fun': budget_constraint}
# Initial value x0 = 0
# Bounds bounds = [(0, None)]
# Solve the optimization problem solution = minimize(objective_function, x0, method='SLSQP', bounds=bounds, constraints=[budget_constraint_eq])
# Display the results
print("Status:", soluti.success and "Optimal Cost" or "No solution found")
print("Number of units that can be purchased:", solution.x[0])
print("Total storage capacity obtained:", solution.x[0] * capacity_per_unit, "liters")
print("Total budget:", solution.x[0] * price_per_unit, "Rupiah")
Here are the results I got:
● Status: Optimal Cost
● Number of units that can be purchased: 1
● Total storage capacity obtained: 1.0 liters
● Total budget: 500000.0 Rupiah
Using the above calculations, we can graph the capacity vs. quantity comparison obtained with the following code:
# Chart List unit_list = [] total_capacity_list = []
# Calculation of Optimal Solution for unit_count in range(int(max_budget / price_per_unit) + 1): total_capacity = unit_count * capacity_per_unit total_cost = unit_count * price_per_unit
if total_capacity > optimal_total_capacity and total_cost <= max_budget: optimal_unit_count = unit_count optimal_total_capacity = total_capacity optimal_total_cost = total_cost
# Adding data to the list for the chart unit_list.append(unit_count) total_capacity_list.append(total_capacity)
# Result Display
print("Number of hydrogen storage units to be purchased:", optimal_unit_count)
print("Total storage capacity:", optimal_total_capacity, "liters")
print("Total cost:", optimal_total_cost, "Rupiah")
# Displaying the Chart
plt.plot(unit_list, total_capacity_list, color='red')
plt.xlabel('Number of Hydrogen Storage')
plt.ylabel('Total Storage Capacity (liters)')
plt.title('Hydrogen Storage Optimization')
plt.grid(True)
plt.show()
The code used above aims to optimize hydrogen storage given the given capacity and budget constraints. The main goal is to determine the number of hydrogen storage units that should be purchased to achieve maximum storage capacity, taking into account the constraints used.
Iterations are performed to retrieve data for each unit purchased and capacity and other identified limits. We test each unit quantity to calculate total storage capacity and total associated costs. The iteration stops when the value exceeds the limit. 800x800px
