Hashing Algorithms

Algorithmes de Hachage (Hashing Algorithms)

• A hashing algorithm is a function that takes data as input and returns a fixed-size hash value.
• This hash value is often called a "hash" or "fingerprint".

Properties of Hashing Algorithms

• Deterministic: the algorithm should always produce the same output for the same input.
• Fast to compute.
• Fixed size: The size of the hash must be fixed regardless of the size of the input.
• Non-reversible (one-way): It should be difficult to find the input from the hash.
• Collision resistance: It should be difficult to find two different inputs that produce the same hash.

Examples of Hashing Algorithms

• MD5 (Message Digest 5).
• SHA-1 (Secure Hash Algorithm 1).
• SHA-256 (Secure Hash Algorithm 256 bits).
• SHA-384 (Secure Hash Algorithm 384 bits).
• SHA-512 (Secure Hash Algorithm 512 bits).
• BLAKE2.
• bcrypt.
• scrypt.
• Argon2.

Applications of Hashing Algorithms

• Verification of data integrity: If the hash of a file before and after a transfer is the same, it shows whether it has been altered.
• Storing passwords: Instead of storing passwords in plain text, their hash is stored. This protects passwords in case the database is stolen.
• Hash tables: Hash tables use hash functions to quickly store and retrieve data.
• Digital signatures: Hash functions can be used to create digital signatures, which verify the authenticity of a document.
• Blockchain algorithms extensively use hashing to ensure integrity and security.

Common Attacks

• Brute force attack: Trying all possible inputs until finding the one that matches the given hash.
• Dictionary attack: Using a list of common words and their hashes to find the word associated with a given hash.
• Collision attack: Finding two different inputs that produce the same hash.

Hashing Example in Python

• Python can be used to calculate SHA-256 hash of a string of characters

import hashlib

## Calculating the SHA-256 hash of a character string
message = "Bonjour le monde !"
hash_object = hashlib.sha256(message.encode('utf-8'))
hex_dig = hash_object.hexdigest()
print(hex_dig)

## Sortie:
## 81c9890c16f22f9492ba1439a2ef9594b4a70a0a99b73d9cb94ff4ff09b241a0

Chemical Engineering Design Project

Introduction

• The project aims to design a heat exchanger for a chemical process
• Heat exchangers are key components in chemical plants as they assist the transfer of heat between fluids for safety, process efficiency and optimal energy use

Process Conditions

• Parameters of the hot oil (Fluid 1):
  • Inlet Temperature: 150°C
  • Outlet Temperature: 80°C
  • Flow Rate: 5 kg/s
  • Specific Heat Capacity: 2.0 kJ/kg.K
• Cooling water (Fluid 2) parameters:
  • Inlet Temperature: 25°C
  • Outlet Temperature: 50°C
  • Flow Rate: To be determined
  • Specific Heat Capacity: 4.186 kJ/kg.K

Design Considerations

• Shell and tube heat exchanger.
• 316 Stainless steel material of construction
• Fouling factors of 0.0002 m².K/W for both fluids
• Maximum allowable pressure drop of 10 kPa for both fluids.

Design Calculations

• Formula to calculate the heat duty (𝑄): $Q = m_1 * Cp_1 * (T_{1,in} - T_{1,out})$
  • $m_1$ is the mass flow rate of Fluid 1
  • $Cp_1$ is the specific heat capacity of Fluid 1
  • $T_{1,in}$ is the inlet temperature of Fluid 1
  • $T_{1,out}$ is the outlet temperature of Fluid 1
• The formula yields $ 700 kW$
• Formula to calculate the flow rate of cooling water ($m_2$) = $\frac{Q}{Cp_2 * (T_{2,in} - T_{2,out})}$
  • $Cp_2$ is the specific heat capacity of Fluid 2
  • $T_{2,in}$ is the inlet temperature of Fluid 2
  • $T_{2,out}$ is the outlet temperature of Fluid 2
• The second formula gives $6.68 \frac{kg}{s}$
• These intermediate parameters are calculated to find Log Mean Temperature Difference (LMTD):
  • $\Delta T_1 = T_{1,in} - T_{2,out} = 150°C - 50°C = 100°C$
  • $\Delta T_2 = T_{1,out} - T_{2,in} = 80°C - 25°C = 55°C$
• The Log Mean Temperature Difference (LMTD) is calculated:
  • $LMTD = \frac{\Delta T_1 - \Delta T_2}{ln(\frac{\Delta T_1}{\Delta T_2})} = \frac{100 - 55}{ln(\frac{100}{55})} = 75.13°C$-
• Assuming a value for shell and tube heat exchangers leads to this Overall Heat Transfer Coefficient (U): $U = 500 \frac{W}{m^2.K}$
• Heat transfer area (A) calculation:
  • $A = \frac{Q}{U * LMTD} = \frac{700,000 W}{500 \frac{W}{m^2.K} * 75.13°C} = 18.63 m^2$

Mechanical Design

• Tube Diameter: 25mm
• Tube Length: 5m
• The specific number of tubes depends on the shell diameter
• Calculation of shell diameter depends on the tube layout and tube quantity
• Baffle Spacing is designed to minimize the pressure drop and boost heat transfer

Cost Estimation

• Material Costs: Stainless steel 316
• Manufacturing Costs: Based on design
• Installation Costs: Depends on location and size of heat exchanger

Safety and Environmental Considerations

• Safety: Proper and regular maintenance is needed to minimize failures and leaks
• Environmental: Proper disposal of waste materials and minimization of energy consumption

Conclusion

• This design targets to provide a safe and efficient solution for heat transfer between hot process oil and cooling water.
• The heat duty, flow rates, temperature variations, and material choices, are all carefully considered during the design process.

Endnote

• D.Q. Kern authored "Process Heat Transfer," McGraw-Hill, 1950.
• Coulson, J.M., and J.F. Richardson authored "Chemical Engineering," Volume 6. Butterworth-Heinemann, 1999.

Appendix

• Contains detailed mechanical drawings
• Contains additional calculations
• Contains material specifications

The Heat Equation

Introduction

• Physics problems often involve transport such as:
  • River pollutant diffusion
  • Solid heat transport.
  • Population diffusion.
• These systems share similar mathematical properties. Heat conduction through solid forms an example of transport.

Fourier's Law of Heat Conduction

• Heat moves from warm to cold regions
• States that the heat flux $\mathbf{q}$ is proportional to the temperature gradient and is in the direction opposite of the heat flow
• The formula is $\mathbf{q} = -k \nabla T$ where:
  • $T(\mathbf{x},t)$ is the temperature.
  • $\mathbf{q}$ is the heat flux.
  • $k$ is the thermal conductivity.

Conservation of Energy

• The heat energy in a solid volume $V$ enclosed by a surface $S$ can be represented as $E = \iiint_V c\rho T dV$ where:
• $c$ is the specific heat of the material
• $\rho$ is the density.
• The rate of change of heat energy is $\frac{\partial E}{\partial t} = \iiint_V c\rho \frac{\partial T}{\partial t} dV$
• Can be rewritten as$\frac{\partial E}{\partial t} = -\iint_S \mathbf{q} \cdot \mathbf{n} dS + \iiint_V Q dV$ Or $\iiint_V c\rho \frac{\partial T}{\partial t} dV = -\iint_S \mathbf{q} \cdot \mathbf{n} dS + \iiint_V Q dV$ Where:
  • $\iint_S \mathbf{q} \cdot \mathbf{n} dS$ Represents the heat flux across the surface $S$.
  • $\iiint_V Q dV$ Represents an external heat source within $V$
  • $Q$ is the heat source density
• Using the divergence theorem
  • Surface integral is written as a volume integral
  • $\iint_S \mathbf{q} \cdot \mathbf{n} dS = \iiint_V \nabla \cdot \mathbf{q} dV$
• Hence the formula becomes:
  • $\iiint_V \left( c\rho \frac{\partial T}{\partial t} + \nabla \cdot \mathbf{q} - Q \right) dV = 0$
• Yielding the relationship $c\rho \frac{\partial T}{\partial t} + \nabla \cdot \mathbf{q} - Q = 0$
• By substituting $\mathbf{q} = -k \nabla T$ heat equation becomes $c\rho \frac{\partial T}{\partial t} - \nabla \cdot (k \nabla T) - Q = 0$ or $\frac{\partial T}{\partial t} = \kappa \nabla^2 T + \frac{Q}{c\rho}$
  • $\kappa = k/c\rho$ is the thermal diffusivity.
• The heat equation or diffusion equation in the absence of sources : $\frac{\partial T}{\partial t} = \kappa \nabla^2 T$

Initial and Boundary Conditions

Initial Conditions

• The heat equation involves specification of an initial condition one time derivative: $T(\mathbf{x},0) = f(\mathbf{x})$

Boundary Conditions

• Two boundary specifications are required given the heat equation involves a second derivative in space
• Dirichlet boundary condition
  • The temperature is specified on the boundary: $T(\mathbf{x},t) = g(\mathbf{x},t), \quad \mathbf{x} \in \partial \Omega$
• Neumann boundary condition
  • The heat flux is specified on the boundary: $-k \nabla T \cdot \mathbf{n} = h(\mathbf{x},t), \quad \mathbf{x} \in \partial \Omega$
  • $\mathbf{n}$ is the outward normal to the boundary.
  • When $h = 0$, the boundary is insulated
• Robin boundary condition
  • A linear combination of the temperature and heat flux is specified on the boundary: $-k \nabla T \cdot \mathbf{n} + \alpha T = \beta(\mathbf{x},t), \quad \mathbf{x} \in \partial \Omega$

Example: Heat Equation in One Dimension

• Constant thermal properties and no internal heat sources:

  • Rod of length $L$
  • temperature is :$\frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial x^2}, \quad 0 < x < L, \quad t > 0$

• Initial condition: $T(x,0) = f(x), \quad 0 \le x \le L$

• Boundary conditions: $T(0,t) = T_1, \quad T(L,t) = T_2, \quad t > 0$

Equilibrium Solution

• Expect a trend to steady state when $t \rightarrow \infty$ I.e $\frac{\partial T}{\partial t} = 0$
• Hence the heat equation states $\frac{\partial^2 T}{\partial x^2} = 0$
• Integration gives $T(x) = Ax + B$
• boundary conditions further reduce to an equilibrium solution $T(x) = \frac{T_2 - T_1}{L}x + T_1$

Homogeneous Boundary Conditions

• Transformation of the boundary conditions required to solve the heat equation
• Let $T(x,t) = U(x,t) + T_E(x)$ where:
  • $T_E(x)$ is the equilibrium solution .
  • $U(x,t)$ is the transient solution
• Transformation by substituting yields:
  • $\frac{\partial U}{\partial t} = \kappa \frac{\partial^2 U}{\partial x^2}$
  • $U(x,0) = f(x) - T_E(x)$
  • $U(0,t) = 0, \quad U(L,t) = 0$

Separation of Variables

• To solve the heat equation with homogeneous boundary conditions, use separation of variables
• The Solution is assumed to be representable as product of single variable functions such as $U(x,t) = X(x) \Theta(t)$
• Further reduced to two ordinary differential equations by substituting for U. These are:
  • $\Theta'(t) = -\kappa \lambda \Theta(t)$
  • $X''(x) = -\lambda X(x)$
• Boundary conditions become $X(0) = 0, \quad X(L) = 0$

Solving the Spatial Equation

• Problematic equation is $X''(x) + \lambda X(x) = 0$
• The general solutions are:
• For $\lambda # Chapter 2: Chemical Equilibrium

2.1 Introduction

• Reactions do not necessarily go to completion.
• Most reactions are reversible; they occur in both directions.
• Reversible reactions reach a state of dynamic equilibrium.
• Forward and reverse reaction rates are equal.
• Concentrations remain constant, but are not necessarily equal.

2.2 The Equilibrium Constant

2.2.1 Expression of the Equilibrium Constant $K_c$

For a general reversible reaction:

$aA + bB \rightleftharpoons cC + dD$

where a, b, c, and d are stoichiometric coefficients.

$K_c = \frac{{\left[ C \right]^c \left[ D \right]^d}}{{\left[ A \right]^a \left[ B \right]^b}}$

• Concentrations are in mol/L (M).
• $K_c$ is unitless.
• $K_c$ depends on temperature.

2.2.2 Interpretation of $K_c$

• $K_c >> 1$: Equilibrium favors products (right).
• $K_c Lecture 25: Hypothesis Testing

1. Basic Concepts

1.1 Definition

• A Hypothesis is a statement about a population parameter.
• Null Hypothesis ($H_0$): Claim presumed true.
• Alternative Hypothesis ($H_1$): Challenges the null hypothesis.

1.2 Example

• Claim: Average height of adult males is 1.75m.
• $H_0: \mu = 1.75$
• $H_1: \mu \neq 1.75$
• Do not aim to "accept" $H_0$, but rather assess if there's enough evidence to reject it

1.3 Hypothesis Testing Framework

1. State Hypotheses ($H_0$, $H_1$)
2. Choose Significance Level ($\alpha$).
3. Compute Test Statistic.
4. Make Decision (Reject $H_0$ or Fail to Reject $H_0$).

1.4 Errors in Hypothesis Testing

• Type I Error: Incorrectly rejecting a true null hypothesis.
• $P(\text{Type I Error}) = \alpha$
• Also known as the significance level.
• Type II Error: Incorrectly accepting a false null hypothesis.
• $P(\text{Type II Error}) = \beta$
• Power of the Test: Probability of correctly rejecting a false null hypothesis.
• $P(\text{Reject } H_0 \mid H_0 \text{ is False}) = 1 - \beta$

1.5 P-value

• Probability of observing a test statistic as extreme, assuming $H_0$ is true.
• Decision Rule:
• If p-value $\leq \alpha$, reject $H_0$.
• If p-value $> \alpha$, fail to reject $H_0$.

2. Hypothesis Tests for Population Mean ($\mu$)

2.1 Test Statistic

• Case 1: Population Variance ($\sigma^2$) Known

  $Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} \sim N(0,1)$

  where:

• $\bar{X}$ is the sample mean.

• $\mu_0$ is the value of $\mu$ under $H_0$.

• $\sigma$ is the population standard deviation.

• $n$ is the sample size.

• Case 2: Population Variance ($\sigma^2$) Unknown

  $t = \frac{\bar{X} - \mu_0}{S / \sqrt{n}} \sim t_{n-1}$

  where:

• $S$ is the sample standard deviation.

• $t_{n-1}$ is the t-distribution with $n-1$ degrees of freedom.

2.2 Example

• Claim: Average score of students is greater than 70.

• $H_0: \mu = 70$

• $H_1: \mu > 70$.

• Suppose $\bar{X} = 72$, $S = 10$, $n = 25$, and $\alpha = 0.05$.

  $t = \frac{72 - 70}{10 / \sqrt{25}} = 1$

  $p\text{-value} = P(t_{24} > 1) = 0.1635$

  Since $0.1635 > 0.05$, we fail to reject $H_0$.

2.3 One-Sided vs Two-Sided Tests

• One-Sided Test: Tests for difference in one direction.

  Example: $H_1: \mu > 70$ or $H_1: \mu < 70$.

• Two-Sided Test: Tests for difference in either direction.

  Example: $H_1: \mu \neq 70$.

3. Hypothesis Tests for Population Variance ($\sigma^2$)

3.1 Test Statistic

$\chi^2 = \frac{(n-1)S^2}{\sigma_0^2} \sim \chi^2_{n-1}$

where:

• $S^2$ is the sample variance.
• $\sigma_0^2$ is the value of $\sigma^2$ under $H_0$.
• $\chi^2_{n-1}$ is the chi-squared distribution with $n-1$ degrees of freedom.

3.2 Example

• Claim: Variance of product weights is less than 4.

• $H_0: \sigma^2 = 4$

• $H_1: \sigma^2 < 4$.

• Suppose $S^2 = 3$, $n = 20$, and $\alpha = 0.05$

  $\chi^2 = \frac{(20-1) \cdot 3}{4} = 14.25$

  $p\text{-value} = P(\chi^2_{19} < 14.25) = 0.2958$

  Since $0.2958 > 0.05$, we fail to reject $H_0$.

Chemical Kinetics - Example Problems

Example 1: Rate Law

• Reaction: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$

• Data at 45°C: | Time (min) | $[N_2O_5]$ (M) | |---|---| | 0 | 0.40 | | 20 | 0.29 | | 40 | 0.20 | | 60 | 0.14 | | 80 | 0.10 | | 100 | 0.07 |

• Several steps are involved in solving, these include

• Determining the rate law.
• Calculating the rate constant
• Calculating the half life
• Calculating $[N_2O_5]$ after 3.0 hours.

Determining the rate law Method

• Plotting [N_2O_5] vs. t, ln[N_2O_5] vs. t, and 1/[N_2O_5] vs. t. The plot of ln[N_2O_5] vs. t is linear, so the reaction is first order.

  • Rate = $k[N_2O_5]$

• Method of Initial Rates (using the integrated rate law): $\ln \frac{[N_2O_5]_t}{[N_2O_5]_0} = -kt$

  $\ln \frac{0.29}{0.40} = -k(20 min)$

  $k = 0.016 min^{-1}$

  $\ln \frac{0.20}{0.29} = -k(20 min)$

  $k = 0.018 min^{-1}$

• Since k remains constant, the reaction is first order with Rate = $k[N_2O_5]$

Rate Constant

Average k = $k = 0.017 min^{-1}$

Half Life Calculation

$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.017 min^{-1}} = 41 min$

Calculation of $[N_2O_5]$ after 3.0 hours

$\ln \frac{[N_2O_5]_t}{[N_2O_5]_0} = -kt$

$\ln \frac{[N_2O_5]_t}{0.40} = -(0.017 min^{-1})(180 min)$

$[N_2O_5]_t = 0.019 M$

Example 2: Activation Energy

• Reaction involves first order decomposition of ethyl iodide:

$CH_3CH_2I(g) \rightarrow CH_2=CH_2(g) + HI(g)$

• Rate constant $1.60 \times 10^{-5} s^{-1}$ at 650 K
• Rate constant $6.36 \times 10^{-3} s^{-1}$ at 725 K

Activation Energy Calculation

• Arrhenius Equation
  • $\ln \frac{k_2}{k_1} = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2})$
• $\ln \frac{6.36 \times 10^{-3}}{1.60 \times 10^{-5}} = \frac{E_a}{8.314 J/mol \cdot K} (\frac{1}{650 K} - \frac{1}{725 K})$
• $E_a = 2.09 \times 10^5 J/mol = 209 kJ/mol$

Explore hashing algorithms, functions that generate fixed-size hash values from data. Key properties include determinism, speed, fixed size, one-way function, and collision resistance. Common examples are MD5, SHA-1, and SHA-256.