Rate Processes: Heat and Fluid Flow

Introduction to Rate Processes

Rate: The amount of a quantity that changes per unit time.
- Represented as the change in 'n' (amount of something) over a change in 't' (time): $\Delta n / \Delta t$ where 'n' is the amount of something.
- Calculus connection: The slope of the curve (change in volume over time) is the derivative, representing the flow rate. $\frac{dV}{dt}$
Flow Rate (r): The amount of a quantity 'n' that flows past a point in a given length of time 't'.
- Example: The volume of water 'V' passing through a screen per unit time 't' is the flow rate 'r'.

Flux

Flux (J): The flow rate 'r' per unit of cross-sectional area 'A'.
- Formula: $J = \frac{r}{A} = \frac{n}{At}$
Illustration: A machine gun firing bullets at a target.
- Target size: $1 \text{ meter} \times 1 \text{ meter}$
- Machine gun fires $5 \text{ bullets per second}$
- Flux of bullets to the target: $5 \frac{\text{bullets}}{\text{s}} \text{ over } 1 \text{ m}^2 = 5 \frac{\text{bullets}}{s \cdot m^2}$

Driving Force

Driving Force: What makes something move or flow.
- It is related to work when force is measured over a distance.
Relationship between Flux and Driving Force: Flux is proportional to the change in driving force over a distance 'x'.
- Initial Proportion: $J \propto \frac{\text{Driving Force}}{x}$
- This proportionality needs a coefficient (k) because different materials exhibit different flow characteristics (e.g., sand vs. water in a pipe).
 - Coefficient's Role: Accounts for material properties (like friction, viscosity, thermal conductivity).
- Final Formula: $J = k \frac{\text{Driving Force}}{x}$
 - Here, 'k' is a proportionality constant.
 - 'x' represents the change in distance: $(t{in} - t{out})$

Heat Flow

Heat: Energy flow resulting from a temperature difference.
Heat Flux ( $J_{heat}$ ): The amount of heat 'Q' that flows per unit cross-sectional area 'A' per unit time 't'.
- Formula: $J_{heat} = \frac{Q}{At}$ (where Q is a quantity of heat, not volume)
- Proportionality: Directly proportional to the temperature difference ( $\Delta T$ ) and inversely proportional to the distance 'x' separating the temperatures.
  - Formula: $J_{heat} = k \frac{\Delta T}{x}$
  - Where 'k' (little k) is the thermal conductivity.
Thermal Conductivity (k):
- Measures how well heat flows through a material.
- Low 'k': Material is an insulator (heat does not flow well).
- High 'k': Material is a conductor (heat flows well).

Heat Flow Problem: Copper Rod

Scenario: Copper rod, $0.5 \text{ m}$ long, $0.025 \text{ m}$ in diameter. One end in boiling water ( $373.15 \text{ K}$ ), the other in an ice bath ( $273.15 \text{ K}$ ). Walls are insulated to prevent heat loss. Calculate the heat flow 'Q' down the rod.
Given:
- Length ( $x$ ) = $0.5 \text{ m}$
- Diameter ( $D$ ) = $0.025 \text{ m}$
- Temperature 1 ( $T_1$ ) = $373.15 \text{ K}$
- Temperature 2 ( $T_2$ ) = $273.15 \text{ K}$
- Thermal conductivity of copper ( $k$ ) = $386 \frac{\text{J}}{\text{s} \cdot \text{m} \cdot \text{K}}$ (a known constant).
Goal: Find the heat flow 'Q' (rate of heat flow, not flux).
Relationship between Flux and Flow: Flux is flow divided by area ( $J = \text{Flow Rate} / A$ ). Therefore, flow rate is flux multiplied by area ( $\text{Flow Rate} = J \cdot A$ ).
Equations Used:
- Heat flux: $J_{heat} = k \frac{\Delta T}{x}$
- Heat flow (rate): $\dot{Q} = J_{heat} \cdot A = k \frac{\Delta T}{x} \cdot A$
- Cross-sectional area of the rod (circular): $A = \pi r^2 = \pi (\frac{D}{2})^2 = \frac{\pi D^2}{4}$
Calculation Steps:
1. Calculate (\Delta T = T1 - T2 = 373.15 \text{ K} - 273.15 \text{ K} = 100 \text{ K}).
2. Calculate area (A = \frac{\pi (0.025 \text{ m})^2}{4} = 0.000490875 \text{ m}^2).
3. Plug values into the heat flow equation:
  $\dot{Q} = (386 \frac{\text{J}}{\text{s} \cdot \text{m} \cdot \text{K}}) \cdot \frac{(100 \text{ K})}{(0.5 \text{ m})} \cdot (0.000490875 \text{ m}^2)$
Result: $\dot{Q} \approx 37.9 \frac{\text{J}}{\text{s}}$ or $37.9 \text{ Watts}$
- Units Check: $(\frac{\text{J}}{\text{s} \cdot \text{m} \cdot \text{K}}) \cdot (\frac{\text{K}}{\text{m}}) \cdot (\text{m}^2) = \frac{\text{J}}{\text{s}} \rightarrow$ which is energy over time, confirming the units for heat flow.

Fluid Flow

Driving Force for Fluids: Pressure difference ( $\Delta P$ ).
Poisson's Equation (for fluid flux through a circular pipe):
- Given fluid volume 'V' through a circular pipe with cross-sectional area 'A' and length 'x'.
- Fluid Flux ( $J{fluid}$ ): $J{fluid} = \frac{1}{8 \mu} \frac{\Delta P}{x} r^2$
 - This formula often appears in slightly different forms, but the core components ( $\Delta P$ , (x), (\mu), (r^2)) are related.
 - Where (\mu) (mu) is the fluid's viscosity.
Viscosity ((\mu)): A measure of the thickness of a fluid; how well a liquid flows (resistance to flow).
Laminar Flow (Non-Turbulent Flow): A non-disrupted, unidirectional stream of flow where fluid molecules move parallel to each other without mixing.
- Poisson's equation is valid only for laminar flow.
Determining Laminar vs. Turbulent Flow: We can use a dimensionless number (Reynolds number concept).
- The equation to determine flow type is not fully given but can be related to the ratio of inertial forces to viscous forces.
- A common criterion: If \frac{\rho v D}{\mu} < 2300 (where (\rho) is fluid density, (v) is fluid velocity, (D) is pipe diameter, (\mu) is viscosity), the flow is laminar.
Superfluid: A fluid with no viscosity ( $\mu = 0$ ). Such a fluid would flow without a pressure difference.

Fluid Flow Problem: Water Flow Rate

Scenario: Calculate the flow rate of water at $300 \text{ K}$ through a pipe $2 \text{ m}$ long with an inner diameter of $0.001 \text{ m}$ . Inlet pressure ( $P{in}$ ) is $1.0153 \times 10^5 \text{ Pa}$ and outlet pressure ( $P{out}$ ) is $1.0133 \times 10^5 \text{ Pa}$ .
Given:
- Length ( $x$ ) = $2 \text{ m}$
- Diameter ( $D$ ) = $0.001 \text{ m}$
- Inlet Pressure ( $P_{in}$ ) = $1.0153 \times 10^5 \text{ Pa}$
- Outlet Pressure ( $P_{out}$ ) = $1.0133 \times 10^5 \text{ Pa}$
- Viscosity of water at $300 \text{ K}$ (not explicitly given in the problem text, but mentioned in the solution part as $86 \times 10^{-5} \text{ Pa} \cdot \text{s}$ or $8.6 \times 10^{-4} \text{ Pa} \cdot \text{s}$ ). Let's use the given value as $86 \times 10^{-5} \text{ Pa} \cdot \text{s}$ for consistency with the provided transcript.
Goal: Find the flow rate of water (volume per unit time).
Flow Rate Equation for Pipe (Poiseuille's Law, derived from flux):
- The volume flow rate ( $\dot{V}$ ) through a circular pipe is: $\dot{V} = \frac{\pi D^4}{128 \mu x} \Delta P$
- The instructor used a slightly different derivation starting from flux then multiplying by area. The final form used was: $\dot{V} = \frac{\pi r^4}{8\mu} \frac{\Delta P}{x}$
  - Substituting $r = D/2$ : $\dot{V} = \frac{\pi (D/2)^4}{8\mu} \frac{\Delta P}{x} = \frac{\pi D^4}{128 \mu x} \Delta P$
Calculation Steps:
1. Calculate (\Delta P = P{in} - P{out} = (1.0153 - 1.0133) \times 10^5 \text{ Pa} = 0.0020 \times 10^5 \text{ Pa} = 200 \text{ Pa}).
2. Diameter ( $D$ ) = $0.001 \text{ m}$ . Therefore, $D^4 = (0.001)^4 = 1 \times 10^{-12} \text{ m}^4$ . (Note: the instructor might have intended to derive the formula slightly differently, but the final plug-in matches the standard Poiseuille's law form given).
3. Viscosity ( $\mu$ ) = $86 \times 10^{-5} \text{ Pa} \cdot \text{s}$ .
4. Plug values into the flow rate equation:
  $\dot{V} = \frac{\pi (0.001 \text{ m})^4}{128 (86 \times 10^{-5} \text{ Pa} \cdot \text{s}) (2 \text{ m})} (200 \text{ Pa})$
Result: $\dot{V} \approx 2.85 \times 10^{-9} \frac{\text{m}^3}{\text{s}}$
- Units Check: $(\frac{\text{m}^4}{\text{Pa} \cdot \text{s} \cdot \text{m}}) \cdot \text{Pa} = \frac{\text{m}^3}{\text{s}}$ which represents volume over time, confirming the units for flow rate.

Electricity (Future Topic)

Will cover basic electrical concepts and circuits in the next session.
Students are encouraged to read the textbook chapter on this topic beforehand.