6.2 Bernoulli Equation and Mechanical Energy Balances

The Ideal Bernoulli Equation

The Bernoulli equation is a simplified form of the mechanical energy balance, representing the conservation of mechanical energy along a streamline. It is derived under four strict assumptions:

1. Steady flow: Velocity and properties at any point do not change with time.
2. Incompressible flow: Fluid density ($\rho$) remains constant.
3. Frictionless (inviscid) flow: Viscous shear stresses are negligible ($\mu = 0$).
4. Flow along a streamline: The energy constant is conserved along specific paths of fluid particles.

Mathematically, the Bernoulli equation is expressed as:

$\frac{P_1}{\rho} + \frac{v_1^2}{2} + g z_1 = \frac{P_2}{\rho} + \frac{v_2^2}{2} + g z_2$

Dividing each term by the gravitational acceleration ($g$) converts the equation into units of 'head' (length, e.g., meters or feet), which represents energy per unit weight of fluid:

$\frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2$

where $\frac{P}{\rho g}$ is the pressure head, $\frac{v^2}{2g}$ is the velocity (or dynamic) head, and $z$ is the elevation (or potential) head.

The Extended Mechanical Energy Balance

In real chemical engineering systems, friction and work interactions cannot be ignored. The engineering Bernoulli equation, or extended mechanical energy balance, accounts for shaft work ($w_s$, energy per unit mass) added by pumps or extracted by turbines, and frictional losses ($h_f$, head loss):

$\frac{P_1}{\rho g} + \alpha_1 \frac{v_1^2}{2g} + z_1 + h_p = \frac{P_2}{\rho g} + \alpha_2 \frac{v_2^2}{2g} + z_2 + h_t + h_f$

where $h_p$ is the head added by a pump, $h_t$ is the head extracted by a turbine, and $h_f$ is the total head loss due to friction. The dimensionless kinetic energy correction factor ($\alpha$) accounts for the velocity profile: $\alpha = 2.0$ for fully developed laminar flow, and $\alpha \approx 1.05 - 1.08$ (often approximated as $1.0$) for turbulent flow.

Frictional Losses in Pipes: Major and Minor Losses

Total head loss ($h_f$) consists of major losses due to skin friction in straight pipes and minor losses due to flow disruptions in valves, fittings, expansions, and contractions:

$h_f = h_{f,\text{major}} + h_{f,\text{minor}}$

Major Losses (Darcy-Weisbach Equation): Major losses are calculated using the Darcy-Weisbach equation:

$h_{f,\text{major}} = f_D \frac{L}{D} \frac{v^2}{2g}$

where $f_D$ is the Darcy friction factor.
CRITICAL EXAM WARNING: The NCEES FE Reference Handbook provides both the Darcy friction factor ($f_D$) and the Fanning friction factor ($f_f$), which are related by:

$f_D = 4 f_f$

For laminar flow ($Re < 2100$), the friction factors are analytically determined as:

$f_D = \frac{64}{Re} \quad \text{and} \quad f_f = \frac{16}{Re}$

Using the wrong friction factor will result in a factor-of-4 error. For turbulent flow, $f_D$ is obtained from the Moody diagram or approximated by the Colebrook equation, depending on the relative roughness ($\epsilon/D$).

Minor Losses: Minor losses are expressed using a loss coefficient ($K_L$):

$h_{f,\text{minor}} = K_L \frac{v^2}{2g}$

Alternatively, minor losses can be represented using an equivalent length of pipe ($L_{\text{eq}}$) added to the physical pipe length in the Darcy-Weisbach equation.

Flow through Packed Beds: The Ergun Equation

Chemical engineers frequently model pressure drop through packed beds (such as catalytic reactors or absorption columns). The Ergun equation calculates the pressure drop ($\Delta P$) over a bed of length $L$ packed with spherical particles of diameter $D_p$ and sphericity $\phi_s$, at a void fraction (porosity) $\epsilon$:

$\frac{\Delta P}{L} = \frac{150 v_o \mu (1-\epsilon)^2}{D_p^2 \phi_s^2 \epsilon^3} + \frac{1.75 \rho v_o^2 (1-\epsilon)}{D_p \phi_s \epsilon^3}$

where $v_o$ is the superficial velocity, defined as the volumetric flow rate divided by the empty bed cross-sectional area ($v_o = Q / A_{\text{bed}}$).

The Ergun equation combines two distinct terms:

1. The Viscous Term (Kozeny-Carman Equation): The first term dominates at low particle Reynolds numbers ($Re_p < 10$), where viscous forces govern the flow. The particle Reynolds number is defined as:

   $Re_p = \frac{D_p v_o \rho}{\mu (1-\epsilon)}$

2. The Inertial Term (Burke-Plummer Equation): The second term dominates at high particle Reynolds numbers ($Re_p > 1000$), where turbulent inertial losses dominate.

Summary of Friction Factors

Worked Example 1: Pump Head in a Piping System

Problem: Water ($\rho = 1000\text{ kg/m}^3$, $\mu = 1.0\text{ cP}$) is pumped from a lower reservoir to an upper reservoir open to the atmosphere. The elevation difference is $15\text{ m}$. The pipe has a diameter of $0.10\text{ m}$, a length of $100\text{ m}$, and a flow velocity of $2.0\text{ m/s}$. The system includes four $90^\circ$ elbows ($K_L = 0.75$ each) and a fully open gate valve ($K_L = 0.15$). The Darcy friction factor is $f_D = 0.02$. Calculate the required pump head ($h_p$) and the pump shaft power if the pump is $75%$ efficient.

Solution:

First, identify states 1 and 2 at the reservoir surfaces: $P_1 = P_2 = P_{\text{atm}}$ (so $\Delta P = 0$) and $v_1 \approx v_2 \approx 0$.

The extended Bernoulli equation simplifies to:

$h_p = (z_2 - z_1) + h_f$

Calculate the major head loss:

$h_{f,\text{major}} = f_D \frac{L}{D} \frac{v^2}{2g} = 0.02 \cdot \frac{100\text{ m}}{0.10\text{ m}} \cdot \frac{(2.0\text{ m/s})^2}{2 \cdot 9.807\text{ m/s}^2} = 20 \cdot 0.2039\text{ m} \approx 4.08\text{ m}$

Calculate total minor loss coefficient:

$\sum K_L = 4 \cdot 0.75 + 0.15 = 3.15$

Compute minor head loss:

$h_{f,\text{minor}} = \sum K_L \frac{v^2}{2g} = 3.15 \cdot \frac{(2.0\text{ m/s})^2}{2 \cdot 9.807\text{ m/s}^2} \approx 0.64\text{ m}$

Total head loss:

$h_f = 4.08\text{ m} + 0.64\text{ m} = 4.72\text{ m}$

Required pump head:

$h_p = 15\text{ m} + 4.72\text{ m} = 19.72\text{ m}$

Calculate the mass flow rate ($\dot{m}$):

$\dot{m} = \rho A_c v = 1000\text{ kg/m}^3 \cdot \left[ \frac{\pi}{4} (0.10\text{ m})^2 \right] \cdot 2.0\text{ m/s} \approx 15.71\text{ kg/s}$

Compute pump shaft power ($\dot{W}_{\text{shaft}}$):

$\dot{W}_{\text{shaft}} = \frac{\dot{m} g h_p}{\eta} = \frac{15.71\text{ kg/s} \cdot 9.807\text{ m/s}^2 \cdot 19.72\text{ m}}{0.75} = \frac{3038.5\text{ W}}{0.75} \approx 4051\text{ W} = 4.05\text{ kW}$

Worked Example 2: Packed Bed Pressure Drop

Problem: Air ($\rho = 1.2\text{ kg/m}^3$, $\mu = 1.8 \times 10^{-5}\text{ Pa}\cdot\text{s}$) flows through a catalyst bed of length $2.0\text{ m}$ packed with spherical beads ($D_p = 3.0\text{ mm}$, $\phi_s = 1.0$) at a superficial velocity of $0.05\text{ m/s}$. The bed porosity is $\epsilon = 0.40$. Assuming laminar flow dominates, calculate the pressure drop across the bed.

Solution:

First, check the particle Reynolds number to confirm laminar dominance:

$Re_p = \frac{D_p v_o \rho}{\mu (1-\epsilon)} = \frac{0.003\text{ m} \cdot 0.05\text{ m/s} \cdot 1.2\text{ kg/m}^3}{1.8 \times 10^{-5}\text{ Pa}\cdot\text{s} \cdot (1 - 0.40)} = \frac{0.00018}{1.08 \times 10^{-5}} = 16.7$

Since $Re_p$ is relatively low, we use the full Ergun equation but note that the viscous term will be dominant. Let's calculate both terms to be precise:

$\text{Viscous term: } \frac{\Delta P_{\text{viscous}}}{L} = \frac{150 v_o \mu (1-\epsilon)^2}{D_p^2 \phi_s^2 \epsilon^3} = \frac{150 \cdot 0.05 \cdot (1.8 \times 10^{-5}) \cdot (1-0.40)^2}{(0.003)^2 \cdot (1.0)^2 \cdot (0.40)^3} = \frac{4.86 \times 10^{-5}}{5.76 \times 10^{-7}} \approx 84.38\text{ Pa/m}$

$\text{Inertial term: } \frac{\Delta P_{\text{inertial}}}{L} = \frac{1.75 \rho v_o^2 (1-\epsilon)}{D_p \phi_s \epsilon^3} = \frac{1.75 \cdot 1.2 \cdot (0.05)^2 \cdot (1-0.40)}{0.003 \cdot 1.0 \cdot (0.40)^3} = \frac{0.00315}{0.000192} \approx 16.41\text{ Pa/m}$

Total pressure drop per unit length:

$\frac{\Delta P}{L} = 84.38 + 16.41 = 100.79\text{ Pa/m}$

Total pressure drop over the 2.0 m bed:

$\Delta P = 100.79\text{ Pa/m} \cdot 2.0\text{ m} \approx 202\text{ Pa}$