Question 3 of 6: Pressure Drop and Power for Gas Flow through a Bed

Question A3: Pressure Drop and Power for Gas Flow through a Bed (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) pressure drop across the bed at the stated gas rate; (b) the power the blower must supply for that flow.

Approach. Compute the bed voidage from the solids volume, the superficial velocity from the gas flow, then the Carman–Kozeny (laminar packed-bed) pressure drop with $K=5$; power is that pressure drop times the volumetric gas flow. Cross-check against the fluidization (buoyant-weight) pressure drop and the minimum-fluidization velocity to confirm the flow regime.

1. Bed voidage. Solids volume $V_s=m_p/\rho_p=1800/2100=0.857\ \text{m}^3$; bed volume $V_b=\tfrac{\pi}{4}(1.0)^2(1.6)=1.257\ \text{m}^3$, so $$\varepsilon=1-\frac{V_s}{V_b}=1-\frac{0.857}{1.257}=\boxed{0.318}.$$
2. Superficial velocity and particle Reynolds number. $A=\tfrac{\pi}{4}(1.0)^2=0.785\ \text{m}^2$, so $u=Q/A=0.2306/0.785=0.294\ \text{m/s}$ and $$Re_p=\frac{\rho u d_p}{\mu}=\frac{1.21(0.294)(0.002)}{1.42\times10^{-5}}=50.0.$$ At $Re_p/(1-\varepsilon)\approx73$ the flow is transitional: the viscous term is the larger part, but inertial losses are no longer negligible (see the Ergun cross-check in step 5). The examiner's explicit Kozeny constant directs the Carman–Kozeny form, so that is the answer reported.
3. (a) Carman–Kozeny pressure drop. For spheres the specific surface is $S=6/d_p$; with Kozeny constant $K=5$, $$\frac{\Delta P}{H}=K\left(\frac{6}{d_p}\right)^2\frac{(1-\varepsilon)^2}{\varepsilon^3}\,\mu u.$$ Substituting $\left(6/0.002\right)^2=9.0\times10^{6}$, $(1-0.318)^2=0.465$, $\varepsilon^3=0.0322$: $$\frac{\Delta P}{H}=5(9.0\times10^{6})\frac{0.465}{0.0322}(1.42\times10^{-5})(0.294)=2716\ \text{Pa/m},$$ so over $H=1.6$ m, $\Delta P=\boxed{4.35\ \text{kPa}}.$
4. (b) Power requirement. Blower power = pressure rise × volumetric flow: $$P=\Delta P\cdot Q=4346(0.2306)=1002\ \text{W}=\boxed{1.0\ \text{kW}}.$$ Keeping the superficial velocity low (below fluidization) is exactly what minimizes this power demand, as the question notes.
5. Regime check (buoyant-weight limit). Were the bed fluidized, its pressure drop would equal the buoyant weight per area, $\Delta P_\text{fl}=\dfrac{m_p g(1-\rho/\rho_p)}{A}=\dfrac{1800(9.81)(1-1.21/2100)}{0.785}=22.5\ \text{kPa}$, reached only at the minimum-fluidization velocity. Setting the Ergun equation equal to the buoyant weight per unit bed volume, $150\dfrac{(1-\varepsilon)^2}{\varepsilon^3}\dfrac{\mu u_{mf}}{d_p^2}+1.75\dfrac{1-\varepsilon}{\varepsilon^3}\dfrac{\rho u_{mf}^2}{d_p}=(1-\varepsilon)(\rho_p-\rho)g$, gives $u_{mf}\approx0.64\ \text{m/s}$. (The purely viscous Kozeny form gives 1.52 m/s, but at that velocity $Re_p\approx260$, far outside its range.) Since the operating $u=0.294\ \text{m/s}\lt u_{mf}$ on either estimate, the bed is not yet fluidized, and the packed-bed value 4.35 kPa is the answer at the stated flow. For reference, the full Ergun equation at $u=0.294$ m/s gives $\Delta P\approx6.7$ kPa (power ≈1.5 kW), which shows the inertial contribution the Kozeny form omits.