# Step-by-Step Solution **Question Recap:** A process stream with 1. mole% hydrocarbon in oil is washed in a scrubber (absorption column) using steam (water vapor) as the stripping agent. - Oil flow to the tower: 30 kmol/h (total inlet flow) - Steam (stripping agent) flow: 90 kmol/h (counter-current) - 90% of hydrocarbon to be stripped - Oil is non-volatile (does not enter vapor phase) - Steam does **not** condense (all vapor phase) - Hydrocarbon equilibrium: \( y = 2.5x \) (where \( y \) = mole fraction in vapor, \( x \) = mole fraction in liquid) You are to determine the number of theoretical stages (plates) needed. **Hint:** Draw a schematic and do a mass balance. --- ## 1. **Draw Schematic and Define Streams** ``` ┌────────────┐ │ SCRUBBER │ └────────────┘ Oil in →| |→ Oil out HC, Oil | | HC (↓) | | | | Steam in →| |→ Steam + HC out (↑) | | ``` Let: - \( L \): Oil flow rate (liquid phase) = **30 kmol/h** (all oil, no steam) - \( V \): Steam flow rate (vapor phase) = **90 kmol/h** (all vapor, no oil) - \( x_{in} \): Hydrocarbon mole fraction in oil in = **.01** - \( x_{out} \): Hydrocarbon mole fraction in oil out = **unknown** - \( y_{in} \): Hydrocarbon mole fraction in vapor in = **** (pure steam in) - \( y_{out} \): Hydrocarbon mole fraction in vapor out = **unknown** --- ## 2. **Do a Hydrocarbon Mass Balance** Let \( F_{HC,in} \) = HC in with oil feed Let \( F_{HC,out,oil} \) = HC out with oil Let \( F_{HC,out,steam} \) = HC out with steam \[ \text{HC in} = \text{HC out with oil} + \text{HC out with steam} \] \[ (30)(.01) = (30)(x_{out}) + (90)(y_{out}) \] \[ .3 = 30x_{out} + 90y_{out} \] --- ## 3. **Apply the 90% Stripping Requirement** 90% of hydrocarbon is stripped = only 10% remains in the oil leaving: \[ (30)(x_{out}) = .1 \times (30)(.01) = .1 \times .3 = .03 \] \[ x_{out} = \frac{.03}{30} = .001 \] --- ## 4. **Find Hydrocarbon in Vapor Out** Plug \( x_{out} \) into the mass balance: \[ .3 = 30(.001) + 90y_{out} \] \[ .3 = .03 + 90y_{out} \] \[ .27 = 90y_{out} \] \[ y_{out} = \frac{.27}{90} = .003 \] --- ## 5. **Plot the Operating Line** For a counter-current stripper: \[ \text{Operating Line:} \quad y_{n+1} = y_{in} + \frac{L}{V}(x_n - x_{out}) \] But with constant flow rates, can use the simplified two-endpoints: - Bottom (oil in): \( x_{in} = .01 \), \( y_{in} = \) - Top (oil out): \( x_{out} = .001 \), \( y_{out} = .003 \) Line passes through: \((x_{out}, y_{in})\) and \((x_{in}, y_{out})\): - Bottom: \( (.001,\, ) \) - Top: \( (.01,\, .003) \) --- ## 6. **Draw the Equilibrium Line** Given: \[ y = 2.5x \] --- ## 7. **Graphical Construction (McCabe-Thiele Method)** - **Plot x-axis:** \( x \) from .001 to .01 - **Plot y-axis:** \( y \) from to .003 - Draw equilibrium line \( y = 2.5x \) - Draw operating line from \( (.001, ) \) to \( (.01, .003) \) - Step off stages (horizontal to equilibrium, vertical to operating line) until you reach \( x_{in} \) --- ## 8. **Estimate the Number of Stages (Analytical Approach)** If you cannot draw, estimate using Kremser Equation for stripping: \[ N = \frac{\ln \left[ \frac{(y_{out} - Sy_{in})}{(y_{eq,out} - Sy_{eq,in})} \right]}{\ln S} \] Where - \( S = \frac{L/V}{K} \) - \( K \) = slope of equilibrium line = 2.5 - \( L = 30 \) - \( V = 90 \) - \( y_{in} = \) - \( y_{out} = .003 \) - \( x_{in} = .01 \) - \( x_{out} = .001 \) First, calculate \( S \): \[ S = \frac{L/V}{K} = \frac{30/90}{2.5} = \frac{1/3}{2.5} = \frac{1}{7.5} = .133 \] Now, for stripping: Let’s use the stripping Kremser equation for the fraction remaining: \[ \text{Fraction remaining} = \frac{S^N - 1}{S - 1} \cdot \frac{x_{out}}{x_{in}} \] But since we know all streams, let's use the stage-to-stage construction for 2-3 steps, or use: \[ N = \frac{\ln \left( \frac{y_{out} - S y_{in}}{y^*_{out} - S y^*_{in}} \right)}{\ln S} \] But \( y_{in} = \), \( y^*_{out} = K x_{out} = 2.5 \times .001 = .0025 \), \( y^*_{in} = K x_{in} = 2.5 \times .01 = .025 \) \[ N = \frac{\ln \left( \frac{.003 - }{.0025 - .133 \times .025} \right)}{\ln(1/.133)} \] Calculate denominator: \[ .133 \times .025 = .003325 \] \[ .0025 - .003325 = -.000825 \] But negative denominator doesn't make sense; so let's directly use McCabe-Thiele (graphical) or try Kremser's for stripping: \[ x_{out} = x_{in} \left( \frac{S^N - 1}{S^{N+1} - 1} \right) \] But this is for absorption, for stripping: \[ \frac{x_{out}}{x_{in}} = S^N \] \[ N = \frac{\ln(x_{out}/x_{in})}{\ln S} \] \[ N = \frac{\ln(.001/.01)}{\ln(.133)} \] \[ N = \frac{\ln(.1)}{\ln(.133)} = \frac{-2.3026}{-2.017} = 1.14 \] So, **number of theoretical stages ≈ 1.14** (so, 2 stages needed, since you can't have a fraction of a stage). --- ## 9. **Summary** ### **Answers** - **Mass balance:** \( x_{out} = .001 \), \( y_{out} = .003 \) - **Operating and equilibrium lines:** Use endpoints and \( y = 2.5x \) - **Number of theoretical stages:** **2 stages** (rounded up from 1.14) --- ## 10. **Schematic Diagram** ``` ┌───────────────────────────┐ Oil in | ABSORPTION TOWER | Oil out x=.01 | | x=.001 30 kmol/h 30 kmol/h | | | | Steam in| | Steam + HC out y= | | y=.003 90 kmol/h| | 90 kmol/h └───────────────────────────┘ ``` --- ## 11. **Steps** 1. **Draw schematic and label streams** 2. **Write mass balances for hydrocarbon** 3. **Use 90% stripping requirement to determine outlet compositions** 4. **Use equilibrium line \( y = 2.5x \) and operating line to construct McCabe-Thiele diagram (or use Kremser equation)** 5. **Calculate or graphically determine number of theoretical stages** 6. **Answer: 2 stages required** --- **If you can draw the diagram and plot the lines, you'll see you need just over 1 stage, so 2 stages are sufficient.**