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It is desired to absorb 95% ammonia from a feed mixture containing 10% ammonia and rest air. The gas enters at rate of 500 kmol/h. If water is used as solvent at the rate of 1.5 times of the minimum calculate (a) NTU (b) L, actual (c) height of the tower if HTU 1m. The equilibrium relationship is given by Y=20x

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my150790

Answer:

To solve this problem, we can use the concept of the Number of Transfer Units (NTU) for absorption towers. NTU is defined as the ratio of the actual mass transfer rate to the maximum possible mass transfer rate.

Given data:

- Gas flow rate (G) = 500 kmol/h

- Ammonia concentration in the feed (Cin) = 10%

- Desired removal efficiency (η) = 95%

- Equilibrium relationship: Y = 20X (where Y is the ammonia concentration in the gas phase and X is the ammonia concentration in the liquid phase)

(a) To calculate NTU:

\[ NTU = -\ln(1-\eta) \]

(b) To calculate L, actual (actual liquid flow rate):

\[ L_{actual} = G \times \frac{(1 - Y_1)}{X_1 - X_2} \]

(c) To calculate the height of the tower:

\[ H_{tower} = L_{actual} \times HTU \]

First, let's find the ammonia concentration at the exit of the absorber (X2) using the equilibrium relationship:

\[ Y1 = 10\% \]

\[ Y2 = 95\% \] (desired removal efficiency)

\[ Y = 20X \]

\[ X2 = Y2 / 20 \]

\[ X2 = 95\% / 20 = 4.75\% \]

(a) Calculate NTU:

\[ NTU = -\ln(1-0.95) \]

\[ NTU ≈ -\ln(0.05) ≈ 2.9957 \]

(b) Calculate L, actual:

\[ L_{actual} = G \times \frac{(1 - Y1)}{X1 - X2} \]

\[ L_{actual} = 500 \, \text{kmol/h} \times \frac{(1 - 0.10)}{0.10 - 0.0475} \]

\[ L_{actual} ≈ 500 \, \text{kmol/h} \times \frac{0.90}{0.0525} \]

\[ L_{actual} ≈ 8561.9 \, \text{kmol/h} \]

(c) Calculate the height of the tower:

Given HTU = 1m

\[ H_{tower} = L_{actual} \times HTU \]

\[ H_{tower} ≈ 8561.9 \times 1 \]

\[ H_{tower} ≈ 8561.9 \, \text{m} \]

So,

(a) NTU ≈ 2.9957

(b) L, actual ≈ 8561.9 kmol/h

(c) Height of the tower ≈ 8561.9 m

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