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Question 3. (30) Points) In a filled tower with a cross-sectional area of 200 cm³ and a height of 180 cm, a gas with a flow rate of 200 kmol/s is absorbed by water. Thus, the concentration of the soluble gas is reduced from 4% to 1% in moles. There is no dissolved gas in the liquid entering the tower. Equilibrium relation When expressed by the equation y=mx, \begin{equation} (NTU) \ BE = \int_{x_a}^{x_b} \frac{dx}{x^* - x} = \frac{1}{\left(\frac{L}{mG} - 1\right)} \text{of} \left[ 1 + r - \left(\frac{mg}{L}\right)r \right] \end{equation} Subtract the equation. is expressed where r is the ratio of the absorbed gas to the gas leaving the tower, and \begin{equation} r = \frac{Lx_b}{Gy_a} \end{equation}

          Question 3. (30) Points) In a filled tower with a cross-sectional area of 200 cm³ and a height of 180 cm, a gas with a flow rate of 200 kmol/s is absorbed by water. Thus, the concentration of the soluble gas is reduced from 4% to 1% in moles. There is no dissolved gas in the liquid entering the tower. Equilibrium relation When expressed by the equation y=mx,
\begin{equation} (NTU) \ BE = \int_{x_a}^{x_b} \frac{dx}{x^* - x} = \frac{1}{\left(\frac{L}{mG} - 1\right)} \text{of} \left[ 1 + r - \left(\frac{mg}{L}\right)r \right] \end{equation} Subtract the equation. is expressed where r is the ratio of the absorbed gas to the gas leaving the tower, and
\begin{equation} r = \frac{Lx_b}{Gy_a} \end{equation}

Question 3. (30) Points) In a filled tower with a cross-sectional area of 200 cm³ and a height of 180 cm, a gas with a flow rate of 200 kmol/s is absorbed by water. Thus, the concentration of the soluble gas is reduced from 4% to 1% in moles. There is no dissolved gas in the liquid entering the tower. Equilibrium relation When expressed by the equation y=mx,

(NTU) BE = ∫xa^xb(dx)/(x^* - x) = (1)/(((L)/(mG) - 1))of[ 1 + r - ((mg)/(L))r ]
Subtract the equation. is expressed where r is the ratio of the absorbed gas to the gas leaving the tower, and

r = (Lxb)/(Gya)

Added by Laura W.

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