Ammonia Absorption Tower Design and Diameter Calculations

Posted on Jun 13, 2026 in Physics

Ammonia Absorption: 1-inch Intalox Packing

Ammonia is being absorbed in a tower using pure water at 25°C and 1.0 atm absolute pressure. The feed rate is 2000 lb/h and contains 2.5 mole % ammonia in air. The process design specifies a liquid-to-gas mass flow rate ratio of 2:1. Using 50% of the flooding velocity and 1-inch Intalox packing, calculate:

• (i) Pressure drop
• (ii) Gas and liquid flow rates
• (iii) Tower diameter

Solution

Given Data:

• Gas rate (G): 2000 lb/h
• Mole fraction of NH₃: 0.025
• Ratio L/G: 2:1
• Packing: 1-inch Intalox packing
• Temperature (T): 25°C (298 K)
• Pressure (P): 1 atm (1.013 bar)
• Operating Velocity: 50% of flooding velocity

Liquid Rate Calculation

L/G = 2
L = 2 × 2000 = 4000 lb/h

The pressure at the tower bottom will be above atmospheric. Since the pressure drop will not be large, we take the properties of air at 1 atm (1.013 bar) and 25°C (298 K).

Average Molecular Weight of Feed Gas

The feed gas consists of 97.5% air and 2.5% NH₃:
M = (0.975)(29) + (0.025)(17) = 28.7 g/mol

Gas Density Calculation

ρ_G = PM / RT
ρ_G = (1.013)(28.7) / (0.08317)(298)
ρ_G = 1.17 kg/m³ = 0.0732 lb/ft³

Liquid Properties

Liquid viscosity (μ_L) is taken as that of water at 25°C: 0.81 cP.
Liquid density (ρ_L): 1000 kg/m³ = 62.4 lb/ft³

Flow Parameter Calculation

F_lv = (L’/G’) (ρ_G/ρ_L)^0.5
F_lv = (4000/2000) (1.17/1000)^0.5 = 0.07

Calculation of Column Diameter

Using Strigle’s version chart for 2nd-generation packing:

• μ_L = 0.81 cP
• Packing factor (F_p) for 1-inch Intalox: 92 per ft
• For F_lv = 0.07 and assumed ΔP/L = 0.2 inch water/ft, the capacity parameter is 0.66 (interpolated).

Capacity Parameter Formula:
C_s(F_p)^0.5 (μ)^0.05 = 0.66
C_s(92)^0.5 (0.81)^0.05 = 0.66
C_s = 0.0695

Superficial Gas Velocity (u_sG):
C_s = u_sG (ρ_G / (ρ_L − ρ_G))^0.5
0.0695 = u_sG ((1.17)/(1000 − 1.17))^0.5
u_sG = 2.029 ft/s = 0.618 m/s

Operating Gas Velocity (u_op):
u_op = (0.618)(0.5) = 0.309 m/s

Volumetric Gas Flow Rate

Convert 2000 lb/h (907.2 kg/h) into m³/s:
Q = (907.2 kg/h) / (1.17 kg/m³ × 3600 s/h) = 0.21 m³/s

Tower Dimensions

Cross-sectional Area (A_c):
A_c = (0.21 m³/s) / (0.309 m/s) = 0.69 m²

Tower Diameter (D_c):
D_c = (4A_c/π)^0.5
D_c = 0.94 m = 3.086 ft ≈ 3 ft

Pressure Drop and Final Flow Rates

Pressure Drop at Flooding:
(ΔP/L)_fl = 0.115(F_p)^0.7 = (0.115)(92)^0.7 = 2.725 inch water/ft

Gas Flow Rate:
ρu_opA_c = (1.17 kg/m³)(0.309 m/s)(0.69 m²) = 0.249 kg/s = 898.04 kg/h

Water Flow Rate:
(1.4)(898.04 kg/h) = 1257.256 kg/h

Ammonia Absorption: 2-inch Berl Saddles

Ammonia is being absorbed in a packed tower using pure water at 28°C and 1.0 atm pressure. The feed rate is 2500 kg/h and contains 4 mole % ammonia in air. The process design specifies a liquid-to-air mass flow rate ratio of 2:1 and the use of 2-inch Berl saddles.

• (i) Calculate the pressure drop in the packing and air mass velocity at flooding condition.
• (ii) Using 60% of the flooding velocity, calculate the pressure drop, gas and liquid flow rate, and tower diameter.

Solution

Given Data:

• Gas flow rate (G): 2500 kg/h
• Mole fraction of NH₃: 0.04
• Liquid to gas ratio (L/G): 2:1
• Packing: 2-inch Berl saddles
• Temperature (T): 28°C (301 K)
• Pressure (P): 1 atm (1.013 bar)
• Operating velocity: 60% flooding velocity

Liquid Flow Rate

L = 2 × 2500 = 5000 kg/h

Average Molecular Weight of Gas Mixture

M = (0.96 × 29) + (0.04 × 17) = 27.84 + 0.68 = 28.52 g/mol

Gas Density

ρ_G = PM / RT
ρ_G = (1.013 × 28.52) / (0.08314 × 301) = 1.15 kg/m³

Liquid Properties (Water at 28°C)

ρ_L = 1000 kg/m³
μ_L = 0.85 cP

Flow Parameter

F_lv = (L/G)(ρ_G/ρ_L)^0.5
F_lv = (5000/2500)(1.15/1000)^0.5 = 2(0.00115)^0.5 = 0.068

Capacity Parameter and Velocity

Packing factor (F_p) for 2-inch Berl saddles: 40 per ft.
From the generalized pressure drop chart for F_lv = 0.068 and assumed ΔP/L = 0.2 inch water/ft:
Capacity parameter = 0.82

Capacity Coefficient:
C_s(F_p)^0.5(μ_L)^0.05 = 0.82
C_s(40)^0.5(0.85)^0.05 = 0.82 ⇒ C_s = 0.128

Flooding Gas Velocity (u_sG):
C_s = u_sG(ρ_G/(ρ_L − ρ_G))^0.5
0.128 = u_sG[(1.15)/(1000 − 1.15)]^0.5
u_sG = 3.77 ft/s = 1.15 m/s (superficial gas velocity at flooding)

Operating Gas Velocity (u_op):
u_op = 0.60 × 1.15 = 0.69 m/s

Tower Dimensions

Volumetric Gas Flow Rate (Q):
Q = 2500 / (1.15 × 3600) = 0.604 m³/s

Cross-sectional Area (A_c):
A_c = 0.604 / 0.69 = 0.875 m²

Tower Diameter (D_c):
D_c = (4 × 0.875 / π)^0.5 = 1.06 m ≈ 3.48 ft

Pressure Drop Calculations

At Flooding:
(ΔP/L)_f = 0.115(F_p)^0.7 = 0.115(40)^0.7 = 1.52 inch water/ft

At 60% Flooding:
ΔP/L = (0.60)² × 1.52 = 0.55 inch water/ft

Final Answers:

• Pressure drop at flooding: 1.52 inch water/ft
• Operating pressure drop: 0.55 inch water/ft
• Flooding gas velocity: 1.15 m/s
• Operating gas velocity: 0.69 m/s
• Gas flow rate: 2500 kg/h
• Liquid flow rate: 5000 kg/h
• Tower diameter: 1.06 m ≈ 3.48 ft

Ammonia Absorption: 1-inch Metal Intalox

Ammonia is being absorbed in a tower using pure water at 25°C and 1.0 atm absolute pressure. The feed rate is 1800 lb/h and contains 4 mole % ammonia in air. The process design specifies a liquid-to-gas mass flow rate ratio of 2:1. Using 1-inch metal Intalox as packing, calculate the pressure drop and gas mass velocity at flooding condition. Also, using 60% of the flooding flow rate, calculate the tower diameter.

Solution

Given Data:

• Gas flow rate (G): 1800 lb/h
• Mole fraction of NH₃: 0.04
• Liquid to gas ratio (L/G): 2:1
• Packing: 1-inch metal Intalox
• Temperature (T): 25°C (298 K)
• Pressure (P): 1 atm (1.013 bar)
• Operating velocity: 60% flooding velocity

Liquid Flow Rate

L = 2 × 1800 = 3600 lb/h

Average Molecular Weight and Density

M = (0.96 × 29) + (0.04 × 17) = 28.52 g/mol
ρ_G = (1.013 × 28.52) / (0.08314 × 298) = 1.17 kg/m³ = 0.073 lb/ft³

Liquid Properties (Water at 25°C)

ρ_L = 1000 kg/m³
μ_L = 0.81 cP

Flow Parameter

F_lv = (3600/1800)(1.17/1000)^0.5 = 2(0.00117)^0.5 = 0.068

Capacity and Velocity

Packing factor (F_p) for 1-inch metal Intalox: 92 per ft.
From chart (F_lv = 0.068, ΔP/L = 0.2): Capacity parameter = 0.66

Capacity Coefficient:
C_s(92)^0.5(0.81)^0.05 = 0.66 ⇒ C_s = 0.0695

Flooding Gas Velocity (u_sG):
0.0695 = u_sG[(1.17)/(1000 − 1.17)]^0.5
u_sG = 2.03 ft/s = 0.618 m/s

Operating Gas Velocity (u_op):
u_op = 0.60 × 0.618 = 0.371 m/s

Tower Dimensions

Volumetric Gas Flow Rate (Q):
1800 lb/h = 816.48 kg/h
Q = 816.48 / (1.17 × 3600) = 0.194 m³/s

Cross-sectional Area (A_c):
A_c = 0.194 / 0.371 = 0.523 m²

Tower Diameter (D_c):
D_c = (4 × 0.523 / π)^0.5 = 0.816 m ≈ 2.68 ft

Pressure Drop at Flooding

(ΔP/L)_f = 0.115(92)^0.7 = 2.725 inch water/ft

Final Answers:

• Pressure drop at flooding: 2.725 inch water/ft
• Flooding gas velocity: 0.618 m/s
• Operating gas velocity: 0.371 m/s
• Tower diameter: 0.816 m ≈ 2.68 ft

Ammonia Absorption: 1-inch Intalox at 3.5% NH3

Ammonia is being absorbed in a tower using pure water at 25°C and 1.0 atm absolute pressure. The feed rate is 2100 lb/h and contains 3.5 mole % ammonia in air. The process design specifies a liquid-to-gas mass flow rate ratio of 2.2:1. Using 60% of flooding and 1-inch Intalox packing, calculate:

• (i) Pressure drop
• (ii) Gas and liquid flow rates
• (iii) Tower diameter

Solution

Given Data:

• Gas flow rate (G): 2100 lb/h
• Mole fraction of NH₃: 0.035
• Liquid to gas ratio (L/G): 2.2:1
• Packing: 1-inch Intalox
• Temperature (T): 25°C (298 K)
• Pressure (P): 1 atm (1.013 bar)
• Operating velocity: 60% flooding velocity

Liquid Flow Rate

L = 2.2 × 2100 = 4620 lb/h

Average Molecular Weight and Density

M = (0.965 × 29) + (0.035 × 17) = 27.985 + 0.595 = 28.58 g/mol
ρ_G = (1.013 × 28.58) / (0.08314 × 298) = 1.17 kg/m³ = 0.073 lb/ft³

Liquid Properties (Water at 25°C)

ρ_L = 1000 kg/m³
μ_L = 0.81 cP

Flow Parameter

F_lv = (4620/2100)(1.17/1000)^0.5 = 2.2(0.00117)^0.5 = 0.075

Capacity and Velocity

Packing factor (F_p) for 1-inch Intalox: 92 per ft.
From chart (F_lv = 0.075, ΔP/L = 0.2): Capacity parameter = 0.67

Capacity Coefficient:
C_s(92)^0.5(0.81)^0.05 = 0.67 ⇒ C_s = 0.0705

Flooding Gas Velocity (u_sG):
0.0705 = u_sG[(1.17)/(1000 − 1.17)]^0.5
u_sG = 2.06 ft/s = 0.628 m/s

Operating Gas Velocity (u_op):
u_op = 0.60 × 0.628 = 0.377 m/s

Tower Dimensions

Volumetric Gas Flow Rate (Q):
2100 lb/h = 952.56 kg/h
Q = 952.56 / (1.17 × 3600) = 0.226 m³/s

Cross-sectional Area (A_c):
A_c = 0.226 / 0.377 = 0.60 m²

Tower Diameter (D_c):
D_c = (4 × 0.60 / π)^0.5 = 0.874 m ≈ 2.87 ft

Pressure Drop Calculations

At Flooding:
(ΔP/L)_f = 0.115(92)^0.7 = 2.725 inch water/ft

Operating Pressure Drop (60% Flooding):
ΔP/L = (0.60)² × 2.725 = 0.981 inch water/ft

Final Answers:

• Pressure drop: 0.981 inch water/ft packed height
• Flooding gas velocity: 0.628 m/s
• Operating gas velocity: 0.377 m/s
• Gas flow rate: 2100 lb/h (952.56 kg/h)
• Liquid flow rate: 4620 lb/h (2095.63 kg/h)
• Tower diameter: 0.874 m ≈ 2.87 ft