Torch design

Come and learn*learn*. I encountered this problem before and didn’t manage to figure it out; I’m hoping for guidance from the experts on the Chemical Engineering forum: http://bbs.hg707.com

We’ve encountered this before; people come to learn how to do it, but we’ll leave it to the manufacturer. We just need to tell the manufacturer the maximum allowable emission levels and the composition of the gases

How are the drag coefficients C and C(Re)2 of droplets in a gas determined?

Reply 1# zhang926: The radius of the barrier zone generally needs to be calculated based on the height of the torch. The thermal radiation radius at any point on the ground is less than 1.5 kj/㎡; taking solar radiation into account, the value used is usually 2.5 kj/㎡.

It’s better to find a professional to handle it; ordinary people can’t do it

It is recommended that LZ refer to HG/T 20570-95.12 on the design of flare systems; I hope this will be helpful to you!

Software calculations are required, such as FlareSIM; generally, the temperature and pressure at the maximum release rate are used (which should correspond to the maximum operating pressure). As for wind speed, it depends on the considerations specific to the project – these three values are not necessarily required.

1. Calculation of calorific value: Air release (1) Original condition exhaust volume 8106.6 Nm3/h Exhaust temperature -165 ℃ 108.15 K Exhaust pressure 1.3 MPa (2) Serial number component content mol% Molecular weight single component low calorific value MJ/Nm3 Single component calorific value MJ Single component density kg/Nm3 Single component weight kg/Nm3 1 H2 0 2.016 10.786 0 0.0899 0 2 CO 0.48 28.0104 12.636 0.060653 1.2506 0.006003 3 CH4 96.99 16.043 35.902 34.82135 0.7174 0.695806 4 O2 0.7 32 0 0 1.5363 0.010754 5 COS 0 60.07 25.218 0 2.6817 0 6 N2 0 28.01 / 0 1.25 0 7 Ar 0 39.948 / 0 1.7834 0 8 CO2 0 44.0098 / 0 1.9771 0 9 C2H6 1.83 30 70.9 1.29747 0.833 0.015244 10 Total 100 16.468 36.179 0.7278 The calculation results are as follows: Average molecular weight 16.468 Mass flow rate kg/h 5900 Low calorific value MJ/Nm3 36.179 Mass flow rate kg/s 1.64 Low calorific value kcal/Nm3 8641.3 Gas weight kg/Nm3 0.7278 (3) Exhaust volume Nm3/h Low calorific value MJ/Nm3 Low calorific value kcal/Nm3 Total calorific value MJ/h Total calorific value kcal/h Total calorific value kW Gas density kg/Nm3 8106.6 36.179 8641.3 293292.5 70051713 81470 0.7278 Note: Xinjiang Guanghui Cold Flare Calculation Book (LNG) 08.12.10 The relevant data to calculate the total calorific value are mainly based on the "Design Code for Fuel Gas Systems and Combustible Gas Emission Systems of Petrochemical Enterprises" (SH3009-2001) 2. 1. Ventilation air calorific value and density exhaust volume Nm3/h Low calorific value MJ/Nm3 Low calorific value kcal/Nm3 Total calorific value MJ/h Average molecular weight and total calorific value kW Gas density kg/Nm3 8106.6 36.179 8641.3 293292.5 16.468 81470 0.7278 2. Adiabatic index and viscosity of vent air (1) Original condition exhaust volume 8106.6 Nm3/h Exhaust temperature -165 ℃ 108.15 K Exhaust pressure 1.3 MPa (2) Serial number Component content mol% Single component adiabatic index Single component index quantity Single component μ 0×106 kg * s/m2 C single component μ ×106 kg * s/m2 1 H2 0 1.4 0 0.86 71.7 0 2 CO 0.48 1.4 0.00672 1.69 118 0.003497 3 CH4 96.99 1.3 1.26087 1.22 225.9 0.440641 4 O2 0.7 1.4 0.0098 1.91 138 0.005566 5 COS 0 1.3 0 1 200 0 6 N2 0 1.4 0 1.69 118 0 7 Ar 0 1.6 0 1.74 114 0 8 CO2 0 1.3 0 1.4 239.7 0 9 CH3OH 0 1.3 0 1 200 0 10 C2H4 1.83 1.4 0.02562 0.94 200 0.006584 11 Total 100 1.303 0.456288 (D.0.3-4) Vs Ka RTM 266.73 1.303 8314 108.15 16.468 5. Find the speed of sound Vs in the gas. Calculate the relevant data Ka-----Adiabatic index Ka of the exhaust gas R-------Gas constant R＝8314N·m/(kg·K) T------Gas temperature K under operating conditions M------Gas average molecular weight Flare head outlet gas flow rate Va=mVs MVK a RT S = Va-----Torch head outlet gas flow rate m/s 0.12 Va M Vs 32.008 0.12 266.73 6. (D.0.3-2) According to the ideal gas state equation, ρ＝ρ0T0 / T ρ ρ0 T0 (℃) T (℃) 1.8382 0.7278 0 -165 AG ρ Va 0.0279 5900 1.8382 32.008 Df (m) A (m2) 0.1883 0.0279 Torch head inner diameter m Torch head D outer mm Wall thickness mm Lining wall thickness mm Effective cross-sectional area of the torch head m2 0.309 0.42 10 50 0.074953 m-----Mach number Assumption m= Vs-----Sound wave propagation speed in gas m/s Flare head outlet diameter Df A-----Torch head effective flow cross-sectional area m2 G-----Mass flow rate of discharge gas kg/h ρ----Density of discharge gas under operating conditions kg/m3 Va----Torch exit gas linear velocity = 32.587m/s Flare head exit effective cross-sectional area A (m2) (D.0.3-1) Flare head exit diameter Df (m) The actual outer diameter wall thickness of the flare head is Φ420×10, lined with 50mm refractory material. Inside the inner diameter D of the torch head outlet (m) Actual outlet flow rate of the flare gas, Mach number a VAG 3600ρ = DA f = 1.128 AVG a 3600ρ = Actual outlet flow rate of the flare gas Va Mass flow rate G (kg/h) Relative density ρ Effective area of the flare head A (m2) Mach number m 11.895 5900 1.8382 0.075 0.0446 Calculation of flame height 7. "Design Specifications for Fuel Gas Systems and Combustible Gas Emission Systems of Petrochemical Enterprises" (SH3009- Appendix D of the Provisions) The calculation of the torch process introduces the Mach number between 0 and 0.12 when h/Df = 270M is converted.: h=270MDf where: h Flame height in meters (inner diameter) hm Df 3.72 0.0446 0.309 3.72 m 8. Design Specifications for Flare High Material Gas System and Combustible Gas Emission System M Mach number Df Flare head outlet diameter m Flame height (SH3009-2001) tff X hhhq Q h = − − φ − φ+ π ε cos 3 sin ) 3 ( 4 2 In the formula, hf ε is the thermal emissivity, which is taken as 0.2 for simplified calculation. Flame tilt angle φ when there is wind φ=Arctg (Vw/Va) Vw——maximum wind speed at the outlet of the torch head Va——gas speed at the outlet of the torch head The design is ht=1.8m Qf The total heat released by the flame (kW) q The thermal radiation intensity of the torch, including sunlight (sunlight is 0.93kW/m2) 1.5kW/m2 is the thermal radiation intensity where people can move freely, 4.7kW/m2 is the thermal radiation intensity where there are no protective measures, but the operator has appropriate protective clothing and needs to stay for a few minutes, a. Take the heat radiation intensity as 1.5kW/m2, then q=1.5-0.93=0.57kW/m2 b. Take the heat radiation intensity as 3.0kW/m2, then q=3.0-0.93=2.07kW/m2 c. Take the heat radiation intensity as 4.7kW/m2, then q=4.7-0.93=3.77kW/m2 d. Taking the thermal radiation intensity as 2.5kW/m2, then q=2.5-0.93=1.57W/m2 Va Vw Φ 11.895 35 71.22872 SIN 71.22872 ＝ 0.946812 ε X q COS 71.22872 ＝ 0.321788 0.2 0 3 22.5897 Flame height = 20 m, torch outlet inner diameter 0.3m, gas flow rate in the flame cylinder is 0.2M. Figure D.0.1 Schematic diagram of the torch. Height of the torch cylinder (m). Substitute all known values into the following formula: hf = … The calculated value for the torch height is: tf = –φ – φ + πε cos(3sinθ)/42. Actual exit velocity of the gas from the torch, Va; mass flow rate, G (kg/h); relative density, ρ; effective area of the torch nozzle, A (m2); Mach number, M. Df(m): 53.347, 5900, 2.3897, 0.013, 0.2000, 0.127898. Actual exit velocity of the gas from the torch, Va; mass flow rate, G (kg/h); relative density, ρ; effective area of the torch nozzle, A (m2); Mach number, M. Df(m): 4.314, 5900, 2.3897, 0.159, 0.0162, 0.45. Water seal tank: D = 1.35 m