How to calculate the safety valve for silicon tetrachloride storage tanks
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I am designing a silicon tetrachloride storage tank, but I’m not sure how to calculate the safety valve for it I earnestly ask netizens to help solve this! Thank you in advance!
I also need to design a storage tank for silicon tetrachloride. Let’s exchange ideas; send me a message on the site
There are specialized safety valve calculation software programs that can be used for this purpose.
Full-valve calculation example I. Safety valve calculation example: In the design and periodic inspection of pressure vessels, it is necessary to select or verify the safe discharge capacity of safety valves. Due to incomplete information in the past, larger sizes were often used in place of smaller ones, resulting in unnecessary waste. It is now proposed to use Appendix B-B5.1 b) of GB150 as a basis for performing selection calculations on the safe discharge capacity of safety valves under different media, pressures, and temperatures. Example 1: There is an air storage tank with a DN of 1000 mm, a volume of V = 5 m³, a maximum operating pressure of 0.8 MPa, and an operating temperature of 30°C. The inlet pipe has a diameter of φ57X3.5. Determine the size of the safety valve.
Solution:
1) Determine the gas condition parameters.
Let Po be the pressure on the outlet side of the safety valve (absolute pressure), which is 0.103 MPa (approximately 0.1 MPa). Then, the discharge pressure of the safety valve (absolute pressure) is given by:
Pd = 1.1Pw + 0.1 + 10%P = 1.068 MPa (according to GB150 Appendix B4.2.1).
When the outlet side of the safety valve is at atmospheric pressure, Po/Pd = 0.103/1.068 = 0.0936.
Meanwhile, (2/(k+1))k/(k-1) = (2/(1.4+1))×1.4/(1.4-1) = 0.53.
Therefore, Po/Pd < (2/(k+1))k/(k-1), indicating that it is in a critical state. The discharge area A of the safety valve is calculated using formula (B5):
A ≥ mm² (per formula B5).
Here, C is a coefficient related to the properties of the gas; it can be obtained from Table B1 or calculated as C = 520√k/(2/(k+1))(k+1)/(k-1)).
K is the rated discharge coefficient of the safety valve; it is typically 0.9 times the actual discharge coefficient (the actual discharge coefficient is provided by the manufacturer, usually around 0.75), or it can be selected according to the relevant provisions in Section 2 of Appendix V of the Pressure Vessel Regulations.
2) Calculation of the safe discharge capacity of the container:
The safe discharge capacity of containers holding compressed gases or water vapor is determined as follows:
a. For compressor storage tanks or containers holding water vapor, the maximum gas production rates of the compressor and water vapor generator respectively are used.
b. The safe discharge capacity of gas storage tanks is calculated using formula (B1):
Ws = 2.83×10⁻³ρυd², in kg/h (per formula B1).
Here, ρ is the density of the gas at the discharge pressure.
ρ = M (molecular weight) × Pw’ (absolute discharge pressure) × 273/(22.4×(273+t)).
For air, M = 28.95, and the absolute discharge pressure Pw’ = 10.68 kg/cm². Substituting these values into the formula gives:
ρ = 28.95×10.68×273/22.4×303 = 12.44 kg/m³.
υ is the flow velocity of the gas in the inlet pipe under operating pressure, in m/s. According to Table 2, υ ranges from 10 to 15 m/s.
Table 2: Fluid name/Transport pressure (MPa) | Flow velocity range (m/s)
Compressed air | 0–0.1, >0.1–<0.6, >0.6–<1.0, >1.0–<2.0, >2.0–<3.0
General gases (at normal pressure) | Oxygen: <0.6, 0.05–0.6
Saturated water vapor (main pipe) | Low-pressure steam: <1.0, 10–15, 10–20, 8–10
Gas (initial pressure) 2 KPa | 3.0–6.0, 10–20, 5.0–10.0, 7.0–8.0
Gas (initial pressure) 6 KPa | 30–40, 20–30, 15–20, 20–40, 40–60
Coal gas | Saturated water vapor (branch pipes)
Ammonia | ≤0.6, 1.0–2.0
Liquid ammonia | Nitrogen
Acetylene | Hydrogen
Tap water (main pipe) | Flammable gases: 40–60, 35–40, 0.75–3.0, 3–12, 10–20, 3.0–8.0, 0.3–1.0, 2.0–5.0, 2.0–8.0, ≤8.0, 1.5–3.5, 1.0–1.5, ≤1.0
Taking υ = 10 m/s, substituting the values of ρ, υ, and d into the formula gives:
Ws = 2.83×10⁻³×12.44×15×50² = 1320.2 kg/h.
Thus, A = 205.4 mm².
If a safety valve with a handle is used, then A = 0.785d₀² = 205.4 mm². Therefore, d₀ = (205.4/0.785)¹/² = 16.2 mm.
According to statistical estimates, the ratio of the throat diameter d₀ to the nominal diameter DN for full-opening safety valves is approximately 0.625, while for slightly opening safety valves, this ratio is approximately 0.8. Therefore, a full-opening safety valve with a handle and a nominal diameter of DN32 is selected.
Table 3: Relationship between nominal pressure and nominal diameter of safety valves
| Nominal pressure (MPa) | Full-opening type | DN | Slightly opening type | DN |
| PN 1.0, 2.5, 4.0, 6.4 | do = 20, 25, 32, 50, 65 | 15, 20, 25, 32, 40, 50, 80, 100 | PN 10.0 | do = 20, 25, 32, 40, 50 | 12, 20 |
| PN 16, 32 | do = 12, 16, 20, 25, 32, 40, 65, 80 | PN 16, 32 | do = 8, 12 | PN 16, 32 | 14/16 |
Example 2: In Example 1, the medium is changed to steam. Solution: In pressure vessels, the ratio of the pressure on the outlet side of the vast majority of safety valves to their discharge pressure, namely Po/pd, is always less than the theoretical value of 0.528. (This value, obtained using air as the test medium with Po/pd=0.528, indicates a critical state.) Pd – the discharge pressure of the safety valve (absolute pressure). Pd = 1.1×P + 0.1 MPa = 1.1×1.1Pw + 0.1 = 1.21×0.8 + 0.1 = 1.068 MPa. The value of ρ is 5.388 Kg/m3. K = 455° (at t = 182°C). Therefore, WS = 2.83×10^-3 × ρ × ν × d^2 = 2.83×10^-3 × 5.388 × 25 × 50^2 = 953 Kg. The minimum discharge area A is given by: A = …, where Z is the compression coefficient of the steam at the operating temperature and pressure. The calculation can be carried out using the formula provided in question 71 of the book \"Safety Technical Issues of Pressure Vessels\" edited by Gao Honghua: (Note 1) Z===0.93. In the formula: R——848 Kg·m/Kmol·K; T——absolute temperature of the steam in K; ν——specific volume of the steam; M——molecular weight. For steam, k=1.135 ∴ A=245.7 mm2; do===17.7 mm. Therefore, a safety valve with DN=32 should be used. (do=20㎜) Note 1: The compressibility coefficient of the medium can be calculated in accordance with Appendix B of GB150. The critical properties of some commonly used media are given in Table 4. Main physical properties of certain gases Table 4 Name Molecular weight Critical temperature t℃ Critical pressure Patm (absolute pressure) K=Cp/Cv Hydrogen H2 2.02 -239.9 12.8 1.407 Oxygen O2 32 -118.8 49.71 1.4 Air 29 -140.8 37.25 1.4 Nitrogen oxide NO 30 -94 67.2 1.4 Carbon dioxide CO2 44 31.1 72.9 1.30 Water vapor H2O 18.2 374.1 225.4 1.3 (superheated) 1.135 Ammonia NH3 17.03 132.4 111.5 1.29 Hydrogen sulfide H2S 34.09 100.4 88.9 1.3 Fluorine-12 CF2Cl2 120.09 111.7 39.6 1.14 Chlorine Cl2 70.91 144.0 76.1 1.36 Propane C3H8 44.09 96.84 42.01 1.133 Butane C4H10 58.12 152.01 37.47 1.094 Benzene C6H6 78.11 287.6 48.7 1.18 Acetylene H2C2 26.04 36.3 61.6 1.238 Isobutane 58.12 58.12 36.00 1.079 The above operating conditions. It can also be calculated using equation (B7). In this formula, the complicated calculation of the coefficient Z is omitted; when Pd≤10 Mpa, A===242.8 mm2 and do===17.6 mm. As is well known, the compression coefficient Z reflects the differences between real gases and ideal gases in terms of the relationship among pressure, temperature, and specific volume. At normal temperatures and not too high pressures, the difference between real gases and ideal gases is minimal. That is, the compression coefficient Z is approximately 1, whereas the adiabatic index K for commonly used diatomic gases such as air, oxygen, nitrogen, hydrogen, and carbon monoxide is 1.4. Therefore, the formula for calculating the discharge capacity of a safety valve is simplified to the following equation: W=27KPdA. Using the conditions from Example 1, substitution yields A==205.4 mm2 and Do===16.2 mm. The results differ very little from those obtained through the detailed calculations in Example 1. On the other hand, it should be noted that for the safety valves of ammonia synthesis circulators, the pressure on the outlet side is very high. Therefore, when the pressure ratio Po/Pd > (2/(k+1))·(k/(k-1)), the system is in a subcritical state, and equation (B6) should be used to calculate the discharge volume of the safety valve. However, the selection and calculation of safety valves for boiler systems must be carried out using the formulas and coefficients provided in the Boiler Code. Example 3: Liquefied petroleum gas storage tank, with an inner diameter Di of 1600 mm, a length of L = 6000 mm, a wall thickness of δn = 16 mm, and a volume of V = 13.3 m3. The head type is elliptical. The composition of the medium is as follows: 50% propylene, 15% propane, 15% n-isobutylene, 15% n-isobutane, and 5% residues. The components of liquefied petroleum gas are shown in Table 5. Single-component components of liquefied petroleum gas and their vaporization latent heat Table 5 Weight percentages: X1 Propane C3H8, Propylene C3H6, n-Isobutylene, i-Isobutane, Residue. Vaporization latent heat at 50°C in kJ/kg: 285.5, 285.96, 343.7, 317.8, 337. Liquefied petroleum gas storage tanks are generally not insulated, and water spray systems are installed for cooling during summer days. Solution: For pressure vessels without an insulating layer made of adiabatic material, the safe discharge volume is calculated according to (B3). W = 2.55×10^5 × FAt × 0.82/q kg/h. Where: F is a coefficient; for containers on the ground, F = 1. At is the heating area of the container – for a horizontal tank with an elliptical head, At = πD0(L + 0.3D0) = 3.14×1.632×(6.916 + 0.3×1.632) = 37.9 m². The vaporization latent heat at 50°C is given by r = ∑Xiri = 285.5×0.15 + 285.96×0.5 + 343.7×0.15 + 317.8×0.15 + 337×0.05 = 301.9 KJ/kg. Therefore, the safe discharge rate W’ = 16640 kg/h. Regarding the discharge capacity of safety valves, when the length of the tank’s cylinder is ≥ 6 meters, two safety valves should be installed. Under normal circumstances, it is more reasonable to calculate by dividing it in half. This prevents the safety valve from being chosen to be overly large, thus avoiding waste. The minimum discharge area A of the safety valve is A = 577.3 mm2. In this formula, Pd = 1.1P + 0.1 = 1.1 × 1.8 + 0.1 = 2.08 MPa; the molecular weight of M is 44 (for the main component, propane). Therefore, do = 27.12 mm. Two full-opening safety valves of type DN40A42H—4.0 were selected. (This safety valve has a minimum nominal diameter of 40.) Gas is delivered from two or more units to a storage tank (centralized tank) in a centralized manner. Or when gas is supplied from one device to several storage tanks (gas distribution tanks) separately. The calculation of the safe vent volume for the storage tank is shown in Example 4. Example 4: Two air compressors supply air simultaneously to a gas collection tank with a volume of V=100 m3. Its gas delivery pressure is Pw=1.0MPa ； t is room temperature. The intake pipe is φ108×4. The safe discharge volume of the storage tank at this time. ∵At Pw=1.0Mpa ； At t=20℃, ρ=12.87 kg/m3. The gas flow velocity in the intake pipe is ν=15 m/s. Therefore, the safe discharge rate of the storage tank is W = W’ = 7.55(ρo)Vd2 = 7.55×1.293×15×1002 = 6.55×103 kg/h. Here, ρo represents the density of the gas under standard conditions, in kg/m3 ； Under standard conditions, the air has a pressure of P=12.87 kg/m3. Pd represents the discharge pressure of the container in MPa (absolute), T represents the discharge temperature of the container in Kelvin, and d represents the inner diameter of the container’s intake pipe in mm. In fact, W’=7.55ρoVd2 is equivalent to W=28×10-3ρVd2; the difference lies in the fact that there is no need to determine the density of the gas at the discharge condition, ρ (kg/m3). In addition to the aforementioned common storage tanks, we have also encountered devices such as evaporators and reactors, where pressure increases due to the heating and evaporation of liquids inside, or because of chemical reactions that cause the medium to vaporize. As the volume increases and internal pressure rises, the safe discharge capacity should be determined based on the heat released by the heat transfer medium supplied, or the maximum amount of gas that may be generated as a result of chemical reactions within the vessel, as well as the time required for these reactions to take place. In addition, JB/T4750‑2003 \"Pressure Vessels for Refrigeration Equipment\", section B.4 on the diameter of safety valves and rupture discs, outlines the method for calculating the diameter of safety valves to be installed on vessels, which we can refer to in our design work. d = C1 (Equation B.1), where C1 = 35 (Equation B.2). D0 – the outer diameter of the container in meters; L – the length of the container in meters; P – the design pressure in MPa. When two or more containers are connected together, the diameter of their safety valves is calculated by substituting the sum of the D0L values of each container into Equation B.1.
Solution:
1) Determine the gas condition parameters.
Let Po be the pressure on the outlet side of the safety valve (absolute pressure), which is 0.103 MPa (approximately 0.1 MPa). Then, the discharge pressure of the safety valve (absolute pressure) is given by:
Pd = 1.1Pw + 0.1 + 10%P = 1.068 MPa (according to GB150 Appendix B4.2.1).
When the outlet side of the safety valve is at atmospheric pressure, Po/Pd = 0.103/1.068 = 0.0936.
Meanwhile, (2/(k+1))k/(k-1) = (2/(1.4+1))×1.4/(1.4-1) = 0.53.
Therefore, Po/Pd < (2/(k+1))k/(k-1), indicating that it is in a critical state. The discharge area A of the safety valve is calculated using formula (B5):
A ≥ mm² (per formula B5).
Here, C is a coefficient related to the properties of the gas; it can be obtained from Table B1 or calculated as C = 520√k/(2/(k+1))(k+1)/(k-1)).
K is the rated discharge coefficient of the safety valve; it is typically 0.9 times the actual discharge coefficient (the actual discharge coefficient is provided by the manufacturer, usually around 0.75), or it can be selected according to the relevant provisions in Section 2 of Appendix V of the Pressure Vessel Regulations.
2) Calculation of the safe discharge capacity of the container:
The safe discharge capacity of containers holding compressed gases or water vapor is determined as follows:
a. For compressor storage tanks or containers holding water vapor, the maximum gas production rates of the compressor and water vapor generator respectively are used.
b. The safe discharge capacity of gas storage tanks is calculated using formula (B1):
Ws = 2.83×10⁻³ρυd², in kg/h (per formula B1).
Here, ρ is the density of the gas at the discharge pressure.
ρ = M (molecular weight) × Pw’ (absolute discharge pressure) × 273/(22.4×(273+t)).
For air, M = 28.95, and the absolute discharge pressure Pw’ = 10.68 kg/cm². Substituting these values into the formula gives:
ρ = 28.95×10.68×273/22.4×303 = 12.44 kg/m³.
υ is the flow velocity of the gas in the inlet pipe under operating pressure, in m/s. According to Table 2, υ ranges from 10 to 15 m/s.
Table 2: Fluid name/Transport pressure (MPa) | Flow velocity range (m/s)
Compressed air | 0–0.1, >0.1–<0.6, >0.6–<1.0, >1.0–<2.0, >2.0–<3.0
General gases (at normal pressure) | Oxygen: <0.6, 0.05–0.6
Saturated water vapor (main pipe) | Low-pressure steam: <1.0, 10–15, 10–20, 8–10
Gas (initial pressure) 2 KPa | 3.0–6.0, 10–20, 5.0–10.0, 7.0–8.0
Gas (initial pressure) 6 KPa | 30–40, 20–30, 15–20, 20–40, 40–60
Coal gas | Saturated water vapor (branch pipes)
Ammonia | ≤0.6, 1.0–2.0
Liquid ammonia | Nitrogen
Acetylene | Hydrogen
Tap water (main pipe) | Flammable gases: 40–60, 35–40, 0.75–3.0, 3–12, 10–20, 3.0–8.0, 0.3–1.0, 2.0–5.0, 2.0–8.0, ≤8.0, 1.5–3.5, 1.0–1.5, ≤1.0
Taking υ = 10 m/s, substituting the values of ρ, υ, and d into the formula gives:
Ws = 2.83×10⁻³×12.44×15×50² = 1320.2 kg/h.
Thus, A = 205.4 mm².
If a safety valve with a handle is used, then A = 0.785d₀² = 205.4 mm². Therefore, d₀ = (205.4/0.785)¹/² = 16.2 mm.
According to statistical estimates, the ratio of the throat diameter d₀ to the nominal diameter DN for full-opening safety valves is approximately 0.625, while for slightly opening safety valves, this ratio is approximately 0.8. Therefore, a full-opening safety valve with a handle and a nominal diameter of DN32 is selected.
Table 3: Relationship between nominal pressure and nominal diameter of safety valves
| Nominal pressure (MPa) | Full-opening type | DN | Slightly opening type | DN |
| PN 1.0, 2.5, 4.0, 6.4 | do = 20, 25, 32, 50, 65 | 15, 20, 25, 32, 40, 50, 80, 100 | PN 10.0 | do = 20, 25, 32, 40, 50 | 12, 20 |
| PN 16, 32 | do = 12, 16, 20, 25, 32, 40, 65, 80 | PN 16, 32 | do = 8, 12 | PN 16, 32 | 14/16 |
Example 2: In Example 1, the medium is changed to steam. Solution: In pressure vessels, the ratio of the pressure on the outlet side of the vast majority of safety valves to their discharge pressure, namely Po/pd, is always less than the theoretical value of 0.528. (This value, obtained using air as the test medium with Po/pd=0.528, indicates a critical state.) Pd – the discharge pressure of the safety valve (absolute pressure). Pd = 1.1×P + 0.1 MPa = 1.1×1.1Pw + 0.1 = 1.21×0.8 + 0.1 = 1.068 MPa. The value of ρ is 5.388 Kg/m3. K = 455° (at t = 182°C). Therefore, WS = 2.83×10^-3 × ρ × ν × d^2 = 2.83×10^-3 × 5.388 × 25 × 50^2 = 953 Kg. The minimum discharge area A is given by: A = …, where Z is the compression coefficient of the steam at the operating temperature and pressure. The calculation can be carried out using the formula provided in question 71 of the book \"Safety Technical Issues of Pressure Vessels\" edited by Gao Honghua: (Note 1) Z===0.93. In the formula: R——848 Kg·m/Kmol·K; T——absolute temperature of the steam in K; ν——specific volume of the steam; M——molecular weight. For steam, k=1.135 ∴ A=245.7 mm2; do===17.7 mm. Therefore, a safety valve with DN=32 should be used. (do=20㎜) Note 1: The compressibility coefficient of the medium can be calculated in accordance with Appendix B of GB150. The critical properties of some commonly used media are given in Table 4. Main physical properties of certain gases Table 4 Name Molecular weight Critical temperature t℃ Critical pressure Patm (absolute pressure) K=Cp/Cv Hydrogen H2 2.02 -239.9 12.8 1.407 Oxygen O2 32 -118.8 49.71 1.4 Air 29 -140.8 37.25 1.4 Nitrogen oxide NO 30 -94 67.2 1.4 Carbon dioxide CO2 44 31.1 72.9 1.30 Water vapor H2O 18.2 374.1 225.4 1.3 (superheated) 1.135 Ammonia NH3 17.03 132.4 111.5 1.29 Hydrogen sulfide H2S 34.09 100.4 88.9 1.3 Fluorine-12 CF2Cl2 120.09 111.7 39.6 1.14 Chlorine Cl2 70.91 144.0 76.1 1.36 Propane C3H8 44.09 96.84 42.01 1.133 Butane C4H10 58.12 152.01 37.47 1.094 Benzene C6H6 78.11 287.6 48.7 1.18 Acetylene H2C2 26.04 36.3 61.6 1.238 Isobutane 58.12 58.12 36.00 1.079 The above operating conditions. It can also be calculated using equation (B7). In this formula, the complicated calculation of the coefficient Z is omitted; when Pd≤10 Mpa, A===242.8 mm2 and do===17.6 mm. As is well known, the compression coefficient Z reflects the differences between real gases and ideal gases in terms of the relationship among pressure, temperature, and specific volume. At normal temperatures and not too high pressures, the difference between real gases and ideal gases is minimal. That is, the compression coefficient Z is approximately 1, whereas the adiabatic index K for commonly used diatomic gases such as air, oxygen, nitrogen, hydrogen, and carbon monoxide is 1.4. Therefore, the formula for calculating the discharge capacity of a safety valve is simplified to the following equation: W=27KPdA. Using the conditions from Example 1, substitution yields A==205.4 mm2 and Do===16.2 mm. The results differ very little from those obtained through the detailed calculations in Example 1. On the other hand, it should be noted that for the safety valves of ammonia synthesis circulators, the pressure on the outlet side is very high. Therefore, when the pressure ratio Po/Pd > (2/(k+1))·(k/(k-1)), the system is in a subcritical state, and equation (B6) should be used to calculate the discharge volume of the safety valve. However, the selection and calculation of safety valves for boiler systems must be carried out using the formulas and coefficients provided in the Boiler Code. Example 3: Liquefied petroleum gas storage tank, with an inner diameter Di of 1600 mm, a length of L = 6000 mm, a wall thickness of δn = 16 mm, and a volume of V = 13.3 m3. The head type is elliptical. The composition of the medium is as follows: 50% propylene, 15% propane, 15% n-isobutylene, 15% n-isobutane, and 5% residues. The components of liquefied petroleum gas are shown in Table 5. Single-component components of liquefied petroleum gas and their vaporization latent heat Table 5 Weight percentages: X1 Propane C3H8, Propylene C3H6, n-Isobutylene, i-Isobutane, Residue. Vaporization latent heat at 50°C in kJ/kg: 285.5, 285.96, 343.7, 317.8, 337. Liquefied petroleum gas storage tanks are generally not insulated, and water spray systems are installed for cooling during summer days. Solution: For pressure vessels without an insulating layer made of adiabatic material, the safe discharge volume is calculated according to (B3). W = 2.55×10^5 × FAt × 0.82/q kg/h. Where: F is a coefficient; for containers on the ground, F = 1. At is the heating area of the container – for a horizontal tank with an elliptical head, At = πD0(L + 0.3D0) = 3.14×1.632×(6.916 + 0.3×1.632) = 37.9 m². The vaporization latent heat at 50°C is given by r = ∑Xiri = 285.5×0.15 + 285.96×0.5 + 343.7×0.15 + 317.8×0.15 + 337×0.05 = 301.9 KJ/kg. Therefore, the safe discharge rate W’ = 16640 kg/h. Regarding the discharge capacity of safety valves, when the length of the tank’s cylinder is ≥ 6 meters, two safety valves should be installed. Under normal circumstances, it is more reasonable to calculate by dividing it in half. This prevents the safety valve from being chosen to be overly large, thus avoiding waste. The minimum discharge area A of the safety valve is A = 577.3 mm2. In this formula, Pd = 1.1P + 0.1 = 1.1 × 1.8 + 0.1 = 2.08 MPa; the molecular weight of M is 44 (for the main component, propane). Therefore, do = 27.12 mm. Two full-opening safety valves of type DN40A42H—4.0 were selected. (This safety valve has a minimum nominal diameter of 40.) Gas is delivered from two or more units to a storage tank (centralized tank) in a centralized manner. Or when gas is supplied from one device to several storage tanks (gas distribution tanks) separately. The calculation of the safe vent volume for the storage tank is shown in Example 4. Example 4: Two air compressors supply air simultaneously to a gas collection tank with a volume of V=100 m3. Its gas delivery pressure is Pw=1.0MPa ； t is room temperature. The intake pipe is φ108×4. The safe discharge volume of the storage tank at this time. ∵At Pw=1.0Mpa ； At t=20℃, ρ=12.87 kg/m3. The gas flow velocity in the intake pipe is ν=15 m/s. Therefore, the safe discharge rate of the storage tank is W = W’ = 7.55(ρo)Vd2 = 7.55×1.293×15×1002 = 6.55×103 kg/h. Here, ρo represents the density of the gas under standard conditions, in kg/m3 ； Under standard conditions, the air has a pressure of P=12.87 kg/m3. Pd represents the discharge pressure of the container in MPa (absolute), T represents the discharge temperature of the container in Kelvin, and d represents the inner diameter of the container’s intake pipe in mm. In fact, W’=7.55ρoVd2 is equivalent to W=28×10-3ρVd2; the difference lies in the fact that there is no need to determine the density of the gas at the discharge condition, ρ (kg/m3). In addition to the aforementioned common storage tanks, we have also encountered devices such as evaporators and reactors, where pressure increases due to the heating and evaporation of liquids inside, or because of chemical reactions that cause the medium to vaporize. As the volume increases and internal pressure rises, the safe discharge capacity should be determined based on the heat released by the heat transfer medium supplied, or the maximum amount of gas that may be generated as a result of chemical reactions within the vessel, as well as the time required for these reactions to take place. In addition, JB/T4750‑2003 \"Pressure Vessels for Refrigeration Equipment\", section B.4 on the diameter of safety valves and rupture discs, outlines the method for calculating the diameter of safety valves to be installed on vessels, which we can refer to in our design work. d = C1 (Equation B.1), where C1 = 35 (Equation B.2). D0 – the outer diameter of the container in meters; L – the length of the container in meters; P – the design pressure in MPa. When two or more containers are connected together, the diameter of their safety valves is calculated by substituting the sum of the D0L values of each container into Equation B.1.
The main issue is that the characteristic parameters of the medium are unknown – things like the latent heat of vaporization, specific heat capacity, etc. Also, it’s not clear what the corrosiveness of this medium is, and whether heat treatment is necessary