HCBBS Forum (English)
Submit Chemical Projects / Find Solutions
Amplify Your Requirements on a Broader Chemical Platform *Engineering · Technology · Equipment · Solutions*
Submit Request

What is the flow rate of 1.4 kg of nitrogen rushing into a tank at atmospheric pressure?

2019-01-11View Original

Thread Content

What is the flow rate of 1.4 kg of nitrogen rushing into a tank at atmospheric pressure? The diameter of the nitrogen pipe is DN25; I need help. Also, how is it calculated?
Reply #22019-01-11
If it is a pressure vessel at normal pressure but the equipment containing nitrogen is at high pressure, I think it would be 340 m/s
Reply #32019-01-11
I’m really interested in understanding the relationship between pressure, flow rate, cross-sectional area, and flow volume. :)
Reply #42019-01-11
Those who don’t understand fluid flow problems generally also don’t understand Ohm’s law for electric current. In Ohm’s law, I=U/R; here, U is generally considered to be voltage, analogous to the pressure in fluids. In fact, junior high school physics deals with voltage, while in high school physics and certainly in university physics, this U is referred to as potential difference, not potential. Therefore, when applied to fluid flow, pressure is actually not useful; it is the difference between two pressures that matters. As for R, it represents the sum of the resistance of the wires, the resistance of the circuit components, and even the internal resistance of the power supply. When considering fluid flow, pipes and valves should also be taken into account comprehensively. In this question, the term “1.4 kg” is an amateurish way of expressing it; one needs to read books instead of relying solely on what experienced craftsmen say, otherwise the result will be something incomplete and unsatisfactory. kg*f: 0.14 – 0 = 0.14 Mpa in terms of pressure difference. If there is no flow-limiting element in between, only the resistance of the wires exists, which results in a short circuit; as a result, the current is very high and the air flow rate is also high. A value of 340 m/S is generally not achievable; it needs to be adjusted accordingly
Reply #52019-01-11
:The handshake is used as a metaphor; potential energy = F*S, where position = (P*S) * position. Kinetic energy = (mv^2)/2. As pressure decreases, speed also decreases; there are too many variables involved
Reply #62019-01-17
If we don’t even know the volume of the atmospheric pressure tank, how can we calculate 340 m/s? You can calculate it by checking the pressure pipeline manual
Reply #72019-01-18
The flow rate will be very high and difficult to control; considering the use of flow-limiting orifice plates. Generally, the required flow rate is specified first, and then the diameter of the flow control orifice plate is calculated based on that flow rate.
Reply #82019-01-19
Discussion on the precision of the question description: 1. \"1.4 kg of nitrogen\" is a unit for measuring the mass of the medium; using the density of nitrogen under normal temperature and pressure, its volume can be determined. 2. To measure flow rate, a time parameter is required, which is not provided in the question, so it is impossible to obtain the answer desired by the poster. 3. Since the time parameter is missing, the flow rate cannot be determined; therefore, no relationship can be established between the pipe specifications given in the question and the mass of the medium. 4. Given that the question is unclear and unresolvable, it would be interesting to know how the data provided by other users was obtained. Please provide a more precise description of the question
Reply #92019-01-19
This post was last edited by lupg on 2019-1-19 21:24. 1. Flow rate / pipe cross-sectional area = velocity; this can be understood by looking at the units: (m3/s) / m2 = m/s. 2. Pressure has no direct relationship with the other three factors. For compressible media such as gases, pressure is related to temperature and volume, which is referred to as the gas law. According to the ideal gas law, given five of the six parameters describing two states of a medium, it is possible to determine the sixth parameter. 3. The ideal gas law is expressed as: P1*V1/T1 = P2*V2/T2. 4. By understanding the gas law, it can be seen that pressure affects volume, and thus it also affects volumetric flow rate. Pressure establishes an indirect relationship with the other three factors. 5. For liquids, the volume change of the medium under pressure is minimal, so it can be ignored in calculations without affecting the flow velocity. 6. At the critical points of phase transition in a medium, changes in pressure cause a change in the phase of the medium – for example, from liquid to gas or vice versa – which has a significant impact on the flow velocity; the extent of this impact depends on the physical properties of the medium

Submit a Project

**Looking for Chemical Technology, Equipment & Solutions?** No Registration Required Broader Platform Exposure | Global Chemical Service Provider Connections

Submit Request — Free Consultation

Disclaimer

This is an automated machine translation of the original thread. Some technical terms may have inaccuracies; the original text shall prevail. Click "View Original" at the top right to access the source page, which supports IP-based automatic real-time language translation. Please watch out for contact details and sales inducements to prevent fraud. All content and translations are for reference only, representing solely the poster's personal views. For enquiries, email service@hcbbs.com.