The calculation of the flow rate of the small hole, and the help for high scores, require a specific

It is known that the general air density is the density at 0 degrees Celsius and 1 standard atmosphere p0=; The radius of the wall hole r=1mm; The velocity of the gas at the junction between the hole and the interior v0=0;T=300K at room temperature, T0=273K corresponding to 0 degrees Celsius; g=；Internal pressure f'=3g(n), external pressure f=1g(n); p0 = corresponding to 1 standard atmosphere.

Let the velocity of the gas at the junction between the hole and the outside be V, the cross-sectional area of the small hole is S, and the internal and external pressures are P', p, the molar mass of the air is u, and the molar mass is n.

Solution: From the equation of state of an ideal gas: pv=nrt, and n=m u; So, p=[(m u) v]rt=(p u)rt, so, pt p=p'*t/p'=p0*t0/p0。

s=πrr=，p'=f'/s=3*，p=f/s=。

p'=p0*t0*p'/p0/t=，p=p0*t0*p/p0/t=p'/3=。

From Bernoulli's equation can be listed: p'+p'*v0 2 2=p+pvv 2, so, vv=2(p'-p)/p，v=(2*(。

Gas flow rate l=psv=.

You go and look at the Principles of Chemical Engineering, the chapter on fluid flow.

Summary. Hello dear, the application condition is that the rotational energy is conserved by the potential energy, and the exit velocity is obtained: v = root number (2gh) = root number (2 * 10 * 5) = 10 m seconds hole cross-sectional area:

square meter flow rate: square meter * 10 m second = cubic meter second The above does not consider the resistance, and the empirical estimate of the actual flow rate is square meter * meter second = cubic meter second.

The flow calculation formula for the free outflow of thin-walled small orifice can still be used for large orifice, and what are the application conditions.

Hello, I am helping you to inquire about the relevant information and will reply to you immediately.

Hello dear, the application condition is that the potential energy is conserved by the rotational energy, and the velocity from the volt to the outlet: v = root number (2gh) = root number (2 * 10 * 5) = 10 m seconds hole cross-sectional area: * square meters flow:

square meters * 10 meters seconds = cubic meters seconds In the morning, without considering the resistance, the actual flow rate is estimated to be square meters * meters seconds = cubic meters seconds.

Equivalent pressure at the outlet position:

2500000+PGH=2500000+617*10*6=2537020 Pa.

Ejection velocity (excluding friction.)

Effect on flow velocity):

i.e. vv=2*2537020 617

v = meter seconds.

The total amount of liquid ammonia flowing out of the pore within 10 minutes is not counted, for the following reasons:

1) The equivalent outlet diameter is not 50mm, but less than 50mm. It is caused by friction, and it is not known for sure.

2) It is not clear whether the small mouth is directly opened on the tank or the inner diameter of a steel pipe connected to the source cover, which has a significant effect on the actual flow rate and the equivalent hall crack and straight dressing path.

3) The outflow velocity in the larger process decreases gradually, which can be calculated integrally.

This can only be used as a reference, waiting for the professionals.

Known: The density of the air generally employed.

It means atmospheric pressure at 0 degrees Celsius and 1 standard.

density p0=; The radius of the wall hole r=1mm; The velocity of the gas at the junction between the hole and the interior v0=0;T=300K at room temperature, T0=273K corresponding to 0 degrees Celsius; g=；Internal pressure f'=3g(n), external pressure f=1g(n); p0 = corresponding to 1 standard atmosphere.

Let the velocity of the gas at the junction between the hole and the outside be V, the cross-sectional area of the small hole is S, and the internal and external pressures are P', p, the molar mass of the air.

is u, and the molar amount is n

by the equation of state of an ideal gas.

Column: pv=nrt, and n=m u; So, p=[(m u) v]rt=(p u)rt, so, pt p=p'*t/p'=p0*t0/p0.

s=πrr=,p'=f'/s=3*,p=f/s=

p'=p0*t0*p'p0 t=mmm), p=p0*t0*p p0 t=p'/3=

Perturbation equation by Bernu.

Column: p'+p'*v0 2 2=p+pvv 2, so, vv=2(p'-p)/p,v=(2*(

Gas flow rate l=psv=.