Pump compresses air into a tank

In summary, the conversation discusses the calculation of the distance at which air will start entering a big tank, given its dimensions and pressure. The problem is solved using the adiabatic process equation and the ideal gas law. The final result is that the air will start entering the tank at a distance of 7.9 cm. Alternative methods, such as using logarithms, are also mentioned.
  • #1
Karol
1,380
22

Homework Statement


A cylindrical pump of length 25[cm] pumps air in atmoshpheric pressure ant temp 270C into a big tank. the manometric pressure in the tank is 4[atm]. at which distance will the air start entering the tank.

Homework Equations


In adiabatic process: ##P_1V_1^\gamma=P_2V_2^\gamma##
γ for air=1.4

The Attempt at a Solution


I assume an adiabatic process. i think that manometric pressure is gauge pressure:
$$P_1V_1^\gamma=P_2V_2^\gamma\rightarrow V_2=\sqrt[\gamma]{\frac{P_1}{P_2}V^{\gamma}}$$
$$\rightarrow \ln V_2=\ln\left(\frac{P_1}{P_2}V_1^{\gamma}\right)^{\frac{1}{\gamma}}=\frac{1}{\gamma}\ln\left(\frac{P_1}{P_2}V_1^{\gamma}\right)$$
$$\ln V_2=\frac{1}{\gamma}\left[\ln\left(P_1V_1^{\gamma}\right)-\ln P_2\right]=\frac{1}{\gamma}\left[\ln P_1+\gamma\ln V_1-\ln P_2\right]$$
The volume is area A times length x. for our data:
$$\ln V_2=\ln A+\ln x=\frac{1}{1.4}\left[\ln 0+1.4\ln (A\cdot 25)-\ln 5\right]=\frac{1}{1.4}\left[1.4(\ln A+\ln 25)-\ln 5\right]$$
$$\ln A+\ln x=\ln A+\ln 25-\ln 5\rightarrow \ln x=\ln\left(\frac{25}{5}\right)=\ln 5\rightarrow x=5[cm]$$
It should be 7.9[cm] before the end of the cylinder.
If i don't assume adiabatic process and only use ##PV=nRT## i have 2 unknowns: the distance and the final temperature.
 
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  • #2
You forgot to divide ln(P2) by 1.4.
Also, P1 = 1, so you shouldn't have ln(0) in the equation.
Also, it was not necessary to work this problem in terms of natural logs.

Chet
 
  • #3
$$\ln A+\ln x=\ln A+\ln 25-\frac{\ln 5}{1.4}\rightarrow x=7.9[cm]$$
Which other way besides natural logs?
 
  • #4
Karol said:
$$\ln A+\ln x=\ln A+\ln 25-\frac{\ln 5}{1.4}\rightarrow x=7.9[cm]$$
Which other way besides natural logs?
[tex]\frac{V_2}{V_1}=\left(\frac{P_1}{P_2}\right)^{1/\gamma}[/tex]
[tex]\frac{x}{25}=0.2^{(1/\gamma)}[/tex]
Chet
 
  • #5
Thanks Chet
 

Related to Pump compresses air into a tank

1. How does a pump compress air into a tank?

A pump compresses air into a tank by using a motor or engine to power a piston or impeller, which forces air into a confined space within the tank. This increases the air pressure, allowing more air to be contained within the tank.

2. What types of pumps are commonly used to compress air into a tank?

The most common types of pumps used to compress air into a tank include reciprocating pumps, rotary pumps, and centrifugal pumps. Each type uses different mechanisms to compress and push air into the tank.

3. Can a pump compress air into a tank indefinitely?

No, a pump has a limited capacity and can only compress air into a tank until the tank reaches its maximum pressure. The pump will continue to run to maintain the pressure, but it will not be able to compress any more air into the tank.

4. What is the purpose of compressing air into a tank?

Compressing air into a tank allows it to be stored for later use. The high pressure air can then be released to power tools, inflate tires, or run pneumatic machinery. It also allows for more efficient use of space as compressed air takes up less volume than uncompressed air.

5. Are there any safety concerns with using a pump to compress air into a tank?

Yes, there are some safety concerns to be aware of when using a pump to compress air into a tank. Over-pressurization of the tank can cause it to rupture, resulting in potential injury or property damage. It is important to follow proper safety procedures and regularly inspect the tank and pump to ensure safe operation.

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