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Chapter 4: Problem 71

The cylindrical pressure vessel contains methane at an initial absolute pressure of \(2 \mathrm{MPa}\). If the nozzle is opened, the mass flow depends upon the absolute pressure and is \(\dot{m}=3.5\left(10^{-6}\right) p \mathrm{~kg} / \mathrm{s},\) where \(p\) is in pascals. Assuming the temperature remains constant at \(20^{\circ} \mathrm{C}\), determine the time required for the pressure to drop to \(1.5 \mathrm{MPa}\).

Short Answer

Expert verified
The time required for the pressure to drop to 1.5 MPa can be calculated using the formula \(t = t2 - t1 = (V/3.5 x 10^{-6}R) * ln(P1/P2)\)

Step by step solution

01

Define Constants

Define the given constants for this problem: initial pressure \(\(P_1 = \(2 MPa\) or \(2 x 10^6 Pa\), final pressure \(P_2 = 1.5 MPa\) or \(1.5 x 10^6 Pa\), and the rate of mass flow \(\(\dot{m} = f(P) = 3.5 x 10^{-6} P\). Also, note that the unit of mass flow is in kg/s and pressure in Pascals.
02

Formulate the Integral Equation

We are looking for the time needed for the pressure to drop from \(P_1\) to \(P_2\). Since \(\(\dot{m} = f(P)\), using this in the elemental time \(\(dt\) results in \(\(dm = f(P).dt\). Since \(P = \(\frac{mRT}{V}\) (from ideal gas equation where V=constant, R= gas constant, T= temperature), we have \(dm = (3.5 x 10^{-6} P).dt => dm = (3.5 x 10^{-6} mRT/V).dt\). This reduces to \(\(dm/m = (3.5 x 10^{-6} RT/V).dt\). Integrate both sides, \(\(\int_{m1}^{m2} \frac{dm}{m} = \int_{t1}^{t2} (3.5 x 10^{-6} RT/V).dt\), where \(m = (P_1V)/(RT)\) to \(m = (P_2V)/(RT)\), and \(t = t1\) to \(t = t2\).
03

Solve the Integral

Perform the integration to obtain: \(\ln(m2/m1) = (3.5 x 10^{-6} RT/V)(t_2 - t_1)\). Substituting \(m = (PV)/(RT)\) and solving for \(t = t2 - t1\) will give the time it requires for the pressure to drop to 1.5 MPa

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Vessels
Pressure vessels are containers designed to hold gases or liquids at a pressure substantially different from the ambient pressure. In fluid mechanics, these vessels play a significant role, especially when dealing with gases like methane under high pressure, as mentioned in our exercise with the cylindrical pressure vessel.
Pressure vessels must be constructed with care to withstand the stress caused by the pressure of the gas or liquid inside. This stress can lead to deformation or even rupture if not properly managed.
  • Design: Vessels are usually cylindrical, like the one in our example, because this shape can efficiently contain high-pressure gases.
  • Materials: Typically made from metal, pressure vessels require strong, durable materials that can endure the internal forces without failing.
  • Safety: Ensuring the vessel's integrity involves regular inspections and sometimes the presence of safety valves to manage unexpected pressure surges.
Understanding these principles can help navigate problems involving pressure vessels with more confidence.
Mass Flow Rate
The mass flow rate is a measure of the mass of fluid passing through a point or area per unit time, typically expressed in kilograms per second (kg/s). In the context of our exercise, we consider how the mass flow rate \(\dot{m}\) depends on the pressure inside the vessel.

In this instance, the mass flow rate is given by the equation \(\dot{m} = 3.5 \times 10^{-6} P\), with pressure \(P\) influencing how quickly the fluid exits the pressure vessel when the nozzle is opened.
  • Variable Dependency: Note that the mass flow rate changes with pressure 鈥� the higher the pressure, the greater the mass flow rate.
  • Impact: This direct relationship means as the pressure drops, the mass flow rate will reduce, ultimately affecting the time it takes for the pressure to decline to a specified level.
Mastering the connection between pressure and mass flow rate is crucial for controlling the discharge of gases from pressure vessels.
Ideal Gas Law
The Ideal Gas Law is foundational in understanding the behavior of gases under various conditions, particularly in pressure vessels. It is expressed as \(PV = nRT\), linking pressure \(P\), volume \(V\), the amount of substance in moles \(n\), the gas constant \(R\), and temperature \(T\).

For our pressure vessel scenario, this equation helps in relating the pressure within the tank to the changes as gas escapes.
  • Simplified Use: Here, it is applied as \(P = \frac{mRT}{V}\) by substituting the mass of gas for moles, visualizing how pressure depends on the mass \(m\) and temperature.
  • Assumptions: Assumes that the gas behaves ideally, with no interactions between molecules which is reasonably accurate under many conditions but less so at very high pressures or low temperatures.
  • Relevance: This relationship is critical for modeling how pressure decreases in the vessel as gas exits, using it to calculate the time necessary for certain pressure drops.
Ultimately, comprehending the Ideal Gas Law aids in predicting how changes in pressure relate to other variables like mass and temperature.
Pressure Drop Calculation
Calculating the pressure drop within a pressure vessel involves understanding how the mass flow rate, governed by the aforementioned relationship with pressure, depletes the mass of the gas over time.

In our exercise, it's essential to integrate over time to find how quickly the pressure decreases from an initial level \(P_1\) to a final level \(P_2\).
  • Integration: The relationship between mass flow rate and pressure is embedded in the integral \(\int_{m1}^{m2} \frac{dm}{m} = (3.5 \times 10^{-6} \frac{RT}{V}) \int_{t1}^{t2} dt\), which ultimately determines the time span for the pressure drop.
  • Rearrangement: By solving this integral, we connect the change in gas mass to elapsed time, linking directly to the watched pressure drop \(\Delta P = P_2 - P_1\).
  • Outcome: This enables the calculation of the specific time required for the system to reach a desired lower pressure, taking into account constants like temperature and the gas constant.
Effectively calculating pressure drops is crucial for designing systems that ensure safe, efficient depressurization in various applications involving gases.

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Most popular questions from this chapter

The unsteady flow of linseed oil is such that at \(A\) it has a velocity of \(V_{A}=\left(0.4 t^{2}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds. Determine the acceleration of a fluid particle located at \(x=0.25 \mathrm{~m}\) when \(t=2 \mathrm{~s}\).

The 2 -m-diameter cylindrical emulsion tank is being filled at \(A\) with cyclohexanol at an average rate of \(V_{A}=4 \mathrm{~m} / \mathrm{s}\) and at \(B\) with thiophene at an average rate of \(V_{B}=2 \mathrm{~m} / \mathrm{s}\). Determine the rate at which the depth of the mixture is increasing when \(h=1 \mathrm{~m} .\) Also, what is the average density of the mixture? Take \(\rho_{c y}=779 \mathrm{~kg} / \mathrm{m}^{3},\) and \(\rho_{t}=1051 \mathrm{~kg} / \mathrm{m}^{3}\)

In a mixing chamber, Carbon dioxide flows into the tank at \(A\) at \(V_{A}=12 \mathrm{~m} / \mathrm{s}\), and nitrogen flows in at \(B\) at \(V_{B}=6 \mathrm{~m} / \mathrm{s}\). Both enter at a gage pressure of \(310 \mathrm{kPa}\) and a temperature of \(260^{\circ} \mathrm{C}\). Determine the average velocity of the mixed gas at \(C\). The mixture has a density of \(\rho=1.546 \mathrm{~kg} / \mathrm{m}^{3}\).

A jet cnginc takes intake of air at \(30 \mathrm{~kg} / \mathrm{s}\) and jet fuel at \(0.25 \mathrm{~kg} / \mathrm{s}\). If the density of the expelled air- fuel mixture is \(1.4 \mathrm{~kg} / \mathrm{m}^{3},\) determine the average velocity of the exhaust relative to the plane. The exhaust nozzle has a diameter of \(0.3 \mathrm{~m}\).

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