Question

A pressure cooker cooks a lot faster than an ordinary pan by maintaining a higher pressure and temperature inside. The lid of a pressure cooker is well sealed, and steam can escape only through an opening in the middle of the lid. A separate metal piece, the petcock, sits on top of this opening and prevents steam from escaping until the pressure force overcomes the weight of the petcock. The periodic escape of the steam in this manner prevents any potentially dangerous pressure buildup and keeps the pressure inside at a constant value. Determine the mass of the petcock of a pressure cooker whose operation pressure is 100 kPa gage and has an opening cross-sectional area of \(4 \mathrm{mm}^{2}\). Assume an atmospheric pressure of \(101 \mathrm{kPa}\), and draw the free-body diagram of the petcock.

Short answer

Answer: The mass of the petcock of the pressure cooker is 0.082 kg.

Step-by-step solution

Convert the given pressure and area values into SI units

The operation pressure is given as 100 kPa gage, and the atmospheric pressure is 101 kPa. Convert them to SI units, by multiplying them with \(10^3\) (since 1 kPa = 1000 Pa). - Operation pressure: \(100 \times 10^3 \ \mathrm{Pa}\) - Atmospheric pressure: \(101 \times 10^3 \ \mathrm{Pa}\) The cross-sectional area of the opening is given as \(4 \mathrm{mm}^{2}\). Convert this to SI units as well, by multiplying with \(10^{-6}\) (since 1 \(\mathrm{mm}^{2}\) = \(1 \times 10^{-6} \mathrm{m}^{2}\)). - Cross-sectional area: \(4 \times 10^{-6} \ \mathrm{m}^{2}\)

Calculate the absolute pressure inside the pressure cooker

The operation pressure is given as gage pressure, which means it is the difference between the absolute pressure inside the cooker and the atmospheric pressure. Find the absolute pressure inside the cooker by adding the gage pressure to the atmospheric pressure: \(P_{\text{abs}} = P_{\text{gage}} + P_{\text{atm}} = 100 \times 10^3 \ \mathrm{Pa} + 101 \times 10^3 \ \mathrm{Pa} = 201 \times 10^3 \ \mathrm{Pa}\)

Determine the force exerted on the petcock due to the pressure inside the pressure cooker

The force exerted on the petcock can be calculated as the product of the absolute pressure inside the pressure cooker and the opening cross-sectional area: \(F_{\text{pressure}} = P_{\text{abs}} \times A = 201 \times 10^3 \ \mathrm{Pa} \times 4 \times 10^{-6} \ \mathrm{m}^{2} = 0.804 \ \mathrm{N}\)

Calculate the mass of the petcock using the gravitational force

The force exerted by the petcock's weight can be calculated as the product of the mass of the petcock and the acceleration due to gravity (\(g = 9.81 \ \mathrm{m/s^2}\)). Since the pressure force and the weight of the petcock are balanced, the force exerted by the weight of the petcock equals the force exerted by the pressure inside the pressure cooker: \(F_{\text{weight}} = m_{\text{petcock}} \times g\) Divide both sides by the acceleration due to gravity to find the mass of the petcock: \(m_{\text{petcock}} = \frac{F_{\text{weight}}}{g} = \frac{0.804 \ \mathrm{N}}{9.81 \ \mathrm{m/s^2}} = 0.082 \ \mathrm{kg}\)

Conclude the answer and draw the free-body diagram of the petcock

The mass of the petcock of the pressure cooker is \(0.082 \ \mathrm{kg}\). The free-body diagram of the petcock will have two forces acting on it: the downward gravitational force (weight) and the upward force due to the pressure inside the pressure cooker, both of which balance each other.

Key concepts

Pressure Gauge

When dealing with pressure in various systems such as a pressure cooker, a pressure gauge is an indispensable tool. It measures the pressure within a container or system and displays it, often relative to atmospheric pressure, which leads to what we term as gauge pressure.

In the case of the pressure cooker, the gauge pressure is given as 100 kPa gage. To comprehend what this means, imagine a hypothetical dial on the pressure cooker that points to '100' when it's at operating pressure. This measurement is actually signifying that the pressure inside the cooker is 100 kPa higher than the outside atmospheric pressure. It's crucial to convert this to absolute pressure to perform any calculations related to force or mass, which is why one needs to add atmospheric pressure to gauge pressure to get the absolute pressure inside the cooker.

Absolute Pressure

Understanding absolute pressure is essential in thermodynamics and fluid mechanics. Absolute pressure is the total pressure exerted within a system, including the atmospheric pressure, if the system is not sealed off from the atmosphere. For applications like our pressure cooker, we equate absolute pressure to the sum of the gauge pressure and atmospheric pressure.

In mathematical terms, if the gauge pressure is the additional pressure within the system and atmospheric pressure is the ambient pressure experienced at sea level, then the formula is: \blockquote{ \[P_{\text{abs}} = P_{\text{gage}} + P_{\text{atm}}\] }To illustrate, in our exercise, the gauge pressure is 100 kPa, and the atmospheric pressure is 101 kPa, making the absolute pressure in the cooker \[201 \times 10^3 \mathrm{Pa}\]. It's essential to present pressure in SI units to align with international standards and ensure clarity in communication.

Free-Body Diagram

A free-body diagram is a simple graphical illustration used to show the relative magnitude and direction of all forces acting upon a single object in a given situation. It's a staple in physics and engineering problem-solving, as it helps to visualize and solve the balance of forces. In our pressure cooker scenario, to draw the free-body diagram for the petcock, one would depict the petcock as a dot or a small box and draw arrows representing the forces acting on it.

An upward arrow representing the force due to the internal pressure (\[F_{\text{pressure}}\]) would be equal and opposite to a downward arrow showing the force of gravity, which is the product of the petcock's mass and the acceleration due to gravity (\[F_{\text{weight}} = m_{\text{petcock}} \times g\]). These diagrams are immensely useful as they allow us to apply Newton's laws of motion to analyze the equilibrium of the object in question.