Question

Steam is leaving a pressure cooker whose operating pressure is 20 psia. It is observed that the amount of liquid in the cooker has decreased by 0.6 gal in 45 minutes after the steady operating conditions are established, and the cross-sectional area of the exit opening is 0.15 in \(^{2}\). Determine \((a)\) the mass flow rate of the steam and the exit velocity, (b) the total and flow energies of the steam per unit mass, and (c) the rate at which energy is leaving the cooker by steam.

Short answer

Based on the given information and calculations: (a) The mass flow rate of the steam is 0.113 lb/min, and the exit velocity is 20.49 in/min. (b) The total energy of the steam per unit mass is 3230 Btu/lb, and the flow energy per unit mass is 2160 Btu/lb. (c) The rate at which energy is leaving the cooker by steam is 364.99 Btu/min.

Step-by-step solution

Calculate the volumetric flow rate of liquid in the pressure cooker

First, we need to convert 0.6 gal to cubic inches: 1 gal = 231 in\(^3\) 0.6 gal = 0.6 * 231 in\(^3\) = 138.6 in\(^3\) Now, we will calculate the decrease in liquid volume per minute: Decrease in liquid volume per minute (Q) = Decrease in volume / time Q = 138.6 in\(^3\) / 45 min = 3.08 in\(^3\)/min

Determine the mass flow rate of the steam

We know that mass flow rate (m) is equal to the volumetric flow rate (Q) multiplied by the density of the substance (ρ). In this case, we are given the density of steam (ρ) as 0.0367 lb/in\(^3\). The mass flow rate of the steam (m) = Q * ρ m = 3.08 in\(^3\)/min * 0.0367 lb/in\(^3\) = 0.113 lb/min

Calculate the exit velocity of the steam

Now, we can use the mass flow rate and the cross-sectional area of the exit opening (A) to calculate the exit velocity of the steam (V). Exit velocity (V) = Mass flow rate (m) / (Density (ρ) * Cross-sectional area (A)) V = 0.113 lb/min / (0.0367 lb/in\(^3\) * 0.15 in\(^2\)) = 20.49 in/min

Calculate the internal energy and flow work per unit mass

To find the total and flow energies per unit mass, we need to determine the internal energy (U) and flow work (W\(_\text{flow}\)) per unit mass. Internal energy per unit mass (U): We know that internal energy is proportional to temperature. We are not given the temperature in the problem, but since it's a pressure cooker at 20 psia, we can assume it's saturated steam at 20 psia. The internally energy per unit mass of saturated steam at 20 psia is approximately 1070 Btu/lb. Flow work per unit mass (W\(_\text{flow}\)): Flow work is the work done to push the fluid by the pressure difference (P). The flow work per unit mass is given by P * specific volume (v). The specific volume of saturated steam at 20 psia is approximately 7.5 ft\(^3\)/lb. W\(_\text{flow}\) = P * v W\(_\text{flow}\) = 20 * 144 lbf/ft\(^2\) * 7.5 ft\(^3\)/lb = 2160 lbf.ft/lb = 2160 Btu/lb Note: We multiplied by 144 to convert psi to lbf/ft\(^2\)

Determine the total and flow energies per unit mass

Now, we can find the total energy per unit mass (E\(_\text{total}\)) and flow energy per unit mass (E\(_\text{flow}\)). Total energy per unit mass (E\(_\text{total}\)) = Internal energy (U) + Flow work (W\(_\text{flow}\)) E\(_\text{total}\) = 1070 Btu/lb + 2160 Btu/lb = 3230 Btu/lb Flow energy per unit mass (E\(_\text{flow}\)) = Flow work (W\(_\text{flow}\)) E\(_\text{flow}\) = 2160 Btu/lb

Calculate the rate at which energy is leaving the cooker by steam

Finally, we can calculate the rate at which energy is leaving the cooker by steam. Energy leaving the cooker by steam = Mass flow rate (m) * Total energy per unit mass (E\(_\text{total}\)) Energy leaving the cooker = 0.113 lb/min * 3230 Btu/lb = 364.99 Btu/min The results are: (a) The mass flow rate of the steam is 0.113 lb/min, and the exit velocity is 20.49 in/min. (b) The total energy of the steam per unit mass is 3230 Btu/lb, and the flow energy per unit mass is 2160 Btu/lb. (c) The rate at which energy is leaving the cooker by steam is 364.99 Btu/min.

Key concepts

Steam Properties

Understanding steam properties is crucial when studying thermodynamics, particularly when dealing with steam systems such as the pressure cooker mentioned in our exercise. Steam can exist in different phases: as water (liquid), steam (vapor), or a mixture of both, where its behavior and properties vary significantly.

The quality of steam, which refers to the ratio of vapor to liquid in a mixture, is determined by pressure and temperature. In the problem, the conditions within the cooker are set to be 20 psia, which suggests it's producing saturated steam, an important term in steam properties that indicates a balance where water and vapor coexist at a particular pressure and temperature.

Key properties that are essential in this exercise include the steam's density and specific volume. Density is the mass per unit volume, which helps to determine mass flow rate when multiplied by volumetric flow rate. Specific volume, conversely, is the volume occupied by a unit mass of steam and is instrumental in calculating flow work. When dealing with energy calculations in a steam system, being familiar with these properties makes it possible to confidently engage with the exercise and find meaningful results through the established relationships between pressure, volume, temperature, and energy.

Energy Conservation

The principle of energy conservation underpins much of thermodynamics and plays a central role in the solution for our pressure cooker scenario. Energy conservation, simply put, states that energy can neither be created nor destroyed, only transformed from one form to another.

In the context of the exercise, this principle allows us to track the energy as it is carried away by the steam exiting the pressure cooker. By calculating the internal energy and flow work, we account for the total energy per unit mass that the steam possesses. Internal energy relates to the molecular energy within the steam, primarily influenced by the temperature, while flow work is the energy necessary to move the steam out of the cooker, which is a product of pressure and specific volume.

The converted energy—stored in the steam and calculated per unit mass—is then easy to manage when multiplied by the mass flow rate, giving us the rate of energy leaving the system. By adhering to energy conservation, we ensure that the results obtained are consistent with the laws of physics, providing reliable conclusions and deeper understanding for solving such thermal-related problems.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. This discipline is the foundation upon which the exercise is based, and its principles allow us to analyze and predict the behavior of systems involving heat transfer and work.

One of the primary laws of thermodynamics that we apply in the provided problem is the first law, or the law of energy conservation. When we calculate the mass flow rate, exit velocity, and energy rates, we are essentially using thermodynamic principles to track how energy is transformed and transported within the steam system of the pressure cooker.

An understanding of these concepts enables the clear interpretation of how energy is converted from the heat supplied into the work done by the steam and how it carries away internal energy. By applying thermodynamics to the exercise, students can understand these processes beyond memorization, by actually engaging with the natural laws that govern energy interactions in practical, everyday systems like a pressure cooker.