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Chapter 4: Problem 12

Liquid propane enters an initially empty cylindrical storage tank at a mass flow rate of \(10 \mathrm{~kg} / \mathrm{s}\). Flow continues until the tank is filled with propane at \(20^{\circ} \mathrm{C}, 9\) bar. The tank is \(25 \mathrm{~m}\) long and has a \(4-\mathrm{m}\) diameter. Determine the time, in minutes, to fill the tank.

Short Answer

Expert verified
It takes approximately 258.19 minutes to fill the tank.

Step by step solution

01

- Calculate the Volume of the Tank

To find the volume of the cylindrical tank, use the formula for the volume of a cylinder: \[ V = \frac{\text{Base Area} \times \text{Height}}\]First, calculate the base area of the cylinder:\[ \text{Base Area} = \text{Area of a Circle} = \pi r^{2} \]Given the diameter is 4 meters, the radius is \[ r = \frac{4}{2} = 2 \text{ meters} \] Therefore:\[ \text{Base Area} = \pi \times (2)^2 = 4\pi \text{ square meters} \] Now multiply by the height (length) of the cylinder, which is 25 meters:\[ V = 4 \pi \times 25 = 100\pi \text{ cubic meters} \]Hence, the volume of the tank is \[ 100\pi \text{ cubic meters} \]
02

- Convert Volume to Mass

To determine the mass of liquid propane needed to fill the tank, use the density of propane at given conditions. We need the density of propane at 20掳C and 9 bar, which is approximately 493 kg/m鲁.\[ \text{Mass} = \text{Density} \times \text{Volume} = 493 \times 100 \pi = 49300\pi \text{ kg} \]
03

- Determine the Time to Fill the Tank

The mass flow rate of liquid propane is given as 10 kg/s.Using the formula for time: \[ \text{Time} = \frac{\text{Mass}}{\text{Flow Rate}} \texpressed as\ Time = \frac{49300 \pi}{10} = 4930\pi \text{seconds} \] Convert the time from seconds to minutes. Since there are 60 seconds in a minute:\[ \text{Time in minutes} = \frac{4930\pi}{60} \text{ minutes} = 82.17 \pi \text{ minutes} 鈮� 258.19 \text{ minutes} \]

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Key Concepts

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

cylindrical volume calculation
Understanding how to calculate the volume of a cylinder is vital in many thermodynamic problems. To start, we need to know the formula, which is given by the base area of the cylinder multiplied by its height. In this exercise, the base is a circle, so its area is calculated using \(\text{Base Area} = \pi r^{2}\) where \( r \) is the radius.

Given the diameter is 4 meters, we can find the radius by dividing the diameter by 2, resulting in \(\text{radius} = 2\text{ meters}\). Next, we can calculate the base area: \( \text{Base Area} = \pi \times (2)^{2} = 4\pi \ text {square meters}\).

Finally, the volume of the cylinder is computed by multiplying this base area by the height (or length) of the tank, which is given as 25 meters: \(V = 4\pi \times 25 = 100 \ pi \ text {cubic meters}\). Thus, the total volume of the tank is \(100\ \ pi \text{ cubic meters})\).
mass flow rate
The mass flow rate is a crucial concept because it tells us how much mass is moving through a given point per unit time. In our example, the mass flow rate is given as 10 kg/s. This means that each second, 10 kilograms of propane enter the tank.

Using the mass flow rate allows us to easily determine the total amount of mass that has entered the tank over a period. Since 10 kg of propane flows each second, knowing the total mass required will help us calculate the time needed to fill the tank, which we discuss in later sections.
density of propane
Density is the mass per unit volume of a substance. For propane at given conditions (20掳C and 9 bar), the density is approximately 493 kg/m鲁. To calculate the mass of propane required to fill a tank, we use the formula: \( \text{Mass} = \text{Density} \times \text{Volume} = 493 \times 100 \pi = 49300\ \ pi \ text{ kg}\).

This means that the propane needed to fill the tank is about 49300\ \ pi \ kilograms. Density thus serves as a bridge between volume and mass, helping us understand material properties and behavior under specified conditions.
time calculation in thermodynamics
Calculating time in thermodynamics problems often means determining how long a process will take. Here, we use the mass flow rate and total mass to find the time required to fill the tank. The formula is: \(\text{Time} = \frac{\text{Mass}}{\text{Flow Rate}}\)

Substituting the values, we get: \( \frac{49300 \ pi}{10 } \ = 4930\text{pi}\ \text{ seconds}\). To convert seconds into minutes, since there are 60 seconds in a minute, we divide: \( \text{Time in minutes} = \frac { 4930 \ pi}{60} = 82.17 \pi 鈮� 258.19 \ \text{minutes}\).

Thus, it will take approximately 258.19 minutes to fill the tank with propane. This step-by-step approach helps ensure a comprehensive understanding of the time calculation process.

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Most popular questions from this chapter

Air enters a diffuser operating at steady state at \(540^{\circ} \mathrm{R}\), 15 lbf/in. \({ }^{2}\), with a velocity of \(600 \mathrm{ft} / \mathrm{s}\), and exits with a velocity of \(60 \mathrm{ft} / \mathrm{s}\). The ratio of the exit area to the inlet area is 8 . Assuming the ideal gas model for the air and ignoring heat transfer, determine the temperature, in \({ }^{\circ} \mathrm{R}\), and pressure, in lbf/in. \({ }^{2}\), at the exit.

Air enters a compressor operating at steady state at \(1.05\) bar, \(300 \mathrm{~K}\), with a volumetric flow rate of \(12 \mathrm{~m}^{3} / \mathrm{min}\) and exits at 12 bar, \(400 \mathrm{~K}\). Heat transfer occurs at a rate of \(2 \mathrm{~kW}\) from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in \(\mathrm{kW}\).

A rigid tank whose volume is \(10 \mathrm{~L}\) is initially evacuated. A pinhole develops in the wall, and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) enters until the pressure in the tank becomes 1 bar. No significant heat transfer between the contents of the tank and the surroundings occurs. Assuming the ideal gas model with \(k=1.4\) for the air, determine (a) the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\), and (b) the amount of air that leaks into the tank, in \(g\).

Steam enters a counterflow heat exchanger operating at steady state at \(0.07 \mathrm{MPa}\) with a specific enthalpy of \(2431.6 \mathrm{~kJ} / \mathrm{kg}\) and exits at the same pressure as saturated liquid. The steam mass flow rate is \(1.5 \mathrm{~kg} / \mathrm{min}\). A separate stream of air with a mass flow rate of \(100 \mathrm{~kg} / \mathrm{min}\) enters at \(30^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). The ideal gas model with \(c_{p}=\) \(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) can be assumed for air. Kinetic and potential energy effects are negligible. Determine (a) the quality of the entering steam and (b) the rate of heat transfer between the heat exchanger and its surroundings, in \(\mathrm{kW}\).

Figure P4.101 shows a pumped-hydro energy storage system delivering water at steady state from a lower reservoir to an upper reservoir using off-peak electricity (see Sec. 4.8.3). Water is delivered to the upper reservoir at a volumetric flow rate of \(150 \mathrm{~m}^{3} / \mathrm{s}\) with an increase in elevation of \(20 \mathrm{~m}\). There is no significant change in temperature, pressure, or kinetic energy from inlet to exit. Heat transfer from the pump to its surroundings occurs at a rate of \(0.6 \mathrm{MW}\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the pump power required, in MW. Assuming the same volumetric flow rate when the system generates on-peak electricity using this water, will the power be greater, less, or the same as the pump power? Explain.
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