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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°C and 9 bar, which is approximately 493 kg/m³.\[ \text{Mass} = \text{Density} \times \text{Volume} = 493 \times 100 \pi = 49300\pi \text{ kg} \]
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:\[ \text{Time in minutes} = \frac{4930\pi}{60} \text{ minutes} = 82.17 \pi \text{ minutes} ≈ 258.19 \text{ minutes} \]

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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°C and 9 bar), the density is approximately 493 kg/m³. 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: \( \text{Time in minutes} = \frac { 4930 \ pi}{60} = 82.17 \pi ≈ 258.19 \ \text{minutes}\).

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

Refrigerant 134 a enters a compressor operating at steady state as saturated vapor at \(0.12 \mathrm{MPa}\) and exits at \(1.2 \mathrm{MPa}\) and \(70^{\circ} \mathrm{C}\) at a mass flow rate of \(0.108 \mathrm{~kg} / \mathrm{s}\). As the refrigerant passes through the compressor, heat transfer to the surroundings occurs at a rate of \(0.32 \mathrm{~kJ} / \mathrm{s}\). Determine at steady state the power input to the compressor, in \(\mathrm{kW}\).

Air modeled as an ideal gas enters a combustion chamber at \(20 \mathrm{lbf} / \mathrm{in}^{2}\) and \(70^{\circ} \mathrm{F}\) through a rectangular duct, \(5 \mathrm{ft}\) by \(4 \mathrm{ft}\). If the mass flow rate of the air is \(830,000 \mathrm{lb} / \mathrm{h}\), determine the velocity, in \(\mathrm{ft} / \mathrm{s}\).

Air enters a compressor operating at steady state with a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) and a temperature of \(70^{\circ} \mathrm{F}\). The volumetric flow rate at the inlet is \(16.6 \mathrm{ft}^{3} / \mathrm{s}\), and the flow area is \(0.26 \mathrm{ft}^{2}\). At the exit, the pressure is \(35 \mathrm{lbf} / \mathrm{in}^{2}\), the temperature is \(280^{\circ} \mathrm{F}\), and the velocity is \(50 \mathrm{ft} / \mathrm{s}\). Heat transfer from the compressor to its surroundings is \(1.0 \mathrm{Btu}\) per \(\mathrm{lb}\) of air flowing. Potential energy effects are negligible, and the ideal gas model can be assumed for the air. Determine (a) the velocity of the air at the inlet, in \(\mathrm{ft} / \mathrm{s}\), (b) the mass flow rate, in \(\mathrm{lb} / \mathrm{s}\), and (c) the compressor power, in Btu/s and hp.

Refrigerant \(134 a\) enters an insulated compressor operating at steady state as saturated vapor at \(-20^{\circ} \mathrm{C}\) with a mass flow rate of \(1.2 \mathrm{~kg} / \mathrm{s}\). Refrigerant exits at 7 bar, \(70^{\circ} \mathrm{C}\). Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in \(\mathrm{m}^{3} / \mathrm{s}\), and (b) the power input to the compressor, in \(\mathrm{kW}\).

Air with a mass flow rate of \(2.3 \mathrm{~kg} / \mathrm{s}\) enters a horizontal nozzle operating at steady state at \(450 \mathrm{~K}, 350 \mathrm{kPa}\), and velocity of \(3 \mathrm{~m} / \mathrm{s}\). At the exit, the temperature is \(300 \mathrm{~K}\) and the velocity is \(460 \mathrm{~m} / \mathrm{s}\). Using the ideal gas model for air with constant \(c_{p}=1.011 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the area at the inlet, in \(\mathrm{m}^{2}\). (b) the heat transfer between the nozzle at its surroundings, in kW. Specify whether the heat transfer is to or from the air.

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