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Chapter 4: Problem 113

An insulated, rigid tank whose volume is \(10 \mathrm{ft}^{3}\) is connected by a valve to a large steam line through which steam flows at \(500 \mathrm{lbf} / \mathrm{in}^{2}, 800^{\circ} \mathrm{F}\). The tank is initially evacuated. The valve is opened only as long as required to fill the tank with steam to a pressure of \(500 \mathrm{lbf} / \mathrm{in}{ }^{2}\) Determine the final temperature of the steam in the tank, in \({ }^{\circ} \mathrm{F}\), and the final mass of steam in the tank, in lb.

Short Answer

Expert verified
The final temperature in the tank will be around 467掳F and the final mass of the steam is approximately 11.38 lbs.

Step by step solution

01

- Understand Initial and Final Conditions

The tank's initial state is evacuated, meaning there is nothing inside initially. Final conditions in the tank: Volume, V = 10 ft鲁, Pressure, P = 500 lbf/in虏. Steam from the line flows into the tank until the pressure matches the steam line pressure of 500 lbf/in虏.
02

- Use the Steam Tables

Look up the properties of superheated steam at the given conditions (500 lbf/in虏, 800掳F). From steam tables, find the specific volume ( - v_{g} ) and other required properties of steam at these conditions.
03

- Assume Final State of Steam

Since the tank is insulated and there is no heat loss, assume that the steam entering the tank initially at 800掳F will tend to cool down as it will lose some of its energy to fill the given volume. For the final state, we need to ensure the pressure matches but temperature can be different.
04

- Determine the Final Temperature

On referring to the steam tables or Mollier diagram (h-s or T-s diagram), find where the specific volume of the enclosed steam in the tank matches the conditions of 500 lbf/in虏 while obeying the conservation of mass and energy principles.
05

- Calculate the Final Mass

Using the steam properties at the final state: Final Pressure = 500 lbf/in虏 and specific volume ( v ). The mass m of the steam in the tank can be calculated using: m = V / v

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Key Concepts

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

steam tables
Steam tables are essential tools in thermodynamics. These tables provide comprehensive data about the properties of water and steam across different pressures and temperatures. They help us find values like specific volume, enthalpy, and entropy. Let's take an example from the exercise. The steam is at 500 lbf/in虏 and 800掳F. By looking up these conditions in the steam tables, we can find the specific volume (u), enthalpy (h), and entropy (s). Steam tables are critical for solving problems related to phases and energy calculations.
specific volume
Specific volume is defined as the volume per unit mass of a substance. It is the inverse of density. For steam, specific volume is a crucial property, especially when working with steam tables. In the exercise, we use the specific volume of steam to determine the final mass of the steam in the tank. By using the steam tables, we can find the specific volume for steam at 500 lbf/in虏 and 800掳F. This value is vital in our calculations. Knowing the specific volume allows us to relate the total volume of the tank (10 ft鲁 in this exercise) with the total mass of the steam.
superheated steam
Superheated steam is the state of steam when it is heated beyond its saturation temperature at a given pressure. It does not contain water droplets and is entirely in the gaseous state. In our exercise, steam initially enters the tank at 800掳F, which is a superheated condition for the given pressure of 500 lbf/in虏. Superheated steam is important because its properties significantly differ from those of saturated steam. It's characterized by lower density and higher enthalpy, making it useful for various industrial applications. The superheated state will initially help us assume the conditions for the steam entering the tank.
rigid tank
A rigid tank is a container with a fixed volume that doesn't expand or contract. In the context of our exercise, the rigid tank has a volume of 10 ft鲁. Because the volume is constant, it simplifies certain calculations, especially when combined with the known pressure. A key feature of a rigid tank is that the internal energy change of the working fluid (steam, in this case) follows conservation laws more straightforwardly. Since the tank is also insulated, we can assume no heat transfer across its boundary. This directly impacts our conservation of energy calculations, simplifying our task by focusing on internal energy changes alone.
conservation of energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transferred or converted. In our exercise, the tank is insulated, meaning no heat is lost. The energy changes inside the tank result from the steam entering it. Specifically, the internal energy of the steam initially entering at a higher temperature will redistribute within the tank, leading to a new equilibrium. This means that although the temperature of the steam may change as it fills the tank, the total energy remains the same. By applying the conservation of energy principle, we can deduce the final temperature of the steam once it reaches the same pressure as the source.

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Most popular questions from this chapter

Refrigerant \(134 \mathrm{a}\) enters a well-insulated nozzle at \(200 \mathrm{lbf} / \mathrm{in} .{ }^{2}, 220^{\circ} \mathrm{F}\), with a velocity of \(120 \mathrm{ft} / \mathrm{s}\) and exits at 20 lbf/in. \({ }^{2}\) with a velocity of \(1500 \mathrm{ft} / \mathrm{s}\). For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in \({ }^{\circ} \mathrm{F}\).

Propane vapor enters a valve at \(1.0 \mathrm{MPa}, 60^{\circ} \mathrm{C}\), and leaves at \(0.3 \mathrm{MPa}\). If the propane undergoes a throttling process, what is the temperature of the propane leaving the valve, in \({ }^{\circ} \mathrm{C}\) ?

Refrigerant \(134 \mathrm{a}\) flows at steady state through a horizontal tube having an inside diameter of \(0.05 \mathrm{~m}\). The refrigerant enters the tube with a quality of \(0.1\), temperature of \(36^{\circ} \mathrm{C}\), and velocity of \(10 \mathrm{~m} / \mathrm{s}\). The refrigerant exits the tube at 9 bar as a saturated liquid. Determine (a) the mass flow rate of the refrigerant, in \(\mathrm{kg} / \mathrm{s}\). (b) the velocity of the refrigerant at the exit, in \(\mathrm{m} / \mathrm{s}\). (c) the rate of heat transfer, in \(\mathrm{kW}\), and its associated direction with respect to the refrigerant.

If a kitchen-sink water tap leaks one drop per second, how many gallons of water are wasted annually? What is the mass of the wasted water, in lb? Assume that there are 46,000 drops per gallon and that the density of water is \(62.3 \mathrm{lb} / \mathrm{ft}^{3}\).

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 \(g\).
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