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Saturated steam at \(100^{\circ} \mathrm{C}\) is heated to \(350^{\circ} \mathrm{C}\). Use the steam tables to determine (a) the required heat input (J/s) if a continuous stream flowing at \(100 \mathrm{kg} / \mathrm{s}\) undergoes the process at constant pressure and (b) the required heat input (J) if \(100 \mathrm{kg}\) undergoes the process in a constant-volume container. What is the physical significance of the difference between the numerical values of these two quantities?

Short Answer

Expert verified
The required heat input under constant pressure and volume can be calculated by applying the principles of conservation of energy and using specific enthalpy and internal energy values obtained from steam tables respectively. The difference in numerical quantities reflects the work done against external forces during the constant pressure process.

Step by step solution

01

Identify Key Data

From the problem statement, there are two key temperatures: the initial temperature of the steam, which is \(100^{\circ}C\) and the final temperature we're heating it to, which is \(350^{\circ}C\). Additionally, the mass flow rate of the steam is given as \(100kg/s\) and the total mass in the constant volume scenario is \(100kg\). The properties of steam at these temperatures will be obtained from the steam tables.
02

Find Enthalpy and Internal Energy at Initial Temperature

Firstly, identify the specific enthalpy (\(h_1\)) and specific internal energy (\(u_1\)) of saturated steam at \(100^{\circ}C\) from steam tables. Note that the values of \(h\) and \(u\) are usually presented in terms of kJ/kg.
03

Find Enthalpy and Internal Energy at Final Temperature

Secondly, identify the specific enthalpy (\(h_2\)) and specific internal energy (\(u_2\)) of steam at \(350^{\circ}C\) from steam tables. The selected values should be for superheated steam since the given temperature is above the saturation temperature.
04

Compute Required Heat for Constant Pressure

The heat required (\(Q_p\)) for a continuous flow at constant pressure can be calculated using the equation \(Q_p = m \cdot (h_2 - h_1)\), where \(m\) is the mass flow rate. This formula is based on the principle of energy conservation applied to an open or flow system and uses enthalpy (which includes the energy required to make room for a system by displacing its environment and establishing its volume and pressure).
05

Compute Required Heat for Constant Volume

The heat required (\(Q_v\)) in a constant volume container can be calculated using the equation \(Q_v = m \cdot (u_2 - u_1)\), where \(m\) is the total mass. This formula is again based on the principle of energy conservation, but for a closed or non-flow system, and uses internal energy.
06

Explain the Physical Significance

The difference between these two quantities can be explained in terms of the different conditions for the two processes – constant pressure vs. constant volume. It shows the additional work done against external forces (or less heat supplied) when the process happens at constant pressure because heating at constant pressure allows for expansion (doing work to push the surroundings), while at constant volume it does not.

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy, heat, and work, and how they interrelate within physical systems. It essentially governs how energy is transferred between systems and their surroundings. In our exercise, we are dealing with a steam heating process, where heat energy is added to steam to increase its temperature. This is a great example of the first law of thermodynamics, which states that energy cannot be created or destroyed, only transformed from one form to another.

In simpler terms, thermodynamics helps dictate how much energy we need to add to transform the steam from its initial to its final state. It also helps us understand why different amounts of energy are needed in processes that are carried out under constant pressure versus constant volume conditions.
Enthalpy
Enthalpy is a fundamental thermodynamic property, representing the total heat content of a system. It is especially useful in processes occurring at constant pressure. When dealing with steam heating, enthalpy quantifies the system's energy considering both internal energy and the work required to "make room" by expanding against the surrounding pressure.

The change in enthalpy (abla h) between two states provides insight into the energy input needed during a heating process at constant pressure. Mathematically, it is expressed as: \( abla h = h_2 - h_1 \). In our exercise, enthalpy change helps calculate the heat input when steam is heated from 100°C to 350°C at constant pressure.
Internal Energy
Internal energy is another key thermodynamic property, representing the energy stored within a system as a result of molecular motion and interactions. Unlike enthalpy, internal energy is particularly invaluable in analyzing processes occurring at constant volumes, where no work is done by expanding against external pressures.

Understanding the change in internal energy (abla u) is important because it indicates how much heat is absorbed by the system itself. It is calculated using \( abla u = u_2 - u_1 \). Hence, for processes in a constant-volume container like in our problem, internal energy is vital in determining the heat energy required to achieve the specified temperature change.
Steam Tables
Steam tables are vital tools in thermodynamics, providing values for different properties of water and steam, such as enthalpy and internal energy, at various temperatures and pressures. By consulting steam tables, engineers and students can determine state properties needed for calculations without performing complex measurements or simulations.

In our problem, steam tables aid in locating the specific enthalpies and internal energies ( abla h) and ( abla u) of steam at initial and final temperatures. They essentially serve as a reliable reference to determine how much energy is involved in heating steam through detailed thermodynamic data.
Constant Pressure and Volume Processes
Processes characterized by constant pressure or constant volume have unique thermodynamic implications.

In constant-pressure processes, as seen in our problem, the steam can expand, performing work. The heat required is calculated using changes in enthalpy, signifying energy involved in both heating and expansion. The formula used is \( Q_p = m imes (h_2 - h_1) \).

Conversely, in constant-volume processes, the system is confined, leading to no work being done against external pressures. Consequently, heat calculations rely on changes in internal energy, expressed as \( Q_v = m imes (u_2 - u_1) \).

This physical distinction underscores the difference in energy requirements between processes taking place under constant-pressure and constant-volume conditions. In essence, it reflects how the system's ability to do work impacts the required heat input, as demonstrated in our exercise.

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Most popular questions from this chapter

Water is to be pumped from a lake to a ranger station on the side of a mountain (see figure). The length of pipe immersed in the lake is negligible compared to the length from the lake surface to the discharge point. The flow rate is to be \(95 \mathrm{gal} / \mathrm{min}\), and the flow channel is a standard 1-inch. Schedule 40 steel pipe (ID \(=1.049\) inch). A pump capable of delivering \(8 \mathrm{hp}\left(=\dot{W}_{\mathrm{s}}\right)\) is available. The friction loss \(\tilde{F}\left(\mathrm{ft} \cdot \mathrm{lb}_{\mathrm{f}} / \mathrm{lb}_{\mathrm{m}}\right)\) equals \(0.041 L,\) where \(L(\mathrm{ft})\) is the length of the pipe. (a) Calculate the maximum elevation, \(z\), of the ranger station above the lake if the pipe rises at an angle of \(30^{\circ}\) (b) Suppose the pipe inlet is immersed to a significantly greater depth below the surface of the lake, but it discharges at the elevation calculated in Part (a). The pressure at the pipe inlet would be greater than it was at the original immersion depth, which means that \(\Delta P\) from inlet to outlet would be greater, which in turn suggests that a smaller pump would be sufficient to move the water to the same elevation. In fact, however, a larger pump would be needed. Explain (i) why the pressure at the inlet would be greater than in Part (a), and (ii) why a larger pump would be needed.

A certain gasoline engine has an efficiency of \(30 \% ;\) that is, it converts into useful work \(30 \%\) of the heat generated by burning a fuel. (a) If the congine consumes \(0.80 \mathrm{L}\) /h of a gasoline with a heating value of \(3.25 \times 10^{4} \mathrm{kJ} / \mathrm{L}\), how much power does it provide? Express the answer both in \(\mathrm{kW}\) and horsepower. (b) Suppose the fuel is changed to include \(10 \%\) ethanol by volume. The heating value of ethanol is approximately \(2.34 \times 10^{4} \mathrm{kJ} / \mathrm{L}\) and volumes of gasoline and ethanol may be assumed additive. At what rate ( \((\text { / } h\) ) does the fuel mixture have to be consumed to produce the same power as gasoline?

A turbine discharges \(200 \mathrm{kg} / \mathrm{h}\) of saturated steam at \(10.0 \mathrm{bar}\) absolute. It is desired to generate steam at \(250^{\circ} \mathrm{C}\) and 10.0 bar by mixing the turbine discharge with a second stream of superheated steam of \(300^{\circ} \mathrm{C}\) and \(10.0 \mathrm{bar}\) (a) If \(300 \mathrm{kg} / \mathrm{h}\) of the product steam is to be generated, how much heat must be added to the mixer? (b) If instead the mixing is carried out adiabatically, at what rate is the product steam generated?

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