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A balloon filled with helium at \(20^{\circ} \mathrm{C}, 1\) bar and a volume of \(0.5 \mathrm{~m}^{3}\) is moving with a velocity of \(15 \mathrm{~m} / \mathrm{s}\) at an elevation of \(0.5 \mathrm{~km}\) relative to an exergy reference environment for which \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar. Using the ideal gas model with \(k=1.67\), determine the specific exergy of the helium, in \(\mathrm{kJ} / \mathrm{kg}\).

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01

- Gather information

List all the given data:- Temperature: ewline - Velocity: \( p_{0} = 1 \text{ bar} \)ewline - Gas constant for helium: \(k=1.67\)ewline - Volume: \( V = 0.5 \text{ m}^3 $$ T_{0} = 20^{\textdegree} \text{C} \)
02

Calculate Mass of Helium

Specific heat ratio for helium is approx: k=1.67Air is made of about 78% nitrogen, 21% oxygen, 0.9% argon and trace amounts of other gases\( P_{0}V=RT_{0}\) where R = specific gas constant, typically in J/(kg*K)0.143J/gC * (491)/2

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Key Concepts

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

Ideal gas model
The ideal gas model is a simplified version of gas behavior and is crucial for calculations in thermodynamics. It assumes that gas molecules move randomly and their interactions are perfectly elastic. This means there are no forces between the molecules except during collisions. The pressure, volume, and temperature of the gas are related by the ideal gas law, which is characterized by the equation: \[ PV = nRT \] This equation tells us how a given amount (n) of gas will behave under different conditions of pressure (P) and temperature (T), where R is the universal gas constant. In our problem, we treat helium as an ideal gas. This allows us to use the ideal gas law to determine various properties, like the specific exergy.
Specific heat ratio
The specific heat ratio (k) is an important property in thermodynamics and is defined as the ratio of the specific heat at constant pressure (Cp) to the specific heat at constant volume (Cv), given by: \[ k = \frac{C_p}{C_v} \] For helium, the specific heat ratio is approximately 1.67. This value is useful for calculating properties like the speed of sound in a gas and for determining specific exergy. Understanding this ratio is key to interpreting how energy transfers within the gas. For helium, having a higher specific heat ratio means it can carry more energy per unit temperature increase at constant volume or constant pressure.
Helium properties
Helium is a noble gas, which means it has a very stable configuration and doesn't easily react with other elements. It's the second lightest and second most abundant element in the observable universe. Because of these properties, helium behaves almost as an ideal gas under a wide range of conditions. Its low density and high thermal conductivity make it useful in various industrial applications. For our calculations, key properties include its molecular mass and specific heat ratio. These properties are essential for determining other thermodynamic quantities, such as specific exergy.
Exergy reference environment
The concept of the exergy reference environment is a fundamental idea in thermodynamics. It represents the state of the surroundings with which the system can interact. For our problem, this reference environment has a temperature (T0) of 20°C and a pressure (P0) of 1 bar. These conditions are crucial because exergy measures the maximum useful work possible during a process that brings the system into equilibrium with the reference environment. By knowing the reference environment, we can calculate how much work can be extracted from the helium-filled balloon.
Thermodynamics equations
Thermodynamics is governed by a set of core equations that relate various physical quantities. These equations allow us to determine important properties like energy, entropy, and exergy. For instance, the exergy of a system can be calculated using the equation: \[ Ex = (h - h_0) - T_0(s - s_0) + \frac{1}{2}v^2 + gz \] Here, h and s are the specific enthalpy and specific entropy of the system, while h0 and s0 are their respective values in the reference environment. The term (1/2)v^2 accounts for kinetic energy, and gz represents potential energy. These thermodynamics equations form the backbone of our calculations, helping to analyze and solve real-world problems efficiently.

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Most popular questions from this chapter

Consider \(100 \mathrm{~kg}\) of steam initially at 20 bar and \(240^{\circ} \mathrm{C}\) as the system. Determine the change in exergy, in \(\mathrm{kJ}\), for each of the following processes: (a) The system is heated at constant pressure until its volume doubles. (b) The system expands isothermally until its volume doubles. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar and ignore the effects of motion and gravity.

Steam at \(450 \mathrm{lbf} / \mathrm{in}^{2}, 700^{\circ} \mathrm{F}\) enters a well-insulated turbine operating at steady state and exits as saturated vapor at a pressure \(p\). (a) For \(p=50 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\), determine the exergy destruction rate, in Btu per lb of steam expanding through the turbine, and the turbine exergetic and isentropic efficiencies. (b) Plot the exergy destruction rate, in Btu per lb of steam flowing, and the exergetic efficiency and isentropic efficiency, each versus pressure \(p\) ranging from 1 to \(50 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) Ignore the effects of motion and gravity and let \(T_{0}=70^{\circ} \mathrm{F}\), \(p_{0}=1 \mathrm{~atm} .\)

At steady state, hot gaseous products of combustion from a gas turbine cool from \(3000^{\circ} \mathrm{F}\) to \(250^{\circ} \mathrm{F}\) as they flow through a pipe. Owing to negligible fluid friction, the flow occurs at nearly constant pressure. Applying the ideal gas model with \(c_{p}=0.3 \mathrm{Btu} / \mathrm{lb} \cdot{ }^{\circ} \mathrm{R}\), determine the exergy transfer accompanying heat transfer from the gas, in Btu per lb of gas flowing. Let \(T_{0}=80^{\circ} \mathrm{F}\) and ignore the effects of motion and gravity.

A system undergoes a refrigeration cycle while receiving \(Q_{\mathrm{C}}\) by heat transfer at temperature \(T_{\mathrm{C}}\) and discharging energy \(Q_{\mathrm{H}}\) by heat transfer at a higher temperature \(T_{\mathrm{H}}\) There are no other heat transfers. (a) Using energy and exergy balances, show that the net work input to the cycle cannot be zero. (b) Show that the coefficient of performance of the cycle can be expressed as $$ \beta=\left(\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}\right)\left(1-\frac{T_{\mathrm{H}} \mathrm{E}_{\mathrm{d}}}{T_{0}\left(Q_{\mathrm{H}}-Q_{\mathrm{C}}\right)}\right) $$ where \(\mathrm{E}_{\mathrm{d}}\) is the exergy destruction and \(T_{0}\) is the temperature of the exergy reference environment. (c) Using the result of part (b), obtain an expression for the maximum theoretical value for the coefficient of performance.

A system consists of \(2 \mathrm{~kg}\) of water at \(100^{\circ} \mathrm{C}\) and 1 bar. Determine the exergy, in kJ, if the system is at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1 \mathrm{bar}\).

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