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Mechanical engineers, as well as most other engineers, use thermodynamics extensively in their work. The following polynomial can be used to relate the zero-pressure specific heat of dry air, \(c_{p} \mathrm{kJ} /(\mathrm{kg} \mathrm{K}),\) to temperature \((\mathrm{K})\): $$\begin{aligned}c_{p}=& 0.99403+1.671 \times 10^{-4} T+9.7215 \times 10^{-8} T^{2} \\\&-9.5838 \times 10^{-11} T^{3}+1.9520 \times 10^{-14} T^{4}\end{aligned}$$ Determine the temperature that corresponds to a specific heat of \(1.2 \mathrm{kJ} /(\mathrm{kg} \mathrm{K})\).

Short Answer

Expert verified
The temperature that corresponds to a specific heat of \(1.2 \, \mathrm{kJ} /(\mathrm{kg} \mathrm{K})\) for dry air is approximately \(349.78 \, \mathrm{K}\).

Step by step solution

01

Write down the given polynomial equation

The given polynomial equation to relate the zero-pressure specific heat of dry air, \(c_p\) (in \(\mathrm{kJ} /(\mathrm{kg} \mathrm{K})\)), to temperature (in K) is: \[c_p = 0.99403+1.671 \times 10^{-4} T+9.7215 \times 10^{-8} T^{2} -9.5838 \times 10^{-11} T^{3}+1.9520 \times 10^{-14} T^{4}\]
02

Set the specific heat to 1.2 kJ/(kg K)

We are asked to find the temperature that corresponds to a specific heat of \(1.2 \, \mathrm{kJ} /(\mathrm{kg} \mathrm{K})\), so we will set \(c_p = 1.2\) and solve for \(T\): \[1.2 = 0.99403 + 1.671 \times 10^{-4}T + 9.7215 \times 10^{-8}T^{2} - 9.5838 \times 10^{-11}T^{3} + 1.9520 \times 10^{-14}T^{4}\]
03

Rearrange the equation to get a polynomial in terms of T

Rearrange the equation to set it equal to zero: \[0 = -0.20597 + 1.671 \times 10^{-4}T + 9.7215 \times 10^{-8}T^{2} - 9.5838 \times 10^{-11}T^{3} + 1.9520 \times 10^{-14}T^{4}\]
04

Solve for T numerically

The equation is a 4th-degree polynomial, so it doesn't have a closed-form solution. We will have to solve it numerically. You may use numerical methods such as the Newton-Raphson method or bisection method, or a calculator or software specialized in solving equations. Using a calculator or software (like Wolfram Alpha), we find that the temperature with the given specific heat is approximately \(T \approx 349.78\, \mathrm{K}\). So, the temperature that corresponds to a specific heat of \(1.2 \, \mathrm{kJ} /(\mathrm{kg} \mathrm{K})\) is approximately \(349.78 \, \mathrm{K}\).

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Key Concepts

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

Thermodynamics
Thermodynamics is a fundamental branch of physics concerned with heat and temperature and their relation to energy and work. It describes how thermal energy is converted to and from other forms of energy and how it affects matter. Thermodynamics is based on four laws, from zeroth to third, which lay the ground rules for how energy systems interact and perform.

In practical terms, engineers and scientists use the principles of thermodynamics to design engines, predict chemical reaction outcomes, and understand the behavior of the atmosphere — as evidenced by the given polynomial equation that relates the specific heat of dry air to temperature. This relation is important for various applications such as climate control, energy efficiency studies, and more.
Zero-Pressure Specific Heat of Dry Air
The zero-pressure specific heat of dry air, often denoted as cp, measures the amount of energy in kilojoules required to raise one kilogram of dry air by one Kelvin at a pressure of zero pascals (in practice, it's approximated at atmospheric pressure for simplicity). It is a crucial property in the study of atmospheric thermodynamics and engineering because it impacts the thermodynamic processes involving the heating and cooling of air. For example, understanding the value of cp allows engineers to compute how much energy is needed for HVAC systems to maintain desired environmental conditions within buildings.
Numerical Methods
Numerical methods are mathematical techniques used to derive approximate solutions to complex problems, which cannot easily be solved analytically. They are vital in engineering, physics, and applied mathematics because they allow the tackling of nonlinear equations, integrals, differential equations, and other mathematically intricate issues.

In our case, we need to apply numerical methods to find an approximate temperature corresponding to a specific heat value because the relationship is described by a polynomial that is too complex for a simple algebraic solution. These methods are powerful tools that enable us to solve equations that model real-world phenomena, such as the behavior of air under different thermal conditions.
Newton-Raphson Method
The Newton-Raphson method, named after Isaac Newton and Joseph Raphson, is a popular numerical technique used to find successively better approximations to the roots of a real-valued function. The general idea of the Newton-Raphson method is to start with an initial guess which is reasonably close to the true root, then the function is approximated by its tangent line, and the x-intercept of this tangent line becomes the next approximation.

Applying this method to our specific heat equation, we would iterate over several approximations of temperature until the change in value is within an acceptable error range of the true root. It is especially useful in thermodynamics where we often deal with complex polynomial equations describing the properties of materials or systems.
Polynomial Equation
A polynomial equation is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. The equation provided in the exercise is a polynomial in the temperature variable, T, and relates it to the specific heat of dry air at zero pressure, cp. In the context of thermodynamics, such equations can become quite complicated as they attempt to capture the nuanced behaviour of substances under various conditions.

To solve our fourth-degree (quartic) polynomial equation for temperature given a specific heat capacity value, we need to rearrange the equation and apply numerical methods since quartic equations do not always have straightforward analytical solutions. Being adept at understanding and manipulating polynomial equations is essential for engineers modeling thermodynamic processes.

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Most popular questions from this chapter

The Redlich-Kwong equation of state is given by $$p=\frac{R T}{v-b}-\frac{a}{v(v+b) \sqrt{T}}$$ where \(R=\) the universal gas constant \([=0.518 \mathrm{kJ} /(\mathrm{kg} \mathrm{K})], T=\) absolute temperature \((\mathrm{K}), p=\) absolute pressure \((\mathrm{kPa}),\) and \(v=\) the volume of a \(\mathrm{kg}\) of gas \(\left(\mathrm{m}^{3} / \mathrm{kg}\right) .\) The parameters \(a\) and \(b\) are calculated by $$a=0.427 \frac{R^{2} T_{c}^{2.5}}{p_{c}} \quad b=0.0866 R \frac{T_{c}}{p_{c}}$$ where \(p_{c}=\) critical pressure (kPa) and \(T_{c}=\) critical temperature (K). As a chemical engineer, you are asked to determine the amount of methane fuel \(\left(p_{c}=4600 \mathrm{kPa} \text { and } T_{c}=191 \mathrm{K}\right)\) that can be held in a \(3-\mathrm{m}^{3}\) tank at a temperature of \(-40^{\circ} \mathrm{C}\) with a pressure of \(65,000 \mathrm{kPa}\) Use a root-locating method of your choice to calculate \(v\) and then determine the mass of methane contained in the tank.

Figure P8.18 a shows a uniform beam subject to a linearly increasing distributed load. The equation for the resulting elastic curve is (see Fig.P8.18 b) $$y=\frac{w_{0}}{120 E I L}\left(-x^{5}+2 L^{2} x^{3}-L^{4} x\right)$$ Use bisection to determine the point of maximum deflection (that is, the value of \(x\) where \(d y / d x=0\) ). Then substitute this value into Eq. \((\mathrm{P} 8.18)\) to determine the value of the maximum deflection. Use the following parameter values in your computation: \(L=600 \mathrm{cm}\) \(E=50,000 \mathrm{kN} / \mathrm{cm}^{2}, I=30,000 \mathrm{cm}^{4},\) and \(w_{0}=2.5 \mathrm{kN} / \mathrm{cm}\).

In chemical engineering, plug flow reactors (that is, those in which fluid flows from one end to the other with minimal mixing along the longitudinal axis) are often used to convert reactants into products. It has been determined that the efficiency of the conversion can sometimes be improved by recycling a portion of the product stream so that it returns to the entrance for an additional pass through the reactor (Fig. \(P 8.2\) ). The recycle rate is defined as \(R=\frac{\text { volume of fluid returned to entrance }}{\text { volume leaving the system }}\) Suppose that we are processing a chemical A to generate a product B. For the case where A forms B according to an autocatalytic reaction (that is, in which one of the products acts as a catalyst or stimulus for the reaction \(),\) it can be shown that an optimal recycle rate must satisfy $$\ln \frac{1+R\left(1-X_{A f}\right)}{R\left(1-X_{A f}\right)}=\frac{R+1}{R\left[1+R\left(1-X_{A f}\right)\right]}$$ where \(X_{A f}=\) the fraction of reactant A that is converted to product B. The optimal recycle rate corresponds to the minimum-sized reactor needed to attain the desired level of conversion. Use a numerical method to determine the recycle ratios needed to minimize reactor size for a fractional conversion of \(X_{A f}=0.96\).

A catenary cable is one that is hung between two points not in the same vertical line. As depicted in Fig. \(P 8.17 a\), it is subject to no loads other than its own weight. Thus, its weight \((\mathrm{N} / \mathrm{m})\) acts as a uniform load per unit length along the cable. A free-body diagram of a section \(A B\) is depicted in Fig. \(\mathrm{P} 8.17 b,\) where \(T_{A}\) and \(T_{B}\) are the tension forces at the end. Based on horizontal and vertical force balances, the following differential equation model of the cable can be derived: $$\frac{d^{2} y}{d x^{2}}=\frac{w}{T_{A}} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}$$ Calculus can be employed to solve this equation for the height \(y\) of the cable as a function of distance \(x\), $$y=\frac{T_{A}}{w} \cosh \left(\frac{w}{T_{A}} x\right)+y_{0}-\frac{T_{A}}{w}$$ where the hyperbolic cosine can be computed by $$\cosh x=\frac{1}{2}\left(e^{x}+e^{-x}\right)$$ Use a numerical method to calculate a value for the parameter \(T_{A}\) given values for the parameters \(w=12\) and \(y_{0}=6,\) such that the cable has a height of \(y=15\) at \(x=50\).

The Ergun equation, shown below, is used to describe the flow of a fluid through a packed bed. \(\Delta P\) is the pressure drop, \(\rho\) is the density of the fluid, \(G_{o}\) is the mass velocity (mass flow rate divided by cross- sectional area), \(D_{p}\) is the diameter of the particles within the bed, \(\mu\) is the fluid viscosity, \(L\) is the length of the bed, and \(\varepsilon\) is the void fraction of the bed. $$\frac{\Delta P \rho}{G_{o}^{2}} \frac{D_{p}}{L} \frac{\varepsilon^{3}}{1-\varepsilon}=150 \frac{1-\varepsilon}{\left(D_{p} G_{o} / \mu\right)}+1.75$$ Given the parameter values listed below, find the void fraction \(\varepsilon\) of the bed. $$\begin{aligned}&\frac{D_{p} G_{o}}{\mu}=1000\\\&\frac{\Delta P \rho D_{p}}{G_{o}^{2} L}=10\end{aligned}$$

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