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Estimate the constant-pressure specific heat and the constant-volume specific heat for R134a at 30 psia and \(100^{\circ} \mathrm{F}\). SOLUTION We write the derivatives in finite-difference form and, using values on either side of \(100^{\circ} \mathrm{F}\) for greatest accuracy, we find $$ \begin{aligned} &C_{p} \equiv \frac{\Delta h}{\Delta T}=\frac{126.39-117.63}{120-80}=0.219 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \\ &C_{v} \equiv \frac{\Delta u}{\Delta T}=\frac{115.47-107.59}{120-80}=0.197 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \end{aligned} $$

Short Answer

Expert verified
\(C_p = 0.219 (Btu/lbm-°F)\)\(C_v = 0.197 (Btu/lbm-°F)\)

Step by step solution

01

Understand the Formulae

To estimate the constant-pressure specific heat (Cp) and the constant-volume specific heat (Cv), use the following finite-difference form formulae:Cp = \(\frac{\Delta h}{\Delta T}\) and Cv = \(\frac{\Delta u}{\Delta T}\), where \(\Delta h\) and \(\Delta u\) are the differences in specific enthalpy and internal energy, respectively, and \(\Delta T\) is the temperature difference.
02

Identify the Given Data

We are given: Specific Enthalpy Values:- At 80°F: 117.63 (Btu/lbm)- At 120°F: 126.39 (Btu/lbm)Specific Internal Energy Values:- At 80°F: 107.59 (Btu/lbm)- At 120°F: 115.47 (Btu/lbm)
03

Calculate \(\Delta h\) and \(\Delta u\)

\(\Delta h\) = 126.39 - 117.63 = 8.76 (Btu/lbm)\(\Delta u\) = 115.47 - 107.59 = 7.88 (Btu/lbm)
04

Calculate \(\Delta T\)

The temperature difference is 120°F - 80°F = 40°F.
05

Calculate \(C_p\) and \(C_v\)

Cp = \(\frac{\Delta h}{\Delta T}\) = \(\frac{8.76}{40}\) = 0.219 (Btu/lbm-°F)Cv = \(\frac{\Delta u}{\Delta T}\) = \(\frac{7.88}{40}\) = 0.197 (Btu/lbm-°F)

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Key Concepts

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

constant-pressure specific heat
Constant-pressure specific heat, denoted by \(C_p\), is an important thermodynamic property. It represents the amount of heat required to raise the temperature of a substance by one degree while maintaining constant pressure. This is crucial in processes where the volume changes, such as in an open system with fluid flow.
Mathematically, \(C_p = \frac{dH}{dT}\), where \(H\) is the enthalpy and \(T\) is the temperature. To estimate \(C_p\), we can use the finite-difference method as follows:
\( C_p \approx \frac{\triangle h}{\triangle T} \),
where \(\triangle h\) is the change in specific enthalpy and \(\triangle T\) is the change in temperature. In the exercise, specific enthalpies are given at 80°F and 120°F, allowing us to calculate \(\triangle h\).
Given the enthalpies:
\( h_{80°F} = 117.63 \text{ Btu/lbm} \)
\( h_{120°F} = 126.39 \text{ Btu/lbm} \)
we find:
\( \triangle h = h_{120°F} - h_{80°F} = 126.39 - 117.63 = 8.76 \text{ Btu/lbm} \).
Using a temperature difference \( \triangle T = 40°F \), we get:
\( C_p \approx \frac{8.76 \text{ Btu/lbm}}{40 \text{ °F}} = 0.219 \text{ Btu/lbm-°F} \).
constant-volume specific heat
Constant-volume specific heat, denoted by \(C_v\), refers to the amount of heat needed to raise the temperature of a substance by one degree while the volume is kept constant. This property is particularly relevant in closed systems.
Mathematically, \(C_v = \frac{dU}{dT}\), where \(U\) is the internal energy and \(T\) is the temperature. Using the finite-difference method, we can approximate \(C_v\) as follows:
\( C_v \approx \frac{\triangle u}{\triangle T} \),
where \(\triangle u\) is the change in specific internal energy. For the given exercise, specific internal energies at 80°F and 120°F facilitate this calculation.
Given the internal energies:
\( u_{80°F} = 107.59 \text{ Btu/lbm} \)
\( u_{120°F} = 115.47 \text{ Btu/lbm} \),
we find:
\( \triangle u = u_{120°F} - u_{80°F} = 115.47 - 107.59 = 7.88 \text{ Btu/lbm} \).
Using a temperature difference \( \triangle T = 40°F \), we get:
\( C_v \approx \frac{7.88 \text{ Btu/lbm}}{40 \text{ °F}} = 0.197 \text{ Btu/lbm-°F} \).
finite-difference method
The finite-difference method is a numerical approach used to approximate derivatives. It's helpful in estimating changes in thermodynamic properties when exact analytical solutions are difficult to obtain.
In the specific heat calculations, the finite-difference method aids in determining \(C_p\) and \(C_v\) by using known values at discrete points. Instead of dealing with complex analytical derivatives, the method uses:
\( \frac{dy}{dx} \approx \frac{\triangle y}{\triangle x} \),
where \(\triangle y\) and \(\triangle x\) are the changes in the respective quantities.
For \(C_p\) and \(C_v\), values of enthalpy and internal energy at different temperatures are used to estimate:
\( C_p \approx \frac{\triangle h}{\triangle T} \) and
\( C_v \approx \frac{\triangle u}{\triangle T} \).
The accuracy of the finite-difference method improves with smaller intervals for \(\triangle T\). However, in many practical cases, limited data points are sufficient for a reasonable estimate.
thermodynamic properties
Thermodynamic properties are characteristics of a system that describe its physical state and behavior under varying conditions. Key properties include:
  • Temperature (T)
  • Pressure (P)
  • Specific volume (v)
  • Enthalpy (H)
  • Internal energy (U)
  • Entropy (S)
These properties are interrelated through various thermodynamic laws and equations.
For instance, specific enthalpy \(H\) (or simply enthalpy) combines internal energy \(U\) and the product of pressure and volume, as in:
\( H = U + PV \).
This linkage explains why both \(C_p\) and \(C_v\) are necessary. They describe heat capacity depending on whether the pressure or volume remains constant during heating.
Understanding and calculating these properties accurately are essential for designing efficient thermal systems and predicting system behaviors.
enthalpy and internal energy
Enthalpy \(H\) and internal energy \(U\) are fundamental thermodynamic concepts.
Internal energy is the total energy contained within a system due to the kinetic and potential energies of its molecules. It's a state function, meaning it depends only on the state of the system and not on the path taken to reach that state.
Enthalpy, on the other hand, is defined as:
\( H = U + PV \),
where \(P\) is pressure, and \(V\) is volume. It represents the total heat content of a system.
For constant-pressure processes, changes in enthalpy (\translate_h) are directly proportional to the heat added or removed. This is why \(C_p\), related to \( \translate_h \), is such a critical measure.
Similarly, for constant-volume processes, changes in internal energy (\translate_u) reflect the heat exchange, making \(C_v\) relevant.
When dealing with specific heats, we analyze energy changes concerning temperature changes. Thus, these properties provide insight into system energy dynamics and help understand heat transfer processes effectively.

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Most popular questions from this chapter

Air travels through the \(4 \times 2 \mathrm{~m}\) test section of a wind tunnel at \(20 \mathrm{~m} / \mathrm{s}\). The gage pressure in the test section is measured to be \(-20 \mathrm{kPa}\) and the temperature \(20^{\circ} \mathrm{C}\). After the test section, a diffuser leads to a 6-m-diameter exit pipe. Estimate the velocity and temperature in the exit pipe. SOLUTION The energy equation (4.72) for air takes the form $$ V_{2}^{2}=V_{1}^{2}+2 C_{p}\left(T_{1}-T_{2}\right)=20^{2}+(2)(1.00)\left(293-T_{2}\right) $$ The continuity equation, \(\rho_{1} A_{1} V_{1}=\rho_{2} A_{2} V_{2}\), yields $$ \frac{P_{1}}{R T_{1}} A_{1} V_{1}=\rho_{2} A_{2} V_{2} \quad \therefore \rho_{2} V_{2}=\left[\frac{80}{(0.287)(293)}\right]\left[\frac{8}{\pi(3)^{2}}\right](20)=5.384 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s} $$ The best approximation to the actual process is the adiabatic quasi- equilibrium process. Using (4.49), letting \(\rho=1 / v\), we have $$ \frac{T_{2}}{T_{1}}=\left(\frac{\rho_{2}}{\rho_{1}}\right)^{k-1} \quad \text { or } \quad \frac{T_{2}}{\rho_{2}^{0.4}}=\frac{293}{[80 /(0.287)(293)]^{0.4}}=298.9 $$ The above three equations include the three unknowns \(T_{2}, V_{2}\), and \(\rho_{2}\). Substitute for \(T_{2}\) and \(V_{2}\) back into the energy equation and find $$ \frac{5.384^{2}}{\rho_{2}^{2}}=20^{2}+(2)(1.00)\left[293-(298.9)\left(\rho_{2}^{0.4}\right)\right] $$ This can be solved by trial and error to yield \(\rho_{2}=3.475 \mathrm{~kg} / \mathrm{m}^{3}\). The velocity and temperature are then $$ V_{2}=\frac{5.384}{\rho_{2}}=\frac{5.384}{3.475}=1.55 \mathrm{~m} / \mathrm{s} \quad T_{2}=(298.9)\left(\rho_{2}^{0.4}\right)=(298.9)(3.475)^{0.4}=492 \mathrm{~K} \quad \text { or } \quad 219^{\circ} \mathrm{C} $$

Water enters a radiator through a 4-cm-diameter hose at \(0.02 \mathrm{~kg} / \mathrm{s}\). It travels down through all the rectangular passageways on its way to the water pump. The passageways are each \(10 \times 1 \mathrm{~mm}\) and there are 800 of them in a cross section. How long does it take water to traverse from the top the bottom of the 60 -cm-high radiator? SOLUTION The average velocity through the passageways is found from the continuity equation, using \(\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^{3}\) : $$ \dot{m}=\rho_{1} V_{1} A_{1}=\rho_{2} V_{2} A_{2} \quad \therefore V_{2}=\frac{\dot{m}}{\rho_{2} A_{2}}=\frac{0.02}{(1000)[(800)(0.01)(0.001)]}=0.0025 \mathrm{~m} / \mathrm{s} $$ The time to travel \(60 \mathrm{~cm}\) at this constant velocity is $$ t=\frac{L}{V}=\frac{0.60}{0.0025}=240 \mathrm{~s} \quad \text { or } \quad 4 \mathrm{~min} $$

The air in the cylinder of an air compressor is compressed from \(100 \mathrm{kPa}\) to \(10 \mathrm{MPa}\). Estimate the final temperature and the work required if the air is initially at \(100^{\circ} \mathrm{C}\). Assume constant specific heats. SOLUTION Since the process occurs quite fast, we assume an adiabatic quasi-equilibrium process. Then $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(373)\left(\frac{10000}{100}\right)^{(1.4-1) / 1.4}=1390 \mathrm{~K} \quad \text { or } \quad 1117^{\circ} \mathrm{C} $$ The work is found by using the first law with \(Q=0\) : $$ w=-\Delta u=-C_{v}\left(T_{2}-T_{1}\right)=-(0.717)(1390-373)=-729 \mathrm{~kJ} / \mathrm{kg} $$ The work per unit mass is calculated since the mass (or volume) was not specified.

Steam with a mass flux of \(600 \mathrm{lbm} / \mathrm{min}\) exits a turbine as saturated steam at 2 psia and passes through a condenser (a heat exchanger). What mass flux of cooling water is needed if the steam is to exit the condenser as saturated liquid and the cooling water is allowed a \(15^{\circ} \mathrm{F}\) temperature rise? SOLUTION The energy equations (4.75) are applicable to this situation. The heat transfer rate for the steam is, assuming no pressure drop through the condenser, $$ \dot{Q}_{s}=\dot{m}_{s}\left(h_{s 2}-h_{s 1}\right)=(600)(94.02-1116.1)=-613.200 \mathrm{Btu} / \mathrm{min} $$ This energy is gained by the water. Hence, $$ \dot{Q}_{w}=\dot{m}_{w}\left(h_{w 2}-h_{w 1}\right)=\dot{m}_{w} C_{p}\left(T_{w 2}-T_{w 1}\right) \quad \frac{613,200}{60}=\dot{m}_{w}(1.00)(15) \quad \dot{m}_{w}=681 \mathrm{lbm} / \mathrm{s} $$

A \(1500-\mathrm{kg}\) automobile traveling at \(30 \mathrm{~m} / \mathrm{s}\) is brought to rest by impacting a shock absorber composed of a piston with small holes that moves in a cylinder containing water. How much heat must be removed from the water to return it to its original temperature? As the piston moves through the water, work is done due to the force of impact moving with the piston. The work that is done is equal to the kinetic energy change; that is, $$ W=\frac{1}{2} m V^{2}=\left(\frac{1}{2}\right)(1500)(30)^{2}=675000 \mathrm{~J} $$ The first law for a cycle requires that this amount of heat must be transferred from the water to return it to its original temperature; hence, \(Q=675 \mathrm{~kJ}\).

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