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Using steam table data, determine the indicated property data for a process in which there is no change in specific entropy between state 1 and state 2 . In each case, locate the states on a sketch of the \(T-s\) diagram. (a) \(T_{1}=40^{\circ} \mathrm{C}, x_{1}=100 \%, p_{2}=150 \mathrm{kPa}\). Find \(T_{2}\), in \({ }^{\circ} \mathrm{C}\), and \(\Delta h\), in \(\mathrm{kJ} / \mathrm{kg}\). (b) \(T_{1}=10^{\circ} \mathrm{C}, x_{1}=75 \%, p_{2}=1 \mathrm{MPa}\). Find \(T_{2}\), in \({ }^{\circ} \mathrm{C}\), and \(\Delta u\), in \(\mathrm{kJ} / \mathrm{kg}\).

Short Answer

Expert verified
For part (a): Given \( T_{1} = 40°C\) and \(x_{1} = 1\): Find \( T_{2} \) and \(Δh. \) For part (b): Given \( T_{1} = 10°C\) and \(x_{1} = 75\text{\text{\text{%}}}\): Find \( T_{2} \) and \(Δu\).

Step by step solution

01

Identify the initial state properties

For part (a): Given - Temperature at state 1: \(T_{1} = 40^{\text{}}^{\text{}} \text{C}\) - Quality at state 1: \(x_{1} = 100\text{\text{%}} (saturated vapor)\) For part (b): Given - Temperature at state 1: \(T_{1} = 10^{\text{}}^{\text{}} \text{C}\) - Quality at state 1: \(x_{1} = 75\text{\text{%}}\)
02

Use steam tables to find specific entropy at state 1

For part (a): - Refer to the steam tables at \(T_{1} = 40^{\circ}\text{C}\) - Since the quality \(x_{1} = 1\), we are at a saturated vapor point. - Find the specific entropy \(s_{1} = s_{g}(40^{\circ}\text{C})\) For part (b): - Refer to the steam tables at \(T_{1} = 10^{\circ}\text{C}\) - Since the quality \(x_{1} = 0.75\), use the equation: \(s_{1} = s_{f} + x_{1}(s_{g} - s_{f})\)
03

Locate specific entropy at state 2

For both parts (a) and (b): - We know specific entropy remains constant: \(s_{2} = s_{1}\) - For part (a): \( p_{2} = 150\text{kPa} \) - For part (b): \( p_{2} = 1\text{MPa} \) - Refer to the steam tables to find the temperature \(T_{2}\) such that \(s_{2} = s_{1} \) at the given pressure.
04

Calculate energy changes

For part (a): - Determine the specific enthalpy at state 1 and state 2. \( h_{1} = h_{g}(40\text{\text{}^{\text{C}}}) \) \( h_{2} \) from the steam tables corresponding to \( s_{2} = s_{1} \) at \( p_{2} = 150\text{kPa} \). - Calculate \( Δh = h_{2} - h_{1} \)For part (b): - Determine the specific internal energy at state 1 and state 2 Use properties and tables to find \(u_{1}\) and \(u_{2}\) - Calculate \( Δu = u_{2} - u_{1} \)
05

Sketch the T-s diagram

Draw the \(T-s\) diagram for both parts (a) and (b) and mark states 1 and 2 along with the isentropic line for reference.

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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 invaluable references in thermodynamics. They provide properties of water and steam for various pressures and temperatures. They include values for specific enthalpy, specific entropy, and specific internal energy. When working with steam cycles or any process involving steam, these tables help you find key properties needed for computations. For example, if you know the temperature and quality (x), you can find specific entropy (\( s \)) and specific enthalpy (\( h \)) directly from the steam tables. Similarly, if you know the pressure and temperature, you can find other properties. Remember, steam tables usually have sections for both saturated and superheated steam.
specific entropy
Entropy measures the randomness or disorder of a system. In thermodynamic processes, specific entropy (\( s \)) is commonly used to describe the state of steam. It is calculated in units of \( \text{kJ/kg·K} \). In an isentropic process, entropy remains constant (\( \Delta s = 0 \)). To find specific entropy at a state, refer to steam tables using your known properties like temperature (\( T \)) and quality (\( x \)). For instance, if you have saturated vapor, the specific entropy can be read directly for that temperature. If you have a mixture, use: \[ s = s_f + x(s_g - s_f) \], where \( s_f \) is the entropy of the saturated liquid and \( s_g \) is the entropy of the saturated vapor. This equation helps you find the entropy of the mixture.
T-s diagram
A \( T-s \) diagram is a temperature-entropy graph used to visualize thermodynamic processes. In this diagram, temperature (\( T \)) is on the vertical axis, and entropy (\( s \)) is on the horizontal axis. The \( T-s \) diagram helps to identify the states and processes of a system. Isentropic processes, where entropy remains constant, appear as vertical lines on this diagram. When solving problems, sketching the \( T-s \) diagram allows you to clearly locate and identify states and visualize energy changes. For instance, a movement from state 1 to state 2 without change in entropy will be a straight vertical line. This visualization aids in understanding the behavior of the system during the transformation.
specific enthalpy
Specific enthalpy (\( h \)) is a measurement of the total heat content of a system, given in \( \text{kJ/kg} \). It combines internal energy and the product of pressure and volume. Specific enthalpy is essential in energy balance calculations, especially in processes involving phase changes. Using steam tables, you can find \( h \) directly if you know the temperature and quality. For example, for a saturated vapor, \( h_g \) gives the specific enthalpy. For mixtures: \[ h = h_f + x(h_g - h_f) \], where \( h_f \) is the enthalpy of saturated liquid and \( h_g \) is the enthalpy of saturated vapor. Calculating changes in specific enthalpy (\( \Delta h \)) involves finding the enthalpy at different states and taking the difference.
specific internal energy
Specific internal energy (\( u \)) represents the energy contained within a system due to molecular motion and interactions. It's measured in \( \text{kJ/kg} \). Like other properties, specific internal energy can be found using steam tables, particularly when looking at superheated or saturated states. For mixtures: \[ u = u_f + x(u_g - u_f) \], where \( u_f \) represents the internal energy at the saturated liquid state, and \( u_g \) represents the saturated vapor state. Specific internal energy is key to understanding energy changes (\( \Delta u \)) within the system as heat is added or removed or work is done by or on the system. For example, changes in specific internal energy can be calculated by finding the internal energies at two states and subtracting them.

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Most popular questions from this chapter

Construct a plot, to scale, showing constant-pressure lines of \(5.0\) and \(10 \mathrm{MPa}\) ranging from 100 to \(400^{\circ} \mathrm{C}\) on a \(T-s\) diagram for water.

Air in a piston-cylinder assembly expands isentropically from \(T_{1}=1800^{\circ} \mathrm{R}, p_{1}=20 \mathrm{lbf} / \mathrm{in} .^{2}\), to \(p_{2}=2000 \mathrm{lbf} / \mathrm{in}^{2}\) Assuming the ideal gas model, determine the temperature at state 2 , in \({ }^{\circ} \mathrm{R}\), using (a) data from Table \(\mathrm{A}-22 \mathrm{E}\), and (b) a constant specific heat ratio, \(k=1.4\). Compare the values obtained in parts (a) and (b) and comment.

A cylindrical copper rod of base area A and length \(L\) is insulated on its lateral surface. One end of the rod is in contact with a wall at temperature \(T_{\mathrm{H}}\). The other end is in contact with a wall at a lower temperature \(T_{\mathrm{C}}\). At steady state, the rate at which energy is conducted into the rod from the hot wall is $$ \dot{Q}_{\mathrm{H}}=\frac{\kappa \mathrm{A}\left(T_{\mathrm{H}}-T_{\mathrm{C}}\right)}{L} $$ where \(\kappa\) is the thermal conductivity of the copper rod. (a) For the rod as the system, obtain an expression for the time rate of entropy production in terms of \(\mathrm{A}, L, T_{\mathrm{H}}, T_{\mathrm{C}}\), and \(\kappa\). (b) If \(T_{\mathrm{H}}=327^{\circ} \mathrm{C}, T_{\mathrm{C}}=77^{\circ} \mathrm{C}, \kappa=0.4 \mathrm{~kW} / \mathrm{m} \cdot \mathrm{K}, \mathrm{A}=0.1 \mathrm{~m}^{2}\), plot the heat transfer rate \(\dot{Q}_{\mathrm{H}}\), in \(\mathrm{kW}\), and the time rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), each versus \(L\) ranging from \(0.01\) to \(1.0 \mathrm{~m}\). Discuss.

Water at 20 bar, \(400^{\circ} \mathrm{C}\) enters a turbine operating at steady state and exits at \(1.5\) bar. Stray heat transfer and kinetic and potential energy effects are negligible. A hard-to-read data sheet indicates that the quality at the turbine exit is \(98 \%\). Can this quality value be correct? If no, explain. If yes, determine the power developed by the turbine, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of water flowing.

Air at \(1 \mathrm{~atm}, 520^{\circ} \mathrm{R}\) enters a compressor operating at steady state and is compressed adiabatically to 3 atm. The isentropic compressor efficiency is \(80 \%\). Employing the ideal gas model with \(k=1.4\) for the air, determine for the compressor (a) the power input, in Btu per lb of air flowing, and (b) the amount of entropy produced, in Btu/ \(/ \mathrm{R}\) per lb of air flowing. Ignore kinetic and potential energy effects.

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