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Chapter 6: Problem 123

Carbon monoxide enters a nozzle operating at steady state at \(5 \mathrm{bar}, 200^{\circ} \mathrm{C}, 1 \mathrm{~m} / \mathrm{s}\) and undergoes a polytropic expansion to 1 bar with \(n=1.2\). Using the ideal gas model and ignoring potential energy effects, determine (a) the exit velocity, in \(\mathrm{m} / \mathrm{s}\). (b) the rate of heat transfer between the gas and its surroundings, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of gas flowing.

Short Answer

Expert verified
The exit velocity is approximately 533.19 m/s. The rate of heat transfer is approximately zero kJ per kg.

Step by step solution

01

Identify the known parameters

Identify and list all the given parameters. Initial pressure, \(P_1 = 5 \text{ bar} = 500 \text{ kPa}\)Initial temperature, \(T_1 = 200^{\circ} \text{C} = 473 \text{ K}\)Initial velocity, \(V_1 = 1 \text{ m/s}\)Final pressure, \(P_2 = 1 \text{ bar} = 100 \text{ kPa}\)Polytropic index, \(n = 1.2\)
02

Apply the polytropic process equation

Use the polytropic process equation to find the final temperature. The equation is \[ P_{1}V_{1}^{n} = P_{2}V_{2}^{n} \] Since \(V \propto \frac{T}{P}\), we can rewrite the equation as: \[ \frac{T_{1}}{P_{1}^{(1-1/n)}} = \frac{T_{2}}{P_{2}^{(1-1/n)}} \] Solve for \( T_{2} \): \[ T_{2} = T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(1-1/n)} = 473\left(\frac{100}{500}\right)^{(1-1/1.2)} = 335.68 \text{ K} \]
03

Use the energy balance equation

Apply the energy balance equation for a control volume (nozzle) with no potential energy effects and steady state conditions. The equation is: \[ h_{1} + \frac{V_{1}^{2}}{2} = h_{2} + \frac{V_{2}^{2}}{2} + q \]Using \[ h = C_{P}T \], write the equation as: \[ C_{P}T_{1} + \frac{V_{1}^{2}}{2} = C_{P}T_{2} + \frac{V_{2}^{2}}{2} + q \]
04

Calculate the specific heat capacity

For carbon monoxide, assume \(C_{P} \approx 1.034 \text{ kJ/kg}\text{路} \text{K}\)
05

Compute the exit velocity

Subtract the enthalpy terms on both sides to isolate \( V_{2} \). That gives: \[ V_2 = \sqrt{ 2(C_P(T_1 - T_2)) + V_1^2 } \]Plugging in the values: \[ V_2 = \sqrt{ 2(1.034 \times 10^3 \times (473 - 335.68)) + 1^2 } \= \sqrt{ 2(1.034 \times 10^3 \times 137.32) + 1 } \= \sqrt{ 284,196 + 1 } = 533.19 \text{ m/s} \]
06

Calculate the rate of heat transfer

Make use of the energy balance equation: \[ h_{1} + \frac{V_{1}^{2}}{2} = h_{2} + \frac{V_{2}^{2}}{2} + q \]Rearrange the equation to solve for \ q : \[ q = C_P T_1 + \frac{V_1^2}{2} - C_P T_2 - \frac{V_2^2}{2} \= 1,000 \approx 0 \text{kJ/kg} . \]

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Key Concepts

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

steady state conditions
In thermodynamics, steady state conditions imply that properties within the system do not change over time, despite mass or energy crossing its boundaries. Everything entering or leaving has a balance. This helps make the equations simpler. For the nozzle problem, it means all entries and exits of mass and energy are constant over time.
  • The nozzle operates consistently without changing its internal state.
  • The entering and exiting mass flow rates are equal.
  • The heat transfer rate is constant.
Assuming steady state helps simplify the energy balance equation. This is because we don't need to account for changes over time. So, it lets us focus on the energy transformations occurring from inlet to outlet.
ideal gas model
The ideal gas model is a mathematical approximation that simplifies the behavior of gases. It relies on the ideal gas law: \[ PV = nRT \] where P is pressure, V is volume, n is the amount of substance (moles), R is the ideal gas constant, and T is temperature. This model assumes:
  • Gases consist of many tiny particles moving randomly.
  • Intermolecular forces are negligible.
  • The volume of individual molecules is minuscule compared to the total volume of gas.
In the nozzle problem, carbon monoxide (CO) is treated as an ideal gas. This means we can use simplified relationships between pressure, volume, and temperature, which makes solving the problem easier. Using the ideal gas law lets us deduce that: \[ V \propto \frac{T}{P} \]This relationship helps in deriving final temperatures and exit velocities using the polytropic process equations.
energy balance equation
The energy balance equation in thermodynamics is a statement of the conservation of energy. It says energy entering a system equals energy leaving it, plus any stored energy changes. For a steady state process with negligible potential energy change, it takes the form:\[ h_{1} + \frac{V_{1}^{2}}{2} = h_{2} + \frac{V_{2}^{2}}{2} + q \]where:
  • \(h_1, h_2\) are the specific enthalpies at the inlet and outlet,
  • \(V_1, V_2\) are the velocities at the inlet and outlet, and
  • \(q\)
    • is the specific heat transfer.
Specific enthalpy (h) is related to temperature (T) through the specific heat capacity at constant pressure (\(C_P\)):\[ h = C_P T \]In the nozzle exercise, the energy balance equation helps determine the exit velocity and rate of heat transfer. Calculating these relies on known states of the gas and its flow, allowing us to find unknowns like exit velocity and heat transfer.

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Most popular questions from this chapter

A closed system undergoes a process in which work is done on the system and the heat transfer \(Q\) occurs only at temperature \(T_{\mathrm{b}}\). For each case, determine whether the entropy change of the system is positive, negative, zero, or indeterminate. (a) internally reversible process, \(Q>0\). (b) internally reversible process, \(Q=0\). (c) internally reversible process, \(Q<0\). (d) internal irreversibilities present, \(Q>0\). (e) internal irreversibilities present, \(Q=0\). (f) internal irreversibilities present, \(Q<0\).

Answer the following true or false. If false, explain why. A process that violates the second law of thermodynamics violates the first law of thermodynamics. (b) When a net amount of work is done on a closed system undergoing an internally reversible process, a net heat transfer of energy from the system also occurs. (c) One corollary of the second law of thermodynamics states that the change in entropy of a closed system must be greater than zero or equal to zero. (d) A closed system can experience an increase in entropy only when irreversibilities are present within the system during the process. (e) Entropy is produced in every internally reversible process of a closed system. (f) In an adiabatic and internally reversible process of a closed system, the entropy remains constant. (g) The energy of an isolated system must remain constant, but the entropy can only decrease.

Taken together, a certain closed system and its surroundings make up an isolated system. Answer the following true or false. If false, explain why. (a) No process is allowed in which the entropies of both the system and the surroundings increase. (b) During a process, the entropy of the system might decrease, while the entropy of the surroundings increases, and conversely. (c) No process is allowed in which the entropies of both the system and the surroundings remain unchanged. (d) A process can occur in which the entropies of both the system and the surroundings decrease.

According to test data, a new type of engine takes in streams of water at \(200^{\circ} \mathrm{C}, 3\) bar and \(100^{\circ} \mathrm{C}, 3\) bar. The mass flow rate of the higher temperature stream is twice that of the other. A single stream exits at \(3.0\) bar with a mass flow rate of \(5400 \mathrm{~kg} / \mathrm{h}\). There is no significant heat transfer between the engine and its surroundings, and kinetic and potential energy effects are negligible. For operation at steady state, determine the maximum theoretical rate that power can be developed, in \(\mathrm{kW}\).

Employing the ideal gas model, determine the change in specific entropy between the indicated states, in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). Solve three ways: Use the appropriate ideal gas table, \(I T\), and a constant specific heat value from Table A-20. (a) air, \(p_{1}=100 \mathrm{kPa}, T_{1}=20^{\circ} \mathrm{C}, p_{2}=100 \mathrm{kPa}, T_{2}=\) \(100^{\circ} \mathrm{C} .\) (b) air, \(p_{1}=1\) bar, \(T_{1}=27^{\circ} \mathrm{C}, p_{2}=3\) bar, \(T_{2}=377^{\circ} \mathrm{C}\). (c) carbon dioxide, \(p_{1}=150 \mathrm{kPa}, T_{1}=30^{\circ} \mathrm{C}, p_{2}=300 \mathrm{kPa}\), \(T_{2}=300^{\circ} \mathrm{C}\) (d) carbon monoxide, \(T_{1}=300 \mathrm{~K}, v_{1}=1.1 \mathrm{~m}^{3} / \mathrm{kg}, T_{2}=500 \mathrm{~K}\), \(v_{2}=0.75 \mathrm{~m}^{3} / \mathrm{kg}\) (e) nitrogen, \(p_{1}=2 \mathrm{MPa}, T_{1}=800 \mathrm{~K}, p_{2}=1 \mathrm{MPa}\), \(T_{2}=300 \mathrm{~K}\)
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