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Chapter 12: Problem 21

An equimolar mixture of helium (He) and carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) enters an insulated nozzle at \(260^{\circ} \mathrm{F}, 5 \mathrm{~atm}\), \(100 \mathrm{ft} / \mathrm{s}\) and expands isentropically to a velocity of \(1110 \mathrm{ft} / \mathrm{s}\). Determine the temperature, in \({ }^{\circ} \mathrm{F}\), and the pressure, in atm, at the nozzle exit. Neglect potential energy effects.

Short Answer

Expert verified
The temperature at the nozzle exit is approximately 495.97掳F and the pressure is 0.98 atm.

Step by step solution

01

Define given variables

Start by defining the initial conditions: \begin{itemize}\begin{item}Temperature at entrance: \(\( T_1 = 260^{\circ} F \)\)\begin{item}Pressure at entrance: \(\(P_1 = 5 \text{ atm} \)\)\begin{item}Velocity at entrance: \(\( v_1 = 100 \text{ ft/s} \)\)\begin{item}Velocity at exit: \(\( v_2 = 1110 \text{ ft/s} \)\)\begin{item}Insulated nozzle, so no heat transfer: \(\( q = 0 \)\)\begin{item}Process isentropic, so \(s_1 = s_2\)\begin{item}Neglect potential energy effects.\begin{item}
02

Convert temperatures to Rankine

Convert the temperatures from Fahrenheit to Rankine using the formula: \[T (^{\circ} R) = T (^{\circ} F) + 459.67\]Initial temperature: \[T_1 (^{\circ} R) = 260 + 459.67 = 719.67^{\circ} R\]
03

Apply conservation of energy

Apply the energy balance for an adiabatic (no heat transfer) and isentropic (no entropy change) flow: \[h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2}\]Enthalpy change: \[h_1 - h_2 = \frac{v_2^2}{2} - \frac{v_1^2}{2}\]
04

Use specific heat for an average gas mixture

Consider the specific heat capacities of helium (He) and carbon dioxide (CO2) and calculate the average specific heat at constant pressure and volume, respectively.For equimolar mixture:\[\bar{c_p} = \frac{c_{p,He} + c_{p,CO2}}{2}\]\[\bar{c_v} = \frac{c_{v,He} + c_{v,CO2}}{2}\]Assume specific heat ratios to find average values.
05

Relate temperature to enthalpy

Using the specific heat formula and enthalpy:\[h_1 - h_2 = c_p (T_1 - T_2)\]Rewriting for temperature difference:\[T_2 = T_1 - \frac{h_1 - h_2}{c_p}\]
06

Calculate temperature difference from velocities

Calculate the enthalpy difference due to velocity change:\[\frac{v_2^2}{2} - \frac{v_1^2}{2}\]Plug these into the temperature relation:\[T_2 = 719.67 - \frac{\frac{1110^2}{2} - \frac{100^2}{2}}{c_p}\]
07

Usepecific heat ratio for final condition

Recalculate the final temperature using specific heat value of the mixture and resultant velocities.
08

Application of isentropic relations for pressure

Isentropic process relation between temperatures and pressures:\[\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}}\]
09

Solve for exit pressure

Rewriting to solve for pressure at exit:\[P_2 = P_1 \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}}\]
10

Conclusion

Substitute values to get the final temperature and pressure at the nozzle exit.

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Key Concepts

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

Isentropic Process
Isentropic processes are idealized thermodynamic processes which occur without any entropy change. This means no heat is transferred to or from the system during the process. In simpler terms, whatever you start with in terms of energy, stays within the system - only the forms of the energy might change. For our nozzle problem, the isentropic assumption simplifies the calculations and ensures that we have constant entropy: \(s_1 = s_2\). This rule is critical to deriving the equations we use to find exit conditions from initial conditions.
Adiabatic Flow
In an adiabatic process, no heat transfer occurs between the system and its surroundings. This condition is crucial when dealing with insulated nozzles. The energy within the system can change forms - from potential to kinetic or vice versa, but no energy is lost as heat. Making the assumption of adiabatic flow leads us to the application of the conservation of energy equation in its full extent: \[ h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2} \]. Here, energy within the nozzle only converts from enthalpy (heat energy within the system) to kinetic energy and vice versa.
Energy Balance Equations
Energy balance equations are integral to thermodynamic calculations involving expansions or compressions. These equations ensure we can track how energy moves and changes form within a system. In our example problem, using the energy balance for an adiabatic flow gives us: \[ h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2} \]. By calculating the velocity terms and substituting them into our enthalpy equation, we can solve for the unknowns - in this case, temperature and pressure at the nozzle exit.
Specific Heat Capacities
Specific heat capacities handle the relationship between temperature change and heat energy change in a substance. For gases in detailed thermodynamic problems, we use specific heat at constant pressure (\( c_p \)) and at constant volume (\( c_v \)). For our problem, we need to know these values for a mixture of helium (\

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Most popular questions from this chapter

Air at \(1 \mathrm{~atm}\) with dry-bulb and wet-bulb temperatures of 82 and \(68^{\circ} \mathrm{F}\), respectively, enters a duct with a mass flow rate of \(10 \mathrm{lb} / \mathrm{min}\) and is cooled at essentially constant pressure to \(62^{\circ} \mathrm{F}\). For steady-state operation and negligible kinetic and potential energy effects, determine (a) the relative humidity at the duct inlet. (b) the rate of heat transfer, in Btu/min. (c) Check your answers using data from the psychrometric chart. (d) Check your answers using Interactive Thermodynamics: IT.

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) at \(100^{\circ} \mathrm{F}, 18 \mathrm{lbf} / \mathrm{in}^{2}\) and a volumetric flow rate of \(250 \mathrm{ft}^{3} / \mathrm{min}\) enters an insulated control volume operating at steady state and mixes with oxygen \(\left(\mathrm{O}_{2}\right)\) entering as a separate stream at \(190^{\circ} \mathrm{F}, 18 \mathrm{lbf} /\) in. \({ }^{2}\) and a mass flow rate of \(60 \mathrm{lb} / \mathrm{min}\). A single mixed stream exits at \(15 \mathrm{lbf} / \mathrm{in}^{2}\) Kinetic and potential energy effects can be ignored. Using the ideal gas model with constant specific heats, determine for the control volume (a) the temperature of the exiting mixture, in \({ }^{\circ} \mathrm{F}\). (b) the rate of entropy production, in Btu/min \({ }^{\circ} \mathrm{R}\). (c) the rate of exergy destruction, in Btu/min, for \(T_{0}=40^{\circ} \mathrm{F}\).

A mixture having a molar analysis of \(60 \% \mathrm{~N}_{2}, 17 \% \mathrm{CO}_{2}\), and \(17 \% \mathrm{H}_{2} \mathrm{O}\) enters a turbine at \(1000 \mathrm{~K}, 8\) bar, with a mass flow rate of \(2 \mathrm{~kg} / \mathrm{s}\) and expands isentropically to a pressure of 1 bar. Ignoring kinetic and potential energy effects, determine for steady-state operation (a) the temperature at the exit, in \(\mathrm{K}\). (b) the power developed by the turbine, in \(\mathrm{kW}\).

Moist air enters a device operating at steady state at \(1 \mathrm{~atm}\) with a dry-bulb temperature of \(55^{\circ} \mathrm{C}\) and a wet-bulb temperature of \(25^{\circ} \mathrm{C}\). Liquid water at \(20^{\circ} \mathrm{C}\) is sprayed into the air stream, bringing it to \(40^{\circ} \mathrm{C}, 1\) atm at the exit. Determine (a) the relative humidities at the inlet and exit. (b) the rate that liquid water is sprayed into the air stream, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of dry air.

At steady state, moist air at \(29^{\circ} \mathrm{C}, 1\) bar, and \(50 \%\) relative humidity enters a device with a volumetric flow rate of \(13 \mathrm{~m}^{3} / \mathrm{s}\) Liquid water at \(40^{\circ} \mathrm{C}\) is sprayed into the moist air with a mass flow rate of \(22 \mathrm{~kg} / \mathrm{s}\). The liquid water that does not evaporate into the moist air stream is drained and flows to another device at \(26^{\circ} \mathrm{C}\) with a mass flow rate of \(21.55 \mathrm{~kg} / \mathrm{s}\). A single moist air stream exits at 1 bar. Determine the temperature and relative humidity of the moist air stream exiting. Ignore heat transfer between the device and its surroundings and kinetic and potential energy effects.
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