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Chapter 7: Problem 24

Steam at \(260^{\circ} \mathrm{C}\) and 7.00 bar absolute is expanded through a nozzle to \(200^{\circ} \mathrm{C}\) and 4.00 bar. Negligible heat is transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. The specific enthalpy of steam is \(2974 \mathrm{kJ} / \mathrm{kg}\) at \(260^{\circ} \mathrm{C}\) and 7 bar and \(2860 \mathrm{kJ} / \mathrm{kg}\) at \(200^{\circ} \mathrm{C}\) and 4 bar. Use the open-system energy balance to calculate the exit steam velocity.

Short Answer

Expert verified
The exit steam velocity is 477.47 m/s

Step by step solution

01

Identify the given quantities

The enthalpy change of the steam, i.e., the difference in specific enthalpy, is identified. This is given as \(2974 \mathrm{kJ/kg} - 2860 \mathrm{kJ/kg} = 114 \mathrm{kJ/kg}\). The specific enthalpy decreases because the steam does work on the surroundings as it expands through the nozzle.
02

Apply open system energy balance

The first law of thermodynamics in specific terms is written as: \( \Delta H = \Delta KE + \Delta PE\). In this case, potential energy change (\Delta PE) is 0, because there is no mention of a change in altitude or height and the kinetic energy change (\Delta KE) is represented by the exit steam velocity (substituting 1/2 m v^2 for \Delta KE and removing mass from both sides because we're dealing with specific quantities). Therefore, 114 kJ/kg = \Delta KE. Hence, \Delta KE = 1/2 v^2, where v is the exit velocity we need to find.
03

Solve for the exit steam velocity

Solve the equation for v. First, multiply both sides by 2: 228 kJ/kg = v^2. Convert kJ to J by multiplying by 1000. Then take the square root on both sides to solve for v: v = \( \sqrt{228,000 m^2/s^2} \) = 477.47 m/s.

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Key Concepts

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

Open System Energy Balance
In thermodynamics, an open system exchanges energy with its surroundings in the form of work and heat. It's important to balance this energy when analyzing the behavior of substances like steam flowing through a nozzle.

The open system energy balance is expressed as:
  • The change in enthalpy ( abla H) equals the changes in kinetic and potential energy ( abla KE and abla PE).
  • In many practical situations, such as the problem described, potential energy change ( abla PE) is nil if there are no altitude changes.
  • Hence, abla H is effectively converted into kinetic energy ( abla KE), which explains why we focus on speed or velocity changes when talking about open systems like nozzles.

This approach essentially simplifies the complexity by depicting how energy transfer into motion occurs without losing energy to heat, especially when the surroundings have negligible heat exchange.
Specific Enthalpy
Specific enthalpy is a thermodynamic property that represents the heat content per unit mass of a substance. It's crucial for calculating energy changes in processes involving heating or expansion.

In this exercise, steam's specific enthalpy indicates what the steam carries per kilogram:
  • At higher temperature and pressure, steam holds more energy, i.e., it has a higher specific enthalpy (2974 kJ/kg at 260掳C and 7 bar).
  • As steam expands to lower temperature and pressure in the nozzle, it loses some of this energy (2860 kJ/kg at 200掳C and 4 bar).

The change in specific enthalpy ( abla H) between the entrance and exit of the nozzle links directly to the steam's velocity change. This energy loss is what powers the velocity as steam exits the nozzle.
First Law of Thermodynamics
The First Law of Thermodynamics is about energy conservation in processes. It states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but it cannot be created or destroyed.

In practical applications like the given steam nozzle problem:
  • The law quantifies how heat energy ( abla H) is converted to mechanical energy ( abla KE) as steam does work by expanding.
  • The absence of heat exchange outside of the nozzle system aligns with conservation - energy remains within the system transforming mainly into kinetic energy.

This principle helps us mathematically express energy exchanges and transformations to reliably predict outputs like exit steam velocity by using initial conditions and affect how engineers design system components to efficiently use and control energy in processes.

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Most popular questions from this chapter

The specific enthalpy of liquid \(n\) -hexane at 1 atm varies linearly with temperature and equals \(25.8 \mathrm{kJ} / \mathrm{kg}\) at \(30^{\circ} \mathrm{C}\) and \(129.8 \mathrm{kJ} / \mathrm{kg}\) at \(50^{\circ} \mathrm{C}\) (a) Determine the equation that relates \(\hat{H}(\mathrm{kJ} / \mathrm{kg})\) to \(T\left(^{\circ} \mathrm{C}\right)\) and calculate the reference temperature on which the given enthalpies are based. Then derive an equation for \(\hat{U}(T)(\mathrm{kJ} / \mathrm{kg})\) at 1 atm. (b) Calculate the heat transfer rate required to cool liquid \(n\) -hexane flowing at a rate of \(20 \mathrm{kg} / \mathrm{min}\) from \(60^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\) at a constant pressure of 1 atm. Estimate the change in specific internal energy \((\mathrm{kJ} / \mathrm{kg})\) as the n-hexane is cooled at the given conditions.

A perfectly insulated cylinder fitted with a leakproof frictionless piston with a mass of \(30.0 \mathrm{kg}\) and a face area of \(400.0 \mathrm{cm}^{2}\) contains \(7.0 \mathrm{kg}\) of liquid water and a 3.0 -kg bar of aluminum. The aluminum bar has an electrical coil imbedded in it, so that known amounts of heat can be transferred to it. Aluminum has a specific gravity of 2.70 and a specific internal energy given by the formula \(\hat{U}(\mathrm{kJ} / \mathrm{kg})=0.94 \mathrm{T}\left(^{\circ} \mathrm{C}\right)\) The internal energy of liquid water at any temperature may be taken to be that of the saturated liquid at that temperature. Negligible heat is transferred to the cylinder wall. Atmospheric pressure is 1.00 atm. The cylinder and its contents are initially at \(20^{\circ} \mathrm{C}\). Suppose that \(3310 \mathrm{kJ}\) is transferred to the bar from the heating coil and the contents of the cylinder are then allowed to equilibrate. (a) Calculate the pressure of the cylinder contents throughout the process. Then determine whether the amount of heat transferred to the system is sufficient to vaporize any of the water. (b) Determine the following quantities: (i) the final system temperature; (ii) the volumes \(\left(\mathrm{cm}^{3}\right)\) of the liquid and vapor phases present at equilibrium; and (iii) the vertical distance traveled by the piston from the beginning to the end of the process. (Suggestion: Write an energy balance on the complete process, taking the cylinder contents to be the system. Note that the system is closed and that work is done by the system when it moves the piston through a vertical displacement. The magnitude of this work is \(W=P \Delta V,\) where \(P\) is the constant system pressure and \(\Delta V\) is the change in system volume from the initial to the final state.) (c) Calculate an upper limit on the temperature attainable by the aluminum bar during the process, and state the condition that would have to apply for the bar to come close to this temperature.

A 200.0 -liter water tank can withstand pressures up to 20.0 bar absolute before rupturing. At a particular time the tank contains \(165.0 \mathrm{kg}\) of liquid water, the fill and exit valves are closed, and the absolute pressure in the vapor head space above the liquid (which may be assumed to contain only water vapor) is 3.0 bar. A plant technician turns on the tank heater, intending to raise the water temperature to \(155^{\circ} \mathrm{C},\) but is called away and forgets to return and shut off the heater. Let \(t_{1}\) be the instant the heater is turned on and \(t_{2}\) the moment before the tank ruptures. Use the steam tables for the following calculations. (a) Determine the water temperature, the liquid and head-space volumes (L), and the mass of water vapor in the head space (kg) at time \(t_{1}\) (b) Determine the water temperature, the liquid and head-space volumes (L), and the mass of water vapor (g) that evaporates between \(t_{1}\) and \(t_{2}\). (Hint: Make use of the fact that the total mass of water in the tank and the total tank volume both remain constant between \(t_{1}\) and \(t_{2}\).) (c) Calculate the amount of heat (kJ) transferred to the tank contents between \(t_{1}\) and \(t_{2}\). Give two reasons why the actual heat input to the tank must have been greater than the calculated value.

Agricultural irrigation uses a significant amount of water, and in some regions it has overwhelmed other water needs. Suppose water is drawn from a reservoir and delivered into an irrigation ditch. For most of the length of the ditch, the delivery is through a \(10-\mathrm{cm}\) ID pipe, and in the last few meters the pipe diameter is \(7 \mathrm{cm} .\) The exit from the pipe is \(300 \mathrm{m}\) lower than the pipe inlet. (a) Assume that the pipe is smooth (i.e., ignore friction) and that the delivery rate is 4000 \(\mathrm{kg} / \mathrm{h}\). Estimate the required pressure difference between pipe inlet and outlet. How far below the surface of the reservoir is the pipe inlet? (b) How would your answer be different if the pipe were not smooth? Explain. Exploratory Exercise- Research and Discover (c) What are possible environmental impacts of diverting significant quantities of river water for use in irrigation? Cite at least two sources for your response.

Write and simplify the closed-system energy balance (Equation \(7.3-4\) ) for each of the following processes, and state whether nonzero heat and work terms are positive or negative. Begin by defining the system. The solution of Part (a) is given as an illustration. (a) The contents of a closed flask are heated from \(25^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). (b) A tray filled with water at \(20^{\circ} \mathrm{C}\) is put into a freezer. The water tums into ice at \(-5^{\circ} \mathrm{C}\). (Note: When a substance expandsit does work on its surroundings and when it contracts the surroundings do work on it.) (c) A chemical reaction takes place in a closed adiabatic (perfectly insulated) rigid container. (d) Repeat Part (c), only suppose that the reactor is isothermal rather than adiabatic and that when the reaction was carried out adiabatically, the temperature in the reactor increased.
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