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Chapter 1: Problem 36

Pressurized water \(\left(p_{\text {in }}=10\right.\) bar, \(\left.T_{\text {in }}=110^{\circ} \mathrm{C}\right)\) enters the bottom of an \(L=10\)-m-long vertical tube of diameter \(D=100 \mathrm{~mm}\) at a mass flow rate of \(\dot{m}=1.5 \mathrm{~kg} / \mathrm{s}\). The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at \(p_{\text {out }}=7\) bar, \(T_{\text {out }}=600^{\circ} \mathrm{C}\). Determine the change in the rate at which the following quantities enter and exit the tube: (a) the combined thermal and flow work, (b) the mechanical energy, and (c) the total energy of the water. Also, (d) determine the heat transfer rate, \(q\). Hint: Relevant properties may be obtained from a thermodynamics text.

Short Answer

Expert verified
To find the change in the rate at which the combined thermal and flow work, mechanical energy, and total energy enter and exit the tube, as well as the heat transfer rate, follow these steps: 1. Obtain relevant properties of water at given pressures and temperatures, specifically specific enthalpy (h), and specific volume (v) for the entry and exit points of the water in the tube. 2. Calculate the water's velocities at the entrance and exit points using the mass flow rate (\(\dot{m}\)) and specific volume (v). 3. Evaluate the change in the combined thermal and flow work by using specific enthalpy values: \(\Delta \dot{W}_{thermal+flow}=\dot{m}(h_{out}-h_{in})\) 4. Assess the change in mechanical energy across the tube using the velocities obtained in Step 2: \(\Delta \dot{W}_{mechanical}=\frac{1}{2} \dot{m} (v_{out}^2-v_{in}^2)\) 5. Determine the total energy change of the water by summing the changes in thermal and flow work and mechanical energy: \(\Delta \dot{W}_{total}=\Delta \dot{W}_{thermal+flow}+\Delta \dot{W}_{mechanical}\) 6. Calculate the heat transfer rate, q, using the First Law of Thermodynamics and the calculated total energy change: \(q=\Delta \dot{W}_{total}+\dot{m}(h_{out}-h_{in})\) By completing these steps, you will have determined the change in the rate of the desired quantities and the heat transfer rate.

Step by step solution

01

Obtain relevant properties of water at given pressures and temperatures

You will need to refer to a thermodynamics text or a steam table to find the specific enthalpy (h), and specific volume (v) for the entry and exit points of the water in the tube. You will have four values: \(h_{in}\), \(v_{in}\), \(h_{out}\), and \(v_{out}\).
02

Calculate the velocity of the water at the inlet and outlet points

Using the specific volume (v) and mass flow rate (\(\dot{m}\)), we can evaluate the water's velocity (\(v_{in}\) and \(v_{out}\)) at the entrance and exit points using the formula: \[v_{water}=\frac{\dot{m}}{\rho A}\] where \(\rho\) is the density, \(A\) is the cross-sectional area of the tube, and is calculated using the given diameter. \[A = \frac{\pi D^2}{4}\] Calculate \(\rho_{in}\), \(v_{in}\), \(\rho_{out}\), and \(v_{out}\).
03

Evaluate the change in the combined thermal and flow work entering and exiting the tube

Using the known specific enthalpies \(h_{in}\) and \(h_{out}\), we can find the change in the combined thermal and flow work using the following formula: \[\Delta \dot{W}_{thermal+flow}=\dot{m}(h_{out}-h_{in})\] Calculate \(\Delta \dot{W}_{thermal+flow}\) using the obtained specific enthalpy values.
04

Assess the change in mechanical energy entering and exiting the tube

Using the velocities from Step 2, we can determine the change in mechanical energy across the tube using the following formula: \[\Delta \dot{W}_{mechanical}=\frac{1}{2} \dot{m} (v_{out}^2-v_{in}^2)\] Calculate \(\Delta \dot{W}_{mechanical}\) using the velocity values obtained in Step 2.
05

Determine the total energy change of the water

Sum the changes in thermal and flow work, and mechanical energy to determine the total energy change of the water: \[\Delta \dot{W}_{total}=\Delta \dot{W}_{thermal+flow}+\Delta \dot{W}_{mechanical}\] Calculate \(\Delta \dot{W}_{total}\) using the values obtained in Steps 3 and 4.
06

Calculate the heat transfer rate, q

Using the First Law of Thermodynamics and the calculated total energy change, we can find the heat transfer rate, q: \[q=\Delta \dot{W}_{total}+\dot{m}(h_{out}-h_{in})\] Calculate q using \(\Delta \dot{W}_{total}\) obtained in Step 5 and the specific enthalpy values. By following these steps, you will have determined the change in the rate at which the combined thermal and flow work, mechanical energy, and total energy enter and exit the tube, as well as the heat transfer rate, q.

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Key Concepts

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

Thermal and Flow Work
Understanding thermal and flow work is pivotal in the realm of thermodynamics, particularly when dealing with scenarios involving fluid flow and heat exchange, such as in the given exercise. Thermal work (also known as 'flow work') is the work needed to push mass into or out of a control volume, and it's intrinsically connected to the pressure and volume of the fluid.

In the context of the exercise, the thermal and flow work is calculated by multiplying the mass flow rate \( \dot{m} \) by the difference in specific enthalpy \( h \) at the inlet and outlet. The specific enthalpy encapsulates the energy needed to create space for the fluid (flow work) in addition to its internal energy (thermal work).

The formula for evaluating this change in thermal and flow work is:
\[ \Delta \dot{W}_{thermal+flow}=\dot{m}(h_{out}-h_{in}) \]
Here, \( h_{in} \) and \( h_{out} \) are determined using a steam table, which lists the specific enthalpy values at the given pressures and temperatures for water and steam.
Mechanical Energy in Thermodynamics
Mechanical energy in the context of thermodynamics, especially within a flowing fluid, includes kinetic and potential energies. The exercise at hand deals predominantly with the kinetic energy part, due to the motion of the water through the tube.

For our purposes, the mechanical energy change is the difference between the kinetic energies at the entrance and exit of the control volume 鈥� in this case, a section of the tube. Since potential energy is not explicity mentioned, we assume its change to be negligible due to the constancy of gravitational acceleration and similar inlet and outlet heights.

The formula used to calculate the change in mechanical energy is:
\[ \Delta \dot{W}_{mechanical}=\frac{1}{2} \dot{m} (v_{out}^2-v_{in}^2) \]
This takes into account the velocities \( v_{in} \) and \( v_{out} \) at the inlet and outlet, which are derived from the mass flow rate and densities corresponding to the inlet and outlet conditions.
Total Energy Change
The total energy change of the system encompasses both thermal (including flow) work and mechanical energy. In the exercise we're examining, this encompasses the enthalpy change due to heat transfer and the kinetic energy change due to velocity variation.

Summing the changes in thermal and flow work with the changes in mechanical energy gives us the total energy change of the water, or any fluid system undergoing similar conditions:
\[ \Delta \dot{W}_{total}=\Delta \dot{W}_{thermal+flow}+\Delta \dot{W}_{mechanical} \]
Such a comprehensive view of energy transformation is fundamental for a thorough analysis of energy conservation in thermodynamic processes. It is essential for students to grasp that the total energy change is the sum total of all forms of energy work - not just the sum of their numerical values but a representation of interconversion of energy forms in compliance with the First Law of Thermodynamics.
Steam Table Usage
When solving thermodynamic problems involving phase changes and properties of water and steam, as in our given problem, steam tables become indispensable tools. They provide crucial data, such as specific enthalpy \( h \) and specific volume \( v \) - necessary metrics in determining energy changes and work.

For accurate heat transfer rate calculations, you locate the properties of water at the specified pressure and temperature conditions (for both inlet and outlet) in these tables. The values found are tailored to the water's state 鈥� whether it's saturated liquid, saturated vapor, or superheated steam, each requiring a careful consultation of the steam table.

Furthermore, in our exercise, to calculate the velocity of water, the specific volume \( v \) from the steam table is converted to density \( \rho \) using its reciprocal, and then employed with the mass flow rate to find the inlet and outlet velocities. Hence, proficiency in using these tables is vital for anyone working on thermodynamics and heat transfer problems.

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Most popular questions from this chapter

A freezer compartment is covered with a 2 -mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

In one stage of an annealing process, 304 stainless steel sheet is taken from \(300 \mathrm{~K}\) to \(1250 \mathrm{~K}\) as it passes through an electrically heated oven at a speed of \(V_{s}=10 \mathrm{~mm} / \mathrm{s}\). The sheet thickness and width are \(t_{s}=8 \mathrm{~mm}\) and \(W_{s}=2 \mathrm{~m}\), respectively, while the height, width, and length of the oven are \(H_{o}=2 \mathrm{~m}\), \(W_{o}=2.4 \mathrm{~m}\), and \(L_{o}=25 \mathrm{~m}\), respectively. The top and four sides of the oven are exposed to ambient air and large surroundings, each at \(300 \mathrm{~K}\), and the corresponding surface temperature, convection coefficient, and emissivity are \(T_{s}=350 \mathrm{~K}, h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(\varepsilon_{s}=0.8\). The bottom surface of the oven is also at \(350 \mathrm{~K}\) and rests on a \(0.5\)-m-thick concrete pad whose base is at \(300 \mathrm{~K}\). Estimate the required electric power input, \(P_{\text {elec }}\), to the oven.

A small sphere of reference-grade iron with a specific heat of \(447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and a mass of \(0.515 \mathrm{~kg}\) is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to \(14^{\circ} \mathrm{C}\) in \(6.35 \mathrm{~s}\). The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of \(1.263 \mathrm{~kg}\). If the same observed temperature change occurs in \(4.59 \mathrm{~s}\), what is the specific heat of the unknown material?

A cartridge electrical heater is shaped as a cylinder of length \(L=200 \mathrm{~mm}\) and outer diameter \(D=20 \mathrm{~mm}\). Under normal operating conditions, the heater dissipates \(2 \mathrm{~kW}\) while submerged in a water flow that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Neglecting heat transfer from the ends of the heater, determine its surface temperature \(T_{s}\). If the water flow is inadvertently terminated while the heater continues to operate, the heater surface is exposed to air that is also at \(20^{\circ} \mathrm{C}\) but for which \(h=50\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding surface temperature? What are the consequences of such an event?
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