Air enters a length of constant area pipe with p1= 200 kPa( abs ), T1=500 K, and V1=400 m / s .

step 1 Let me wait to this going. Air enters length of constant area pipe with a P1, 200 khal's absolut

Air enters a length of constant area pipe with p1= 200 kPa(...

Air enters a length of constant area pipe with $p_{1}=$ $200 \mathrm{kPa}(\text { abs }), T_{1}=500 \mathrm{K},$ and $V_{1}=400 \mathrm{m} / \mathrm{s} .$ If $500 \mathrm{kJ} / \mathrm{kg}$ of en ergy is removed from the air by frictionless heat transfer between sections (1) and $(2),$ determine $p_{2}, T_{2},$ and $V_{2} .$ Sketch a temperature-entropy diagram for the flow between sections and

Air enters a length of constant area pipe with $p_{1}=$ $200 \mathrm{kPa}(\text { abs }), T_{1}=500 \mathrm{K},$ and $V_{1}=400 \mathrm{m} / \mathrm{s} .$ If $500 \mathrm{kJ} / \mathrm{kg}$ of en
ergy is removed from the air by frictionless heat transfer between sections (1) and $(2),$ determine $p_{2}, T_{2},$ and $V_{2} .$ Sketch a temperature-entropy diagram for the flow between sections and

Key Concepts

Temperature-Entropy (T-s) Diagram

A Temperature-Entropy diagram is a graphical tool used in thermodynamics to represent the state changes of a system and to visualize the effects of heat transfer and process reversibility. It plots temperature on the vertical axis and entropy on the horizontal axis, portraying how a fluid’s state path changes in processes like heating or cooling. In reversible processes with heat transfer, the diagram can illustrate the decrease in entropy corresponding to heat removal, and it helps in understanding the thermodynamic efficiency and the nature of the process.

Conservation of Mass in Open Systems

The conservation of mass in open systems states that the mass flow rate remains constant along a streamline, especially in a constant-area pipe. This principle, often described by the continuity equation, ensures that the product of cross-sectional area, density, and velocity is the same at different sections in the absence of mass sources or sinks, thereby providing an important constraint when solving for unknown flow variables.

Steady Flow Energy Equation

The steady flow energy equation is a form of the first law of thermodynamics applied to open systems. It accounts for the energy balance between the enthalpy, kinetic energy, and potential energy of a flowing fluid, along with the work and heat interactions across a control volume. In problems involving heat transfer with a compressible fluid, this equation is used to relate the changes in temperature, pressure, and velocity as the fluid flows through a duct or a pipe.

Ideal Gas Thermodynamic Relations

For fluids like air that can be modeled as an ideal gas, the relationship between thermodynamic properties is governed by the ideal gas law along with specific heat capacities. These relations allow for the connection between temperature and enthalpy changes (by using cp for constant pressure processes), and they are essential when computing changes in state variables such as pressure, temperature, and density resulting from energy exchange in the system.

Transcript

0:00 Let me wait to this going.

00:01 Air enters length of constant area pipe with a p1, 200 khal's absolute, a t1, a 500 kelvin, and an initial velocity of 400 meters per second.

00:12 If a 500 kilojoules per kilogram of energy is removed from the air by frictionless heat transfer between sections 1 and 2, what am i supposed to find there? determine p2, t2, and v2, and sketch the temperature.

00:42 Time.

00:43 Okay, now we've got it.

00:47 Okay, so let's start with the heat discharge equation, and that'll be plus t -o -1.

01:11 We need the mock value for this section, and m -a -1 will equal v -1 over c -1, which equal v1 over square root of r -t1, k, and this will equal 400 to divided by the square root of 286 .9 times 500 times 1 .4.

01:55 And this will equal 0 .89.

02:00 So from that value, we're going to get the ratio from 0 .89.

02:10 We're going to get t1 over t01, equal 0 .899.

02:18 And t01 will equal 0 .899.

02:22 And t01 will equal 500 divided by 0 .99 equals 579 .8 kelvin and then t02 equal 500 divided by 1 .004 kelvin plus 580 kelvin and that'll equal 81 .99 kelvin.

03:08 That's my t2.

03:12 Okay.

03:13 And then from figure d3 we're getting the values for the next list.

03:17 81 .37.

03:36 Okay, so let me get this going here.

03:44 Then from figure d2, d3, next list of ratios, t1 over ta equals 1 .02, p1 over pa is 1 .14, v1 over va equals 0 .9, and t01 over t0a ,000.

04:13 Equals 0 .99.

04:17 Then we can evaluate find ma2 t02 t01 t o1 t oa and this will equal 82 times 99.

04:54 99.

04:57 Divided by 580 equals 0 .139.

05:05 So the mock number from a we can look at the book and we end up with mock a2 is equal to 0 .177...