Assignment 6 - Part a

# Assignment 6 - Part a - Assignment 5 The rate of doing work...

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Unformatted text preview: Assignment 5 The rate of doing work term in the Energy Balance requires special consideration in the development of the forms useful in solving macroscopic problems. This is covered here {see Welty et. al., pgs. 67-68, and Fox et. at, pgs. 134-136 for different approaches). Show all steps in the following development of the rate of doing work term in the Macroscopic Energy Balance [i.e., the First Law of Thermodynamics applied to a CV). First: What is work? Work is the energy required to move a body over a distance under the action of a force. In rigid body mechanics, work equals force time distance. Hence, the rate of doing work is force times velocity. We are interested in developing the relation for W term in the First Law of Thermodynamic applied to a control volume involving flow (or the MEB). ' Use the definition of the stress in a flowing fluid to show that dF=(s'I)dA (1) Then explain why the rate of doing work associated with the force acting on a control volume surface is W: l3-(s-Z‘MA (2) (.2‘ where 2 is the velocity at the surface. Now the velocity at a surface can be divided into normal and tangential components, i.c., v = vng+ v 4/1. as shown in the ﬁgure. Surface dz! With this velocity, show that K'(ﬂ'l’)=VnT rm HAT"; (3) The stress can be divided into its pressure and deviatoric parts: I = ‘Pﬂ + z (4) Show that TM 2 p + rm (5) if}; = 1'" A (6) and 245-1") = (-17 + rm. m + a,sz - (7] or from (2): W = I[(—p + rm)“, + rum ]dA (8) cs For the CV shown on the next page, there are three types of surfaces that make up the CS, the inlet ports A the solid stationary surfaces A punk 3 3m: 3 and the moving surfaces AWM. On the stationary solid surfaces: v” = 0, \$2,, = 0. Also, if the ﬂow is uniform at the inlet and exit ports, there is no ﬂuid deformation and rm 2 0 on these surfaces. Using this information show that rate of work of the ﬂuid on the CS (and on the external environment) is W A; _ I pvndA + deﬂ + mlwr AM, where Wm}; = I [(FP + Tm. M + m ‘5: (10) Am.” “given! = I V]! raid}! I A mm Explain why Pam = 0 if the port surface orientations are normal to the velocity at the port sections. ...
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## This note was uploaded on 10/09/2011 for the course ENGINEERIN 3133 taught by Professor Sarkozi during the Spring '11 term at Texas A&M.

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Assignment 6 - Part a - Assignment 5 The rate of doing work...

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