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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. 6768, and Fox et. at, pgs. 134136 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(sZ‘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 2451") = (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.
 Spring '11
 Sarkozi

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