=
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= 0 1 1 1
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14
However = 1 0 over the opening A1 and zero
elsewhere around the surface
of the vessel. Therefore if the coolant density at the opening
is p1, we get
The solid
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.
=
=1
+
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(4.33)
19
The momentum law, applied to the control volume, accounts
for the rate of momentum change because of both the
accumulation of momentum due to net influx and
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13
Consider loss of coolant from a pressure vessel. The process
can be described by observing the vessel as a control volume
or the coolant as a control mass. Assume the coolant
=
+ (4.3)
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7
The right-hand side of Equation 4.3 describes the rate of change
of the variable in Eulerian coordinates, and the left-hand side
describes the time rate of change in
, =
,
+
,
(4.12)
, =
(4.13)
,
=
,
, =
, +
,
(4.14)
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=
(4.15)
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11
Where = relative velocity of the material with respect to the
surface of the control volume:
T
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2. If valve C is opened just until the
pressure in the two tanks is equalized and
is then closed, what are the final
temperature and pressure in tanks A and
B? Assume no transfer
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= =1 +
.
.
(4.39)
+
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+
20
The energy equation (first law of thermodynamics) applied to a
control volume takes into consideration the rate of change of
energy in the co
=
0
= 0.5
3.5
Solve for T1.
0
1
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Solve for n2/n, and you also know n1/n.
The energy loss from n1 shall equal energy gain of
n2. Therefore,
1 0 1 = 2 2 0
Solve for T2.
0
We assume t
Chapter 3 Reactor Energy
Distribution
Xue Yang, PhD
Assistant Professor of Mechanical Engineering
Texas A&M University-Kingsville
9/8/2016
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1
3.4 Energy generation parameters
Example 3.2: He