Consider the fuel element of Example 5.11, which operates at a uniform volumetric generation rate of q ˙ = 10 7 W / m 3 , until the generation rate suddenly changes to q ˙ = 2 × 10 7 W / m 3 . Use the Finite-Difference Equations, One-Dimensional, Transient conduction model builder of IHT to obtain the implicit form of the finite-difference equations for the 6 nodes, with 2 mm, as shown in the example. (a) Calculate the temperature distribution 1.5 s after the change in operating power, and compare your results with those tabulated in the example. (b) Use the Explore and Graph opt ions of IHT to calculate and plot temperature histories at the midplane (00) and surface (05) nodes for 0 ≤ t ≤ 400 s . What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?
Consider the fuel element of Example 5.11, which operates at a uniform volumetric generation rate of q ˙ = 10 7 W / m 3 , until the generation rate suddenly changes to q ˙ = 2 × 10 7 W / m 3 . Use the Finite-Difference Equations, One-Dimensional, Transient conduction model builder of IHT to obtain the implicit form of the finite-difference equations for the 6 nodes, with 2 mm, as shown in the example. (a) Calculate the temperature distribution 1.5 s after the change in operating power, and compare your results with those tabulated in the example. (b) Use the Explore and Graph opt ions of IHT to calculate and plot temperature histories at the midplane (00) and surface (05) nodes for 0 ≤ t ≤ 400 s . What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?
Solution Summary: The author calculates the energy balance to the center of the rod using one dimensional conduction with uniform heat generation.
Consider the fuel element of Example 5.11, which operates at a uniform volumetric generation rate of
q
˙
=
10
7
W
/
m
3
, until the generation rate suddenly changes to
q
˙
=
2
×
10
7
W
/
m
3
. Use the Finite-Difference Equations, One-Dimensional, Transient conduction model builder of IHT to obtain the implicit form of the finite-difference equations for the 6 nodes, with 2 mm, as shown in the example.
(a) Calculate the temperature distribution 1.5 s after the change in operating power, and compare your results with those tabulated in the example.
(b) Use the Explore and Graph opt ions of IHT to calculate and plot temperature histories at the midplane (00) and surface (05) nodes for
0
≤
t
≤
400
s
. What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?
Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a
mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and
(y2), respectively.
Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s].
Givens:
y1 = 4.112 m
y2 =
0.387 m
b = 0.942 m
Answers:
( 1 ) 1880.186 lit/s
( 2 ) 4042.945 lit/s
( 3 ) 2553.11 lit/s
( 4 ) 3130.448 lit/s
Problem (14): A pump is being used to lift water from an underground
tank through a pipe of diameter (d) at discharge (Q). The total head
loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h
where (V) is the flow velocity in the pipe. The elevation difference
between the pump and tank surface is (h).
Given the values of h [cm], d [cm], and K [-], calculate the maximum
discharge Q [Lit/s] beyond which cavitation would take place at the
pump entrance. Assume Turbulent flow conditions.
Givens:
h = 120.31 cm
d = 14.455 cm
K = 8.976
Q
Answers:
(1) 94.917 lit/s
(2) 49.048 lit/s
( 3 ) 80.722 lit/s
68.588 lit/s
4
Problem (13): A pump is being used to lift water from the bottom
tank to the top tank in a galvanized iron pipe at a discharge (Q).
The length and diameter of the pipe section from the bottom tank
to the pump are (L₁) and (d₁), respectively. The length and
diameter of the pipe section from the pump to the top tank are
(L2) and (d2), respectively.
Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m],
calculate total head loss due to friction (i.e., major loss) in the
pipe (hmajor-loss) in [cm].
Givens:
L₁,d₁
Pump
L₂,d2
오
0.533 lit/s
L1 =
6920.729 m
d1 =
1.065 m
L2 =
70.946 m
d2
0.072 m
Answers:
(1)
3.069 cm
(2) 3.914 cm
( 3 ) 2.519 cm
( 4 ) 1.855 cm
TABLE 8.1
Equivalent Roughness for New Pipes
Pipe
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Equivalent Roughness, &
Feet
Millimeters
0.003-0.03 0.9-9.0
0.001-0.01 0.3-3.0
0.0006-0.003 0.18-0.9
0.00085
0.26
0.0005
0.15
0.045
0.000005
0.0015
0.0 (smooth) 0.0 (smooth)
Commercial steel or wrought iron 0.00015
Drawn…
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