EBK 3N3-EBK: INDUSTRIAL MOTOR CONTROL
7th Edition
ISBN: 9780176919962
Author: Herman
Publisher: VST
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Textbook Question
Chapter 40, Problem 5RQ
Refer to the circuit shown in Figure 40–5. When the Start button is pressed, the motor does not start. Which of the following could not cause this problem?
- a. The control transformer fuse is blown.
- b. Overload contact 2 is open.
- c. TR1 timer coil is open.
- d. Control relay coil CR is open.
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Students have asked these similar questions
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…
Chapter 40 Solutions
EBK 3N3-EBK: INDUSTRIAL MOTOR CONTROL
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- The flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forwardAn experimental setup is being built to study the flow in a large water main (i.e., a large pipe). The water main is expected to convey a discharge (Qp). The experimental tube will be built at a length scale of 1/20 of the actual water main. After building the experimental setup, the pressure drop per unit length in the model tube (APm/Lm) is measured. Problem (20): Given the value of APm/Lm [kPa/m], and assuming pressure coefficient similitude, calculate the drop in the pressure per unit length of the water main (APP/Lp) in [Pa/m]. Givens: AP M/L m = 590.637 kPa/m meen Answers: ( 1 ) 59.369 Pa/m ( 2 ) 73.83 Pa/m (3) 95.443 Pa/m ( 4 ) 44.444 Pa/m *******arrow_forwardFind the reaction force in y if Ain = 0.169 m^2, Aout = 0.143 m^2, p_in = 0.552 atm, Q = 0.367 m^3/s, α = 31.72 degrees. The pipe is flat on the ground so do not factor in weight of the pipe and fluid.arrow_forward
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