Answer: Why is even here then.
Explanation:
Drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been
removed.
Answer:
no it has to be removed
Explanation:
It is completely inappropriate to mention that the drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been removed. Therefore, the statement given above is false.
What is the significance of drum brakes?Drum brakes can be referred to or considered as the types of brakes that are useful in application of brakes to an object, such as wheels, in motion. To understand better, it can be stated that the system of braking under drum brakes is completely in contrast to that of disc brakes.
Drum brakes have a hydraulic pressure, which means that if the condition of lining is to be checked, removal of drums becomes essential. If the drums are not removed, correction or alignment of wheels cannot be performed.
Therefore, the significance regarding drum brakes has been aforementioned, and the statement given above with respect to their removal also holds false.
Learn more about drum brakes here:
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64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump efficiency of 74 percent, determine the required power input to the pump. Disregard frictional losses in the pipes, and assume the geothermal water at 200m depth to be exposed to the atmosphere.
Answer:
835,175.68W
Explanation:
Calculation to determine the required power input to the pump
First step is to calculate the power needed
Using this formula
P=V*p*g*h
Where,
P represent power
V represent Volume flow rate =0.3 m³/s
p represent brine density=1050 kg/m³
g represent gravity=9.81m/s²
h represent height=200m
Let plug in the formula
P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m
P=618,030 W
Now let calculate the required power input to the pump
Using this formula
Required power input=P/μ
Where,
P represent power=618,030 W
μ represent pump efficiency=74%
Let plug in the formula
Required power input=618,030W/0.74
Required power input=835,175.68W
Therefore the required power input to the pump will be 835,175.68W
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A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly Problems 49 CH001.qxd 2/24/11 12:03 PM Page 49 insulated. What is the minimum thickness of styrofoam insulation (k 0.030 W/m K) that must be applied to the top and side walls to ensure a heat load of less than 500 W, when the inner and outer surfaces are 10 and 35 C
Answer:
30 mm is the minimum thickness that must be applied.
Explanation:
Given the data in the question;
Using Fourier's equation. the heat rate is
q = kA(ΔT/Δx)
where
A is the surface area, we must consider all surfaces through which the heat can dissipate through
i.e 2×2 for one wall gives you 4m²,
there are 5 walls, so we will have 20m² for surface area.
k is thermal conductivity of the styrofoam ( 0.030 W/m K)
q is the heat loss (500 W )
ΔT is the Temperature difference ( 35 - 10) = 25°C
Δx = ?
So we substitute
500 = (0.030)(20)(25/Δx)
500 = 0.6 (25/Δx)
500 = 15 / Δx
Δx = 15 / 500
Δx = 0.03 m = 30 mm
Therefore, 30 mm is the minimum thickness that must be applied.
Please calculate the energy stored in a flying wheel. The steel flying wheel is 10 metric tons, in adiameter of 10m, and it rotates at a speed of 1000 rpm. The steel density is 8050 kg/m^3. When energyis fully released, the flying wheel stops its rotation.
Answer:
685.38 MJ
Explanation:
Given that:
mass = 10 tons = 1.0 × 10 ⁴ kg
diameter D = 10 m
radius R = 5 m
speed N = 1000 rpm
Using the formula for K.E = [tex]\dfrac{1}{2}I \omega^2[/tex] to calculate the energy stored
where;
[tex]= \dfrac{2 \pi \times N}{60}[/tex]
[tex]= \dfrac{2 \pi \times 1000}{60}[/tex]
= 104.719 rad/s
Hence, the energy stored is;
[tex]= \dfrac{1}{2}\times (\dfrac{MR^2}{2}) \times \omega^2[/tex]
[tex]= \dfrac{1}{2}\times (\dfrac{10^4\times 5^2}{2}) \times 104.719^2[/tex]
= 685379310.1
= 685.38 MJ
PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the following: a PWM signal source with 1 kHz base frequency and adjustable pulse-width modulation(PWM source), an analog filter with time constant of 0.01 s. Simulate the input and output of the low pass filter for a PWM duty cycle of 25, 50, and 100%
Answer:
Attached below
Explanation:
PWM signal source has 1 KHz base frequency
Analog filter : with time constant = 0.01 s
low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]
PWM duty cycle is a constant block
Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
