Answer:
a). 139498.24 kg
b). 281.85 ohm
c). 10.2 ohm
Explanation:
Given :
Diameter, d = 22 m
Linear strain, = 3%
= 0.03
Young's modulus, E = 30 GPa
Gauge factor, k = 6.9
Gauge resistance, R = 340 Ω
a). Maximum truck weight
σ = Eε
σ =
= 342119.44 N
For the four sensors,
Maximum weight = 4 x P
= 4 x 342119.44
= 1368477.76 N
Therefore, weight in kg is
m = 139498.24 kg
b). Change in resistance
, since
Ω
For 4 resistance of the sensors,
Ω
c).
If linear strain,
, where k = 1
Ω
Answer:
The cathode ray oscilloscope (CRO) consists of a set of blocks. Those are vertical amplifier, delay line, trip circuit, time base generator, horizontal amplifier, cathode ray tube (CRT) and power supply. The CRO block diagram is shown in attached figure.
The function of each CRO block is mentioned below,
Vertical amplifier amplifies the input signal, which will be displayed on the CRT screen.
Delay line provides a certain amount of delay to the signal, which is obtained at the output of the vertical amplifier. This delayed signal is then applied to the CRT vertical deflection plates.
Trigger circuit produces a trigger signal to synchronize the horizontal and vertical deviations of the electron beam.
Time base generator produces a sawtooth signal, which is useful for horizontal deviation of the electron beam.
Horizontal amplifier amplifies the sawtooth signal and then connects it to the CRT horizontal deflection plates.
Power supply produces high and low voltages. The high negative voltage and the low positive voltage apply to CRT and other circuits respectively.
Cathode ray tube (CRT)
it is the main important block of CRO and consists mainly of four parts. Those are electronic guns, vertical deflection plates, horizontal deflection plates and fluorescent display.
The electron beam, which is produced by an electron gun, is deflected both vertically and horizontally by a pair of vertical deflection plates and a pair of horizontal deflection plates, respectively. Finally, the deflected beam will appear as a point on the fluorescent screen.
In this way, CRO will display the input signal applied on the CRT screen. So, we can analyze the signals in the time domain using CRO.
Explanation:
The oscilloscopes which is widely used for analysis purpose of circuits is divided into four main groups: the horizontal and vertical controls, the input controls and the activation controls.
Found in the front panel section marked Horizontal, the oscilloscope's horizontal controls allow users to adjust the horizontal scale of the screen. This section includes the control of the horizontal delay (displacement), as well as the control that indicates the time per division on the x-axis. The first control allows users to scan through a time range, while the latter allows users to approach a particular time range by decreasing the time per division.
Meanwhile, the oscilloscope's vertical controls are usually found in a section specifically marked as Vertical. The controls found in this section allow users to adjust the vertical appearance of the screen and include the control that indicates the number of volts per division on the axis and the grid of the screen. Also in this section is the control of the vertical displacement of the waveform, which translates the waveform up or down on the screen.
Signal activation helps provide a usable and stable display and allows users to synchronize the oscilloscope acquisition in the waveform of interest. The oscilloscope trigger controls allow users to choose the vertical trigger level, as well as the desired trigger capability. Common types of activation include fault activation, edge activation and pulse width activation.
Useful for identifying random errors or failures, the activation of faults allows users to fire at a pulse or event whose width is less than or greater than a specific period of time. This activation mode allows users to capture errors or technical problems that do not occur very frequently, which makes them very difficult to see.
The most famous trigger mode, edge tripping occurs when the voltage exceeds a set threshold value. This mode allows users to choose between shooting on a falling or rising edge.
Although pulse width activation is comparable to fault activation when users search for pulse width, it is, however, more general since it allows users to fire pulses of specified width. Users can also select the polarity of the pulses to be activated and set the horizontal position of the trigger. This allows users to see what happened during pre-shot or post-shot.
The input panels of an oscilloscope usually include two or four analog channels. They are usually numbered and have a button associated with each channel that allows users to activate and deactivate them. This section may also include a selection that allows users to specify the DC or AC coupling. Selecting the DC coupling implies that the entire signal will be input. The AC pairing, on the other hand, blocks the DC component and focuses the waveform around zero volts. Operators can also identify the probe impedance of the channels through a selection button. In adding, the input panels permit users to select the type of sampling to be used.
Answer:
The heat loss rate through one of the windows made of polycarbonate is 252W. If the window is made of aerogel, the heat loss rate is 16.8W. If the window is made of soda-lime glass, the heat loss rate is 1190.4W.
The cost associated with the heat loss through the windows for an 8-hour flight is:
For aerogel windows: $17.472 (most efficient)
For polycarbonate windows: $262.08
For soda-lime glass windows: $1,238.016 (least efficient)
Explanation:
To calculate the heat loss rate through the window, we can use a model of heat transmission by conduction throw flat wall. Using unidimensional Fourier law:
In this case:
If we replace the data provided by the problem we get the heat loss rate through one of the windows of each material (we only have to change the thermal conductivities).
To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).
To calculate the cost associated with the heat loss through the windows for an 8-hour flight we use this formula (using the heat loss rate calculated in each case):
