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3 years ago

Consider a standard room thermostat. Determine the sensor, transducer, output, and control stages for this measurement system.

Engineering

1 answer:

Anni [7]3 years ago

3 0

Answer:

Sensor/transducer: bimetallic thermometer

Output: displacement of thermometer tip

Control Tstages: mercury contact switch (open:furnace off; closed:furnace on

Explanation:

for a standard room thermostat : This is the device that sets/determines the temperature of an enclosure.

Sensor/transducer: bimetallic thermometer: Bimetalic thermometer are used for measuring the temperature of the ambient air . bimetallic thermometer actually contains two metals. they undergo linear expansivity as the temperature of the room changes.in other words, they experience contraction and expansion with increase or decrease in temperature.The sensor is basically coupled with a transducer which turns the measured variable(Temperature) into something else, such as a movement on a dial or an electrical signal

Output: displacement of thermometer tip

Controller: mercury contact switch (open:furnace off; closed:furnace on)

once the contact switch is open the furnace can go off. when the contact switch is closed, the furnace will come up.

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simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

We can see the pic shown in order to understand the question.

We apply

∑MB = 0 (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6 (↑)

Then, we apply

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$

We get the inertia as follows

$I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}$

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

$y=\frac{h}{2} =\frac{0.3m}{2}=0.15m$

If M = Mmax, we have

$(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}$

8 0

3 years ago

A satellite would have a mass of 270 kg on the surface of Mars. Determine the weight of the satellite in pounds if it is in orbi

koban [17]

Answer:

26 lbf

Explanation:

The mass of the satellite is the same regardless of where it is.

The weight however, depends on the acceleration of gravity.

The universal gravitation equation:

g = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))

M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)

d: distance to that body

15000 miles = 24140 km

The distance is to the center of Earth.

Earth radius = 6371 km

Then:

d = 24140 + 6371 = 30511 km

g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2

Then we calculate the weight:

w = m * a

w = 270 * 0.43 = 116 N

116 N is 26 lbf

8 0

3 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

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jeka94

Answer:

why you doin this

Explanation:

is this so we get free points?

5 0

3 years ago

Read 2 more answers

State five applications of thermochromic materials

rusak2 [61]

Explanation:

The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.

8 0

3 years ago