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2 years ago

What term describes a guardrail system

Engineering

2 answers:

Ymorist [56]2 years ago

7 0

Answer:

The ( Fall Protection in Construction ) is the term which relates to these mechanisms , i.e , guardrail system, a positioning system and a personal fall arrest system.

Explanation:

thanks me later

Nitella [24]2 years ago

5 0

Answer:

The ( Fall Protection in Construction ) is the term which relates to these mechanisms , i.e , guardrail system, a positioning system and a personal fall arrest system.

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A thermometer requires 1 minute to indicate 98% of the response to a unit step input. Assuming the thermometer to be a first ord

Rama09 [41]

Answer:

Time constant = 15.34 seconds

The thermometer shows an error of 0.838°

Explanation:

Given

t = 1 minute = 60 seconds

c(t) = 98% = 0.98

According to the question, the thermometer is a first order system.

The first order system transfer function is given as;

C(s)/R(s) = 1/(sT + 1).

To calculate the time constant, we need to calculate the step response.

This is given as

r(t) = u(t) --- Take Laplace Transformation

R(s) = 1/s

Substitute 1/s for R(s) in C(s)/R(s) = 1/(sT + 1).

We have

C(s)/1/s = 1/(sT + 1)

C(s) = 1/(sT + 1) * 1/s

C(s) = 1/s - 1/(s + 1/T) --- Take Inverse Laplace Transformation

L^-1(C(s)) = L^-1(1/s - 1/(s + 1/T))

Since, e^-t <–> 1/(s + 1) --- {L}

1 <–> 1/s {L}

So, the unit response c(t) = 1 - e^-(t/T)

Substitute 0.98 for c(t) and 60 for t

0.98 = 1 - e^-(60/T)

0.98 - 1 = - e^-(60/T)

-0.02 = - e^-(60/T)

e^-(60/T) = 0.02

ln(e^-(60/T)) = ln(0.02)

-60/T = -3.912

T = -60/-3.912

T = 15.34 seconds

Time constant = 15.34 seconds

The error signal is given as

E(s) = R(s) - C(s)

Where the temperature changes at the rate of 10°/min; 10°/60 s = 1/6

So.

E(s) = R(s) - 1/6 C(s)

Calculating C(s)

C(s) = 1/s - 1/(s + 1/T)

C(s) = 1/s - 1/(s + 1/15.34)

Remember that R(s) = 1/s

So, E(s) becomes

E(s) = 1/s - 1/6(1/s - 1/(s + 1/15.34))

E(s) = 1/s - 1/6(1/s - 1/(s + 0.0652)

E(s) = 1/s - 1/6s + 1/(6(s+0.0652))

E(s) = 5/6s + 1/(6(s+0.0652))

E(s) = 0.833/s + 1/(6(s+0.0652)) ---- Take Inverse Laplace Transformation

e(t) = 1/6e^-0.652t + 0.833

For a first order system, the system attains a steady state condition when time is 4 times of Time constant.

So,

Time = 4 * 15.34

Time = 61.36 seconds

So, e(t) becomes

e(t) = 1/6e^-0.652t + 0.833

e(t) = 1/(6e^-0.652(61.36)) + 0.833

e(t) = 0.83821342824942664566211

e(t) = 0.838 --- Approximated

Hence, the thermometer shows an error of 0.838°

4 0

2 years ago

Read 2 more answers

Refrigerant-134a enters a diffuser steadily as saturated vapor at 600 kPa with a velocity of 160 m/s, and it leaves at 700 kPa a

Zina [86]

Answer:

a) V_2 = 82.1 m/s

b) m = 0.298 Kg/s

Explanation:

from A-11 to A-13 we have the following data

P_1 = 600 kpa

V_1 = 0.033925 m^3/kg

h_1 = 262.52 kJ/kg

P_2 = 700 kpa

V_2 = 0.0313 m^3/kg

T_2 = 40°C = 313K

h_2 = 278.66 kJ/kg

Now, from the conversation of mass,

A_2*V_2/u_2 = A_1*V_1/u_1

V_2 = A_1/A_2*u_2/u_1*V_1

V_2 = A_1/1.8*A_1 * 0.0313 /0.033925*160

V_2 = 82.1 m/s

now from the energy balance equation

E_in = E_out

Q_in + m(h_1 + V_1^2/2) = m(h_2 + V_2^2/2)

m = 0.298 Kg/s

4 0

2 years ago

Read 2 more answers

2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to

Korolek [52]

Answer:

Work input =283.47 KJ

Explanation:

Given that

$P_1=150\ KPa$

$P_2=600\ KPa$

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=mRT\ln \dfrac{P_1}{P_2}$

So now putting the values

$W=mRT\ln \dfrac{P_1}{P_2}$

$W=2.5\times 0.287\times 285\ln \dfrac{150}{600}$

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

8 0

2 years ago

During an experiment conducted in a room at 25°C, a laboratory assistant measures that a refrigerator that draws 2 kW of power h

zvonat [6]

Answer:

Not reasonable.

Explanation:

To solve this problem it is necessary to take into account the concepts related to the performance of a reversible refrigerator. The coefficient of performance is basically defined as the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. For a reversible refrigerator the coefficient is given by

$COP_{R,rev} = \frac{1}{\frac{T_1}{T_2}-1}$

Where,

$T_1 =$ High temperature

$T_2 =$ Low Temperature

With our values previous given we can find it:

$T_2 = -30\°C = (-30+273)$

$T_2 = 243K$

$T_1 = 25\°C = (25+273)$

$T_1 = 298K$

With these values we can now calculate the coefficient of performance:

$COP_{R,rev} = \frac{1}{\frac{298}{243}-1}$

$COP_{R,rev} = 4.42$

At the same time we can calculate the work consumption of the refrigerator, this is

$W = \dot{W}\Delta t$

Where,

$\dot{W} =$ Required power input

t = time to remove heat from a cool to water medium

$W = 2kJ/s * 20 min$

$W = 2kJ/s * 1200s$

$W = 2400kJ$

In this way we can calculate the coefficient of the refrigerator directly:

$COP_R = \frac{Q_L}{W}$

Where,

Q = Amoun of heat rejected

$COP_R = \frac{30000}{2400}$

$COP_R = 12.5$

Comparing the values of both coefficients we have that the experiments are NOT reasonable, because the coefficient of a refrigerator is high compared to coefficient of reversible refrigerator.

5 0

2 years ago

Write SQL queries to answer the following questions: What are the names of the course(s) that student Altvater took during the s

amm1812

Answer:

See explanations

Explanation:

Consider the following Venn diagram to retrieve the name of course that student Altvater took during semester I-2015.

• In above Venn diagram inner Ellipse represent the subquery part, this subquery part select the Student ID from STUDENT table.

• Second sub query is used to determine the SectionNo of all student whose studentID retrived in the first subquery

• Finally the main query displays the CourseName from COURSE table.

6 0

2 years ago