Please upload a photo of a damper that you have come across in your daily life activities.Explain the type and nature of dampin

In which of the following branches of engineering is the practice not restricted?

iogann1982 [368]

a) civil engineering.

6 0

2 months ago

Suppose we include the lead resistance in the calculation of temperature for a class A RTD. If R3 = 1000 ohms, Ra = 18 ohms, V0

alex41 [359]

Respuesta:

La temperatura máxima que se puede medir (en °C) es 14170.27°C

Explicación:

Los RTDs son termómetros compuestos de metales cuya resistencia aumenta con la temperatura.

Para un RTD de Clase A, l, Alpha = 0.00385.

La fórmula para el RTD es

Rt = Ro ( 1 + alpha x t)

Donde

Rt es la resistencia a la temperatura t°C,

Ro es la resistencia a 0°C

Alpha es un coeficiente de temperatura constante para un RTD de clase A.

Aquí, Rt = 1000ohms,

Ro se considera como Ra = 18Ohms

Por lo tanto,

1000 = 18 ( 1 + 0.00385t)

Dividiendo ambos lados por 18

1 + 0.00385t = 1000/18

0.00385t = 55.55 - 1

0.00385t = 54.55

t = 54.55/0.00385

t = 14170.27°C

5 0

2 months ago

The rigid beam is supported by a pin at C and an A992 steel guy wire AB of length 6 ft. If the wire has a diameter of 0.2 in., d

Mrrafil [318]

Answer:

Change in length = 0.0913 in

Explanation:

Given data:

Length = 6 ft

Diameter = 0.2 in

Load w = 200 lb/ft

Solution:

We start by applying the equilibrium moment about point C, expressed as

∑M(c) = 0.............1

This can be used to find the force in AB.

10× 200 × ( 5) - (T cos(30)) × 10 = 0

Solving gives us

Tension in wire T(AB) = 1154.7 lb

We also know the modulus of elasticity for A992 is

E = 29000 ksi

And the area will be

Area = $\frac{\pi }{4}\times 0.2^2$

The change in length is expressed as

Change in length = $\frac{PL}{AE}$ .........2

Substituting values results in

Change in length = $\frac{1154.7 \times 6 \times 12}{\frac{\pi }{4}\times 0.2^2 \times 29000 \times 1000}$

Change in length = 0.0913 in

8 0

3 months ago

Design a digital integrator using the impulse invariance method. Find and give a rough sketch of the amplitude response, and com

mote1985 [299]

50 μsec.

8 0

2 months ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

Mrrafil [318]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Inertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy =?

(b) rotor acceleration =?

(c) change in torque angle =?

(c) rotor speed =?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Calculate the rotor's stored energy at synchronous speed.

The stored energy is represented as

$E = G \times H$

Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.

The formula for rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is defined as

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.

The change in torque angle is expressed as

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$