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3 years ago

Why do the pipes in a steam-heating system need to be insulated?

Physics

1 answer:

qaws [65]3 years ago

3 0

Answer:

1. Increase Efficiency: Insulating the condensate or hot water return lines reduces heat loss from the water returning to the boiler. The hotter the water returning to the boiler is, the quicker it is to convert back to steam, which takes less energy to accomplish.

Explanation:

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

What happens when two forces act in the same direction?

Wewaii [24]

Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.

5 0

3 years ago

Read 2 more answers

What would the velocity of an object be if it was on a 1.75m

JulijaS [17]

Answer:

T = 4 sec / 2 = 2 sec period of revolution

S = 2 pi R = 2 * pi * 1.75 m = 11 m

V = S / T = 11 m / 2 sec = 5.5 m/s speed of object

4 0

3 years ago

What must be the acceleration of a train in order for it to stop 12 m/s in a distance if 541 m

earnstyle [38]

Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

Lets explain how to solve the problem

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

We want to find its acceleration

The acceleration will be negative quantity because the train reduced its

velocity from 12 m/s to zero

We need rule contains velocity, acceleration and distance

So we will use ⇒<em> v² = u² + 2as</em>, where v is the final velocity, u is the

initial velocity, a is the acceleration and s is the distance

v = 0, u = 12 m/s, s = 541 m

Substitute these values in the rule

(0)² = (12)² + 2(a)(541)

0 = 144 + 1082 a

Subtract 144 from both sides

-144 = 1082 a

Divide both sides by 1082

- 0.133 = a

<em>The acceleration of the train must be - 0.133 m/s²</em>

3 0

3 years ago

Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0

3 years ago