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2 years ago

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Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

1 answer:

vichka [17]2 years ago

4 0

Answer:

Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

updated information on model changes that do not affect the technician. Who is correct? the answer is c

You might be interested in

1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn

emmasim [6.3K]

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

1.

Given:

height of the window pane, $h=2\ m$

width of the window pane, $w=1\ m$

thickness of the pane, $t=5\ mm= 0.005\ m$

thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

temperature of the inner surface, $T_i=15^{\circ}C$

temperature of the outer surface, $T_o=-20^{\circ}C$

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

2.

air spacing between two glass panes, $dx=0.01\ m$

area of each glass pane, $A=2\times 1=2\ m^2$

thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

temperature difference between the surfaces, $dT=25^{\circ}C$

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$

5 0

2 years ago

A cylindrical drum (2 ft. dia ,3 ft height) is filled with a fluid whose density is 40 lb/ft^3. Determine (a. the total volume o

Ksivusya [100]

Answer:

a) $V=9.42\ ft^3$

b)Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c) $v=0.025\ ft^3/lb$

d) $w=1276 \ lb/ft.s^2$

Explanation:

Given that

d= 2 ft

r= 1 ft

h= 3 ft

Density

$\rho = 40\ lb/ft^3$

a)

We know that volume V given as

$V=\pi r^2 h$

$V=\pi \times 1^2\times 3$

$V=9.42\ ft^3$

b)

Mass = Density x volume

$mass =40\times 9.42\ lb$

mass= 376.8 lb

We know that

1 lb = 0.031 slug

So 376.8 lb= 11.71 slug

Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)

we know that specific volume(v) is the inverse of density.

$v=\dfrac{1}{\rho}\ ft^3/lb$

$v=\dfrac{1}{40}\ ft^3/lb$

$v=0.025\ ft^3/lb$

d)

Specific weight(w) is the product of density and the gravity(g).

w= ρ X g

w = 40 x 31.9

$w=1276 \ lb/ft.s^2$

8 0

2 years ago

6. You are an electrician working in an industrial plant. You are building a motor control cabinet that contains six motor start

Sveta_85 [38]

Answer:

It's indeed safer to suggest a 150 VA transformer. Following table however is the clarification given.

Explanation:

For 2 motors with 0.1 A, the Power will be:

P = $2\times 120\times 0.1$

= $24 \ W$

For 4 motors with 0.18 A, the Power will be:

P = $4\times 120\times 0.18$

= $86.4 \ W$

As we know, for 6 pilot lamps, the power is "5 W".

So,

The total power will be:

⇒ P = $24+86.4+5$

= $115.4 \ W$

Now,

Consider the power factor to be "0.95"

VA of transformer is:

= $PF\times Power$

= $115.4\times 0.9$

= $109.63 \ VA$

8 0

2 years ago

A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula

makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0

2 years ago

2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam

Natasha2012 [34]

Answer:

$\eta = 91.7$ %

Explanation:

Determine the initial velocity

$v_1 = \frac{\dot v}{A_1}$

$= \frac{0.1}{\pi}{4} 0.08^2$

= 19.89 m/s

final velocity

$v_2 =\frac{\dot v}{A_2}$

$= \frac{0.1}{\frac{\pi}{4} 0.12^2}$

=8.84 m/s

total mechanical energy is given as

$E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

$\dot v = \dot m v$ $( v =v_1 =v_2)$

$E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}$

$= mv\Delta P + \dot m \frac{v_2^2 -v_1^2}{2}$

$= \dot v \Delta P + \dot v \rho \frac{v_2^2 -v_1^2}{2}$

$= 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}$

$E_{mech} = 36.34 W$

Shaft power

$W = \eta_[motar} W_{elec}$

$=0.9\times 44 =39.6$

mechanical efficiency

$\eta{pump} =\frac{ E_{mech}}{W}$

$=\frac{36.34}{39.6} = 0.917 = 91.7$ %

8 0

2 years ago