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3 years ago

Which two states about electric motor are true

Engineering

1 answer:

kap26 [50]3 years ago

3 0

Answer:

theres not statements

Explanation:

...........

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A duct for an air conditioning system has a rectangular cross section of 2ft by 9in. The duct is fabricated from galvanized iron

qwelly [4]

Answer:

The answer is ' $2.9584 \ \frac{lbf}{ft^2}$ '

Explanation:

Given:

$\mu=3.96 \times 10^{-7} \ \frac{lbfs}{ ft^2}\\\\\gamma=0.0709 \ \frac{lfb}{ft^3}\\\\v= 1.8 \times 10^{-4} \ \frac{ft^2}{s}$

energy balance equation:

$\ P_1 - P_2 = \rho gh_{L}$

$V= \frac{\ Q}{ A}\\\\V= \frac{\frac{5000}{60}}{2 \times \frac{9}{12}}\\\\V= 55.55 \frac{ft}{s}$

calculating $D_h$ :

$D_h= \frac{4 \times \ scetor \ area }{ wetted \ permetor}\\\\D_h=\frac{4 \times 2 \times \frac{9}{12}}{2(2+\frac{9}{12})}\\\\D_h= 1.0909 ft\\\\$

calculating $R_e$ :

$R_e=\frac{VD_h}{V}\\\\R_e=\frac{55.55 \times 1.0909}{1.8 \times 10^{-4}}\\\\R_e=3.366 \times 10^5\\\\$

calculating relative vovghness:

$\frac{k_s}{D_h}= \frac{\frac{0.006}{12}}{1.0909}\\\\$

= 0.0006875

If $R_e= 3.366 \times 10^5 \ and \ \ \frac{K_s}{D_h} = 0.0006875 \ \ \ so, \ \ f= 0.019$

Calculating $H_f$ :

$H_f= \frac{f \ l\ v^2}{ 2 \ g \ D_h}\\\\$

$=\frac{0.019 \times 50 \times 55.55^2}{2\times 32.2\times 1.0909}\\\\= 41.727 \ ft$

Calculating pressure drop:

$P_1-P_2= \gamma h_{L}\\\\$

$= 0.0709 \times 41.727\\\\= 2.9584 \ \frac{lbf}{ft^2}$

4 0

4 years ago

The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) a bx cx2 , where T is in degrees Cel

Oksi-84 [34.3K]

Answer:

the rate of heat transfer into the wall is $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

the rate of heat output is $\mathbf{q_{out} =182 \ W/m^2}$

the rate of change of energy stored by the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

the convection coefficient is h = 4.26 W/m².K

Explanation:

From the question:

The temperature distribution across the wall is given by :

$T(x) = ax+bx+cx^2$

where;

T = temperature in ° C

and a, b, & c are constants.

replacing 200° C for a, - 200° C/m for b and 30° C/m² for c ; we have :

$T(x) = 200x-200x+30x^2$

According to the application of Fourier's Law of heat conduction.

$q_x = -k \dfrac{dT}{dx}$

where the rate of heat input $q_{in} = q_k$ ; Then x= 0

So:

$q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}$

$q_{in}= -1 (-200+60x)_{x=0}$

$\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

Thus , the rate of heat transfer into the wall is $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The rate of heat output is:

$q_{out} = q_{x=L}$ ; where x = 0.3

$q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}$

replacing T with $200x-200x+30x^2$ and k with 1 W/m.K

$q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}$

$q_{out} = -1 (-200+60x)_{x=0.3}$

$q_{out} = 200-60*0.3$

$\mathbf{q_{out} =182 \ W/m^2}$

Therefore , the rate of heat output is $\mathbf{q_{out} =182 \ W/m^2}$

Using energy balance to determine the change of energy(internal energy) stored by the wall.

$\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\$

$\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

Thus; the rate of change of energy stored by the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

We all know that for a steady state, the heat conducted to the end of the plate must be convected to the surrounding fluid.

So:

$q_{x=L} = q_{convected}$

$q_{x=L} = h(T(L) - T _ \infty)$

where;

h is the convective heat transfer coefficient.

Then:

$Replacing \ 182 W/m^2 \ for \ q_{x=L} , (200-200x +30x \ for \ T(x) \ , 0.3 m \ for \ x \ and \ 100^0 C for \ T$ We have:

182 = h(200-200×0.3 + 30 ×0.3² - 100 )

182 = h (42.7)

h = 4.26 W/m².K

Thus, the convection coefficient is h = 4.26 W/m².K

6 0

3 years ago

Technicians are normally paid in all of the following methods except:

yawa3891 [41]

Answer:

you didnt list the following methods

Explanation:

7 0

3 years ago

Read 2 more answers

In the truss below, the vertical reaction force at A is

Montano1993 [528]

It’s c trust just trust

7 0

3 years ago

In a slowly cooled hypereutectoid iron-carbon steel, the pearlite colonies are

jek_recluse [69]

In a slowly cooled hypereutectoid iron-carbon steel, the pearlite colonies are normally separated from each other by a more or less continuous boundary layer of cementite done by Slower cooling reasons coarse Pearlite, even as rapid cooling reasons first-rate pearlite to form.

<h3>What levels is in Hypereutectoid metal?</h3>

Hypoeutectoid steels can, upon preliminary cooling from the austenite single segment field, exist as extraordinary levels, eutectoid ferrite and austenite, every with extraordinary carbon contents.

At room temperature, hypereutectoid steels have a pearlitic primary microstructure (ferrite grains with embedded cementite lamellae) with moreover induced cementite on the grain boundaries! The micrograph under suggests a hypereutectoid metal with 1.0 Carbon (C100).