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Montano1993 [528]

4 years ago

11

An_______ employee is always prompt and on time.

1 answer:

Paha777 [63]4 years ago

5 0

Do u won’t to go out

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A cold storage room is used to keep the temperature inside the room maintain at low temperature.

lianna [129]

Answer:

The power of the brick wall it may be how the soiled ness of the wall too keep in the cold

Explanation:

5 0

3 years ago

Read 2 more answers

Giải giúp mình câu này được không ạ.em cảm ơn

sp2606 [1]

Answer:

çâdßèöñbvjsjaushdhshs

4 0

3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

A force of 250 N is acting on an area of 2 m2. The pressure is therefore:

lianna [129]

Answer:

125 Pascal

Explanation:

We know that

$Pressure=\frac{Force}{Area}$

Applying given values we get

$Pressure=\frac{250N}{2m^{2}}$

$Pressure=125\frac{N}{m^{2}}$

Pressure = 125 Pascal

7 0

3 years ago

A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to

Aliun [14]

Answer:

The right answer is "2.2099 m³".

Explanation:

Given:

Mass,

m = 5 kg

Temperature,

T = 35℃

or,

= 35 + 273

Pressure,

P = 200 kPa

Gas constant,

R = 0.2870 kj/kgK

By using the ideal gas equation,

The volume will be:

⇒ $PV=mRT$

or,

⇒ $V=\frac{mRT}{P}$

By substituting the values, we get

$=\frac{5(0.2870)(35+273)}{200}$

$=\frac{441.98}{200}$

$=2.2099 \ m^3$

8 0

3 years ago