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Svetllana [295]
3 years ago
6

In an RC parallel circuit, E T = 240 V, R = 1000 Ω, and X C = 1500 Ω. What is the apparent power?

Engineering
1 answer:
denis-greek [22]3 years ago
5 0

Answer:

69.23VA

Explanation:

Z=R//Z_{c}=\left ( \frac{1}{R}+\frac{1}{Z_{c}} \right )^{-1}\\ R=1000\Omega\\ X_{c}=1500\Omega\\ Z=R//Z_{c}=\left ( \frac{1}{1000}+\frac{1}{-j1500} \right )^{-1}=832,05\angle-33,69\Omega

Like is an parallel circuit, we can calculate across the voltage and impedance

S=V.I=\frac{V^2}{Z} \\ S=\frac{240^2}{832,05}\LARGE S=69.23VA

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The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large
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Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

C(d) = 220 / (1.221 * 625)

C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

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3 years ago
Jeff is assembling a rear axle unit. He has all
BartSMP [9]

Answer:

He should read the directions

Explanation:

Reading the directions will tell him what he needs to do in order to assemble the rear axle unit. I hope this helped.

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A slip and fall prevention component should have which of the following elements?
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Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

\frac{P_{1} }{\gamma } +\frac{V_{1}^{2}  }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2}  }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}

For the pipe 1, the flow velocity is:

V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4

\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

h_{1} =\frac{V_{1}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m

For the pipe 2, the flow velocity is:

V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s

The Reynold´s number is:

Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4

\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087

The head of pipe 1 is:

h_{2} =\frac{V_{2}^{2}  }{2g} (k_{L}+\frac{fL}{D}  )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m

The required pumping power is:

P=Q\rho *g*h_{pump}  =0.018*999.1*9.8*1344.55=236965.16W=236.96kW

8 0
3 years ago
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