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Gennadij [26K]
3 years ago
7

The weatherman states that the current temperature is 95 F and the dew point is 74 F. What is the current relative humidity?

Engineering
1 answer:
spayn [35]3 years ago
3 0

Answer:

Current Relative Humidity is 29.623

Given:

Current Temperature, T_{c} = 95 F = 35^{\circ}C

Dew point temperature, T_{d} = 74 F = 23.34^{\circ}C

Solution:

Now, in order to calculate the Relative Humidity, RH, we use the given formula:

T_{d} =100( \frac{\frac{aT_{c}}{b + T}b + lnRH}{a - \frac{aT_{c}}{b + T} + lnRH})

where

a = 17.625

b = 237.7

Now, using the above formula and given values:

23.34 = (237.7\frac{\frac{17.625\times 35}{ 237.7 + 35} + lnRH}{17.625 - \frac{17.625\times 35}{237.7 + 35} + lnRH})

23.34(17.625 - 2.26 + lnRH) = (237.7\times 2.26 + lnRH)

358.62 + 23.34lnRH) = (537.20 + lnRH)

358.62 + 23.34lnRH) = (537.20 + lnRH)

On solving the above eqn, we get:

RH = 29.623

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Which of these are not included as part of a drivetrain
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transmission, driveshafts, differential and axles

Explanation:

The powertrain consists of the prime mover (e.g. an internal combustion engine and/or one or more traction motors) and the drivetrain - all of the components that convert the prime mover's power into movement of the vehicle (e.g. the transmission, driveshafts, differential and axles); whereas the drivetrain does not.

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A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli
Amanda [17]

Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3}  } =926.2A

X_{s} =0.85\frac{13.85}{926.2} =12.7ohm

The impedance Zn is

\sqrt{0.56^{2}+4^{2}  } =4.03ohm

The voltage drop is

I_{a} *Z_{n} =926.2*4.03=3732.58V

r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2}    } =\sqrt{8.97^{2}+12.7^{2}  } =15.55ohm

The induced voltage for leading power factor is

E_{F} ^{2} =OB^{2} +(BC-CD)^{2}

if cosθ = 0.5

E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2}   } =11937.51V

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

\frac{E_{F}-V_{t}  }{V_{t} } *100

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

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Befor operating any plant,machinery or equipment,you should​
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Answer:

Always make sure that you are properly protected and that it is all clear to be operated on.

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