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2 years ago

What are the three main sections of a nuclear power plant?

Engineering

1 answer:

strojnjashka [21]2 years ago

7 0

The reactor, generator, and the cooling towers

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In highways the far left lane is usually the _____

Ivan

Fastest

(Known as the fast lane)

8 0

2 years ago

Read 2 more answers

The amplitudes of the displacement and acceleration of an unbalanced motor were measured to be 0.15 mm and 0.6*g, respectively.

ehidna [41]

Answer:

The speed of shaft is 1891.62 RPM.

Explanation:

given that

Amplitude A= 0.15 mm

Acceleration = 0.6 g

we can say that acceleration= 0.6 x 9.81

$acceleration,a=5.88\ \frac{m}{s^2}$

We know that

$a=\omega ^2A$

So now by putting the values

$a=\omega ^2A$

$5.88=\omega ^2 \0.15\times 10^{-3}$

$\omega =198.09\ \frac{rad}{s}$

We know that

ω= 2πN/60

198.0=2πN/60

N=1891.62 RPM

So the speed of shaft is 1891.62 RPM.

4 0

2 years ago

To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws

ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

$E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\$

E=52000Hp.h

$E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\$

E=38724920Wh

$E=52000Hph(\frac{1977378.4 ft lb}{1Hph}$

E=1.028x10^11 ftlb

3 0

3 years ago

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o

jekas [21]

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

$\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

$\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

$H_m=(\frac{200*10^3}{1000*9.8}-10)$

$H_m=10.39m$

Generally the equation for Power is mathematically given by

$P=\rho gQH_m$

Therefore

$Q=\frac{P}{\rho g H_m}$

$Q=\frac{4*10^4}{1000*9.81*10.9}$

$Q=0.03935m^3/sec$

Since

$Q=AV'$

Where

$A=\pi r^2\\A=3.142 (0.05)^2$

$A=7.85*10^{-3}$

Therefore

$V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

$V'=5m/s$

5 0

2 years ago

River models are used to study many different types of flow situations. A certain small river has an average width and depth of

slavikrds [6]

Answer: 7ft x21 I’d be right but yes I am

Explanation: because it is Welty

4 0

2 years ago