It is possible to have liquid water at 200°C. a)-True b)- False

Re armature of a 4 pole DC generator is required to generate an emf of 520v on open circuit when revolving at a speed of 660rpm.

bija089 [108]

Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

Given the following data:

Emf = 520 Volts

Speed = 660 r.p.m

Number of armature conductors = 144 slots

Number of poles = 4 poles

Number of parallel paths = 2

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

$E = \frac{\theta ZN}{60}$ × $\frac{P}{A}$

Where:

E is the electromotive force in the DC generator.

Z is the total number of armature conductors.

N is the speed or armature rotation in r.p.m.

P is the number of poles.

A is the number of parallel paths in armature.

Ф is the magnetic flux.

First of all, we would determine the total number of armature conductors:

$Z = 144$ × $2$ × $3$

Z = 864

Substituting the given parameters into the formula, we have;

$520 = \frac{\theta (864)(660)}{60}$ × $\frac{4}{2}$

$520 = \theta (864)(11)$ × $2$

$520 = 19008 \theta \\\\\Theta = \frac{520}{19008}$

Magnetic flux = 0.0274 Weber.

Therefore, the magnetic flux per pole is 0.0274 Weber.

5 0

3 years ago

To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01

liq [111]

Answer:

138.9 °C

Explanation:

The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C

3 0

3 years ago

How large a force is required to accelerate a 1300 kg car from rest to a speed of 20 m/s in a distance of 80 m?

topjm [15]

F=m*a

F=80*20

F =1600 ans"

7 0

3 years ago

Read 2 more answers

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

$\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})$

where

'H' is the initial depth of the layer

$C_c$ is the Compression index

$e_o$ is the inital void ratio

$\bar{\sigma_o}$ is the initial effective stress at the depth

$\Delta \bar{\sigma_o}$ is the change in the effective stress at the given depth

Applying the given values we get

$\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04$

3 0

3 years ago

A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60

Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

T_2 = T_1 * r^(k-1/k)

T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

T_7 = T_6 * r^(-k-1/k)

T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

flow(m) = 7.941 lbm/s

- The heat input is as follows:

Q_in = c_p*(T_6 - T_5)

Q_in = 0.24*(1400 - 1022.84)

Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

Q_out = c_p*(T_10 - T_1)

Q_out = 0.24*(711.744 - 520)

Q_out = 56.01856 Btu/lbm

5 0

3 years ago