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Kay [80]
2 years ago
15

The modifications of superheat and reheat for a vapor power plant are specifically better for the operation which of the followi

ng components. Pick one and briefly explain.
a.Condenser
b.Boiler
c.Open feedwater heater
d.Turbine
e.Electric generator
Engineering
1 answer:
MatroZZZ [7]2 years ago
7 0

The modifications of superheating and reheat for a vapor power plant are specifically better for the operation which of the following components b.Boiler.

<h3>What are the primary additives in the vapour strength cycle?</h3>

There are 5 steam strength cycles: The Carnot cycle, the easy Rankine cycle, the Rankine superheat cycle, the Rankine reheat cycle and the regenerative cycle.

  1. Central to expertise the operation of steam propulsion is the primary steam cycle, a method wherein we generate steam in a boiler, increase the steam via a turbine to extract work, condense the steam into water, and sooner or later feed the water again to the boiler.
  2. Reheat now no longer best correctly decreased the penalty of the latent warmness of vaporization in steam discharged from the low-stress quit of the turbine cycle, however, it additionally advanced the first-rate of the steam on the low-stress quit of the mills via way of means of decreasing condensation and the formation of water droplets inside the turbine.

Read more about the Boiler:

brainly.com/question/17362931

#SPJ1

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
Should aircraft wings have infinite stiffness?
Colt1911 [192]

Answer:

No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.

3 0
2 years ago
Read 2 more answers
A 1 turn coil carries has a radius of 9.8 cm and a magnetic moment of 6.2 X 10 -2 Am 2. What is the current through the coil?
Alexus [3.1K]

Answer:

The current through the coil is 2.05 A

Explanation:

Given;

number of turns of the coil, N = 1

radius of the coil, r = 9.8 cm = 0.098 m

magnetic moment of the coil, P = 6.2 x 10⁻² A m²

The magnetic moment is given by;

P = IA

Where;

I is the current through the coil

A is area of the coil = πr² = π(0.098)² = 0.03018 m²

The current through the coil is given by;

I = P / A

I = (6.2 x 10⁻² ) / (0.03018)

I = 2.05 A

Therefore, the current through the coil is 2.05 A

6 0
3 years ago
Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes
Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

  At the inlet of turbine P=10 MPa  ,T=500 C

 AT the exit of turbine  P=10 KPa   ,x=0.9

 Required power=5 MW

From steam table

<u> At 10 MPa and 500 C:</u>

  h=3374 KJ/Kg  ,s=6.59 KJ/Kg-K  (Super heated steam table)

<u>At 10 KPa:</u>

h_g=2675.1 KJ/Kg, h_f=417.51  KJ/Kg

s_g= 7.3  KJ/Kg-K                ,s_f=1.3   KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= h_f+x(h_g- h_f)

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34  KJ/Kg

Lets take m is the mass flow rate of steam

So 5\times 10^3=m\times (3374-2449.34)

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0
3 years ago
Please help i give brainliest​
Mazyrski [523]

Answer:

A mock-up

Explanation:

It is made of cheap and easy to access parts.

5 0
2 years ago
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