1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Zinaida [17]
3 years ago
7

Steam enters an insulated turbine at 100 bar, 400oC. At the exit, the pressure and quality are 200 kPa and 0.47, respectively. D

etermine the power produced (kW) by the turbine if the mass flow rate is 1.99 kg/s.
Engineering
1 answer:
OverLord2011 [107]3 years ago
8 0

Answer:

power produced = 3098.52 kW

Explanation:

given data

insulated turbine = 100 bar

temperature = 400°C

pressure = 200 kPa

mass flow rate = 1.99 kg/s

solution

we use here steam table for At 100 bar and 400°C  

h1 = 3096.5 KJ/Kg

and  

at P2 = 200 Kpa

h2 = hf + 0.47 hg

h2 = 504.7 + 0.47 ×  2201.6  

h2 = 1539.452 KJ/Kg

so here

power produced  is express as

power produced = m × (h1 - h2)    .................1

power produced = 1.99 × ( 3096.5 - 1539.452 )

power produced = 3098.52 kW

You might be interested in
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 5208C.
algol [13]

This question is incomplete, the complete question is;

Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.

Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.

Answer:

- the quality of the steam exiting the second stage of the turbine is 0.9329  

- the thermal efficiency is 36.05%  

Explanation:

get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .

Specific enthalpy h1= 3192.3 kJ/kg

Specific entropy s1 = 5.9566 kJ/kg.K  

Process 1 to 2s is isentropic expansion process in the turbine

S1 = S2s

get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K

h2s = 2822.2 kJ/kg

get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)

0.78 = (3192.3 - h2) / (3192.3 - 2822.2)

h2 = 2903.6 kJ/kg

get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C

h3 = 3422.2 kJ/kg

s3 = 6.8803 kJ/kg.K

Process 3 to 4s is isentropic expansion process in the turbine

S3 = S4s

get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K

h4s = 2118.8 kJ/kg

get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)

0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )

h4 = 2405.5 kJ/kg

get the properties at pressure, p5 = 6 kPa

h5 = hf

= 151.53 kJ/kg

v5 = Vf  

= 0.0010064 m³/kg  

get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at

p6 = p1 = 28 MPa

np = v5( p6 - p5) / (h6 - h5)

0.82 =  ((0.0010064)( 28000 - 6)) / (h6 - 151.53)

h6 = 185.89 kJ/kg  

Now to find the quality of the steam at the exit of the second stage of the turbine

At stat4, p4 = 6kPa  

h4f = 151.53 kJ/kg

h4fg = 2415.9 kJ/kg  

h4 = h4f + x4h4fg

2405.5 = 151.53 + (x4 (2415.9))

x4 = 0.9329  

the quality of the steam exiting the second stage of the turbine is 0.9329  

Also to find the efficiency of the power plant, we use the following equation;

n = Wnet / Qin  

= (Wt1 + Wt2 - Wp) / (Q61 + Q23)

=  [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]

[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]

= 0.3605

n = 36.05%  

therefore the thermal efficiency is 36.05%  

3 0
3 years ago
A. 50
Mademuasel [1]

Answer:ahahahahahah

Explanation:boyboy

6 0
3 years ago
A particular cloud-to-ground lightning strike lasts 500 µµsec and delivers 30 kA across a potential difference of 100 MV. Assu
sasho [114]

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

c)

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

d)

the residential retail value of the energy delivered by the strike is $ 40.83

e)

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

a)

we know that; I = Q/t

so Q = I × t

we substitute

Q =  30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

b)

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

c)

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹²  × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

d)

given that;  1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

e)

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.  

7 0
3 years ago
Technician A says that the most effective way to diagnose wheel speed sensor circuit faults is to utilize the live data function
Marianna [84]

Answer:

Both Technicians A & B are correct

Explanation:

In diagnosing wheel systems, both steps said by technicians A & B are correct because it's the normal procedure to diagnose wheel speed sensor circuit faults.

7 0
3 years ago
Smoke rises from a chimney on a day with variable winds and produces a visible pattern. Does this pattern represent a pathline o
jeyben [28]

Answer:

Pathline

Explanation:

- The smoke coming out of the chimney on a day with variable winds would most closely relate to a path-line. Consider a smoke particle " grey " color particle that is injected into air at every instant in time and follows a continuous "path" along the flow of the of the fluid.

- A streakline concentrates on fluid particles that have gone through a fixed station or point. At some instant of time the position of all these particles are marked and a line is drawn through them.

- The streakline and pathline can coincide on the conditions of flow to be steady that is the rate of change of flow of the air is continuous and does not change with time in a day i.e the variation in the winds or any discontinuity is removed.

5 0
4 years ago
Other questions:
  • . A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The allowable shear stress for the material of the shaft is 42 MPa.
    7·1 answer
  • A circuit has a 24 volt battery and a resistor that is 8 ohms. How many amps are running through the circuit?
    6·1 answer
  • In an RC parallel circuit, E T = 240 V, R = 1000 Ω, and X C = 1500 Ω. What is the apparent power?
    6·1 answer
  • Fin efficiency is defined in relation to the maximum possible heat transfer rate from the fin for given convective conditions. T
    10·1 answer
  • Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: L
    5·1 answer
  • 4. A hydropower installation is to be located where the downstream water-surface elevation is 150 m below the water-surface elev
    14·1 answer
  • PLSSSSS Help !!!!!!!!!!!. It's due today. <br> I will give brainliest to correct answer
    10·2 answers
  • What the advantages and disadvantages of steam distillation??
    7·1 answer
  • How can any student outside apply for studying engineering at Cambridge University​
    7·1 answer
  • A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R1
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!