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
Harlamova29_29 [7]
3 years ago
11

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit

conditions are 10 kPa, 92% quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s.
Determine:
(a) the change in kinetic energy
(b) the power output
(c) the turbine inlet area
Engineering
1 answer:
8090 [49]3 years ago
3 0

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

You might be interested in
n the Spring of 2015, three utility companies in the Ukraine received email purporting to come from Ukraine's parliament, the Ra
Tresset [83]

Answer:

Trojan horse

Explanation:

A trojan horse attack is a type of malware that misleads users, as it appears unsuspicious at first, but actually presents a threat to the user. A common example is that of an email that contains a malicious attachment. Another common example is that of a fake advertisement. The name comes from the Greek story of the Trojan horse that led to the fall of the city of Troy.

5 0
3 years ago
Only answer this if your name is riley
Sati [7]

Answer:

hey im like kinda riley

Explanation:

y u wanna talk to moi

3 0
2 years ago
Read 2 more answers
A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of
Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end  + area of the curve surface

=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}

= 2.136 *10^-4 m^{2}

Heat flux =\frac{0.6}{2.136 * 10^{-4} } = 2808.99 W/m^{2}

fraction of heat dissipated from the top and bottom surface

=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100}  )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175

=11.75%

8 0
3 years ago
Read 2 more answers
Ext
Galina-37 [17]
Engineering is the technical
8 0
1 year ago
Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces app
stiks02 [169]

Answer:

a) Fb= 275.77 lb   Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

   Fay = 337.75 lb

   Fbx = 195 lb

   Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution:

<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>

<u />

a) For no translation condition

∑ F_{x} = 0      &     ∑F_{y} = 0

Hence,

F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0

F_{A}sin(30) - F_{B} sin(45) = 0

and

F_{A} = 390 lb

Inserting the value of F_{A} and solving the remaining equations simultaneously yields (magnitudes),

F_{B} = 275.77 lb\\F_{C} = 142.75 lb

b) Summing up moments

M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))

M = -779.97 lb.ft (i.e. 779.97 lb.ft clockwise)

c)

F_{Ax} = 390 sin(30)  = 195 lb

F_{Ay} = 390 cos(30) = 337.75lb\\

F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb

8 0
3 years ago
Other questions:
  • An air-standard Otto cycle has a compression ratio of 6 and the temperature and pressure at the beginning of the compression pro
    13·1 answer
  • Fictional Corp is looking at solutions for their new CRM system for the sales department. The IT staff already has a fairly heav
    10·1 answer
  • A(n)___ branch circuit supplies two or more receptacles or outlets for lighting and appliances
    10·1 answer
  • A gear motor can develop 2 hp when it turns at 450rpm. If the motor turns a solid shaft with a diameter of 1 in., determine the
    7·1 answer
  • Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo
    10·2 answers
  • Alcohol consumption tends to cause more ___________ behavior.
    14·1 answer
  • A long bone is subjected to a torsion test. Assume that the inner diameter is 0.375 in. and the outer diameter is 1.25 in., both
    14·1 answer
  • To do you blur text in google docs
    10·1 answer
  • What do you understand by the term phase angle?<br>​
    13·1 answer
  • What is the answer of this question please tell me <br><br>dadada please please​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!