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Fudgin [204]
3 years ago
6

A 450 MWt combined cycle plant has a Brayton cycle efficiency of 24% and a Rankine cycle efficiency of 29% with no heat augmenta

tion. Determine the net electric output from the plant.
Engineering
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0

Answer:

Output of plant=207.18 MW

Explanation:

Given input energy Q=450 MW

Brayton cycle efficiency\eta_1=24%

Rankine cycle efficiency \eta_2=29%

Combined efficiency for two cycle given as follow

\eta _b=\eta _1+\eta _2-\eta _1\eta _2

\eta _b=0.24+0.29-0.24\times 0.29

\eta_b=0.4604

We know that efficiency \eta =\dfrac{out\ put}{in\ put}

out put of plant=450\times 0.4604

So output of plant=207.18 MW

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List the thermal conductivities of five
ICE Princess25 [194]

Answer:

Sorry I don't understand

Explanation:

8 0
3 years ago
Consider a vortex filament of strength in the shape of a closed circular loop of radius R. Obtain an expression for the velocity
zysi [14]

Answer:

<em>v</em><em> </em>= T/(2R)

Explanation:

Given

R = radius

T = strength

From Biot - Savart Law

d<em>v</em> = (T/4π)* (d<em>l</em> x <em>r</em>)/r³

Velocity induced at center

<em>v </em>= ∫  (T/4π)* (d<em>l</em> x <em>r</em>)/r³

⇒   <em>v </em>= ∫  (T/4π)* (d<em>l</em> x <em>R</em>)/R³  (<em>k</em>)        <em>k</em><em>:</em> unit vector perpendicular to plane of loop

⇒   <em>v </em>= (T/4π)(1/R²) ∫ dl

If l ∈  (0, 2πR)

⇒   <em>v </em>= (T/4π)(1/R²)(2πR)  (<em>k</em>)    ⇒   <em>v </em>= T/(2R)  (<em>k</em>)  

3 0
3 years ago
A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
Gekata [30.6K]

Answer:

a) \dot m_{3} = 135\,\frac{lbm}{s}, b) h_{3}=168.965\,\frac{BTU}{lbm}, c) T = 200.829\,^{\textdegree}F

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

\dot m_{1} + \dot m_{2} - \dot m_{3} = 0

The mass flow rate exiting the tank is:

\dot m_{3} = \dot m_{1} + \dot m_{2}

\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}

\dot m_{3} = 135\,\frac{lbm}{s}

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}

Properties of water are obtained from tables:

h_{1}=180.16\,\frac{BTU}{lbm}

h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)

h_{2}=29.032\,\frac{BTU}{lbm}

The specific enthalpy at outlet is:

h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }

h_{3}=168.965\,\frac{BTU}{lbm}

c) After a quick interpolation from data availables on water tables, the final temperature is:

T = 200.829\,^{\textdegree}F

8 0
3 years ago
Read 2 more answers
You leave your house at 5:02 PM and run 20 yards down the street. You don't realize that you forgot your Wallet back at home and
Sidana [21]

Answer:

The average velocity is 0.203 m/s

Explanation:

Given;

initial displacement, x₁ = 20 yards = 18.288 m

final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m

change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s

The average velocity is given by;

V = change in displacement / change in time

V = (x₂ - x₁) / Δt

V = (18.288 - 6.096) / 60

V = 0.203 m/s

Therefore, the average velocity is 0.203 m/s

7 0
3 years ago
What is the hardest part of engineering?
Vikki [24]

ANSWER:

Aerospace Engineering. ...

Chemical Engineering. ...

Biomedical Engineering.

EXPLANATION:

This is all i know but ... I hope this helps~

7 0
2 years ago
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