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ki77a [65]
4 years ago
10

When the transportation of natural gas in a pipeline is not feasible for economic reasons, it is first liquefied using nonconven

tional refrigeration techniques and then transported in super-insulated tanks. In a natural gas liquefaction plant, the liquefied natural gas (LNG) enters a cryogenic turbine at 30 bar and –160°C at a rate of 20 kg/s and leaves at 3 bar. If 120 kW power is produced by the turbine, determine the efficiency of the turbine. Take the density of LNG to be 423.8 kg/m3.
Engineering
1 answer:
Charra [1.4K]4 years ago
3 0

Answer:

the   isentropic efficiency of turbine is 99.65%

Explanation:

Given that:

Mass flow rate of LNG  m = 20 kg/s

The pressure at the inlet P_1 =30 \ bar  = 3000 kPa

turbine temperature at the inlet T_1 = -160^0C = ( -160+273)K = 113K

The pressure at the turbine exit P_2 = 3 bar = 300 kPa

Power produced by the turbine  W = 120 kW

Density of LNG \rho = 423.8 \ kg/m^3

The formula for the workdone by an ideal turbine can be expressed by:

W_{ideal} = \int\limits^2_1 {V} \, dP

W_{ideal} ={V} \int\limits^2_1  \, dP

W_{ideal} ={V} [P]^2_{1}

W_{ideal} ={V} [P_1-P_2]

We all know that density = mass * volume i.e \rho= m*V

Then ;

V = \dfrac{m}{\rho}

replacing it into the above previous derived formula; we have:

W_{ideal} ={ \dfrac{m}{\rho}} [P_1-P_2]

W_{ideal} ={ \dfrac{20}{423.8}} [3000-300]

W_{ideal} ={ \dfrac{20}{423.8}} [2700]

W_{ideal} =0.04719*[2700]

W_{ideal} =127.42 kW

However ; the isentropic efficiency of turbine is given by the relation:

n_{isen} =\dfrac{W}{W_{ideal}}

n_{isen} =\dfrac{120}{120.42}

n_{isen} =0.9965

n_{isen} = 99.65%

Therefore, the   isentropic efficiency of turbine is 99.65%

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An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec
Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

4 0
3 years ago
1.Which thematic group uses technology to direct the behavior of dynamical systems, ensuring that they behave in a predictable m
dsp73

The thematic group governing the behavior of dynamic system is control system, and the one packaging the miniature components to conduct electricity is electronic system. Thus, option E is correct.

<h3>What are thematic groups?</h3>

The thematic groups comprises of group of people that work over the same idea or the concept. The thematic groups working on different projects and ideas comprised to work with different systems.

The control of dynamic system in order to produce the desirable outcome has been the role of the control system.

The packaging of the electronic miniatures in system that performs the role of direction of electricity is the role of electronic system thematic group. Thus, option E is correct.

Learn more about thematic group, here:

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8 0
2 years ago
A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli
Amanda [17]

Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3}  } =926.2A

X_{s} =0.85\frac{13.85}{926.2} =12.7ohm

The impedance Zn is

\sqrt{0.56^{2}+4^{2}  } =4.03ohm

The voltage drop is

I_{a} *Z_{n} =926.2*4.03=3732.58V

r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2}    } =\sqrt{8.97^{2}+12.7^{2}  } =15.55ohm

The induced voltage for leading power factor is

E_{F} ^{2} =OB^{2} +(BC-CD)^{2}

if cosθ = 0.5

E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2}   } =11937.51V

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

\frac{E_{F}-V_{t}  }{V_{t} } *100

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

8 0
4 years ago
You want to improve your grades so that you make all A's next 6 weeks. List examples of quantitative and qualitative data you sh
Naddika [18.5K]

Explanation:

z3d33sxurljt 36f

3fभथठभदाफमदखज्ञफादफज्ञादफज्ञिलफ इऋबिअऋब

6 0
3 years ago
Use phasor techniques to determine the impedance seen by the source given that R = 4 Ω, C = 12 μF, L = 6 mH and ω = 2000 rad/sec
Zielflug [23.3K]

Answer:

Z = 29.938Ω ∠22.04°

I = 2.494A

Explanation:

Impedance Z is defined as the total opposition to the flow of current in an AC circuit. In an R-L-C AC circuit, Impedance is expressed as shown:

Z² = R²+(Xl-Xc)²

Z = √R²+(Xl-Xc)²

R is the resistance = 4Ω

Xl is the inductive reactance = ωL

Xc is the capacitive reactance =

1/ωc

Given C = 12 μF, L = 6 mH and ω = 2000 rad/sec

Xl = 2000×6×10^-3

Xl = 12Ω

Xc = 1/2000×12×10^-6

Xc = 1/24000×10^-6

Xc = 1/0.024

Xc = 41.67Ω

Z = √4²+(12-41.67)²

Z = √16+880.31

Z = √896.31

Z = 29.938Ω (to 3dp)

θ = tan^-1(Xl-Xc)/R

θ = tan^-1(12-41.67)/12

θ = tan^-1(-29.67)/12

θ = tan^-1 -2.47

θ = -67.96°

θ = 90-67.96

θ = 22.04° (to 2dp)

To determine the current, we will use the relationship

V = IZ

I =V/Z

Given V = 12V

I = 29.93/12

I = 2.494A (3dp)

7 0
3 years ago
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