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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

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Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B =  164.6 < 85.42°Ω  

C =  2.061 * 10^-3 < 90.32° s

D =  0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i  + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = \sqrt{\frac{z}{y} }  =  \sqrt{\frac{0.03 i  + j 0.35}{j4.4*10^-6 } }    

    = \sqrt{79837.128< 4.899^o}   =  282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate  gl

gl = \sqrt{zy} * d  

 d = 500

 z = 0.03 i  + j 0.35

 y = j4.4*10^-6 S/km

gl =  \sqrt{0.03 i  + j 0.35*  j4.4*10^-6}  * 500

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gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit  =  0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω  = 164.6 < 85.42°Ω  ( as calculated )

C = 1/Zc  sin h (gl) s  =  2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where :  cos h (gl)  = \frac{e^{gl} + e^{-gl}  }{2}

             sin h (gl) = \frac{e^{gl}-e^{-gl}  }{2}

     

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bazaltina [42]
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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be
cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, \zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )

Creep rate at 800⁰C, \zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )

\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\

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Creep rate at 700⁰C

\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )

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Divide equation (1) by equation (2)

\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\

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3 0
3 years ago
A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu
bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

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a.

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

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C = 1.0-nF = 1 * 10^-12F

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d = 1.64 * 10^-3m

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d = 12 * 10^-3m

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A ∝ 1/K

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The dielectric constant of a Teflon is 2.1

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