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aivan3 [116]
3 years ago
5

Estimate the luminosity of a 3 -solar-mass main-sequence star; of a 9 -solar-mass main-sequence star. Can you easily estimate th

e luminosity of a 3 -solar-massred giant star?
Engineering
1 answer:
Oxana [17]3 years ago
6 0

Answer:

a) 46.76 Lsun

b) 2187 Lsun

c ) You can easily estimate the luminosity of a 3-solar massed giant star because its luminosity is close to the luminosity of the sun

Explanation:

To estimate the luminosity of a celestial body in terms of solar luminosity  we use the expression below

= \frac{L}{Lsun}  = (\frac{M}{Msun} )^{3.5}

a) Estimate the luminosity of a 3 solar mass main sequence star

= \frac{L}{Lsun}  = (\frac{3M}{Msun} )^{3.5}

Hence  ; L(luminosity of a 3 solar mass star ) = 46.76  Lsun

b) Estimate the luminosity of a 9-solar-mass main sequence star

= \frac{L}{Lsun}  = (\frac{9M}{Msun} )^{3.5}

hence L( luminosity of a 9-solar mass star ) =  2187 Lsun

c ) You can easily estimate the luminosity of a 3-solar massed giant star because its luminosity is close to the luminosity of the sun

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saveliy_v [14]

Answer:

a)  The slip coefficient is 0.9

b) Blade tip speed is 345.57 m/s

c) Stagnation temperature exiting the impellor is 416.84 k

d) Exit flow velocity is 84.88 m/s

e) Exit flow angle is 76.2°

f) Exit static temperature is 353.84 k

g) Impellor exit Mach number is 0.943

Explanation:

Flow at entry is axial α₁ = 0

Tagential velocity at entry V_{w1}= 0

Blade at exit is radial  β₂ = 0

μ₂ = V_{w2}

3 0
3 years ago
A convergentâdivergent nozzle has an exit area to throat area ratio of 4. It is supplied with air from a large reservoir in whic
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Answer:

Angle of discharge make at the edge of tube=64.9 degrees.

5 0
3 years ago
Convert the following pairs of voltage and current waveforms to phasor form. Each pair of waveforms corresponds to an unknown el
exis [7]

Answer:

a) V = 20 ∠30⁰    ,    I = 4 ∠-210⁰    Z inductive    L = 0,0125 H

b) V = 9∠-60⁰      ,    I = 4 ∠ 190⁰    Z capacitive C = 4,94 *10⁻⁴ F

c) V = 13 ∠240⁰   ,    I = 7 ∠ 150⁰    Z Inductive  L = 0,0074 H

Explanation:

a) v(t) = 20 cos (400*t + 30 )

Phasor form    V = 20 ∠30⁰

i(t) = 4 sin (400*t - 120)

First we need to transform 4sin( 400t - 120 ) as  function cosine

we know that  sin ( x + 90 )  =  cos x

Then  sin ( 400*t -120 )  = cos ( 400*t  - 120 -90 )  = cos ( 400t - 120 - 90)

Phasor form  I = 4 ∠-210⁰

To have the impedance nature we compute

Z = V / I      ⇒  Z = 20 ∠30⁰ / 4  ∠-210⁰    Z = 5 ∠-180⁰

We notice that  voltage advances the current then we are in presence of an inductive impedance

5 = wl      ⇒  5  = 400 *L       ⇒  L  =    0,0125 H        

b) v(t) = 9 cos ( 900t - 60 )

V = 9∠-60⁰

i(t)  = 4 sin ( 900t + 280 )    ⇒  i(t) = 4 cos ( 900t + 280 - 90)

i(t) = 4 cos (900t + 190 )    ⇒  I = 4 ∠ 190⁰

Z = V/I    ⇒  Z = 9∠-60⁰ / 4  ∠ 190⁰    Z = 2,25 ∠-250

In this case the current advances the voltage. Impedance capacitive

1/wc  = 1/ 900*C       1/wc = Z   ⇒ 2,25 = 1/ 900*C

2,25*900 = 1/C     ⇒  2025 =1/C     ⇒  C = 4,94 *10⁻⁴ F

c) v(t) = - 13 cos ( 250t + 60 )

v(t) = 13 cos ( 250t + 60 +180 )    ⇒ v(t) = 13 cos ( 250t +240)

Phasor Form

V = 13 ∠240⁰

i(t) = 7 sin (250t + 240 - 90)  ⇒  i(t) = 7 sin (250t + 150)

Phasor Form  I = 7  ∠150⁰

Z = 13∠240⁰ / 7 ∠150⁰    ⇒  Z = 1,86 ∠ 90⁰

Voltage advances the current then the impedance is inductive

wl = 250L     250 L = 1,86     L  = 1,86/250     L = 0,0074 H

7 0
3 years ago
A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The
attashe74 [19]

Answer:

q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C

Explanation:

R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W

So, heat rate can be calculated as follows:

q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW

Surface temperatures can be calculated as follows:

T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C

6 0
4 years ago
Hỗ trợ mình với được không các bạn
Leya [2.2K]

Answer:

Explanation:

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6 0
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