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aivan3 [116]
2 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]2 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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A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe
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Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

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Using Table ( saturated water - pressure table);

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v_f = 0.001057 m³/kg

v_g = 1.0037 m³/kg

u_f = 486.82 kJ/kg

u_g 2524.5 kJ/kg

h_g = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V_f/v_f + V_g/v_g

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v_g

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m_{out = m₁ - m₂

m_{out = 1.89414  - 0.003985

m_{out = 1.890155 kg

so, Initial internal energy will be;

U₁ = m_fu_f + m_gu_g

U₁ = (V_f/v_f)u_f  + (V_g/v_g)u_g

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E_{in -  E_{out = ΔE_{sys

QΔt - m_{outh_{out = m₂u₂ - U₁

QΔt - m_{outh_g = m₂u_g - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

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