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kondaur [170]
3 years ago
11

A 500-m^3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 20% of the mass is liquid and the 80%

of the mass is vapor, the total mass in the tank is: (a) 635 kg (b) 2809 kg (c) 258 kg (d) 500 kg (e) 705 kg
Engineering
1 answer:
guapka [62]3 years ago
3 0

Answer:

(a) 705 kg

Explanation:

it is given that volume =500m^3

pressure = 200 kPa

it is given that 80% mass is vapor so dryness fraction =x=0.8

from the standard table at 200 kPa

v_f=0.001061 \frac{m^3}{kg}

v_g=0.88578\frac{m^3}{kg}

specific volume v=v_f+x\left ( v_g-v_f \right )

v=0.00106+0.8\left ( 0.88578-0.00106 \right )=0.7088\frac{m^3}{kg}

we know that specific\ volume =\frac{volume}{mass}

0.7088=\frac{500}{mass}

mass=\frac{500}{0.7088}=705.456 \ kg

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Answer:

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Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

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As we know,

The shape factor will be:

⇒ SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}

By putting the value, we get

⇒       =\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}

⇒       =3.7258 \ l

hence,

The heat loss will be:

⇒ Q=SF\times K(T_2-T_1)

       =3.7258\times 1\times 1.8\times (500-300)

       =3.7258\times 1.8\times (200)

       =1341.288 \ W/m

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2 years ago
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Answer:

576.21kJ

Explanation:

#We know that:

The balance mass m_{in}+m_{out}=\bigtriangleup m_{system}

so, m_e=m_1-m_2

Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1

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#Replace known values in the equation above;h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg

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m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg

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Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ

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5 0
3 years ago
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Answer:

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Explanation:

3 0
2 years ago
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

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answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

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where ;

ε = 0.95

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As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

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Qconv = h* As ( ΔT )

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Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

5 0
3 years ago
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