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CaHeK987 [17]
3 years ago
5

A 200‑m rigid vessel contains a saturated liquid‑vapor mixture with a vapor quality of 75%. The temperature of the vessel is mai

ntained at 393°K. What is the total mass of the liquid‑vapor mixture inside the vessel?
Engineering
1 answer:
DerKrebs [107]3 years ago
6 0

Answer:

Given,

Temperature;

T = 393;;K

Convert to Celcius;

T = (393-273) degrees

T = 120°C

Using Table A-4 (Saturated water - Temperature table), at T = 120 C;

vf = 0.001060 m³/kg

vg = 0.89133 m³/kg

Quality is given as;

75% = 0.75

Specific volume is given as;

v = vf + x (vg - vf) = 0.001060 + 0.75(0.89133 _ 0.001060)

v= 0.66876 m³/kg

We know;

v = V/m

0.66876  = 100/m

m = 149.53 kg

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1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
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Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

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\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

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n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

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\Delta t_{min} = \frac{l}{v}

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