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Mnenie [13.5K]
3 years ago
15

The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

and the atmospheric pressure is 97 kPa.
Engineering
1 answer:
s2008m [1.1K]3 years ago
7 0

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325  P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.2 + 273.15) K = 301.15 K  

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K  

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

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A 0.39 percent Carbon hypoeutectoid plain-carbon steel is slowly cooled from 950 oC to a temperature just slightly below 723 oC.
belka [17]

Answer:

53%

Explanation:

To find the weight % in proeutectoid ferrite we use:

Wt % in proeutectoid ferrite=

[(0.80 - 0.39) / (0.80 - 0.02)] * 100=

= (0.41 /0.78) * 100 =

52.56%

which is approximately 53%

8 0
3 years ago
Zack's new home is progressing well, and the foundation work is finished. The general contractor stops by Zack's present home to
iogann1982 [59]

Answer:

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5 0
2 years ago
How much energy in joule is added to a 12 g of sample of aluminum (c=0.897 J/g ◦C) to raise the temperature from 20 ◦C to 45 ◦C?
MArishka [77]

Answer:

269.1J

Explanation:

m = 12g

c = 0.897J/g°C

∆T = 45 - 20 = 25°C

H = mc∆T = 12 × 0.897 × 27 = 269.1J

8 0
3 years ago
Three return steam lines in a chemical processing plan enter a collection tank operating at a steady state at 1 bar. Steam enter
gavmur [86]

Answer:

a. 4kg/s

b. 99.97 °C

Explanation:

Hello,

a. The resulting mass balance turns out into:

F_1+F_2+F_3=F_{out}\\F_{out}=0.8kg/s+2kg/s+1.2kg/s=4kg/s

b. Now, the energy balance is:

F_1h_1+F_2h_2+F_3h_3-Q_{out}=F_{out}h_{out}

In such a way, the first enthalpy is taken as a liquid-vapor mixture at 1 bar and 0.9 quality, it means:

h_1=hf(1bar)+xhfg(1bar)\\h_1=419.06kJ/kg+0.9*2256.5kJ/kg=2449.91kJ/kg

Second enthalpy is taken by identifying that stream as an overheated vapor at 1 bar and 200 °C, thus, the resulting enthalpy is:

h_2=2875.5kJ/kg

Then, the third enthalpy is taken by considering that at 95°C and 1 bar the water is a saturated liquid, thus:

h_3=hf(95^0C)=398.09kJ/kg.

Now, by solving for h_{out}, we've got:

h_{out}=\frac{0.8kg/s*2449.91kJ/kg+2kg/s*2875.5kJ/kg+1.2kg/s*398.09kJ/kg-40kW}{4kg/s} \\h_{out}=\frac{8148.612kJ/s}{4kg/s} \\h_{out}=2037.153kJ/kg

Finally, by searching for that value of enthalpy, one sees that at 1 bar, the exiting stream is a liquid-vapor mixture that is at 99.97 °C and has a 72%- quality.

(NOTE: all the data was extracted from Cengel's book 7th edition).

Best regards.

4 0
3 years ago
Water from an upper tank is drained into a lower tank through a 5 cm diameter iron pipe with roughness 2 mm. The entrance to the
nignag [31]

Answer:

Relative roughness = 0.04

Explanation:

Given that:

Diameter = 5 cm

roughness = 2 mm

At inlet:

Minor coefficient loss k_{L1} = 0.4

At exit:

Minor coefficient loss k_{L2} = 1

Height h = 4m

Length = 5 m

To find the relative roughness:

Relative roughness is a term that is used to describe the set of irregularities that exist inside commercial pipes that transport fluids. The relative roughness can be evaluated by knowing the diameter of the pipe made with the absolute roughness in question. If we denote the absolute roughness as e and the diameter as D, the relative roughness is expressed as:

e_r = \dfrac{e}{D}

e_r = \dfrac{0.2 }{5}

\mathbf{e_r = 0.04}

5 0
3 years ago
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