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2 years ago

7

What computer program can you use to publish and share a research project with others?

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

Word Online is the program

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An isothermal CSTR with a first order irreversible reaction A â> B (and rA[mol/(ft3*min)] = - 0.5 CA) has a constant flow rat

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2 years ago

Which company introduced the Windows operating system correct answer plz <br>

grigory [225]

Answer:

I think Microsoft Corporation

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2 years ago

A rigid tank contains 3 kg of water initially at 43.97% quality and at a temperature of 120°C. The water is heated until it reac

makkiz [27]

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3 years ago

Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,

Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take $p_o$ = 880 kg/m³ and $p_{glycerol$ = 1260 kg/m³

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

Explanation:

Given that;

Inlet velocity of Glycerin, $V_A$ = 6 m/s

Inlet velocity of oil, $V_B$ = 3 m/s

Density velocity of glycerin, $p_{glycerol$ = 1260 kg/m³

Density velocity of glycerin, Take $p_o$ = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, $d_A$ = 100 mm = 0.1 m

Diameter of oil pipe, $d_B$ = 80 mm = 0.08 m

Diameter of outlet pipe $d_C$ = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

$Q_A + Q_B = Q_C$

$A_Av_A + A_Bv_B = A_Cv_C$

π/4 × ( $d_A$ )² $v_A$ + π/4 × ( $d_B$ )² $v_B$ = π/4 × ( $d_C$ )² $v_C$

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)² $v_C$

0.04712 + 0.0150796 = 0.0113097 $v_C$

0.0621996 = 0.0113097 $v_C$

$v_C$ = 0.0621996 / 0.0113097

$v_C$ = 5.5 m/s

Now we apply the mass flow rate condition

$m_A + m_B = m_C$

$p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C$

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p

72.6412 = 0.06220335p

p = 72.6412 / 0.06220335

p = 1167.8 kg/m³

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

4 0

3 years ago

The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 whi

coldgirl [10]

Answer:

The initial temperature will be "385.1°K" as well as final will be "128.3°K".

Explanation:

The given values are:

Helium's initial volume, v₁ = 6 m³

Mass, m = 1.5 kg

Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, $W=p(v_{2}-v_{1})$

On putting the estimated values, we get

⇒ $=200000(2-6)$

⇒ $=200000\times (-4)$

⇒ $=800,000 \ N.m$

Now,

Gas ideal equation will be:

⇒ $pv_{1}=mRT_{1}$

On putting the values. we get

⇒ $200000\times 6=1.5\times 2077\times T_{1}$

⇒ $T_{1}=\frac{1200000}{3115.5}$

⇒ $=385.1^{\circ}K$ (Initial temperature of helium)

and,

⇒ $pv_{2}=mRT_{2}$

On putting the values, we get

⇒ $200000\times 2=1.5\times 2077\times T_{2}$

⇒ $T_{2}=\frac{400000}{3115.5}$

⇒ $=128.3^{\circ}K$ (Final temperature of helium)

3 0

3 years ago