50.38

Neon is compressed from 100 kPa and 20◦C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and

PIT_PIT [208]

Answer:

The specific volume is reduced in 80 per cent due to isothermal compression.

Specific enthalpy remains constant.

Explanation:

Let suppose that neon behaves ideally, the equation of state for ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in cubic meters.

$n$ - Molar quantity, measured in kilomoles,

$T$ - Temperature, measured in kelvins.

$R_{u}$ - Ideal gas constant, measured in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$ .

On the other hand, the molar quantity ( $n$ ) and specific volume ( $\nu$ ), measured in cubic meter per kilogram, are defined as:

$n = \frac{m}{M}$ and $\nu = \frac{V}{m}$

Where:

$m$ - Mass of neon, measured in kilograms.

$M$ - Molar mass of neon, measured in kilograms per kilomoles.

After replacing in the equation of state, the resulting expression is therefore simplified in term of specific volume:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$P\cdot \nu = \frac{R_{u}\cdot T}{M}$

Since the neon is compressed isothermally, the following relation is constructed herein:

$P_{1}\cdot \nu_{1} = P_{2}\cdot \nu_{2}$

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$\nu_{1}$ , $\nu_{2}$ - Initial and final specific volume, measured in cubic meters per kilogram.

The change in specific volume is given by the following expression:

$\frac{\nu_{2}}{\nu_{1}} = \frac{P_{1}}{P_{2}}$

Given that $P_{1} = 100\,kPa$ and $P_{2} = 500\,kPa$ , the change in specific volume is:

$\frac{\nu_{2}}{\nu_{1}} = \frac{100\,kPa}{500\,kPa}$

$\frac{\nu_{2}}{\nu_{1}} = \frac{1}{5}$

The specific volume is reduced in 80 per cent due to isothermal compression.

Under the ideal gas supposition, specific enthalpy is only function of temperature, as neon experiments an isothermal process, temperature remains constant and, hence, there is no change in specific enthalpy.

Specific enthalpy remains constant.

8 0

3 years ago

Who wants me to make a device to resurrect Juice Wrld

Aleksandr-060686 [28]

Lolololollolollolollolololo

5 0

3 years ago

The flow rate in the pipe system below is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. Point 1 is 0.60 m higher

DedPeter [7]

Answer:

Explanation:

The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.

The answer to the above question is

The pressure at point 2 = 75.959 kPa

Explanation:

Bernoulli's equation with losses gives

hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)

Between points 1 and 2, z₁ = z₃ + 0.6 m therefore

hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

But v = Q/A

or since A = π×D²/4 we have

A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²

Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃ = 1.315 m/s from there we find the friction coefficient from Moody Diagram as follows

ε = $\frac{Roughness _. value}{ Diameter}$ Which gives

the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175

Substituting he above values into the $h_{l}$ equation we get $h_{l}$ = 19.761 m

Combined head loss = 19.761 m

Hence 19.743 m = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)

or 260 kPa-18.82 m × 9.81 m/s²×ρ= P₃

Where ρ = density of water, we have

260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa

6 0

3 years ago

Which excerpt from "The Chrysanthemums' best reveals that Elisa is proud of her

Alisiya [41]

Scissors to avoid any rough edges

4 0

3 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$