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3 years ago

For H2O, determine the specified property at the indicated state.

Engineering

1 answer:

zheka24 [161]3 years ago

7 0

Answer:

3.613 bar

1.0290 m3/kg

0.038378 m3/kg

1.363 m3/kg

0.012166 m3/kg

0.002083 m3/kg

0.001177 m3/kg

Explanation:

part a

Table A-3: vf = 1.0435 x 10-3 m 3/kg , vg = 1.673 m 3 /kg. Since vf < v < vg, the state is in the two phase liquid-vapor region, as shown.

From Table A-3, the pressure is the saturation is pressure at 140 C: p = 3.613 bar.

Answer: p = 3.613 bar

part b

The pressure is higher than the critical pressure, as shown on the diagram. Hence, the state is in the compressed liquid region.

From Table A-5: v = 1.0290 m3 /kg.

Answer: v = 1.0290 m3 /kg

part c

Since the temperature is higher than Tsat at 100 bar, the state is super-heated vapor. T sat = 311.1 C

Interpolating in Table A-4, we get

v = 0.038378 m3/kg

Answer: v = 0.038378 m3/kg

part d

vx = vf + x(vg – vf)

v = 1.0291 x 10-3 + (0.4)(3.407 - 1.0291 x 10-3 )

v = 1.363 m3/kg

Answer: v = 1.363 m3/kg

a ) T = 440 C , p = 20 MPa

Tsat(@20 MPa) = 365.75 C

T > Tsat(@20 MPa) hence, super-heated steam

Interpolate the results

v = 0.012166 m3/kg

Answer: v = 0.012166 m3/kg

b) T = 160 C , p= 20 MPa

Tsat(@20 MPa) = 365.75 C

T < Tsat(@20 MPa) hence, compressed liquid region

v = 0.002038 m3/kg

Answer: v = 0.002038 m3/kg

c) T = 40 C , p = 2 MPa

Tsat(@2 MPa) = 212.38 C

T < Tsat(@2 MPa) hence, compressed liquid region

v = 0.001177 m3/kg

Answer: v = 0.001177 m3/kg

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A 2-kW pump is used to pump kerosene ( rho = 0. 820 kg/L) from a tank on the ground to a tank at a higher elevation. Both tanks

Alborosie

The maximum volume flow rate of kerosene is 8.3 L/s

<h3>What is the maximum volume flow rate?</h3>

In fluid dynamics, the maximum volume flow rate (Q) is the volume or amount of fluid flowing via a required cross-sectional area per unit time.

In fluid mechanics, using the following relation, we can determine the maximum volume flow rate of kerosene.

Power = mass flow rate(m) × specific work(w) --- (1)
Specific work = acceleration due to gravity (g) × head (h) ---- (2)
Mass flow rate (m) = density (ρ) × volume flow rate (Q) --- (3)

By combining the three equations together, we have:

The power gained through the fluid pump to be:

P = ρ × Q × g × h

Making Q the subject, we have:

$\mathbf{Q = \dfrac{P}{\rho \times g \times h}}$

where:

P = 2 kW = 2000 W
ρ = 0.820 kg/L
g = 9.8 m/s
h = 30 m

$\mathbf{Q = \dfrac{2000 \ W }{820 \ kg/m^3 \times 9.8 m/s \times 30 \ m}}$

Q = 0.008296 m³/s

Q ≅ 8.3 L/s

Learn more about the maximum volume flow rate here:

brainly.com/question/19091787

8 0

2 years ago

An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?

kifflom [539]

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = $\frac{NE}{t}$

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = $\frac{NE}{t}$ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$ [t = 1 s for per second]

550 × 10³ = $\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}$

550 = N × 4903.24 × 10⁻³⁴

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

7 0

3 years ago

One more because i am running out of points! Have a lovely rest of your guyses day! :)

VMariaS [17]

Answer:

thank you

Explanation:

so much for that

8 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago

If an object moves 2 meters to the left in 1 second, what is the velocity of the object?

tresset_1 [31]

Answer:

Explanation:

possibly -4

7 0

3 years ago