A storage tank has a volume of 3,000 liters. The tank is originally filled to a pressure of 21.1 megapascals with an ideal gas while the temperature is maintained at 28 degrees Celsius. The tank is heated so the temperature increases to a final temperature of 58 degrees Celsius. What will be the new pressure in the tank in units of megapascals?

Answer:

The pump delivers 32.737 kilowatts to the water.

Explanation:

We can describe the system by applying the Principle of Energy Conservation and the Work-Energy Theorem, the pump system, which works at steady state and changes due to temperature are neglected, is represented by the following model:

[tex]\dot W_{in} + \dot m \cdot g \cdot (z_{1}-z_{2}) + \frac{1}{2}\cdot \dot m \cdot (v_{1}^{2}-v_{2}^{2})+\dot m \cdot [(u_{1}+P_{1}\cdot \nu_{1})-(u_{2}+P_{2}\cdot \nu_{2})]-\dot E_{losses} = 0[/tex] (Eq. 2)

Where:

[tex]\dot m[/tex] - Mass flow, measured in kilograms per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final flow speeds at pump nozzles, measured in meters per second.

[tex]u_{1}[/tex], [tex]u_{2}[/tex] - Initial and final internal energies, measured in joules per kilogram.

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.

[tex]\nu_{1}[/tex], [tex]\nu_{2}[/tex] - Initial and final specific volumes, measured in cubic meters per kilogram.

Then, we get this expression:

[tex]\dot W_{in} + \dot m \cdot g \cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot \dot m \cdot (v_{1}^{2}-v_{2}^{2}) +\dot m\cdot \nu \cdot (P_{1}-P_{2})-\dot E_{losses} = 0[/tex]  (Ec. 3)

We note that specific volume is the reciprocal of density:

[tex]\nu = \frac{1}{\rho}[/tex] (Ec. 4)

Where [tex]\rho[/tex] is the density of water, measured in kilograms per cubic meter.

The initial pressure of water ([tex]P_{1}[/tex]), measured in pascals, can be found by Hydrostatics:

[tex]P_{1} = P_{atm} + \rho\cdot g \cdot \Delta z[/tex] (Ec. 5)

Where:

[tex]P_{atm}[/tex] - Atmospheric pressure, measured in pascals.

[tex]\Delta z[/tex] - Depth of the entrance of the inlet pipe with respect to the limit of the water reservoir.

If we know that [tex]p_{atm} = 101325\,Pa[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta z = 6\,m[/tex], then:

[tex]P_{1} = 101325\,Pa+\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}})\cdot (6\,m)[/tex]

[tex]P_{1} = 160167\,Pa[/tex]

And the specific volume of water ([tex]\nu[/tex]), measured in cubic meters per kilogram, is: ([tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex])

[tex]\nu = \frac{1}{1000\,\frac{kg}{m^{3}} }[/tex]

[tex]\nu = 1\times 10^{-3}\,\frac{m^{3}}{kg}[/tex]

The power losses due to friction is found by this expression:

[tex]\dot E_{losses} = \dot m \cdot g\cdot h_{losses}[/tex]

Where [tex]h_{losses}[/tex] is the total friction head loss, measured in meters.

The mass flow is obtained by this:

[tex]\dot m = \rho \cdot \dot V[/tex] (Ec. 6)

Where [tex]\dot V[/tex] is the volumetric flow, measured in cubic meters per second.

If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex] and [tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], then:

[tex]\dot m = \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(0.061\,\frac{m^{3}}{s} \right)[/tex]

[tex]\dot m = 61\,\frac{kg}{s}[/tex]

Then, the power loss due to friction is: ([tex]h_{losses} = 5\,m[/tex])

[tex]\dot E_{losses} = \left(61\,\frac{kg}{s}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (5\,m)[/tex]

[tex]\dot E_{losses} = 2991.135\,W[/tex]

Now, we calculate the inlet and outlet speed by this formula:

[tex]v = \frac{\dot V}{\frac{\pi}{4}\cdot D^{2} }[/tex] (Ec. 7)

Inlet nozzle ([tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], [tex]D = 0.12\,m[/tex])

[tex]v_{1} = \frac{0.061\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.12\,m)^{2} }[/tex]

[tex]v_{1} \approx 5.394\,\frac{m}{s}[/tex]

Oulet nozzle ([tex]\dot V = 0.061\,\frac{m^{3}}{s}[/tex], [tex]D = 0.05\,m[/tex])

[tex]v_{2} = \frac{0.061\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.05\,m)^{2} }[/tex]

[tex]v_{2} \approx 31.067\,\frac{m}{s}[/tex]

([tex]\dot m = 61\,\frac{kg}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2} = 2\,m[/tex], [tex]z_{1} = -6\,m[/tex], [tex]v_{2} \approx 31.067\,\frac{m}{s}[/tex], [tex]v_{1} \approx 5.394\,\frac{m}{s}[/tex], [tex]P_{2} = 101325\,Pa[/tex], [tex]P_{1} = 160167\,Pa[/tex], [tex]\dot E_{losses} = 2991.135\,W[/tex])

[tex]\dot W_{in} = \left(61\,\frac{kg}{s}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [2\,m-(-6\,m)]+\frac{1}{2}\cdot \left(61\,\frac{kg}{s}\right) \cdot \left[\left(31.067\,\frac{m}{s} \right)^{2}-\left(5.394\,\frac{m}{s} \right)^{2}\right] +\left(61\,\frac{kg}{s}\right)\cdot \left(1\times 10^{-3}\,\frac{m^{3}}{kg} \right)\cdot (101325\,Pa-160167\,Pa)+2991.135\,W[/tex]

[tex]\dot W_{in} = 32737.518\,W[/tex]

The pump delivers 32.737 kilowatts to the water.

Answer:

the first one is the correct answer

Answer:

the first one would be correct

Explanation:

Answer:

  a. vehicle color

Explanation:

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  a. vehicle color

__

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