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.
2
A spring balance pulls with 5 N on a beam of 0.5 m.
What is the torque at the end of the beam?
Answer:
The torque at the end of the beam is 2.5 Nm
Explanation:
Given;
length of beam, r = 0.5 m
applied force, F = 5 N
The torque at the end of the beam is given by;
τ = F x r
where;
τ is the torque
F is applied force
r is length of the beam
τ = 5 x 0.5
τ = 2.5 Nm
Therefore, the torque at the end of the beam is 2.5 Nm
Think about a good game story that made you feel a mix of positive and negative emotions. What was the story, what emotions did you feel, and how did it make you feel them? Why did those emotions draw you into the story?
Which of the following best describes empathy?
the understanding of the feelings and beliefs of others
the lack of pride or boastfulness
the courage to speak up with one’s ideas
the possession of honesty and high morals
Answer:
the first one is the correct answer
Answer:
the first one would be correct
Explanation:
3.94 x 105) + (2.04 x 105)
Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is −5°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 = 10 W/m2⋅K and h2 = 25 W/m2⋅K, respectively, and disregard any heat transfer by radiation.
Answer:
The steady rate of heat transfer through the glass window is 707.317 watts.
Explanation:
A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:
[tex]\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}[/tex] (Eq. 1)
Given that window is represented as a flat element, we can expand (Eq. 1) as follows:
[tex]\dot Q_{total} = \frac{T_{i}-T_{o}}{R}[/tex] (Eq. 2)
Where:
[tex]T_{i}[/tex], [tex]T_{o}[/tex] - Indoor and outdoor temperatures, measured in Celsius.
[tex]R[/tex] - Overall thermal resistance, measured in Celsius per watt.
Now, we know that glass window is configurated in series and overall thermal resistance is:
[tex]R = R_{cond} + R_{conv, in}+R_{conv, out}[/tex] (Eq. 3)
Where:
[tex]R_{cond}[/tex] - Conductive thermal resistance, measured in Celsius per watt.
[tex]R_{conv, in}[/tex], [tex]R_{conv, out}[/tex] - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.
And we expand the expression as follows:
[tex]R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}[/tex]
[tex]R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)[/tex] (Eq. 4)
Where:
[tex]w[/tex] - Width of the glass window, measured in meters.
[tex]d[/tex] - Length of the glass window, measured in meters.
[tex]l[/tex] - Thickness of the glass window, measured in meters.
[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius.
[tex]h_{i}[/tex], [tex]h_{o}[/tex] - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.
If we know that [tex]w = 2.4\,m[/tex], [tex]d = 1.5\,m[/tex], [tex]l = 0.006\,m[/tex], [tex]k = 0.78\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex] and [tex]h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex], the overall thermal resistance is:
[tex]R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)[/tex]
[tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex]
Now, we obtain the steady rate of heat transfer from (Eq. 2): ([tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex], [tex]T_{i} = -5\,^{\circ}C[/tex], [tex]T_{o} = 24\,^{\circ}C[/tex])
[tex]\dot Q_{total} = \frac{24\,^{\circ}C-(-5\,^{\circ}C)}{0.041\,\frac{^{\circ}C}{W} }[/tex]
[tex]\dot Q_{total} = 707.317\,W[/tex]
The steady rate of heat transfer through the glass window is 707.317 watts.