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mr_godi [17]
3 years ago
9

When we utilize a visualization on paper/screen, that visualization is limited to exploring: Group of answer choices Relationshi

p among an infinite number of variables Relationship between a dependent and an independent variable As many variables as we can coherently communicate in 2 dimensions Relationship between two variables, x
Engineering
1 answer:
Mila [183]3 years ago
6 0

Answer:

As many variables as we can coherently communicate in 2 dimensions

Explanation:

Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.

When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).

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A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a
almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water (\dot m), in kilograms per second, is determined by following formula:

\dot m = \frac{\eta \cdot \dot W}{g\cdot H} (1)

Where:

\dot W - Pump power, in watts.

\eta - Efficiency, no unit.

g - Gravitational acceleration, in meters per square second.

H - Hydrostatic column, in meters.

If we know that \eta = 0.72, \dot W = 5000\,W, g = 9.807\,\frac{m}{s^{2}} and H = 25\,m, then the mass flow rate of water is:

\dot m = 14.683\,\frac{kg}{s}

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump (\Delta P), in pascals, is determined by this equation:

\Delta P = \rho\cdot g\cdot H (2)

Where \rho is the density of water, in kilograms per cubic meter.

If we know that \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and H = 25\,m, then the pressure difference is:

\Delta P = 245175\,Pa

The pressure difference across the pump is 245.175 kilopascals.

4 0
3 years ago
Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear
Rus_ich [418]
The shear stress at any given point y1 along the height of the cross section is calculated by: where Ic = b·h3/12 is the centroidal moment of inertia of the cross section. The maximum shear stress occurs at the neutral axis of the beam and is calculated by: where A = b·h is the area of the cross section.
3 0
4 years ago
A __________ is a single lane used by drivers to enter and exit a freeway. a) Climbing lane b) Weave lane c) Thru lane d) Offset
Rufina [12.5K]

A Weave lane is a single lane used by drivers to enter and exit a freeway.

<h3>What is this lane about?</h3>

Weave lanes are known to be lanes that acts as an entrance and exits for a lot of cars in highways.

Hence, one can say that A Weave lane is a single lane used by drivers to enter and exit a freeway.

Learn more about Weave lane from

brainly.com/question/10828527

#SPJ1

6 0
2 years ago
A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th
Inessa [10]

Answer:

attached below

Explanation:

4 0
4 years ago
Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.
djyliett [7]

Solution :

The isentropic efficiency of the turbine is given as :

$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$

  $=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$

  $=\frac{h_1-h_2}{h_1-h_{2s}}$

The entropy relation for the isentropic process is given by :

$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$

$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$

$ \frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$

$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$

Now obtaining the properties from the ideal gas properties of air table :

At $T_1 = 1600 \  K,$

$P_{r1}=791.2$

$h_1=1757.57 \ kJ/kg$

Calculating the relative pressure at state 2s :

$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$

$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$

$P_{r2}=63.296$

Obtaining the properties from Ideal gas properties of air table :

At $P_{r2}=63.296$,  $T_{2s}\approx 860 \ K$

Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:

  $\eta=\frac{h_1-h_2}{h_1-h_{2s}}$

$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$

$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$

$0.9=\frac{1600-T_2}{1600-860}$

$T_2= 938 \ K$

So, at $T_2= 938 \ K$, $h_2=975.66 \ kJ/kg$

Now calculating the work developed per kg of air is :

$w=h_1-h_2$

  = 1757.57 - 975.66

  = 781 kJ/kg

Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.

4 0
3 years ago
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