All categories

3 years ago

What Degree Do You Need To Become a Solar Engineer? (2 or more sentences please)

Engineering

1 answer:

Nataliya [291]3 years ago

7 0

Answer: A bachelor's degree in mechanical engineering or electrical engineering.

Explanation:

A bachelor's degree in mechanical engineering or electrical engineering is typically required for solar engineering positions. In some areas, more advanced certification could be required. Degrees in industrial engineering, chemical engineering, and computer software engineering may also be helpful.

Hope this will help you!

You might be interested in

The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%

Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf) ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

quality = 0.4698

so quality is 47 %

SO OPTION C IS CORRECT

4 0

4 years ago

In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical

stealth61 [152]

Answer:

Explanation:

We know that Drag force $F_D$

$F_D=\dfrac{1}{2}C_D\rho AV^2$

Where

$C_D$ is the drag force constant.

A is the projected area.

V is the velocity.

ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

6 0

4 years ago

Consider a fluid with mean inlet temperature Ti flowing through a tube of diameter D and length L, at a mass flow rate m'. The t

EleoNora [17]

Answer:

$T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))$

Explanation:

Our data given are:

$T_i =$ Mean temperature (inlet)

D = Diameter

L = Length

$\dot{m}=$ Mass flow rate

Equation to surface flow as,

$q(x) = a+bsin(x\pi/L)$

We need to consider the perimeter of tube (p) to apply the steady flow energy balance to a tube , that is

$q(x)pdx=\dot{m}c_p dT_m$

Where $p=\pi D$

Re-arrange for $dT_m,$

$dT_m = \frac{q(x)pdx}{\dot{m}c_p}$

Integrating from 0 to x (the distance intelt of pipe) we have,

$\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x q(x)dx$

Replacing the value of q(x)

$\int\limit^{T_m(x)}_T_i dT_m = \frac{p}{mc_p}\int_0^x (a+bsin(x\pi/L))dx$

$T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}\int_0^x(a+bsin(x\pi/L))dx$

$T_m(x)-T_i = \frac{\pi D}{\dot{m}c_p}(ax-\frac{bL}{\pi}cos(\frac{x\pi}{L}))^x_0$

$T_m(x) = T_i+\frac{\pi D}{\dot{m}c_p}(ax+\frac{bL}{\pi}-\frac{bL}{\pi}cos(\frac{x\pi}{L}))$

8 0

3 years ago

Using the Breguet range and endurance equation, estimate the amount of kerosene

love history [14]

The amount of kerosene fuel needed for an aircraft weighing 10 metric tons to fly from Boston to Los Angeles, assuming a distance of 5,000 [km] is 10000 kg.

<h3>What is Breguet equation?</h3>

Breguet equation is used to determine the range of airplane fly in some specified set of parameters.

$R=\dfrac{h_f}{g}\dfrac{L}{D}\times \eta \ln \left(\dfrac{W_{initial}}{W_{final}}\right)$

Here, (L/D) is lift to drag ratio,<em> g</em> is gravitation acceleration,<em> h </em>is height and <em>W</em> is weight.

The weight of aircraft is 10 metric tons (this is the dry weight that includes passengers and cargo) The distance of 5,000 [km], with velocity at 300 [m/s] from Boston to Los Angeles has to be cover by the airplane.

The initial weight is 10 metric tons or 10000 kg. Thus, the final weight with total fuel burn is,

$W_{final}=W_{initial}-W_{fuel}\\W_{final}=1-\dfrac{W_{fule}}{W_{initial}}$

The lift-to-drag ratio is 15 and the overall efficiency is 0.3, The standard air density at an altitude of approximately 6,000 [m] is half.

Thus, put the values, in above formula,

$5000=\dfrac{6000}{9.81}(15)\times(0.3) \ln \left(\dfrac{10000}{1-\dfrac{W_{fuel}}{10000}}}\right)\\$

$\ln \left(\dfrac{10000}{1-\dfrac{W_{fuel}}{10000}}}\right)=1.8167\\\ln \left(\dfrac{1}{10000-W_{fuel}}}\right)=1.8167$

Solving this equation, we get,

$W_{fuel}=10000\rm\; kg\\$

Thus, the amount of kerosene fuel needed for an aircraft weighing 10 metric tons to fly from Boston to Los Angeles, assuming a distance of 5,000 [km] is 10000 kg.

Learn more about the Breguet equation here:

brainly.com/question/15218094

#SPJ1

7 0

2 years ago

Pendulum impacting an inclined surface of a block attached to a spring-Dependent multi-part problem assign all parts NOTE: This

Art [367]

Answer:

vA = -2.55 m/s

vB = 0.947 m/s

Explanation:

Given:-

- The initial angle of rope, α = 30°

- The angle of rope just before impact or wedge angle, θ = 20°

- The weight of sphere, Ws = 1-lb

- The initial position velocity, vi = 4 ft/s

- The coefficient of restitution, e = 0.7

- The weight of the wedge, Ww = 2-lb

- The spring constant, k = 1.8 lb/in

- The length of rope, L = 2.6 ft

Find:-

Determine the velocities of A and B immediately after the impact.

Solution:-

- We can first consider the ball ( acting as a pendulum ) to be isolated for study.

- There are no unbalanced fictitious forces acting on the sphere ball. Hence, we can reasonably assume that the energy is conserved.

- According to the principle of conservation for the initial point and the point just before impact.

Let,

vA : The speed of sphere ball before impact

Change in kinetic energy = Change in potential energy

ΔK.E = ΔE.P

0.5*ms* ( uA^2 - vi^2 ) = ms*g*L*( cos ( θ ) - cos ( α ) )

uA^2 = 2*g*L*( cos ( θ ) - cos ( α ) ) + vi^2

uA = √ [ 2*32*2.6*( cos ( 20 ) - cos ( 30 ) ) + 4^2 ] = √28.25822

uA = 5.316 ft/s

- The coefficient of restitution (e) can be thought of as a measure of the extent to which mechanical energy is conserved when an object bounces off a surface:

e^2 = ( K.E_after impact / K.E_before impact )

- The respective Kinetic energies are:

K.E_after impact = K.E_sphere + K.E_block

= 0.5*ms*vA^2 + 0.5*mb*vB^2

K.E_before impact = K.E = Ws*L*( cos ( θ ) - cos ( α ) )

= 1*2.6*( cos ( 20 ) - cos ( 30 ) )

= 0.1915 J

32*2*0.1915*0.7^2 = Ws*vA^2 + Wb*vB^2

6.00544 = vA^2 + 2*vB^2 ... Eq1

- From conservation of linear momentum we have:

vB = e*( uA - uB )*cos ( 20 ) + vA

vB = 0.7*( 5.316 - 0 )*cos ( 20) + vA

vB = 3.49678 + vA .... Eq 2

- Solve two equation simultaneously:

6.00544 = vA^2 + 2*(3.49678 + vA)^2

6.00544 = 3vA^2 + 13.98*vA + 24.455

3vA^2 + 14.8848*vA + 18.4495 = 0

vA = -2.55 m/s

vB = 0.947 m/s

5 0

4 years ago