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anygoal [31]
3 years ago
14

Where Does a Solar Engineer Work? (2 sentences or more please)

Engineering
2 answers:
Phoenix [80]3 years ago
5 0
Where Does a Solar Engineer Work? Most solar engineers work in offices, but may also travel frequently to different work sites, including overseas locations. They often must climb ladders onto rooftops to inspect installations, and may need to carry heavy loads for short distances.
Vsevolod [243]3 years ago
3 0

Answer:

in a service mask...they also work there to mount mask and take care of the sites

You might be interested in
Question 64 (1 point)
Xelga [282]

Answer: c fine sand aggregate, portland cement,fine sand

Explanation:

7 0
3 years ago
An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t
solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the w_{p} to h₁, where  w_{p}  is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute  w_{p} To computer  

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water v_{f} = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ =  0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

    = 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

    =  417.5 + 0.001043 (14900)

    = 417.5 + 15.5407

    = 433.04 kJ/kg

Find h₃  

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = s_{f} + x_{4} s_{fg}

x_{4} = s_{4} -s_{f} /s_{fg}

x_{4} = 6.14 -  2.45 / 3.89

x_{4}   = 0.9497

The enthalpy h₄:

h₄ = h_{f} +x_{4} h_{fg}

    = 908.4 + 0.9497(1889.8)

    =  908.4 + 1794.7430

    = 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅  = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to  So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = s_{f} + x_{6} s_{fg}

x_{6} = s_{6} -s_{f} /s_{fg}

x_{6} = 7.29 - 1.3028 / 6.0562

x_{6}   = 0.988

The enthalpy h₆:

h₆ = h_{f} +x_{6} h_{fg}

    = 417.51 + 0.988 (2257.5)

    = 417.51 + 2230.41

  h₆ =  2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

P_{p} is found by using:

mass flow rate = m =  1.74 kg/s

Volume of water = v₁ =  0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

P_{p}  = ( m ) ( v₁ ) ( p₂ - p₁ )

     = (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

     = (1.74 kg/s) (0.001043 m³/kg) (14900)

     = 27.04

P_{p} = 27 kW

Compute heat added q_{a} and heat rejected q_{r}  from boiler using computed enthalpies:

q_{a} = ( h₃ - h₂ ) + ( h₅ - h₄ )

      = ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

      = 2726 + 655

      = 3381  kJ/kg

q_{r} =  h₆ - h₁

  = 2648 kJ/kg - 417.5 kJ/kg

  = 2232 kJ/kg

Compute net work

W_{net} = q_{a} - q_{r}

       = 3381  kJ/kg - 2232 kJ/kg

       = 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m =  1.74 kg/s

W_{net} = 1150 kJ/kg

P = m * W_{net}

  = 1.74 kg/s * 1150 kJ/kg

  = 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

   =  1.74 kg/s * 655

   =  1140 kW

Compute Thermal efficiency of this system

μ_{t} = 1 - q_{r} /  q_{a}

   = 1 - 2232 kJ/kg / 3381  kJ/kg

   = 1 - 0.6601

   = 0.34

   = 34%

7 0
3 years ago
500 flights land each day at San Jose’s airport. Assume that each flight has a 5% chance of being late, independently of whether
BARSIC [14]

Answer:

a.0.0199

b.0.1765

c.0.0785

d.0.1268

e.Yes

Explanation:

It is given that   X follows a  Binomial distribution with (n= 500, p = 0.05)

The  probabilities  are computed using the EXCEL .

a) The required probability here is:

P(X less of equal to  15)

= binom.dist(15,500,0.05,TRUE)

=0.0199

Therefore the probability is 0.0199 .

b) The required probability here is:

P(X greater or equal to 30) = 1 - P(X less or equal to  29)  

=1 - binom.dist(29,500,0.05,TRUE)

=0.1765

Therefore the probability is 0.1765

c) P(X = 26 )

= binom.dist(26,500,0.05,FALSE)  

=0.0785

Therefore the probability is 0.0785

d) The required probability here is computed as:

P(10 less or equal to X less or equal to 20 ) = P(X less or equal to 19) - P(X less or equal to 10)

= binom.dist(19,500,0.05,TRUE) - binom.dist(10,500,0.05,TRUE)

=0.1268

Therefore the probability 0.1268

e) Yes . Therefore the probability because that is the assumption used to apply binomial distribution .

6 0
3 years ago
Compare electromagnets and solenoids
natta225 [31]

An electromagnet is a made coil associated with a ferromagnetic core. This way, the strength of the magnet is controlled by the input current. A solenoid is a simple shape used in magnetostatics or magnetics. ... A solenoid is a cylindrical coil of wire whose diameter is small compared to its length.

8 0
3 years ago
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat
zlopas [31]

Answer: 10.631\times 10^3\ N/m^2

Explanation:

Given

Discharge is Q=12.5\ L

Diameter of pipe d=150\ mm

Distance between two ends of pipe L=800\ m

friction factor f=0.008

Average velocity is given by

\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s

Pressure difference is given by

\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow  \Delta P=10.631\ kPa

8 0
2 years ago
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