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Sladkaya [172]
2 years ago
12

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Engineering
1 answer:
vodka [1.7K]2 years ago
5 0

Answer:

(⌐■-■) (⌐■-■) (⌐■-■) (⌐■-■)

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garri49 [273]
Concentrating solar power (CSP) plants use mirrors to concentrate the sun's energy to drive traditional steam turbines or engines that create electricity. The thermal energy concentrated in a CSP plant can be stored and used to produce electricity when it is needed, day or night. Today, roughly 1,815 megawatts (MWac) of CSP plants are in operation in the United States.

Parabolic Trough
Parabolic trough systems use curved mirrors to focus the sun’s energy onto a receiver tube that runs down the center of a trough. In the receiver tube, a high-temperature heat transfer fluid (such as a synthetic oil) absorbs the sun’s energy, reaching temperatures of 750°F or higher, and passes through a heat exchanger to heat water and produce steam. The steam drives a conventional steam turbine power system to generate electricity. A typical solar collector field contains hundreds of parallel rows of troughs connected as a series of loops, which are placed on a north-south axis so the troughs can track the sun from east to west. Individual collector modules are typically 15-20 feet tall and 300-450 feet long.

Compact Linear Fresnel Reflector
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3 0
3 years ago
Read 2 more answers
Can i join three 12 volts batteriesto give me 24 volts output​
bulgar [2K]

Answer:

YES

Explanation:

If we connect batteries in series then the output voltage is the sum of the individual voltage of each battery i.e if you connect three 12 volts batteries in series then their output voltage will be 12+12+12=36 volts, but the current rating of the batteries in series will be same of the individual battery rating in 'mah'.

But when we connect the batteries in parallel their voltage is not added  but their current rating in mah is addition of their individual rating.

So, If you want 24 volts from three 12 volts battery then you can connect two of them in series and the other one in parallel with them this will give 24 volts and the current will be addition of the two series batteries and the third which is in parallel with them. You can use this configuration if current value is not a big factor.

8 0
3 years ago
How can any student outside apply for studying engineering at Cambridge University​
telo118 [61]
Admission to the Engineering course at Cambridge is highly competitive, both in terms of the numbers and quality of applicants. In considering applicants, Colleges look for evidence both of academic ability and of motivation towards Engineering. There are no absolute standards required of A Level achievement, but it should be noted that the average entrant to the Department has three A* grades. You need to get top marks in Maths and Physics.All Colleges strongly prefer applicants for Engineering to be taking a third subject that is relevant to Engineering.
Hope that helps and good luck if you are applying. Can you please mark this as brainliest and press the thank you button and if you have any further questions please let me know!!
3 0
3 years ago
2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compres
lutik1710 [3]
You can see and download from the link

https://tlgur.com/d/GYYVL5lG

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5 0
2 years ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
3 years ago
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