Answer:
A key element is powering economies with clean energy, replacing polluting coal - and gas and oil-fired power stations - with renewable energy sources, such as wind or solar farms. This would dramatically reduce carbon emissions. Plus, renewable energy is now not only cleaner, but often cheaper than fossil fuels
Explanation:
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Given Information:
Initial temperature of aluminum block = 26.5°C
Heat flux = 4000 w/m²
Time = 2112 seconds
Time = 30 minutes = 30*60 = 1800 seconds
Required Information:
Rise in surface temperature = ?
Answer:
Rise in surface temperature = 8.6 °C after 2112 seconds
Rise in surface temperature = 8 °C after 30 minutes
Explanation:
The surface temperature of the aluminum block is given by

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.
After t = 2112 sec:

The rise in the surface temperature is
Rise = 35.1 - 26.5 = 8.6 °C
Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.
After t = 30 mins:

The rise in the surface temperature is
Rise = 34.5 - 26.5 = 8 °C
Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.
Answer:
a) 3581.15067 kw
b) 95.4%
Explanation:
<u>Given data:</u>
compressor efficiency = 85%
compressor pressure ratio = 10
Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K
At turbine inlet : pressure = 950 kPa, temperature = 1400k
Turbine efficiency = 88% , exit pressure of turbine = 100 kPa
A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW
attached below is a detailed solution to the given question
Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557