Answer:
12.332 KW
The positive sign indicates work done by the system ( Turbine )
Explanation:
Stagnation pressure( P1 ) = 900 kPa
Stagnation temperature ( T1 ) = 658K
Expanded stagnation pressure ( P2 ) = 100 kPa
Expansion process is Isentropic, also assume steady state condition
mass flow rate ( m ) = 0.04 kg/s
<u>Calculate the Turbine power </u>
Assuming a steady state condition
( p1 / p2 )^(r-1/r) = ( T1 / T2 )
= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )
= ( 9 )^0.285 = 658 / T2
∴ T2 = 351.22 K
Finally Turbine Power / power developed can be calculated as
Wt = mCp ( T1 - T2 )
= 0.04 * 1.005 ( 658 - 351.22 )
= 12.332 KW
The positive sign indicates work done by the system ( Turbine )
When an object has the same number of positive and negative charges, its electrical charge will become neutral.
What is an electric charge?
When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.
Now when the two equal magnitude charges with opposite natures come together they become neutral.
To know more about charges follow
brainly.com/question/24391667
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Hauling Vehicles that include a semitrailer manufactured prior to or in the model year of 2024, and registered in Illinois prior to January 1, 2025, having 5 axles with a distance of 42 feet or less between extreme axles, may not exceed the following maximum weights: 20,000 pounds on a single axle; 34,000 pounds
Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:
![q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}](https://tex.z-dn.net/?f=q%3D0.149h_%7Bfg%7D%20%5Crho%20_%7Bv%7D%20%28%5Cfrac%7B%5Csigma%20%28%5Crho%20_%7BL%7D-%5Crho%20_%7Bv%7D%29%7D%7B%5Crho%20_%7Bv%7D%5E%7B2%7D%20%7D%20%20%29%5E%7B1%2F4%7D%20%3D0.149%2A2257%2A0.596%2A%28%5Cfrac%7B58.9x10%5E%7B-3%7D%2A%28957.9-0.596%29%20%7D%7B0.596%5E%7B2%7D%20%7D%20%29%5E%7B1%2F4%7D%20%3D18703.42W%2Fm%5E%7B2%7D)
For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²
![q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4](https://tex.z-dn.net/?f=q_%7Bn%7D%20%3DuK%28%5Cfrac%7Bg%28%5Crho_%7BL%7D-%5Crho%20_%7Bv%7D%29%20%20%20%20%20%7D%7B%5Csigma%20%7D%29%5E%7B1%2F2%7D%20%28%5Cfrac%7Bc_%7BpL%7D%2AdeltaT%20%7D%7Bc_%7Bfg%7Dh_%7Bfg%7DPr%20%20%7D%20%5C%5C16833.078%3D279x10%5E%7B-6%7D%20%2A2257x10%5E%7B3%7D%20%28%5Cfrac%7B9.8%2A%28957.9-0.596%29%7D%7B0.596%7D%20%29%5E%7B1%2F2%7D%20%2A%28%5Cfrac%7B4.127x10%5E%7B3%7D%2Adelta-T%20%7D%7B0.0128%2A2257x10%5E%7B3%7D%2A1.76%20%7D%20%29%5E%7B3%7D%20%5C%5Cdelta-T%3D20.4)
The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C