Answer:
Enthalpy, hsteam = 2663.7 kJ/kg
Volume, Vsteam = 0.3598613 m^3 / kg
Density = 2.67 kg/ m^3
Explanation:
Mass of steam, m = 1 kg
Pressure of the steam, P = 0.5 MN/m^2
Dryness fraction, x = 0.96
At P = 0.5 MPa:
Tsat = 151.831°C
Vf = 0.00109255 m^3 / kg
Vg = 0.37481 m^3 / kg
hf = 640.09 kJ/kg
hg = 2748.1 kJ/kg
hfg = 2108 kJ/kg
The enthalpy can be given by the formula:
hsteam = hf + x * hfg
hsteam = 640.09 + ( 0.96 * 2108)
hsteam = 2663.7 kJ/kg
The volume of the steam can be given as:
Vsteam = Vf + x(Vg - Vf)
Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)
Vsteam = 0.3598613 m^3 / kg
From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as
Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that
Now the pressure of helium by using the Adiabatic pressure temperature is
Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that
Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
<u>Determine the design flow ins ft^3/s </u>
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
