Answer:
folding plans neatly after use
Answer:
The maximum power that can be generated is 127.788 kW
Explanation:
Using the steam table
Enthalpy at 20 bar = 2799 kJ/kg
Enthalpy at 2 bar = 2707 kJ/kg
Change in enthalpy = 2799 - 2707 = 92 kJ/kg
Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s
Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW
Answer:
2074.2 KW
Explanation:
<u>Determine power developed at steady state </u>
First step : Determine mass flow rate ( m )
m / Mmax = ( AV )₁ P₁ / RT₁ -------------------- ( 1 )
<em> where : ( AV )₁ = 8.2 kg/s, P₁ = 0.35 * 10^6 N/m^2, R = 8.314 N.M / kmol , </em>
<em> T₁ = 720 K . </em>
insert values into equation 1
m = 0.1871 kmol/s ( mix )
Next : calculate power developed at steady state ( using ideal gas tables to get the h values of the gases )
W( power developed at steady state )
W = m [ Yco2 ( h1 - h2 )co2
Attached below is the remaining part of the detailed solution
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:

Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:

Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:

Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero