Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation
under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q = 
Q = 
Q= 4.6 × 10⁻³ m³/s
we know that
hence , actual velocity will be equal to 8.39 m/s
Answer:
The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V
Explanation:
As the complete question is not given the complete question is found online and is attached herewith.
By applying KCL at node 1
Also
Now applying KVL on loop 1 as indicated in the attached figure
Similarly for loop 2
So the system of equations become
Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as
The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V
Answer:
The electrical work for the process is 256.54 Btu.
Explanation:
From the ideal gas equation:
n = PV/RT
n is the number of moles of air in the tank
P is initial pressure of air = 50 lbf/in^2 = 50 lbf/in^2 × 4.4482 N/1 lbf × (1 in/0.0254m)^2 = 344736.2 N/m^2
V is volume of the tank = 40 ft^3 = 40 ft^3 × (1 m/3.2808 ft)^3 = 1.133 m^3
T is initial temperature of air = 120 °F = (120-32)/1.8 + 273 = 321.9 K
R is gas constant = 8.314 J/mol.k
n = 344736.2×1.133/8.314×321.9 = 145.94 mol
The thermodynamic process is an isothermal process because the temperature is kept constant.
W = nRTln(P1/P2) = 145.94×8.314×321.9×ln(50/25) = 145.94×8.314×321.9×0.693 = 270669 J = 270669 J × 1 Btu/1055.06 J = 256.54 Btu
Answer:
Stress corrosion cracking
Explanation:
This occurs when susceptible materials subjected to an environment that causes cracking effect by the production of folds and tensile stress. This also depends upon the nature of the corrosive environment.
Factors like high-temperature water, along with Carbonization and chlorination, static stress, and material properties.
