Answer:
Qx = 9.10
m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × 
Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 
)²× 9 × sin18 × cos18
Qd = 94.305 × 
m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × 
× π × 85 × ( 9 )³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × - 85.2 ×
Qx = 9.10 m³/s
Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
Answer:
Power = 371.28 kW
Explanation:
Initial pressure, P1 = 5 bar
Final pressure, P2 = 1 bar
Initial temperature, T1 = 320°C
Final temperature, T2 = 160°C
Volume flow rate, V = 0.65m³/s
From steam tables at state 1,
h1 = 3105.6 kJ/kg, s1 = 7.5308 kJ/kgK
v1 = 0.5416 m³/kg
Mass flow rate, m = V/v1
m = 1.2 kg/s
From steam tables, at state 2
h2 = 2796.2 kJ/kg, s2 = 7.6597 kJ/kgK
Power developed, P = m(h1 - h2)
P = 1.2(3105.6-2796.2)
P = 371.28 kW
Answer:
#include <iostream>
using namespace std;
void PrintPopcornTime(int bagOunces) {
if(bagOunces < 3){
cout << "Too small";
cout << endl;
}
else if(bagOunces > 10){
cout << "Too large";
cout << endl;
}
else{
cout << (6 * bagOunces) << " seconds" << endl;
}
}
int main() {
PrintPopcornTime(7);
return 0;
}
Explanation:
Using C++ to write the program. In line 1 we define the header "#include <iostream>" that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.
Answer:
power = 49.95 W
and it is self locking screw
Explanation:
given data
weight W = 100 kg = 1000 N
diameter d = 20mm
pitch p = 2mm
friction coefficient of steel f = 0.1
Gravity constant is g = 10 N/kg
solution
we know T is
T = w tan(α + φ ) 
...................1
here dm is = do - 0.5 P
dm = 20 - 1
dm = 19 mm
and
tan(α) = 
...............2
here lead L = n × p
so tan(α) = 
α = 3.83°
and
f = 0.1
so tanφ = 0.1
so that φ = 5.71°
and now we will put all value in equation 1 we get
T = 1000 × tan(3.83 + 5.71 ) 
T = 1.59 Nm
so
power = 
.................3
put here value
power = 
power = 49.95 W
and
as φ > α
so it is self locking screw
