Answer:0.1898 Pa/m
Explanation:
Given data
Diameter of Pipe
Velocity of water in pipe
We know viscosity of water is
Pressure drop is given by hagen poiseuille equation
We have asked pressure Drop per unit length i.e.
Substituting Values
=0.1898 Pa/m
Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb
1 answer:
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Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.
Answer:
a)120C
b)29kg
Explanation:
Hello!
To solve this exercise follow the steps below
1. we will call 1 the initial state, 2 the steam that enters and 3 the final state
2. We find the quality of the initial state, dividing the mass of steam by the total mass.
3 Find the internal energy in the three states using thermodynamic tables
note:Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
u1=IntEnergy(Water;x=0,6(quality);P=200kPa) =1719KJ/kg
u2=IntEnergy(Water;t=350;P=5000kPa) =2808KJ/kg
u3=IntEnergy(Water;x=1;P=200kPa) =2529KJ/kg
4. use the internal energy and pressure to find the temperature in state 3, using thermodynamic tables
T3=Temperature(Water;P=200kPa;u=u3=2529KJ/kg)=120C
5. Use the first law of thermodynamics in the system, it states that the initial energy in a system must be equal to the final
m1u1+m2u2=(m1+m2)u3
where
m1=inital mass=10kg
m2=the mass of the steam that has entered.
solve for m2
(m1)(u1-u3)=(m2)(u3)-(m2)(u2)