Answer:
irreversible head loss is 38.51 ft
mechanical power 6.55 hp
Explanation:
Given data:
Pump Power 12 hp = 8948.39 watt = 6600 lbs ft/sec
Q = 1.5 ft^3/s
Pump actually delivers
Power that water gains
hence Power lost = 6600 - 2995.2 3604.8 lbs ft/sec = 6.55 hp
head loss
hl = 38.51 ft
Answer:288 pm
Explanation:
Number of atoms(s) for face centered unit cell -
Lattice points: at corners and face centers of unit cell.
For face centered cubic (FCC), z=4.
- whereas
For an FCC lattices √2a =4r =2d
Therefore d = a/√2a = 408pm/√2a= 288pm
I think with this step by step procedure the, the answer was clearly stated.
Answer:
The currents becomes 0
Explanation:
when the fuse burns out due to a sudden surge of current and becomes an open switch (with a resistance of Infinity ∞) this automatically reduces the currents through it to zero
Answer:
(a) ΔU = 125 kJ
(b) ΔU = -110 kJ
Explanation:
(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?
The work is done to the system so w = 150 kJ.
The heat is released by the system so q = -25 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -25 kJ + 150 kJ = 125 kJ
(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?
The work is done by the system so w = -100 kJ.
The heat is released by the system so q = -10 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -10 kJ - 100 kJ = -110 kJ
Answer:
t = 12.03
t = 81.473
velocity of the fluid decreases with level of fluid. Hence, in later stages the time taken is greater than in earlier stages
Explanation:
Given:
- The half angle θ = 30°
- The diameter of the small hole d = 6.25 mm
- The flow rate out of the funnel Q = A*√ 2gy
- The volume of frustum of cone is given by:
Where,
D: Is the larger diameter of the frustum
d: Is the smaller diameter of the frustum
y: The height of the liquid free surface from small diameter d base.
Find:
- Obtain an expression for the time, t, for the funnel to completely drain, and evaluate.
- Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm) and from 150 mm to completely empty (also a change in depth of 150 mm)
- Can you explain the discrepancy in these times?
Solution:
- We will use rate of change analysis by considering the rate of change of volume, and then apply the chain rule.
- The Volume of the frustum is a function of d , D , y. V = f ( d , D , y ). In our case the diameter of base d remains constant. Then we are left with:
V = f ( D , y )
- Determine a relationship between y and d. We will use half angle θ to determine the relationship between D and y by applying trigonometric ratios:
tan ( θ ) = D / 2*y
D = 2*y*tan ( θ )
- We developed a relationship between D and y in terms of half angle which remains constant. So the Volume is now only a function of one variable y:
V = f ( y )
- The volume of frustum of the cone can be written as:
Substituting the relationship for D in terms of y we have:
- Now by rate of change of Volume analysis we have:
dV / dt = [dV / dy] * [dy / dt]
- Computing dV / dy, where V = f(y) only:
- Where, dV/dt = Volume flow rate:
- Then from Chain rule we have:
[dy / dt] = [dV / dt] / [dV / dy]
- Separate variables:
- Integrate both sides:
- Evaluate @ t = 0 , y = 0.3 m
- Time taken from y = 300 to 150 mm:
- Time taken from y = 150 to 0 mm:
t = t_(0-300) - ( t_(0-150) = 90.871587 - 12.0347055 = 81.4721 s
- The discrepancy is time can be explained by the velocity of fluid coming out of bottom is function of y. The velocity of the fluid decreases as the level of fluid decreases hence the time taken from y =150 mm to 0 mm is larger than y = 300 mm to y = 150 mm.
The moment of force at the given slope about point C is 210 ft-lb.
The given parameters:
The magnitude of the two moments are in the following simple ratio;
330:420 = 11:14 (divide through by 30)
The line of action of the force passes line AB at the final following coordinates;
total ratio of 11:14 = 11 + 14 = 25
The position of C = (3, 1)
The resultant position of point C = (3 - 2.2, 2.8-1) = (0.8, 1.8)
The moment of force at the given slope about point C is calculated as;
3(1.8) + 2(0.8) = 7
Recall that this is the simplest form of the moment produced by the force.
Moment about C = 7 x 30 = 210 ft-lb
Thus, the moment of force at the given slope about point C is 210 ft-lb.
Learn more about moment of force here: brainly.com/question/6278006
Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm