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.
Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.
A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.
The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.
The schematic design consists of a rough sketch with markings and measurements.
Therefore, the correct option is B. schematic design.
To learn more about schematic design, refer to the link:
#SPJ5
Answer:
B. schematic design
Explanation:
This correct for Plato/edmentum
Explanation:
Step1
Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.
The expression for absolute pressure is given as follows:
Here, is absolute pressure,
is gauge pressure and
is atmospheric pressure.
Step2
Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.
Fluorescent lamps
Mercury-containing lamps
All of the above
Answer: D all above
Explanation:
Jus done it
Answer:
The amount of heat transferred to the air is 340.24 kJ
Explanation:
From P-V diagram,
Initial temperature T1 = 27°C
Initial pressure P1 = 100 kPa
final pressure P3 = P2 = 300 kPa
volume at point 2, V2 = V1 = 0.4 m³
final temperature T2 = T3 = 1200 K
To determine the final pressure V3, use ideal gas equation
PV = mRT
Where R is the specific gas constant = 0.2870 KPa m³ kg K
But,
from initial condition, mass m = PV/RT
m = (P1*V1)/R*T1
T1 = 27+273 = 300K
m = (100*0.4)/(0.2870*300) = 0.4646 kg
Then;
Final volume V3 = mRT3/P3
V3 = (0.4646*0.2870*1200)/300
V3 = 0.5334 m³
Total work done W is determined where there is volume change which from point 2 to 3.
W = P3*(V3-V2)
W = 300*(0.5334-0.4) = 40.02 kJ
To get the internal energy, the heat capacity at room temperature Cv is 0.718 kJ/kg K
∆U = m*Cv*(T2-T1)
∆U = 0.4646*0.718(1200-300)
∆U = 300.22 kJ
The heat transfer Q = W + ∆U
Q = 40.02 + 300.22 = 340.24 kJ
Determine the amount of heat transferred to the air, in kJ, while increasing the temperature to 1200 K is 340.24 kJ
The attached file shows the Pressure - Volume relationship (P -V graph)
Answer:
A) approximate alkalinity = 123.361 mg/l
B) exact alkalinity = 124.708 mg/l
Explanation:
Given data :
A) determine approximate alkalinity first
Bicarbonate ion = 120 mg/l
carbonate ion = 15 mg/l
Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61
= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3
B) calculate the exact alkalinity of the water if the pH = 9.43
pH + pOH = 14
9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57
[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l
[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l
[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l
therefore the exact alkalinity can be calculated as
= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )
= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )
= 124.708 mg/l
A. The heat transfer rate from natural gas is 2105.26 MW
B. The heat transfer rate to river is 1305.26 MW
Efficiency = (power output / power input) × 100
Power input = Power input / efficiency
Power input = 800 / 38%
Power input = 800 / 0.38
Power input = 2105.26 MW
Thus, the heat transfer from natural gas is 2105.26 MW
Heat to the river = 2105.26 – 800
Heat to the river = 1305.26 MW
Learn more about efficiency:
Answer:
heat transfer from natural gas is 2105.26 MW
heat transfer to river is 1305.26 MW
Explanation:
given data
power output Wn = 800 MW
efficiency = 38%
solution
we know that efficiency is express as
......................1
put here value we get
38% =
Qin = 2105.26 MW
so heat supply is 2105.26
so we can say
Wn = Qin - Qout
800 = 2105.26 - Qout
Qout = 2105.26 - 800
Qout = 1305.26 MW
so heat transfer from natural gas is 2105.26 MW
and heat transfer to river is 1305.26 MW