Explanation:
The main difference between surface erosion and bulk degradation is that in surface erosion as the name implies that the material or polymer begins to erode or degrade only from the external surface until it reaches the interior of the material. However, in bulk degradation, the material undergoes complete degradation from within and outside the material.
Surface erosion is best applied in pharmaceuticals in terms of drug delivery into the body, while bulk degradation is best applied in plastics which may degrade in any form.
Service entrance conductors must be kept a minimum of ____ feet from the sides and bottom of windows that can be opened
A heat exchanger is to heat water (cp = 4.18 kJ/kg·°C) from 25°C to 60°C at a rate of 0.5 kg/s. The heating is to be accomplished by geothermal water (cp = 4.31 kJ/kg·°C) available at 140°C at a mass flow rate of 0.3 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit tempera
Answer:
73.15 kW, 196.6°C
Explanation:
Energy in - Energy out = change in energy
[tex]E_{in}-E_{out}=\Delta E\\\\\Delta E=0\\\\E_{in}-E_{out}=0\\\\E_{in}=E_{out}\\\\Q_{in}+\dot mh_1=\dot mh_2\\\\Q_{in}=\dot mh_2-\dot mh_1\\\\Q_{in}=\dot m c_p(T_2-T_1)[/tex]
The rate of heat transfer to cold water is given as:
[tex]Q_{in}=\dot m c_p(T_2-T_1)\\\\\dot m =0.5\ kg/s,c_p=4.18\ kJ/kg.^oC, T_2=60^oC,T_1=25^oC\\\\Q_{in}=(0.5 *4.18)(60-25)=73.15\ kW[/tex]
For the geothermal water:
[tex]Q_{in}=\dot m c_p(T_2-T_1)\\\\\dot m =0.3\ kg/s,c_p=4.31\ kJ/kg.^oC, ,T_1=140^oC\\\\T_2=\frac{Q}{\dot m c_p} +T_1=\frac{73.15}{0.3*4.31}+140=196.6\\ \\T_2=196.6^oC[/tex]
In a two dimensional flow, the component of the velocity along the X-axis and the Y-axis are u = ax2 + by + cy2 and v = cxy. What should be the condition for the flow field to be incompressible
Answer:
The condition for the flow field to be incompressible is independent of a and c
Explanation:
We are given the component of the velocity along the X-axis and the Y-axis as;
u = ax² + by + cy²
v = cxy
Now, the condition for the flow to be incompressible is;
du/dx + dv/dy = 0
Now,
du/dx = 2ax
dv/dy = cx
Thus;
2ax + cx = 0
(2a + c)x = 0
Thus,the condition for the flow field to be incompressible is independent of a and c
A rectangular channel with a width of 2 m is carrying 15 m3/s. What are the critical depth and the flow velocity
Answer:
The critical depth of the rectangular channel is approximately 1.790 meters.
The flow velocity in the rectangular channel is 4.190 meters per second.
Explanation:
From Open Channel Theory we know that critical depth of the rectangular channel ([tex]y_{c}[/tex]), measured in meters, is calculated by using this equation:
[tex]y_{c} = \sqrt[3]{\frac{\dot V^{2}}{g\cdot b^{2}} }[/tex] (Eq. 1)
Where:
[tex]\dot V[/tex] - Volume flow rate, measured in cubic meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]b[/tex] - Channel width, measured in meters.
If we know that [tex]\dot V = 15\,\frac{m^{3}}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]b = 2\,m[/tex], then the critical depth is:
[tex]y_{c} = \sqrt[3]{\frac{\left(15\,\frac{m^{3}}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (2\,m)^{2}} }[/tex]
[tex]y_{c} \approx 1.790\,m[/tex]
The critical depth of the rectangular channel is approximately 1.790 meters.
Lastly, the flow velocity ([tex]v[/tex]), measured in meters, is obtained from this formula:
[tex]v = \frac{\dot V}{b\cdot y_{c}}[/tex] (Eq. 2)
If we know that [tex]\dot V = 15\,\frac{m^{3}}{s}[/tex], [tex]b = 2\,m[/tex] and [tex]y_{c} = 1.790\,m[/tex], then the flow velocity in the rectangular channel is:
[tex]v = \frac{15\,\frac{m^{2}}{s} }{(2\,m)\cdot (1.790\,m)}[/tex]
[tex]v = 4.190\,\frac{m}{s}[/tex]
The flow velocity in the rectangular channel is 4.190 meters per second.
Assume that the density of fresh water is 998 kg/m3 , and the density of salt water is 1025 kg/m3 . The gate weighs 0.448 kN/m (into the page). The gate is 5 m wide, into the page. Determine the height of the fresh water at which the gate opens. (15 points)
Answer:
hello your question lacks the required diagram attached below is the diagram showing the resultant forces at the gate
answer: 1.975 m
Explanation:
Given data :
density of fresh water = 998 kg/m^3
density of salt water = 1025 kg/m^3
gate weight = 0.448 kN/m ≈ 100 Ib/m
attached below is the detailed solution required
What is another example of radiation? Warm air risingrisingHeat Heat from an old-fashioned radiatorradiatorBatteriesTouching Touching a stove and burning bur-
ning your hand hand
Answer: what’s the answer
Explanation:
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kehrgpiyfealuydvasljfnVjlehgrPIYgefioyGlihv
Answer:
Batteries
Explanation:
They include a lot of radiation in them
It is known that the kinetics of recrystallization for some alloy obey the Avrami equation, and that the value of n in the exponential is 2.4. If, at some temperature, the fraction recrystallized is 0.30 after 100 min, determine the rate of recrystallization at this temperature.
Answer:
rate = 7.580 × [tex]10^{-3}[/tex] /min
Explanation:
given data
value of n = 2.4
fraction recrystallized = 0.30
time = 100 min
solution
we will get here rate of recrystallization at this temperaturewe use here avrami equation that is
y = 1 - exp (-k [tex]t^n[/tex] ) ................1
we get hete first k
k = [tex]-\frac{ln (1-y)}{t^n}[/tex]
put heer value n is 2.4 and y = 0.30 and t is 100
k = [tex]-\frac{ln (1-0.30)}{100^{2.4}}[/tex]
k = 5.65 × [tex]10^{-6}[/tex]
now we get here [tex]t^{0.5}[/tex]
value of t at y 0.5
[tex]t^{0.5}[/tex] = [tex][\frac{-ln(1-y)}{k}]^{1/n}[/tex]
[tex]t^{0.5}[/tex] = [tex][\frac{-ln(1-0.5)}{5.65\times 106{-6}}]^{1/2.4}[/tex]
[tex]t^{0.5}[/tex] = 131.923 min
and
rate = 1 ÷ [tex]t^{0.5}[/tex]
rate = 1 ÷ 131.923
rate = 7.580 × [tex]10^{-3}[/tex] /min
At 7:00PM the temperature was 40 degrees F. If the temperature dropped steadily at a rate of 6 degrees per hour, what was the temperature at 11:00 PM? Do you know what it is
Answer:
Temperature at 11:00 PM = 16 degree
Explanation:
Given:
Temperature at 7:00 PM = 40 degree
Decrease rate = 6 degree per hour
Find:
Temperature at 11:00 PM
Computation:
Total time = 11 - 7 = 4 hour
Total decrease in temperature = 4 x 6 = 24 degree
Temperature at 11:00 PM = 40 - 24
Temperature at 11:00 PM = 16 degree
A high strength waste having an ultimate CBOD of 1,000 mg/L is discharged to a river at a rate of 2 m3 /s. The river has an ultimate CBOD of 10 mg/L and is flowing at a rate of 8 m3 /s. Assuming a reaction rate coefficient of 0.1/day, calculate the ultimate and 5-day CBOD of the waste at the point of discharge (0 km) and 20 km downstream. The river is flowing at a velocity of 10 km/day.
Answer:
Attached below is the detailed solution
answer:
The Ultimate CBOD of waste water at (0) km and 20 Km
= 81.84 mg/l , 67 mg/l
Explanation:
Given Data :
UBOD of waste stream = 1000 mg/l
Volumetric flow rate of waste water = 2 m^3 / s
UBOD of River = 10 mg/l
Volumetric flow rate of river = 8 m^3 / s
reaction rate coefficient = 0.1/day
Velocity of river = 10km/day
The Ultimate CBOD of waste water at (0) km and 20 Km
= 81.84 mg/l , 67 mg/l
A single crystal of copper yields under a shear stress of 0.62 MPa. The shear modulus of copper is 7.9106 psi. Calculate the approximate ratio of the theoretical to the experimental shear strengths.
Answer:
6370
Explanation:
Given data:
shear stress = 0.62 MPa
shear modulus of copper = 7.9 GPa
Determine the approximate ratio of the theoretical to the experimental shear strengths
lets assume the theoretical yield shear stress = 1/2 shear modulus
theoretical yield shear stress = 7.9 / 2 = 3.95 GPa ≈ 3950 MPa
required ratio = [tex]\frac{3950}{0.62}[/tex] ≈ 6370
What type of engineer constructs the infrastructure necessary for roads airfields and water ports
Answer:
Civil Engineer
Explanation:
Structural engineers and Civil engineers are the ones who design, build, and maintain the foundation for our modern society.
Like our roads and bridges, drinking water and energy systems, sea ports and airports, and the infrastructure for a cleaner environment and many more.
