Answer:
The resulting pressure of the gas when its volume decreases is 300 kN/m².
Explanation:
Given;
initial volume of the gas, V₁ = 80 L
number of moles of the gas, n = 0.5 moles
initial pressure of the gas, P₁ = 150 kN/m² = 150 kPa
Determine the constant temperature of the gas using ideal gas equation;
PV = nRT
where;
R is ideal gas constant = 8.315 L.kPa/K.mol
T is the constant temperature
[tex]T = \frac{P_1V_1}{nR} \\\\T = \frac{150.kPa \ \times \ 80 .L}{0.5 .mol \ \times \ 8.315(L.kPa/mol.K)} \\\\T = 2,886.35 \ K[/tex]
When the gas is compressed to half of its volume;
new volume of the gas, V₂ = ¹/₂ V₁
= ¹/₂ x 80L = 40 L
The new pressure, P₂ is calculated as;
[tex]P_2V_2 = nRT\\\\P_2 = \frac{nRT}{V_2} \\\\P_2 = \frac{0.5 \times 8.315\times 2886.35}{40} \\\\P_2 = 300 \ kPa = 300 \ kN/m^2[/tex]
Therefore, the resulting pressure of the gas when its volume decreases is 300 kN/m².
A closed vessel of volume 80 litres contain gas at a gauge pressure of 150 kPa. If the gas is compressed isothermally to half its volume, determine the resulting pressure.
Answer:
The resulting pressure is 300 kilopascals.
Explanation:
Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:
[tex]P_{1}\cdot V_{1} = P_{2} \cdot V_{2}[/tex] (1)
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressure, measured in kilopascals.
[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Initial and final volume, measured in litres.
If we know that [tex]\frac{V_{1}}{V_{2}} = 2[/tex] and [tex]P_{1} = 150\,kPa[/tex], then the resulting pressure is:
[tex]P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}[/tex]
[tex]P_{2} = 300\,kPa[/tex]
The resulting pressure is 300 kilopascals.
write to change past tense
Initially, 1 lbm of water is at rest at 14.7 psia and 70 F. The water then undergoes a process where the final state is 30 psia and 700 F with a velocity of 100 ft/s and an elevation of 100 ft above the starting location. Determine the increases in internal energy, potential energy, and kinetic energy of the water in Btu/lbm. Compare the increases in potential energy and kinetic energy individually to the change in the internal energy.
Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]\text{The mass (m) = 1 lbm}[/tex]
Suppose: g = 32.2 ft/s²
At the inlet conditions:
[tex]\text{mass (m) = 1 \ lbm water} \\ \\ P_1 = 14.7 \ psia \\ \\ T_1 = 70 F \\ \\ z_1 = 0 \ ft[/tex]
At the outlet conditions:
[tex]P_2 = 30 \ psia \\ \\ T_2 = 700\ F \\ \\ v_2 =100 \ ft/s\\ \\ z_2 = 100 \ ft[/tex]
[tex]\text{Using the information obtained from saturated water table at P1 = 14.7 \ psia \ and \ T1 = 70 F }[/tex]
[tex]u_1 =u_f = 38.09 \ Btu/lbm[/tex]
[tex]\text{Applying informations from superheated water vapor table:}[/tex]
[tex]P_2 = 30 \ psia \ and \ T_2 = 700 \ F \\ \\ u_ = 1256.9 \ kJ/kg[/tex]
The change in the internal energy is:
[tex]\Delta U = U_2 -U_1 \\ \\ \Delta U = 1256.9 -38.09 \\ \\ \Delta U = 1218.81 \ Btu/lbm[/tex]
For potential energy (P.E):
Initial P.E = mgz
P.E = 1 × 32.2 × 0 = 0 ft²/s²
Final P.E = mgz
P.E = 1 × 32.2 × 100 = 3220 ft²/s²
The change in the potential energy = PE₂ - PE₁
ΔPE = (3220 - 0) ft²/s²
ΔPE = 3220 ft²/s²
ΔPE = (3220 × 3.9941 × 10⁻⁵) Btu/lbm
ΔPE =0.12861 Btu/lbm
Initial Kinetic energy (K.E)
[tex]KE_1 = \dfrac{1}{2}mV_1[/tex]
[tex]KE_1 = \dfrac{1}{2}(1)(0) = 0 \ lbm \ ft^2/s^2[/tex]
FInal K.E
[tex]KE_2= \dfrac{1}{2}mV_2[/tex]
[tex]KE_2= \dfrac{1}{2}(1)(100)^2_2 = 50000 \ lbm \ ft^2/s^2[/tex]
Change in K.E [tex]\Delta K.E[/tex] = [tex]KE_2-KE_1[/tex]
[tex]\Delta K.E = 50000 -0 = 50000 \ lbm.ft^2/s^2[/tex]
[tex]\mathbf{\Delta K.E = 0.199 \ Btu/lbm}[/tex]
Many households in developing countries prepare food over indoor cook stoves with no ducting system to exhaust the combustion products outside. A woman in a household lights a fire in her cookstove at 4 pm. The fire emits benzo(a)pyrene, a carcinogenic compound, at a rate of 0.01 ng/min. The house has a size of 40 m3 and a ventilation rate of 1 m3/h. The ambient indoor air concentration of benzo(a)pyrene was 0.2 ng/m3 when she started the fire, but no significant concentration of benzo(a)pyrene is found in the outdoor air. What is the indoor benzo(a)pyrene concentration in the household at 6 pm?
Answer:
C = 0.22857 ng / m³
Explanation:
Let's solve this problem for part the total time in the kitchen is
t = 2h (60 min / 1h) = 120 min
The concentration (C) quantity of benzol pyrene is the initial quantity plus the quantity generated per area minus the quantity eliminated by the air flow. The amount removed can be calculated assuming that an amount of extra air that must be filled with the pollutant
amount generated
C = co + time_generation rate / (area_house + area_flow)
C = 0.2 + 0.01 120 / (40+ 2)
C = 0.22857 ng / m³
Shorter lines are faster than longer lines is an example of an algorithm.
Answer:
Programmers count the number of lines of code in an algorithm.
Programmers count the number of lines of code in an algorithm. Thus, shorter lines are faster than longer lines is an example of an algorithm.
What do you mean by algorithm?An algorithm is a finite sequence of exact instructions that is used in mathematics and computer science to solve a class of particular problems or carry out a computation.
For performing calculations and processing data, algorithms are employed as specifications. Conditionals can be used by more sophisticated algorithms to divert code execution along several paths and draw reliable inferences, ultimately leading to automation.
Alan Turing was the first to use terminology like "memory," "search," and "stimulus" to describe human traits as metaphorical descriptions of machines.
A heuristic, on the other hand, is a method for addressing problems that may not be fully articulated or may not provide accurate or ideal solutions, particularly in problem domains where there isn't a clearly defined proper or ideal conclusion.
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In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.8) is 1.2 and there may be a 2 mm crack present in a bone specimen, what would be the maximum tensile stress that can be applied before fracture occurs?
A detailed image of a brain scan with height, width, and depth is an example of a(n) 3D _________ model.
Answer:
is a mathematical representation of something three-dimensional.
Explanation:The typical base of a the model is a 3D mesh; the structural build consists of polygons.
(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring cup and flows into the downsprue. The cross section of the sprue is round, with a diameter at the top = 3.4 cm. If the sprue is 20 cm long, determine the proper diameter at its base so as to main- tain the same volume flow rate.
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top [tex]d_{top}[/tex] = 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
[tex]V_{base}[/tex] = √2gh
we substitute
[tex]V_{base}[/tex] = √(2 × 9.81 m/s² × 0.2 m )
[tex]V_{base}[/tex] = 1.98091 m/s
[tex]V_{base}[/tex] = 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π[tex]d_{bottom}^2[/tex]/4) × [tex]V_{base}[/tex]
[tex]d_{bottom}[/tex] = √(4Q/π[tex]V_{base}[/tex])
we substitute our values into the equation;
[tex]d_{bottom}[/tex] = √(4(400 cm³/s) / (π×198.091 cm/s))
[tex]d_{bottom}[/tex] = 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
What's the ampacity of a No. 10 type TW copper Wire in a raceway containing six wires and located in an area where the ambient Temperature is 50 C
A. 15.1 A
B. 8.6 A
C. 13.9 A
D. 10.8 A
Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon alloy that initially contains 0.10 wt% C. The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be conducted at 1100°C. Use the diffusion data for γ-Fe.
Answer:
the carburizing time necessary to achieve a carbon concentration is 31.657 hours
Explanation:
Given the data in the question;
To determine the carburizing time necessary to achieve the given carbon concentration, we will be using the following equation:
(Cs - Cx) / (Cs - C0) = ERF( x / 2√Dt)
where Cs is Concentration of carbon at surface = 0.90
Cx is Concentration of carbon at distance x = 0.30 ; x in this case is 4 mm = ( 0.004 m )
C0 is Initial concentration of carbon = 0.10
ERF() = Error function at the given value
D = Diffusion of Carbon into steel
t = Time necessary to achieve given carbon concentration ,
so
(Cs - Cx) / (Cs - C0) = (0.9 - 0.3) / (0.9 - 0.1)
= 0.6 / 0.8
= 0.75
now, ERF(z) = 0.75; using ERF table, we can say;
Z ~ 0.81; which means ( x / 2√Dt) = 0.81
Now, Using the table of diffusion data
D = 5.35 × 10⁻¹¹ m²/sec at (1100°C) or 1373 K
now we calculate the carbonizing time by using the following equation;
z = (x/2√Dt)
t is carbonizing time
so we we substitute in our values
0.81 = ( 0.004 / 2 × √5.35 × 10⁻¹¹ × √t)
0.81 = 0.004 / 1.4628 × 10⁻⁵ × √t
0.81 × 1.4628 × 10⁻⁵ × √t = 0.004
1.184868 × 10⁻⁵ × √t = 0.004
√t = 0.004 / 1.184868 × 10⁻⁵
√t = 337.5903
t = ( 337.5903)²
t = 113967.21 seconds
we convert to hours
t = 113967.21 / 3600
t = 31.657 hours
Therefore, the carburizing time necessary to achieve a carbon concentration is 31.657 hours
Please help fast
What would happen if the air outside the hot air balloon were as hot as the air inside.
In a foundry, metal castings are cooled by quenching in an oil bath. Typically, a casting weighting 20 kg and at a temperature of 450 oC is cooled by placing it in a 150 kg in volatile oil bath initially at 50 oC. If the specific heat capacity of the metal is 0.5 J/kg K, and that of the oil is 2.6 J/kg K, determine the entropy change in this process and the common final temperature of the oil and the casting after quenching. Assume that there are no heat losses
Answer:
4.18 J/KgK
Explanation:
Equilibrium point is reached when
m₁*c₁(T₁-T) = m₂*c₂(T -t₂)
m1 = 20
c1 = 0.5
T1 = 450
m2 = 150 kg
c2 = 2.6
T2 = 50
putting these values into the formula
20*0.5 (450-T) = 150*2.6(T - 50)
4500 - 10T = 390T - 19500
4500 + 19500 = 390T + 10T
24,000 = 400T
T = 24000/400
= 60⁰C
ΔSmetal = m1*c1In[t + 273]/[T1+273]
= 20*0.5 In (60+273)/450+273
= 10 ln(333/723)
= 10 * -0.7752
= -7.752
ΔS/oil =
m2*s2(60 + 273)/50 + 273)
= 150*2.6ln(333/323)
= 390 * 0.03048
= 11.88j/KgK
Δtotal = -7.7+11.8
= 4.18J/KgK
this is the enthropy change
The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphere. An orifice of diameter D with a smooth entrance (i.e., negligible losses) is open at the bottom. Develop a relation for the time required for the tank (a) to empty halfway (5-point) and (b) to empty completely (5-point).
Answer:
a. The time required for the tank to empty halfway is presented as follows;
[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]
b. The time it takes for the tank to empty the remaining half is presented as follows;
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The total time 't', is presented as follows;
[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]
Explanation:
a. The diameter of the tank = D₀
The height of the tank = H
The diameter of the orifice at the bottom = D
The equation for the flow through an orifice is given as follows;
v = √(2·g·h)
Therefore, we have;
[tex]\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}[/tex]
[tex]\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}[/tex]
Where;
P₁ = P₂ = The atmospheric pressure
z₁ - z₂ = dh (The height of eater in the tank)
A₁·v₁ = A₂·v₂
v₂ = (A₁/A₂)·v₁
A₁ = π·D₀²/4
A₂ = π·D²/4
A₁/A₂ = D₀²/(D²) = v₂/v₁
v₂ = (D₀²/(D²))·v₁ = √(2·g·h)
The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;
dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh
We have;
[tex]dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh[/tex]
The time for the tank to drop halfway is given as follows;
[tex]\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]
[tex]t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]
[tex]t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]
[tex]t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]The time required for the tank to empty halfway, t₁, is given as follows;
[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]
(b) The time it takes for the tank to empty completely, t₂, is given as follows;
[tex]\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]
[tex]t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)[/tex]
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The time it takes for the tank to empty the remaining half, t₂, is presented as follows;
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The total time, t, to empty the tank is given as follows;
[tex]t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}[/tex]
[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]
A fast moving hurricane produces a constant rainfall intensity of 110 mm/h for 45 min. The area consists almost entirely of sandy loam, which typically has a saturated hydraulic conductivity of 9.1 mm/h, average suction head of 110 mm, porosity of 0.45, field capacity of 0.190, wilting point of 0.085, and depression storage of 4 mm. Calculate the cumulative infiltration after 30 min, assuming ponding is continuous from the beginning of the storm. Assume the initial water content is midway between the field capacity and wilting point..
Answer:
hello some part of your question is missing ( The table ) attached below is the missing detail related to your question
answer: cumulative infiltration after 30 mins = 6.7
Explanation:
Calculate the cumulative infiltration after 30 min
formula for calculating cumulative infiltration
= 11( t - tp + t' ) = F - 34.3 In ( 1 + 0.0291 * F )
Attached below is a detailed solution to the problem
Choose two other elements from the periodic table that you predict should react to form something like table salt
Please pleassssss helppp
I give branlistttttt
Find the minimum stages at total reflux for separating ethanol-acetone mixture to achieve 95 mol% purity of acetone in the distillate and 95 mol% purity of ethanol in the bottoms product. Use equilibrium data from problem 4D7. x y Equilibrium data for acetone (A) and ethanol at 1 atm (Perry et al., 1963, pp. 13-4) are
0.100.150.200.250.300.350.40 0.50 0.60 0.70 0.80 0.90 0.262 0.348 0.417 0.478 0.524 0.566 0.605 0.674 0.739 0.80 20.865 0.929
Answer:
minimum number of stages ≈ 9 ( from Image number 1 )
Explanation:
Attached below is a detailed solution to the question
Total reflux = R Tending to infinity
we can determine the operation equation by stripping and rectifying section in total reflux condition
minimum number of stages ≈ 9 ( from Image number 1 )
Write a python program to get the following output. 1-----99 2-----98 3-----97 . . . . . . 98-----2 99-----1
Answer:
i dont know th answer can u help ?
Explanation:
The output of a first order transducer is to be connected to a signal conditioner which also has first order dynamic characteristics. The transducer has known time constant of x milliseconds and static sensitivity of xx V/kPa while the signal conditioner has a time constant of y milliseconds and a static sensitivity of yy V/V. Determine the steady state response of this measurement system to an input signal of the form: P(t)=(A+B*sin(w*t) ) kPa.
Answer:
hello your question is poorly written attached below is the correct question
answer : y(t) = 1.5 + 0.025 sin( 600t - 0.259 ) v
Explanation:
Given data:
Time constant = 5msec
static sensitivity = 0.05 v/°c
f = 100 hz
ε = 0.8
T(t) = ( 30 + 2.5 sin600t )°C
attached below is a detailed solution to the question above
If an improvement creates no significant change in a product’s performance, then it is a(n) design improvement.
Answer:
Following are the responses to these question:
Explanation:
They might believe that it was an enhanced layout because the quality is not updated. For instance, its new XS Max iPhone does have a better display than the iPhone X, however, the performance wasn't enhanced. It also has the same processor or graphic cards however a bigger pixel every centimeter ratio. When its output AND is not altered, the layout doesn't change, basically the very same item.
Answer:
If an Improvement creates no significant change in a product’s performance, then it is a(n) Superficial design improvement.
Explanation:
DIFFERENT BREED!
Water is the working fluid in a Rankine cycle with reheat. The turbine and the pump have isentropic efficencies of 80%. Superheated vapor enters the turbine at 10 MPa, 480 °C, and the condenser pressure is 6 kPa. Steam expands through the first-stage turbine to 0.7 MPa and then is reheated to 480°C. Determine for the cycle
(a) the rate of heat addition, in kJ per kg of steam entering the first-stage turbine.
(b) the thermal efficiency.
(c) the rate of heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per kg of steam entering the first-stage turbine.
Answer:
A) 3783.952 kJ/kg
B) 34.5%
C) 2476.67 kJ/kg
Explanation:
A) Determine the rate of heat addition entering the first-stage turbine
Qin ( 1st stage ) = ( h1 - h6 ) + ( h3 - h2 )
= ( 3321.4 - 164.07 ) + ( 3438.566 - 2811.944 )
= 3783.952 kJ/kg
h values are gotten from super heated table
B) Determine the thermal efficiency
n = 34.5%
attached below
C) Rate of heat transfer from working fluid passing through the condenser to the cooling water
Qout = h4 - h5
= 2628.2 - 151.53
= 2476.67 kJ/kg
What is the uppermost part of the tree referred to as?
Answer:
canopy I believe it is called
Answer:
Canopy
Explanation:
The canopy of a tree is sometimes also called its “crown.” This part of the tree is the uppermost part, made up of branches, stems, and leaves for a deciduous tree. ... For evergreen trees, the canopy would include the branches, stems, and needles.
Dry air does NOT contain
Explanation:
Dry air doesn't contain water vapor .
Technologies that allow for instant worldwide communication include
high-speed trains and naval ships.
mobile phones and Internet access.
airplanes and container shipping.
outsourcing and new trade markets.
Answer:
mobile phones and internet access
Explanation:
I got it right on my quiz
Technologies that allow for instant worldwide communication include mobile phones and Internet access.
Communication technologies and globalizationThe feasibility for sending and receiving information is a fundamental issue in the emergence of the process known as globalization.
The major communication technologies that have greatly improved the speed in the emission of messages include telephone, radio, television, and internet.For example, the internet can be used to share information and data almost instantly, allowing communication much better than in the past.In conclusion, technologies that allow for instant worldwide communication include mobile phones and Internet access.
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A three-story structure is to be constructed over an 8000-m2 site. The initial subsurface exploration indicates the presence of sinkholes and voids due to dissolution of the limestone formation. The predominant soil type is a silty fine sand grading to a fine sand with seams of sandy clay. The design indicates that shallow foundations can be used for this project provided the soils were made more homogeneous as far as load support and no voids were present within the depth up to 7.6 m below the ground surface. Assume groundwater is not a concern. Dynamic compaction is proposed to improve the ground. The local contractor doing dynamic compaction has a 15-ton tamper with the diameter of 2.0 m and the height of 1.4 m. You are requested to conduct the preliminary design for this dynamic compaction project including drop height, spacing, number of drops, number of passes, estimated crater depth, and settlement
Answer:
a) 24.07 m
b) 4 m
c) 14 number of drops
d) p = number of passes
e) Dcd = 2.27
0.69 m
Explanation:
Given data:
Depth ( D )= 7.6 m below ground surface
dynamic compaction ( w ) = 15-ton , diameter of tamper = 2.0 m , thickness = 1.4 m
Determine :
A) drop height ( H )
D = n √wH
therefore H = 361 / 15 = 24.07 m
where : D = 7.6 m , n = 0.4 , w = 15
B) Drop spacing
drop spacing = average of ( 1.5 to 2.5 ) * diameter of tamper
= 2 * 2.0m = 4 m
C) number of drops
since the applied energy for fine grained soils and day fills range from 250 - 350 kj/m^2 the number of drops can be calculated using the relation below
AE = [tex]\frac{NWHP}{SPACING ^2}[/tex]
w = 15, H = 24.07 , Np = ? , AE = 300 kj/m^2
∴ Np = 4800 / 361.05 = 13.3
the number of drops at one pass = 14
D) number of passes
p = number of passes
E) estimated crater depth and settlement
crater depth ( Dcd ) = 0.028 [tex]N_{d} ^{0.55} \sqrt{wtIt}[/tex]
Nd = 14 , wt = 15, It = 24.07
therefore : Dcd = 2.27
estimate settlement is within 3 to 5% therefore the improved settlement
= 2.27 * 0.04 * 7.6 = 0.69 m
Determine the maximum shear stress in the beam and the minimum yield strength that should be considered to obtain a minimum factor of safety of 2 based on the distortion-energy theory. The maximum shear stress in the beam is 1875 Numeric ResponseEdit Unavailable. 1875 correct.psi. The minimum yield strength based on the distortion-energy theory is
Answer: hello some parts of your question is missing attached below is the missing part
max shear stress = 1875 psi
minimum yield strength = 14911.76 psi
Explanation:
Since The maximum shear stress in the Beam is already given as 1875 psi , I will calculate for the minimum yield strength
Determine minimum yield strength
attached below is the detailed solution
minimum yield strength = 14911.76 psi
We intend to measure the open-loop gain (A open ) of an actual operational amplifier. The magnitude of A_open is in the range of 10^6 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the A_open?
Answer:
circuit diagram is attached below
Explanation:
Magnitude of A_ open = 10^6 v/v
Vp = Vin / 1001
where: Vin = 1 mV.
hence Vp = 1 mV / 1001 ≅ 1 μ V
while Vout = Aopen( Vp )
∴ Aopen = Vout / Vp
since Vp = 1 μV then we can can measure Aopen with Vout ranging up to
10 V
attached below is the measuring circuit for measuring open loop gain of amplifier
the uniform beam has a mass of 50 kg determine the reaction at the support
Answer:
1.4KN 4KN - m C 3KN 0.6m A 0 B 0.6m R 0.6m 0.6m 30° R Resolving Forces Horizontally, we get X=0 Rx
Resistors are used to reduce current flow, adjust signal levels to divide voltages, bias active elements and terminate transmission line.true or false
Answer:
True
Explanation:
Those are the exact uses of a resistor
Consider the formation of p-nitrophenol from p-nitrophenyl trimethyl acetate. The process is known as enzymatic hydrolysis and it occurs in the presence of the enzyme elastase. Along with the formation of p-nitrophenol, trimethyl acetic acid is also formed which is an undesired byproduct. p-nitrophenol is an important intermediate in the manufacture of several pharmaceuticals. Your role as a Chemical Engineer is to maximize the production of p-nitrophenol. The reactions can be denoted as:
E+S → P+ES R1
ES+PE+A R2
where e denotes the enzyme elastase, denotes the substrate p-nitrophenyl trimethyl acetate, es denotes enzyme-substrate intermediate and A denotes the trimethyl acetic acid. The rate of the reactions 1 and 2 are given by:
kg Cs KM + C r2 = kxCp
where Cs and Cp denote the concentrations of the substrate and the product, k, and ky are the rate constants given by 0.015 s' and 0.0026 s. Ky is the Michaelis - Menten constant and is given by 5.53 mol/m!. All the reactants and products are in the liquid phase. The initial concentrations of S and E are 0.5 mol/m3 and 0.001 mol/m..Consider the above reaction to occur in a batch reactor for 15 minutes.
a. Plot the concentration profiles of S, P and A as a function of time in a single figure.
b. Plot the selectivity of P with respect to 5 as a function of time b.
Solution :
cs=zeros(9001);
ca=zeros(9001);
cp=zeros(9001);
psi=zeros(9001);
t=[0:0.1:900];
cs(1)=0.5;
ce(1)=0.001;
cp(1)=0;
ca(1)=0;
psi(1)=0;
for i=1:1:9000
cs(i+1)=cs(i)-0.1*((0.015*cs(i))/(5.53+cs(i)));
cp(i+1)=cp(i)+0.1*((0.015*cs(i))/(5.53+cs(i))-0.0026*cp(i));
ca(i+1)=ca(i)+0.1*0.0026*cp(i);
psi(i+1)=((cp(i+1)-cp(i)))/((cs(i)-cs(i+1)));
end
plot(t,cs,t,cp,t,ca);
plot(t,psi);
The daily average concentration of pollutants in a stream follows a lognormal distribution with a mean of 50 mg/l and a standard deviation of 12 mg/l. It is desired to calculate the following probabilities:
1.) That the average concentration of pollutants in the stream on a given day will exceed 85 mg/l; and
2.) That a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days). Assume that the pollutant concentrations between days are statistically independent.
Answer:
1) probability that the average concentration of pollutants in the stream on a given day will exceed 85 mg/l is 0.0018
2) probability that a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days) is 0.9993
Explanation:
Given the data in the question;
mean μ = 50 mg/l
standard deviation σ = 12 mg/l
we know that; x-score = x-μ / σ
1) probability that the average concentration of pollutants in the stream on a given day will exceed 85 mg/l
p( x > 85 ) = P( Z > 85-50 / 12 )
= P( Z > 35/12 )
= P( Z > 2.9166)
= P( Z > 2.92)
= 1 - P( Z > 2.92)
from z-score table; P( Z > 2.92) = 0.9982
= 1 - 0.9982
p( x > 85 ) = 0.0018
Therefore, probability that the average concentration of pollutants in the stream on a given day will exceed 85 mg/l is 0.0018
2) probability that a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days). Assume that the pollutant concentrations between days are statistically independent.
p( x > 95 ) = p( Z > 95-50 / 12 )
= p( Z > 45 / 12 )
= p( Z > 3.75 )
= 1 - p( Z > 3.75 )
from z-score table; p( Z > 3.75 ) = 0.9999
= 1 - 0.9999
= 0.0001
Now; p = 0.0001 and n = ( week) = 7
x = number of days × exceeds 99 mg/l
x ¬ Binomial ( n =7 p = 0.0001 )
p(x ≤ 1) = p(x=0) + (p=1)
= ¹∑[tex]_{x=0}[/tex] [tex]^7C_x ( 0.0001)^x[/tex] [tex](0.9999)^{7-x}[/tex]
= [tex]^7C_0 ( 0.0001)^0[/tex] [tex](0.9999)^{7-0}[/tex]
= 7!/(0!(7-0)!) [tex]( 0.0001)^0[/tex] [tex](0.9999)^{7-0}[/tex]
= (1) ( 1 ) ( 0.9993 )
= 0.9993
Therefore, probability that a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days) is 0.9993