Answer:
see explaination
Explanation:
The statistical procedure for comparing the 3 groups was the F test with 2 degrees of freedom in the numerator (groups - 1) and 23 df in the denominator (N total - groups)
It is calculated as SSB/2/(SSW/23) = 10.22754
As Fcrit =
from Excel, finv(.05,2,23) = 3.4221, we can reject the null hypothesis of no difference between groups.
SSW = the sum of std i ^ 2 * (n i - 1)
SSB = the sum of ni (mean i - mean)^2
2. From Excel, we get the pvalue from fdist(10.22754,2,23) = .000664
3. For LSD, we calculate (mean 1 - mean 2)/(s * sqrt(1/n1+1/n2))
This is the pooled s = sqrt(SSW/23)
Then, we found t crit from tinv(.05,23) = 2.068
Making use of the chart i made, We found significant differences between std and lac as well as std and veg, but no significant difference between lac and veg.
When considering free energy change, biochemists usually define a standard state, the biochemical standard state, which is modified from the chemical standard state to fit biochemical applications. Determine which of the phrases describe the biochemical standard state, the chemical standard state, or both.
Answer:
Maximum work under this condition (∆G) = Maximum work under Standard Condition (∆G°) + Activities defining this condition
Explanation:
In this equation, the term DGo provides us with a value for the maximal work we could obtain from the reaction starting with all reactants and products in their standard states, and going to equilibrium. The term DG' provides us with a value for the maximal work we could obtain under the conditions defined by the activities in the logarithmic term. The logarithmic term can be seen as modifying the value under standard conditions to account for the actual conditions. In describing the work available in metabolic processes, we are concerned with the actual conditions in the reaction medium (whether that is a test-tube, or the cell cytoplasm); the important term is therefore DG'. If we measure the actual activities (in practice, we make do with concentrations), and look up a value for DGo in a reference book, we can calculate DG' from the above equation.
Values for DGo provide a useful indication through which we can compare the relative work potential from different processes, because they refer to a standard set of conditions.
Therefore both phrases describe the Biochemical and Chemical Standard State
The annual storage in Broad River watershed is 0 cm/y. Annual precipitation is 100 cm/y and evapotranspiration is 50 cm/y. The sandy loam soil in the watershed has an infiltration rate of 5 × 10-7 cm/s. The area of the watershed is 40 km2. Calculate the annual flow at the outlet of Broad River. What is the runoff coefficient for this watershed?
Answer:
0.34232
Explanation:
See attachment
Answer:
The annual flow at the outlet of Broad River is 13688480 m³
The runoff coefficient for the watershed is 0.342212
Explanation:
Here, we have
The annual storage in the watershed = 0 cm/y
Annual precipitation = 100 cm/y
Annual evapotranspiration = 50 cm/y
infiltration rate = 5 × 10-7 cm/s
Watershed area = 40 km²
From the question, annual infiltration rate is given by
Annual infiltration rate = 5 × 10-7 cm/s × 31557600 s/year = 15.7788 cm/y
Therefore, annual flow =
Annual precipitation - Annual evapotranspiration - Annual infiltration rate - Annual storage in the watershed
Annual flow = 100 cm/y - 50 cm/y - 15.7788 cm/y - 0 cm/y = 34.2212 cm/y
The area of the watershed = 40 km²
Therefore, the volume of the annual flow is given by;
Height of annual flow × area of the watershed
Volume of annual flow = 34.2212 cm × 40 km² = 1368.848 cm·km²
= 13688480 m³
The runoff coefficient = [tex]\frac{Amount \hspace{0.1cm} of \hspace{0.1cm}runoff}{Amount \hspace{0.1cm} of \hspace{0.1cm} precipitation \hspace{0.1cm} received}[/tex] =
The amount of precipitation received is given by
100 cm × 40 km² = 4000 cm·km² = 40000000 m³
Runoff coefficient = [tex]\frac{13688480 m^3}{40000000 m^3}[/tex] = 0.342212.
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) and width b = 12 mm (0.47 in.); the distance between support points is 43 mm (1.69 in.).(a) Compute the flexural strength if the load at fracture is 292 N (66 lbf). (b) The point of maximum deflection, ?y, occurs at the center of the specimen and is described by where E is the modulus of elasticity and I is the cross-sectional moment of inertia. Compute ?y at a load of 266 N (60 lbf). Assume that the elastic modulus for the glass is 60 GPa. What is the flexural strength of this sample in MPa if the load in part (a) is applied? What is I, the moment of inertia, in m4? What is the deflection, ?y, in mm, at the load given in the problem?
Answer:
5.21e-2mm
Explanation:
Please see attachment
The flexural strength of the glass specimen under the given load is approximately 84.4 MPa. The cross-sectional moment of inertia is about 0.655 *10^-12 m^4. When subjected to a load of 266N, the deflection at the center of the specimen is approximately 0.87 mm.
Explanation:The subject matter of the question refers to concepts from material science and mechanical engineering, specifically pertaining to the calculation of flexural strength, the moment of inertia, and deflection.
Flexural strength or 'Modulus of rupture' is calculated with the formula σf = 3FL / 2bd^2, where F is the fracture load, L the distance between support points, b the specimen's width, and d its height. Substituting the given values, we get σf = (3*292 * 43*10^-3) / (2 * 12 *10^-3 * (5.4 *10^-3)^2), which gives us about 84.4 MPa.
The moment of inertia for a rectangular cross-section is given by I = b*d^3 / 12. Substituting the given values, we get I = 12*10^-3 * (5.4 *10^-3)^3 / 12, which gives approximately 0.655 *10^-12 m^4.
Deflection, typically denoted by the Greek letter δ (or in this case, Υ), is computed with the formula Υ = FL^3 / 48EI. Given that the modulus of elasticity (E) for glass is 60 GPa or 60*10^9 Pa and we already calculated I, we substitute all values to get Υ = (266 * (43*10^-3)^3) / (48 * 60*10^9 * 0.655*10^-12) which gives us approximately 0.87 mm.
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Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power plant has an efficiency of 3 percent and a net power output of 150 kW, determine the average value of the required solar energy collection rate, in Btu/h.
Answer:
17060700 btu/h
Explanation:
Power output from power plant = 150 kW.
This power output is at 3% efficiency, this means power available in pond is 0.30P
Now, 150 kW = 0.03P
P = 150/0.03 = 5000 kW of power in the pond.
From basic conversion,
1 kW = 3412.14 btu/h
5000 kW = 5000 x 3412.14
average value of the required solar energy collection rate equal
17060700 btu/h
Solar pond power plant with 3% efficiency and 150 kW output needs ~17,060,000 Btu/h solar energy collection rate.
To find the average value of the required solar energy collection rate in Btu/h, we need to calculate the total solar energy collected per unit time and then convert it into Btu/h.
First, let's find the total solar energy collected per unit time:
Given:
- Net power output = 150 kW
- Efficiency = 3%
We know that efficiency is the ratio of output power to input power, so:
Efficiency = (Net Power Output / Solar Energy Input) * 100
Rearranging the equation to find the solar energy input:
Solar Energy Input = (Net Power Output / Efficiency) * 100
Now, let's substitute the given values:
Solar Energy Input = (150 kW / 3%) * 100
Solar Energy Input = (150,000 W / 0.03) * 100
Solar Energy Input = 5,000,000 W
Now, we convert this into Btu/h. To do this, we'll use the conversion factor:
1 watt = 3.412 Btu/h
So, the solar energy input in Btu/h will be:
Solar Energy Input (Btu/h) = (Solar Energy Input in watts) * (3.412 Btu/h per watt)
Solar Energy Input (Btu/h) = 5,000,000 W * 3.412 Btu/h per watt
Solar Energy Input (Btu/h) ≈ 17,060,000 Btu/h
Therefore, the average value of the required solar energy collection rate is approximately 17,060,000 Btu/h.
A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed of 320 km/h in air at 15 C. The student decided to build a 1/50 scale model and test it in a water tunnel at the same temperature.
(a) What conditions are required to ensure the similarity of scale model and prototype.
(b) Find the water speed required to model the prototype.
(c) If the drag force on the model is 5 kN, determine the drag force on the prototype
Answer:
b. 1232.08 km/hr
c. 1.02 kn
Explanation:
a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.
See attachment for the remaining.
The Canadair CL-215T amphibious aircraft is specially designed to fight fires. It is the only production aircraft that can scoop water, at up to 6120 gallons in 12 seconds, from any lake, river, or ocean. Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.
Answer:
Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.= 132.26∪
Explanation:
check attached files for explanation
For some material, the heat capacity at constant volume Cv at 34 K is 0.81 J/mol-K, and the Debye temperature is 306 K. Estimate the heat capacity (in J/mol-K) (a) at 44 K, and (b) at 477 K.
Answer:
a. Heat Capacity = 1.756J/mol-K
b. Heat Capacity = 24.942J/mol-k
Explanation:
Given
Constant volume Cv = 0.81J/mol-k
T1 = 34K
Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K
First, The value of the temperature-independent constant.
Using Cv = AT³
Make A the subject of formula
A = Cv/T³
Substitute each values
A = 0.81/34³
A = 0.000020608589456543
A = 2.061 * 10^-5J/mol-k
The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature
Cv = AT³
So, The heat capacity when T = 44k is then calculated as
Cv = 2.061 * 10^-5 * 44³
Cv = 1.755522084266232
Cv = 1.756J/mol-K
(b) at 477 K.
Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:
Cv = 3R
Where R = universal gas constant
R = 8.314J/mol-k
Cv = 3 * 8.314
Cv = 24.942J/mol-k
Answer:
a) 1.75 J/mol-k
b) 24.94 J/mol-k
Explanation:
We are given:
Cv = 0.81J/mol-k
Debye temperature = 306k
Heat capacity varies with temperature and the relationship between them is given as:
Cv = AT³
To find temperature independent constant A, we have:
[tex] A = \frac {C_v}{T^3} [/tex]
[tex] A = \frac{0.81J/mol-k}{34^3} [/tex]
[tex] A = 2.06*10^-^5 [/tex]
a) for temperature at 44k
Cv = AT³
[tex] C_v = 2.06*10^-^5 * 44^3 [/tex]
Cv = 1.75 J/mol-k
b) for T at 477k
Since the temperature 447k is higher than Debye's temperature, the specific heat will be given as:
Cv = 3R
Where R = universal gas constant = 8.314 J/mol-k
Therefore,
Cv = 3 * 8.314 J/mol-k
Cv = 24.94 J/mol-k
A student wants to determine experimentally, without disconnecting any wires in the circuit, the DC current moving through a copper wire. Which of the following items of laboratory equipment would be sufficient to make the necessary measurements for this determination?(A) Magnetic field sensor only (B) Magnetic field sensor and meterstick (C) Bar magnet and meterstick (D) Stopwatch and meterstick (E) Voltmeter and current sensor
Answer:
A student wants to determine experimentally, without disconnecting any wires in the circuit, the DC current moving through a copper wire. Which of the following items of laboratory equipment would be sufficient to make the necessary measurements for this determination?
(E) Voltmeter and current sensor
Explanation:
DC (direct current) is the unidirectional stream or development of electric charge transporters (which are generally electrons). The power of the current can shift with time, yet the general direction of development remains the equivalent consistently. As a descriptor, the term DC is utilized in reference to voltage whose extremity never turns around. In a DC circuit, electrons rise up out of the negative, or less, shaft and move towards the positive, or additionally, post. All things considered, physicists characterize DC as making a trip from in addition to short.
A voltmeter is an instrument utilized for estimating electrical potential contrast between two focuses in an electric circuit. Simple voltmeters move a pointer over a scale in relation to the voltage of the circuit; advanced voltmeters give a numerical presentation of voltage by utilization of a simple to computerized converter. A voltmeter in a circuit outline is spoken to by the letter V around. Voltmeters are made in a wide scope of styles. Instruments for all time mounted in a board are utilized to screen generators or other fixed mechanical assembly. Versatile instruments, typically prepared to likewise gauge flow and obstruction as a multimeter, are standard test instruments utilized in electrical and hardware work. Any estimation that can be changed over to a voltage can be shown on a meter that is reasonably adjusted; for instance, pressure, temperature, stream or level in a compound procedure plant.
A current sensor is a gadget that identifies electric current in a wire and creates a sign relative to that current. The produced sign could be simple voltage or current or even a computerized yield. The produced sign can be then used to show the deliberate current in an ammeter, or can be put away for additional investigation in an information securing framework, or can be utilized with the end goal of control. Current sensors are either open-or shut circle. Open-circle current sensors measure air conditioning and DC currents and give electrical confinement between the circuit being estimated and the yield of the sensor (the essential current is estimated without electrical contact with the essential circuit, giving galvanic detachment). More affordable than their shut circle cousins, open-circle current sensors are commonly favored in battery-controlled circuits given their low-working force prerequisites and little impression highlights.
A non-profit foundation is hosting a fundraising dinner. A few days ahead of the event you need to notify the caterer the quantity of meals you want served. The cost to cater each meal is $25. Each guest donates $100 to the non-profit foundation to attend the dinner event. Any leftover meals not eaten have no residual value. If a donor arrives at the event wanting to donate and no meals are left (i.e., you under ordered), the disappointed, hungry donor will i) not donate $100 at this event, and ii) be less likely to donate in the future, costing the foundation $75 (goodwill cost).a) (4 pts) What is the underage cost? What is the overage cost? What is the target service level? Cu = 150 Co = 25 Target Service Level = .8571 or 85.71%b) (4 pts) This is the first year of your event, so the only historical data you have is from the caterer who says that from her past experience with other foundations, the number of donors who will attend is well-approximated by a normal distribution with a mean of 100 and a standard deviation of 9.How many meals should you order? What is your effective service level? (NOTE: It might help you to know that for a standard normal distribution: z.8824=1.19, z.8571=1.07, z.90=1.28, z.95=1.64) Order 110 meals Effective Service Level = 85.71%For the remainder of the questions: It is now 10 years later (assume all revenue and costs are the same as in the original question), and you believe the historical data from your past fundraisers will more accurately predict the amount of meals you should order. The following table contains historical donor attendance for your various past fundraisers:Number of Donors. 104 105 106 107% of fundraisers 63% 20% 10% 7%c) Using the historical data, now how many meals should you order? Justify. Q* = 106 mealsd) What is your effective service level? Service Level = 93%e) The president of your foundation comes to you and says that he wants you to ensure that you won’t run out of meals at least 95% for all future fundraisers. How many dinners do you need to buy in order to meet that service level? 107 Meals
Answer:
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Explanation:
An insulated air nozzle with an inlet pressure of 10 bar operates with a mass flow rate of 1.2 kg s−1 . The inlet temperature is 600 K and the outlet velocity is 100 m s−1 . The inlet diameter is 0.5 m and the outlet diameter is 0.05 m. Determine the inlet velocity and the outlet temperature. What is the change in enthalpy across the nozzle? Assume air is an ideal gas. [0.95 m s−1 , 595.2 K, −5 kJ kg−1 ]
Answer:
Given mass flow rate =1.2 kg/s
mass flow rate=density*A*V
Area=pi(douter^2-dinner^2)/4
Area=0.194m^2
The velocity is given by
velocity=2.876 m/s
Between 1975 and 1985, the volume of all iron and steel in a given automobile model decreased from 0.165 m3m3 to 0.118 m3m3 . In the same time frame, the volume of all aluminum alloys increased from 0.012 m3m3 to 0.023 m3m3 . Using the densities of pure FeFe and AlAl, estimate the mass reduction resulting from this trend in materials substitution.
Answer: 340.19kg
Explanation:
The reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. It is referred to as a quantity which allows the two-body problem to be solved as if it were a one-body problem. You should note, however, that the mass determining the gravitational force is not reduced. In the computation or calculation, one mass can be replaced with the reduced mass, if this is compensated by replacing the other mass with the sum of both masses.
It has the dimensions of mass, and SI unit kg.
Given two bodies, first with mass m1 and the other with mass m2, the equivalent one-body problem, with the position of one body with respect to the other referred to as the unknown, is that of a single body of mass.
The density of Iron (Fe) = 7.87Mg/m^3
The volume of Iron decreased from= (0.165 - 0.118) m^3 =0.047m^3
Therefore,
The weight of Iron decreased= (0.047m^3) x (7.87Mg/m^3) = 0.36989Mg
The density of aluminium= 2.70Mg/m^3
The volume of aluminum increased= (0.023 - 0.012)m^3= 0.011m^3
The weight of aluminium is= (0.011m^3) × (2.70Mg/m^3) = 0.0297Mg
Therefore,
The mass reduction is:
The weight of Iron decreased minus the weight of Aluminium:
0.36989Mg - 0.0297Mg= 0.34019Mg
0.34019Mg × 1000
= 340.19Kg
Therefore, the mass reduction resulting from the trend is= 340.19kg
A 280-km-long pipeline connects two pumping stations. If 0.56m3/s are to be pumped through a 0.62-m-diameter line, the discharge station is 250m lower in elevation than the upstream station, and the discharge pressure is to be maintained at300,000Pa,determinethepowerrequiredtopumptheoil.The oilhas akinematic viscosity of4.510"6m2/s anda density of 810kg/m3.Thepipeisconstructedofcommercialsteel.Theinlet pressure may be taken as atmospheric.
The question asks for pump pressure calculations in a pipeline, requiring applications of fluid dynamics principles, such as Bernoulli's equation, and considering factors like elevation changes, viscosity, and pipe characteristics.
Explanation:The student's question relates to the field of fluid dynamics, specifically the calculation of required pump pressure in a pipeline system using principles such as Bernoulli's equation and concepts like pressure, flow rate, pipe diameter, density, viscosity, and elevation change. To solve this type of problem, you typically apply the Bernoulli equation, along with any additional head losses due to viscosity, known as the head loss due to friction, which can be calculated using the Darcy-Weisbach equation or other empirical formulas. You must then calculate the total head needed, convert this to pressure, taking into account the fluid's density, gravitational acceleration, and add any other relevant pressures, such as atmospheric pressure if the pipeline opens to the atmosphere.
Use the indirect pattern when you need to soften or delay bad news until after an explanation is given. Understanding the four components of the indirect pattern will help you craft messages that convey empathy, present reasons, cushion bad news, and close pleasantly. What buffering technique are you using if you show in your opening that you care and are concerned
Answer:
The general method for indirect strategy is Start with a neutral buffer.
Explanation:
The general method for indirect strategy goes like this:
-Start with a neutral buffer, you should never start with good news because it will give the reader false hope that more good news will come. So a neutral “buffer” or a show of appreciation for your business is a good way to start. You are not apologizing for the bad news that is coming, you are simply preparing the reader for it.
-The next part is where you give reasons. There are many studies on the effectiveness of reasons in communication: people like to know why things are the way they are. By offering reasons, you will make the bad news easier to accept and once you have prepared the reader, you give the bad news.
The idea is also not to give infinite turns to the subject in general, since the information and the objective of the message can be lost. It is important not to spend too much time on this: the buffer should be short, so as not to make the moment tedious and lose the attention of the public before reaching the main topic.
-Finally, in your conclusion, divert attention from the bad news. Don't talk about it anymore, be nice, focus your final efforts on future opportunities and recover goodwill.
an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required to provide 1.6 A of current through the 75 ohm resistance
Answer:
The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.
Explanation:
Given;
Resistance, R₁ = 50Ω
Resistance, R₂ = 75Ω
Total resistance, R = (R₁R₂)/(R₁ + R₂)
Total resistance, R = (50 x 75)/(125)
Total resistance, R = 30 Ω
According to ohms law, sum of current in a parallel circuit is given as
I = I₁ + I₂
[tex]I = \frac{V}{R_1} + \frac{V}{R_2}[/tex]
Voltage across each resistor is the same
[tex]1.6 = \frac{V}{R_2}[/tex]
V = 1.6 x R₂
V = 1.6 x 75
V = 120 V
Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.
This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.
Answer:
120 volts
Explanation:
Since the two resistances are connected in parallel across the voltage source, the effective resistance of the circuit can be obtained by using the formula.
[tex]\frac{1}{R_{eq}}=\frac{1}{R_{1}}+ \frac{1}{R_{2}}[/tex]
given that [tex]R_{1} and R_{2}[/tex] are 50 and 75 ohms respectively, we have the equivalent resistance as:
[tex]\frac{1}{R_{eq}}=\frac{1}{50}+ \frac{1}{75}=\frac{1}{30}[/tex]
hence,
[tex]R_{eq}= 30\Omega[/tex]
From Ohm's law, voltage = current X resistance.
given that the current through the 75 ohm resistor is 1.6 A
[tex]V= I\times R[/tex]
[tex]V= 1.6 \times 75\Omega[/tex]
voltage = 120 Volts.
Because the resistors are connected in parallel, it means that they are connected to the same voltage source.
Hence, the voltage source for the 75 Ohm resistance = 120 volts. This is same for the 50 Ohm resistor.
Water at 20 bar, 400oC enters a turbine operating at steady state and exits at 1.5 bar. Neglect heat transfer, kinetic energy and potential energy effects. Someone states that the vapor quality at the turbine exit is 98%. (a) (35%) Find the entropy generation based on this statement. (b) (5%) Is this statement possible?
Answer:
a) [tex]s_{gen} = -0.0219\,\frac{kJ}{kg\cdot K}[/tex], b) No.
Explanation:
The turbine is modelled after the Second Law of Thermodynamics, which states:
[tex]s_{in} - s_{out} + s_{gen} = 0[/tex]
The entropy generation per unit mass is:
[tex]s_{gen} = s_{out} - s_{in}[/tex]
The specific entropy for steam at entrance and exits are obtained from property tables:
Inlet (Superheated Steam)
[tex]s_{in} = 7.1292\,\frac{kJ}{kg\cdot K}[/tex]
Outlet (Liquid-Vapor Mixture)
[tex]s_{out} = 7.1073\,\frac{kJ}{kg\cdot K}[/tex]
[tex]s_{gen} = 7.1073\,\frac{kJ}{kg\cdot K} - 7.1292\,\frac {kJ}{kg\cdot K}[/tex]
[tex]s_{gen} = -0.0219\,\frac{kJ}{kg\cdot K}[/tex]
b) It is not possible, as it contradicts the Kelvin-Planck and Claussius Statements, of which is inferred that entropy generation can only be zero or positive.
A cyclist rode the 1st 20 mile portion of his workout, at a constant speed. For the 12 mile cool down portion of his workout, he reduced his speed by 4 miles per hour. Each portion of the workout took the same time. Find the cyclist speed during the first portion, and find his speed during the cool down portion
Answer:
10 mph
Explanation:
20 / x = 12 / (x - 4)
Cross multiply
12x = 20 (x - 4)
12x = 20x - 80
80 = 20x - 12x
80 = 8x
divide through by '8'
10 = x
x = 10 mph
Answer:
16
Explanation:
Air passing through an array of thin copper tubes submerged in a large ice/water bath is used for air conditioning in the summer. Each tube has diameter D = 50 mm, and the ice bath maintains its inner surface temperature at a uniform, constant 0oC. Air enters each tube at a mean temperature of Tmi 32oC and a flow rate of m 0.01 kg/s, and it is required for the mean temperature of the air to be at or belovw Tmo-22°C by the end of the tube. a) At what temperature do you want to calculate all the air properties (like viscosity and density)? b) Calculate the minimum tube length required to provide an exit temperature of Tmo 22°C. c) Sketch the mean temperature along the pipe, Tmx). d) Calculate the log-mean temperature difference ΔTlm between the pipe surface and the mean air temperature, and use it to determine the total heat transfer rate conv out of the air into the ice bath. I am assuming that you are ignoring the entry region, and assuming the air profile is fully-developed throughout the length of the pipe to complete your analysis. Let's now challenge that assumption e) What is the length of the entry region before the flow is thermally fully-developed? What percentage is that of the total length of pipe you calculated in (a)? Because of the entry region, is your result for the design length of pipe in (a) conservative or not? That is, if your analysis were to account for the entry region, do you expect that the actual temperature of air by the exit would be below 22oC, or would you have to make your pipe longer than what you calculated in (a)? Explain
Answer:
Explanation:
The answers and step by step to the solution Can be found in the attached files, please kindly go through them.
Hydroelectric power plants convert the gravitational potential energy of falling water into electrical power, typically by allowing the water to flow through a pipe called a penstock to rotate a generator located below it. Let the bottom of the penstock be the origin of a Cartesian coordinate system and the point at which the gravitational potential energy is zero.
a) Consider a penstock that is vertical and has a height of h = 61 m. How long, t in seconds, does it take water to fall from the top of the penstock to the bottom? Assume the water starts at rest.
Answer:
3.527 seconds
Explanation:
The height of the falling water, assuming no friction or air resistance, is given by ...
h(t) = -(1/2)gt² +61
where g is the standard gravity value, 9.80665 m/s².
The the time required for h(t) = 0 is ...
1/2gt² = 61
t² = 2·61/g
t = √(2·61/9.80665) ≈ 3.527 . . . . seconds
Use the following information to complete the concrete mixture designs for the following question.
The following materials are available for a concrete mixture design:
ASTM C 150 Type I/II Cement
Relative density of 3.15
Coarse Aggregate: well-graded 3⁄4 in. maximum-size aggregate (MSA), crushed limestone, angular
Oven-dry specific gravity: 2.6
Absorption: 0.72%
Oven-dry rodded bulk density: 102 lb/ft3 Moisture content of coarse stockpile: 0.6%
Fine Aggregate: well-graded natural sand Oven-dry specific gravity: 2.5 Absorption: 1.3%
Moisture content of fine stockpile: 3.6% Fineness Modulus: 2.2
An air entraining admixture will provide enough air for the design mixture at a dosage rate of 3 oz/ft3
QUESTION:
Concrete is required for an 8 in thick exterior concrete slab in central Texas. A specified compressive strength, f’c , of 5000 psi is required at 28 days using an ASTM C 150 Type II portland cement. The design calls for a minimum of 2 in. of concrete cover over the reinforcing steel. The minimum distance between reinforcing bars is 3 in. A slump of 3 inches should be the target. The sidewalk is located in an environment that does not require air entraining. The concrete is required to have low permeability when exposed to water and moderate sulfates. The concrete will be exposed water soluble sulfates in soil at 0.14%. No statistical data on previous mixes are available.
Determine:
a. The theoretical mixture proportions of all concrete constituents (fine aggregate, coarse aggregate, water, cement, chemical admixtures) based on 1 yd3 of concrete on an oven dry basis for aggregates.
b. The as-batched mixture proportions based on the moisture contents provided in the materials available section.
c. How much of each material would be required to make a 20 ft long by 15 ft slab from the concrete mixture described above?
Answer:
a. 147lb/dt^3
b. 148lb/dt^3
c. 7318.52lb
Explanation:
Please see attachments
11.4 Compute the volume percent of graphite VGr in a 3.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
The volume percent of graphite in a 3.5 wt% C cast iron is 11%.
Explanation:
Consider 100 grams 3.5% C cast iron.
Mass of carbon in 100 grams of cast iron = [tex]m_1=100 g\times \frac{3.5 }{100}=3.5 g[/tex]
Mass of iron in cast iron =[tex]m_2[/tex] = 100 g - 3.5 g = 96.5 g
Density of graphite = [tex]d_1=2.3g/cm^3[/tex]
Volume of the graphite = [tex]v_1[/tex]
[tex]v_1=\frac{m_1}{d_1}=\frac{3.5 g}{2.3g/cm^3}=1.52 cm^3[/tex]
Density of iron= [tex]d_2=7.9 g/cm^3[/tex]
Volume of the iron = [tex]v_2[/tex]
[tex]v_2=\frac{m_2}{d_2}=\frac{96.5 g}{7.9 g/cm^3}=12.22 cm^3[/tex]
Total volume of the cast iron ,V=[tex]v_1+v_2=1.52 cm^3+12.22 cm^3=13.74 cm^3[/tex]
Volume percent of the graphite :
[tex]=\frac{v_1}{V}\times 100[/tex]
[tex]=\frac{1.52 cm^3}{13.74 cm^3}\times 100=11.06\%\approx 11\%[/tex]
The volume percent of graphite in a 3.5 wt% C cast iron is 11%.
The volume percent of graphite VGr in a 3.5 wt% C cast iron is;
V_gr = 11.08%
We are given;
Density of Ferrite; ρₐ = 7.9 g/cm³
Density of Graphite; [tex]\rho _{g}[/tex] = 2.3 g/cm³
Weight percent of C cast iron; C₀ = 3.5%
Now the formula for the mass fraction of ferrite is;
Wₐ = [tex]\frac{C_{g} - C_{0}}{C_{g} - C_{a}}[/tex]
If we consider 100 g of 3.5 wt% C cast iron, then [tex]C_{g}[/tex] = 100 and Cₐ = 0
Thus;
Wₐ = [tex]\frac{100 - 3.5}{100 - 0}[/tex]
Wₐ = 0.965
Thus, mass fraction of graphite is;
[tex]W _{g}[/tex] = 1 - Wₐ
[tex]W _{g}[/tex] = 1 - 0.965
[tex]W _{g}[/tex] = 0.035
Formula for the volume percent of graphite is;
[tex]V_{gr} = \frac{\frac{W_{g}}{\rho_{g}}}{\frac{W_{a}}{\rho_{a}} + \frac{W_{g}}{\rho_{g}}} * 100%[/tex]
Plugging in the relevant values gives;
V_gr = [(0.035/2.3)/((0.965/7.9) + (0.035/2.3))]
V_gr = 0.01521739/(0.1221518987 + 0.01521739)
V_gr = 11.08%
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A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of heat transfer to the river water. Will the actual heat transfer rate be higher or lower than this value
Answer:
A The heat transfer to the river would be 900 MW
B. The actual heat transfer would be lower than the theoretical heat transfer.
Explanation:
Part A
For a steam power plant, the process is continuous and the law of thermodynamic is conserved. The steam power plant consists of the cold and hot reservoir and the rate of heat transfer in the steam power plant can be gotten with the expression below.
W = [tex]Q_{H}[/tex] - [tex]Q_{c}[/tex] ......................1
where W is the work output of the steam power plant
[tex]Q_{H}[/tex] is the heat transfer at the hot reservoir
[tex]Q_{c}[/tex] is the heat transfer at the cold reservoir( heat transfer to the nearby river)
The efficiency of the steam power plant would be used to obtain the heat transfer at the hot reservoir ([tex]Q_{H}[/tex]). The efficiency can be obtained with equation 2;
e = [tex]\frac{W}{Q_{H} }[/tex]
e is the thermal efficiency of the steam power plant = 40 % = 0.4
W is the work output of the steam power plant = 600 MW
[tex]Q_{H}[/tex] is the heat transfer at the hot reservoir
Rearranging equation 2 to make [tex]Q_{H}[/tex] the subject formula we have;
[tex]Q_{H}[/tex] = W/e
substituting the values we have;
[tex]Q_{H}[/tex] = 600 MW/ 0.4
[tex]Q_{H}[/tex] = 1500 MW
The heat transfer at the hot reservoir is 1500 MW and putting it into equation 1 to find the heat transfer to the river;
[tex]Q_{c}[/tex] = [tex]Q_{H}[/tex] - W
[tex]Q_{c}[/tex] = 1500 MW - 600 MW
[tex]Q_{c}[/tex] = 900 MW
Therefore the heat transfer to the river would be 900 MW
Part B
The actual heat transfer would be lower than the theoretical heat transfer. Losses dues to friction, head losses and heat loss by evaporation to the surroundings account for the lower value.
The actual heat transfer rate is higher than the power delivered by the steam power plant.
The heat transfer rate released to the river ([tex]\dot Q_{L}[/tex]), in megawatts, is derived from the definition of energy efficiency ([tex]\eta[/tex]), no unit, that is to say:
[tex]\eta = \frac{\dot W}{\dot Q_{L}+\dot {W}}[/tex]
[tex]\eta \cdot \dot Q_{L} + \eta \cdot \dot W = \dot W[/tex]
[tex](1-\eta)\cdot \dot W = \eta \cdot \dot Q_{L}[/tex]
[tex]\dot Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot \dot W[/tex] (1)
Where [tex]\dot W[/tex] is the power delivered by the steam power plant, in megawatts.
If we know that [tex]\eta = 0.4[/tex] and [tex]\dot W = 600\,MW[/tex], then the heat transfer rate released to the river is:
[tex]\dot Q_{L} = \left(\frac{1}{0.4}-1 \right)\cdot (600\,MW)[/tex]
[tex]\dot Q_{L} = 900\,MW[/tex]
Thus, the actual heat transfer rate is higher than the power delivered by the steam power plant.
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A large plate is at rest in water at 15?C. The plate is suddenly translated parallel to itself, at 1.5 m/s. The resulting fluid movement is not exactly like that in a b.l. because the velocity profile builds up uniformly, all over, instead of from an edge. The governing transient momentum equation, Du/Dt = ?(?2u/?y2), takes the form 1 ? ?u ?t = ?2u ?y2
`Determine u at 0.015 m from the plate for t = 1, 10, and 1000 s. Do this by first posing the problem fully and then comparing it with the solution in Section 5.6. [u 0.003 m/s after 10 s.]
Answer:
Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s
Explanation:
du/u² = dt = ∫du/2.3183 = ∫dt
0.4313 u = t + c
(a) t = 0, u= 15m/s, c = 0.647
u = t+c/0.4313 = t + 0.647/0.4313
(a) when t= 1 u = 1+ 0.647/0.4313 = 3.8187m/s
(b) when t= 10 u = 10 + 0.647/0.4313 = 24.0858m/s
(c)when t= 1000 u = 1000 + 0.647/0.4313 = 3220.071m/s
In this lab, you will be creating a class that implements the Rule of Three (A Destructor, A Copy Constructor, and a Copy Assignment Operator). You are to create a program that prompts users to enter in contact information, dynamically create each object, then print the information of each contact to the screen. Some code has already been provided for you. To receive full credit make sure to implement the following:
Default Constructor - set Contact id to -1
Overloaded Constructor - used to set the Contact name, phoneNumber and id (Should take in 3 parameters)
Destructor
Copy Constructor
Copy Assignment Operator
Any other useful functions (getters/setters)
Main.cpp
#include
#include
#include "Contact.h"
using namespace std;
int main() {
const int NUM_OF_CONTACTS = 3;
vector contacts;
for (int i = 0; i < NUM_OF_CONTACTS; i++) {
string name, phoneNumber;
cout << "Enter a name: ";
cin >> name;
cout << "Enter a phoneNumber; ";
cin >> phoneNumber;
// TODO: Use i, name, and phone number to dynamically create a Contact object on the heap
// HINT: Use the Overloaded Constructor here!
// TODO: Add the Contact * to the vector...
}
cout << "\n\n----- Contacts ----- \n\n";
// TODO: Loop through the vector of contacts and print out each contact info
// TODO: Make sure to call the destructor of each Contact object by looping through the vector and using the delete keyword
return 0;
}
Contact.h
#ifndef CONTACT_H
#define CONTACT_H
#include
#include
using std::string;
using std::cout;
class Contact {
public:
Contact();
Contact(int id, string name, string phoneNumber);
~Contact();
Contact(const Contact& copy);
Contact& operator=(const Contact& copy);
private:
int *id = nullptr;
string *name = nullptr;
string *phoneNumber = nullptr;
};
#endif
Contact.cpp
#include "Contact.h"
Contact::Contact() {
this->id = new int(-1);
this->name = new string("No Name");
this->phoneNumber = new string("No Phone Number");
}
Contact::Contact(int id, string name, string phoneNumber) {
// TODO: Implement Overloaded Constructor
// Remember to initialize pointers on the heap!
}
Contact::~Contact() {
// TODO: Implement Destructor
}
Contact::Contact(const Contact ©) {
// TODO: Implement Copy Constructor
}
Contact &Contact::operator=(const Contact ©) {
// TODO: Implement Copy Assignment Operator
return *this;
}
Answer:
Explanation:
============== main.cpp =======================
#include <iostream>
using namespace std;
#include <iostream>
#include <vector>
#include "Contact.h"
using namespace std;
int main() {
const int NUM_OF_CONTACTS = 3;
vector<Contact *> contacts;
for (int i = 0; i < NUM_OF_CONTACTS; i++) {
string name, phoneNumber;
cout << "Enter a name: ";
cin >> name;
cout << "Enter a phoneNumber: ";
cin >> phoneNumber;
// Use i, name, and phone number to dynamically create a Contact object on the heap
// HINT: Use the Overloaded Constructor here!
Contact * newContact = new Contact(i+1,name,phoneNumber);
// Add the Contact * to the vector...
contacts.push_back(newContact);
}
cout << "\n\n----- Contacts ----- \n\n";
// Loop through the vector of contacts and print out each contact info
std::vector<Contact *>::iterator itrContact;
for(itrContact = contacts.begin(); itrContact !=contacts.end(); itrContact++)
{
cout<< " Id : " << (*itrContact)->getId();
cout<< " Name : " << (*itrContact)->getName();
cout<< " Phone-Number : " << (*itrContact)->getPhoneNumber() <<endl;
}
// Make sure to call the destructor of each Contact object by looping through the vector and using the delete keyword
for(int i = 0; i < NUM_OF_CONTACTS; i++)
{
delete contacts[i];
}
return 0;
}
================== contact.h =====================
#ifndef CONTACT_H
#define CONTACT_H
#include <string>
#include <iostream>
using std::string;
using std::cout;
class Contact {
public:
Contact();
Contact(int id, string name, string phoneNumber);
~Contact();
Contact(const Contact& copy);
Contact& operator=(const Contact& copy);
// getters
int getId() const;
string getName() const;
string getPhoneNumber() const;
//setters
void setId(int id);
void setName(string name);
void setPhoneNumber(string phoneNumber);
private:
int *id = nullptr;
string *name = nullptr;
string *phoneNumber = nullptr;
};
#endif
================= contact.cpp ========================
#include "Contact.h"
Contact::Contact() {
this->id = new int(-1);
this->name = new string("No Name");
this->phoneNumber = new string("No Phone Number");
}
Contact::Contact(int id, string name, string phoneNumber) {
// Implement Overloaded Constructor
// Remember to initialize pointers on the heap!
this->id = new int(id);
this->name = new string(name);
this->phoneNumber = new string(phoneNumber);
}
Contact::~Contact() {
// Implement Destructor
if(this->id != nullptr)
delete this->id;
if(this->name != nullptr)
delete this->name;
if(this->phoneNumber != nullptr)
delete this->phoneNumber;
}
Contact::Contact(const Contact ©) {
// Implement Copy Constructor
this->id = new int(copy.getId());
this->name = new string(copy.getName());
this->phoneNumber = new string(copy.getPhoneNumber());
}
Contact &Contact::operator=(const Contact ©) {
// Implement Copy Assignment Operator
if(this == ©) // Checks for self Assignment
return *this;
this->id = new int(copy.getId());
this->name = new string(copy.getName());
this->phoneNumber = new string(copy.getPhoneNumber());
return *this;
}
// getters
int Contact::getId() const
{
return *(this->id);
}
string Contact::getName() const
{
return *(this->name);
}
string Contact::getPhoneNumber() const
{
return *(this->phoneNumber);
}
//setters
void Contact::setId(int id)
{
*(this->id) = id;
}
void Contact::setName(string name)
{
*(this->name) = name;
}
void Contact::setPhoneNumber(string phoneNumber)
{
*(this->phoneNumber) = phoneNumber;
}
====================Output =========================
is attached below
A rectifier charges a battery bank in a substation. The bank rated dc voltage is 48 V. The required charging current is 25 A. The available ac supply is 120 V. The internal resistance of the battery is 2.5 Ω. (a) Analyze the operating conditions of the charger. Plot the ac and dc voltage and current, and determine the feasibility of delay angle control. (b) Calculate the delay angle needed to maintain the 25 A charging current. (c) Calculate the power and power factor at the ac side.
Answer:
See explaination
Explanation:
A constant voltage source is called a DC Voltage with a voltage that varies periodically with time is called an AC voltage. Voltage is measured in volts, with one volt being defined as the electrical pressure required to force an electrical current of one ampere through a resistance of one Ohm.
Please check the attached file.
A pipe of 10 cm inner diameter is used to send crude oil over distance of 400 meters. The entire pipe was laid horizontal. The viscosity of the oil is 10 cp. The density of oil is 800 kg/m3 . a. (20 pts) The desired volumetric flow rate is 0.1 m3 /min. What is the Reynolds number of this flow? Is the flow laminar or turbulent? What is the pressure difference needed to generate this flow rate? b. (15 pts) Three month later, the operator found that they had to triple the pressure difference to maintain the flow rate at 0.1 m3 /min. The operator thought that wax deposition from the oil had reduced the inner diameter of the tube. Based on this assumption, can you estimate the reduced inner diameter of the tube? What is the Reynolds number of the flow?
Answer:
See explaination
Explanation:
Looking at Reynolds number, we can go ahead and describe the Reynolds number as a dimensionless value that is used to determine whether the fluid is exhibiting laminar flow (R less than 2300) or turbulent flow (R greater than 4000). Laminar flow is when a fluid moves smoothly and is predictable.
Check attachments
How many D-cell batteries would it take to power a human for 1 day?
All numbers must be entered as 5000 or 5e3 or 5.0e3 and not with commas as in "5,000" and not as fractions as in "3/4" and not as percentages as in "70%".
Estimate the recomended daily food intake (in Food Calories).
Estimate the daily energy (in joule) a human needs.
Estimate the voltage (in volt) of one D cell battery.
Estimate the charge (in amp-hours) of one D cell battery.
Estimate the energy of one D cell battery (in watt-hour).
Estimate the number of D-cell batteries it takes to power a human for 1 day.
Answer:
it would take approximately 232 to 258 D cell batteries to power a human for 1 day.
Explanation:
Estimate the recommended daily food intake (in Food Calories).
An average adult man requires between 2000 to 3000 calories per day.
Estimate the daily energy (in joule) a human needs.
As we know 1 food calorie is equal to 4.184 Joules of energy
2000*4.184 to 3000*4.184
8368 to 12552 Joules
But for engineering calculations 1 food calorie is equal to 1000 engineering calories, so
8368*1000 to 12552*1000
8368000 to 12552000 Joules
Estimate the voltage (in volt) of one D cell battery.
The voltage of a D cell battery is around 1.5 Volts
Estimate the charge (in amp-hours) of one D cell battery.
The amp-hours of a D cell battery varies with the manufacturing company, the typical amp-hours are in the range of 6 to 10 amp-hours.
Estimate the energy of one D cell battery (in watt-hour).
Energy in watt-hour is given by
voltage*amp-hour
1.5*6 to 1.5*10
9 to 15 watt-hour
Estimate the number of D-cell batteries it takes to power a human for 1 day.
First let us calculate the energy in a D cell battery,
1 watt-hour is equal to 3600 Joules
9*3600 to 15*3600
32400 to 54000 Joules
The number of D cell batteries required is found by dividing the energy need of a human by the energy stored in a D cell battery.
12552000/54000 to 8368000/32400
232 to 258 batteries
Therefore, it would take approximately 232 to 258 D cell batteries to power a human for 1 day.
A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flowing in the supply line at 600 kPa and 22°C. The valve is opened, and air is allowed to enter the tank until the pressure in the tank reaches the line pressure, at which point the valve is closed. A thermometer placed in the tank indicates that the air temperature at the final state is 77°C. 1) Starting with the most general form of the appropriate mass balance equation, determine the mass of air that has entered the tank (4 points). 2) Starting with the most general form of the appropriate energy balance equation, determine the amount of heat transferred and whether it was heat transferred in or out (8 points). 3) List at least 3 assumptions needed to complete this problem (3 points).
Answer:
Check the explanation
Explanation:
First of all the initial or primary and final masses can be calculated with the use of the ideal gas relations.
The net It transfer is determined from the energy balance. The initial and final internal energies and the enthalpy of the air in the supply line are obtained from A-I] for the given temperatures.
kindly check the attached image below to see working.
- Suppose we want to find the largest entry in an unsorted array of n entries. Algorithm A searches the entire array sequentially and records the largest entry seen so far. Algorithm B sorts the array into descending order and then reports the first entry as the largest. Compare the time efficiency of the two approaches. Coding is not required but highly recommended.
Answer:Coding is required in other to compare the time efficiency between two different search approaches.
Explanation
Assuming a linear search approach is to be conducted to find the highest entry in a sequential or descending order, there is need to write a programing code for the two approaches, then compare the time taken for each task to be completed . After, one can now tell the approach that is most time efficient.
This code initially states that n=10 and f=n. Once the code enters the while loop, it stays in the while loop until the condition n>1 is not true. n is subtracted by 1 every iteration. This means that f = 10*9*8*7*6*4*3*2*1. With 9,8,7,6,5,4,3,2, and 1 being repeated n values. The disp (f) command displays the value of f after the while loop, which is equal to (n!).
Answer:
n = 10; f = n; while(n > 1) n = n - 1; f = f * n; end disp(f);Explanation:
The solution code is written in Matlab.
Firstly, we initialize n to 10 and set n to f variable (Line 1-2).
Next, create a while loop that will continue to loop until n is equal or smaller than one. In the loop, decrement n by one and multiply n with current f_variable.
At last display value of f. The output is 3628800.
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1 for $76,900, and Machine 1's adjusted basis was $40,950 at the time of the exchange. Machine 2's seller purchased it for $64,050 and Machine 2's adjusted basis was $55,950 at the time of the exchange. What is Koch's adjusted basis in machine 2 after the exchange?
Answer:
Koch's adjusted basis in machine 2 after the exchange is $60,000
Explanation:
given data
fair market value = $60,000
originally purchased Machine 1 = $76,900
Machine 1 adjusted basis = $40,950
Machine 2 seller purchase = $64,050
Machine 2 adjusted basis = $55,950
solution
As he exchanged machine for another at $60,000
and this exchanged in fair market
so adjusted basis = $50,000
Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell
Adjusted basis increases when a person deducts expenses from factor taxes and operating statements
so Koch's adjusted basis in machine 2 after the exchange is $60,000