Answer:
C) Prospective Cohort study
Explanation:
prospective cohort study can be regarded as longitudinal cohort study that comes up with periods of time when group of individuals that are different in terms of some factors that are undergoing some study, so that how theses factors influence rates of some particular outcomes can be known.
A 4-m-high and 6-m-long wall is constructed of twolarge 2-cm-thick steel plates (k 5 15 W/m·K) separated by1-cm-thick and 20-cm wide steel bars placed 99 cm apart. Theremaining space between the steel plates is filled with fiber-glass insulation (k 5 0.035 W/m·K). If the temperature dif-ference between the inner and the outer surfaces of the wallsis 22°C, determine the
Answer:
fart
Explanation:
Research a machine product or component that we will not cover in the class. The component must require engineering analysis to select the component from the manufacturer of the component. If it is an engineered component, the manufacturer will provide a website/catalog that explains what calculations to perform in order to select a product from their list. (i.e., not the McMaster-Carr catalog.)
(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are held constant at 300oC and 100oC, respectively.(b) Determine the heat loss per hour if the cross-sectional area of the sheet is 0.25 m2.(c) What will be the heat loss per hour if a sheet of soda-lime glass is used instead
Answer:
do the wam wam
Explanation:
The heat flux is =1038kW/m² , the heat lost per hour is =259.5 kW, the heat lost per hour using a sheet of soda- lime glass.
Calculation of heat fluxThe thickness of steel( t) = 10mm = 10× 10^-³m
The temperature difference on both sides = 300-100
∆T = 200°C
But the formula for heat flux = q = k∆T/t
Where K = thermal conductivity for steel = 51.9W/mK.
Substitute the variables into the formula for heat flux;
q = 51.9 × 200/10 × 10-³
q = 10380 × 10³/10
q = 10380000/10
q = 1038000 W/m² = 1038kW/m²
To calculate the heat lost per hour if the cross sectional area is = 0.25 m2 use the formula q × A
= 1038kW/m² × 0.25 m2
= 259.5 kW.
Learn more about heat flux here:
https://brainly.com/question/15217121
Cooling water for a chemical plant must be pumped from a river 2500 ft from the plant site. Preliminary design cans for a flow of 600 gal/min and 6-in. Steel pipe. Calculate the pressure drop and the annual pumping cost if power cost 3 cents per kilowatt-hour. Would the use of an 8-in. Pipe reduce the power cost enough to offset the increased pipe cost? Use $15/ft of length for the installed cost of 6-in. Pipe and $20/ft for 8-in. Pipe. Annual charges are 20 percent of the installed cost.
Answer:
The power cost savings for the 8 inches pipe offsets the increased cost for the pipe therefore to save costs the 8-inch pipe should be used
Explanation:
The given parameters in the question are;
The distance of the river from the the site, d = 2,500 ft.
The planned flow rate = 600 gal/min
The diameter of the pipe, d = 6-in.
The pipe material = Steel
The cost of pumping = 3 cents per kilowatt-hour
The Bernoulli's equation is presented as follows;
[tex]\dfrac{P_a}{\rho} + g\cdot Z_a + \dfrac{V^2_a}{2} + \eta\cdot W_p = \dfrac{P_b}{\rho} + g\cdot Z_b + \dfrac{V^2_b}{2} +h_f +W_m[/tex]
[tex]{P_a} = {P_b} = Atmospheric \ pressure[/tex]
[tex]Z_a = Z_b[/tex]
Vₐ - 0 m/s (The river is taken as an infinite source)
[tex]W_m[/tex] = 0
The head loss in 6 inches steel pipe of at flow rate of 600 gal/min = 1.19 psi/100 ft
Therefore; [tex]h_f[/tex] = 1.19 × 2500/100 = 29.75 psi
[tex]\eta\cdot W_p = \dfrac{V^2_b}{2} +h_f[/tex]
[tex]V_b[/tex] = Q/[tex]A_b[/tex] = 600 gal/min/(π·(6 in.)²/4) = 6.80829479 ft./s
[tex]V_b[/tex] ≈ 6.81 ft./s
The pressure of the pump = P = 62.4 lb/ft³× (6.81 ft./s)²/2 + 29.75 psi ≈ 30.06 psi
The power of the pump = Q·P ≈ 30.06 psi × 600 gal/min = 7,845.50835 W
The power consumed per hour = 7,845.50835 × 60 × 60 W
The cost = 28,243.8301 kW × 3 = $847.31 per hour
Annual cost = $847.31 × 8766 = $7,427,519.46
Pipe cost = $15/ft × 2,500 ft = $37,500
Annual charges = 20% × Installed cost = 0.2 × $37,500 = $7,500
Total cost = $37,500 + $7,500 + $7,427,519.46 = $7,475,519.46
For the 8-in pipe, we have;
[tex]V_b[/tex] = Q/[tex]A_b[/tex] = 600 gal/min/(π·(8 in.)²/4) = 3.83 ft./s
[tex]h_f[/tex] = 1.17 ft/100 feet
Total head loss = (2,500 ft/100 ft) × 1.17 ft. = 29.25 ft.
[tex]h_p = \dfrac{V^2_b}{2 \cdot g} +h_f[/tex]
∴ [tex]h_p[/tex] = (3.83 ft./s)²/(2 × 32.1740 ft/s²) + 29.25 ft. ≈ 29.5 ft.
The power of the pump = ρ·g·h × Q
Power of pump = 62.4 lb/ft³ × 32.1740 ft/s² × 29.5 ft.× 600 gal/min = 3,363.8047 W
The cost power consumed per annum = 3,363.8047 W × 60 × 60 × 3 × 8766 = $3,184,608.1
The Cost of the pipe = $20/ft × 2,500 ft. = $50,000
The total cost plus charges = $3,184,608.1 + $50,000 + $10,000 = $3,244,608.1
Therefore it is more affordable to use the 8-in pipe which provides substantial savings in power costs
For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.6 MN/mm per bolt. The bolts are preloaded to 75 percent of proof strength. Assume the external load is equally distributed to all the bolts. The bolts are M6 × 1 class 5.8 with rolled threads. A fluctuating external load is applied to the entire joint with Pmax = 60 kN and Pmin = 20 kN.(a) Determine the yielding factor of safety.(b) Determine the overload factor of safety.(c) Determine the factor of safety based on joint separation.(d) Determine the fatigue factor of safety using the Goodman criterion.
Answer:
a) 0.978
b) 0.9191
c) 1.056
d) 0.849
Explanation:
Given data :
Stiffness of each bolt = 1.0 MN/mm
Stiffness of the members = 2.6 MN/mm per bolt
Bolts are preloaded to 75% of proof strength
The bolts are M6 × 1 class 5.8 with rolled threads
Pmax =60 kN, Pmin = 20kN
a) Determine the yielding factor of safety
[tex]n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }[/tex] ------ ( 1 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
Input the given values into the equation above
equation 1 becomes ( np ) = [tex]\frac{380*20.1}{0.277*7500*5728.5}[/tex] = 0.978
note : values above are derived values whose solution are not basically part of the required solution hence they are not included
b) Determine the overload factor of safety
[tex]n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}[/tex] ------- ( 2 )
Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N
input values into equation 2 above
hence : [tex]n_{L}[/tex] = 0.9191[tex]n_{L} = 0.9191[/tex]
C) Determine the factor of safety based on joint separation
[tex]n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }[/tex]
Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,
input values into equation above
Hence [tex]n_{0}[/tex] = 1.056
D) Determine the fatigue factor of safety using the Goodman criterion.
nf = 0.849
attached below is the detailed solution .
A benefit to using the medium the author used in "Great Rock and Roll
Pauses" is that the audience:
Incomplete question. The options read;
A. can change the story's ending
B. listens to the dialogue
C. hears the rock songs
D. feels more connected to the text.
Answer:
D. feels more connected to the text.
Explanation:
Remember, the underlying motive of an author when writing any text is to make his readers or audience more connected to the material been read or heard.
Hence, in "Great Rock and Roll Pauses" we can conclude that the author's choice of medium was motivated by the same desire of making the audience feel more connected to the text.
Answer:
The audience can actually hear the music.
Explanation: dont listen to below average iq people :0
A simple Rankine cycle uses water as the working fluid. The boiler operates at 6000 kPa and the condenser at 50 kPa. At the entrance of the turbine the temperature is 450 deg C. The isentropic efficiency of the turbine is 94 percent, pressure and pump losses are negligible, and the water in the condenser is subcooled by 6.3 degC. The boiler is sized for a mass flow rate of 20 kg/s. Determine the rate at which heat is added in the boiler, the power required to operate the pumps, the net power produced by the cycle, and the thermal efficiency.
Answer:
the rate at which heat is added in the boiler = 59597.4 kW
the power required to operate the pumps = 122.57 kW
The net power produced by the cycle = 17925 kW.
The thermal efficiency = 30%.
Explanation:
The specific enthalpy of saturated liquid is equal to the enthalpy of the first point which is equal to 314 kJ/ kg.
The second enthalpy is calculated from the pump work. Therefore, the second enthalpy = first enthalpy point + specific volume of water [ the pressure of the boiler - the pressure of the condenser].
The second enthalpy = 314 + 0.00103 [ 6000 - 50 ] = 320.13 kJ/kg.
The specific enthalpy for the third point = 3300 kJ/kg.
Therefore, the rate at which heat is added in the boiler = 20 × [3300 - 320.13] = 59597.4 kW.
The rate at which heat is added in the boiler = 59597.4 kW.
Also, the power required to operate the pumps = 20 × 0.00103 [6000 - 50] = 122.57 kW.
The power produced by the turbine = 20 [ 300 - ( the fourth enthalpy value)].
The fourth enthalpy value = 3300 - 0.94 [ 3300 - 2340] = 2397.6 kJ/kg
Thus, the power produced by the turbine = 20 [ 300 - 2397.6] = 18048 kW.
The power produced by the turbine = 18048 kW.
The net power produced = 18048 + 122.57 = 17925 kW.
The thermal efficiency = [net power produced] / [the rate at which heat is added in the boiler].
The thermal efficiency = 17925/ 59597.4 = 30%.
A landowner and a contractor entered into a written contract under which the contractor agreed to build a building and pave an adjacent sidewalk for the landowner for $200,000. Later, while construction was proceeding, the landowner and the contractor entered into an oral modification under which the contractor was not obligated to pave the sidewalk but still would be entitled to $200,000 upon completion of the building. The contractor completed the building. The landowner, after discussions with his landscaper, demanded that the contractor pave the adjacent sidewalk. The contractor refused.
Has the contractor breached the contract?
A reservoir rock system located between a depth of 2153m and a depth of
2383m , as the pressure at these depths is 18.200 MPa , 19.643 MPa
respectively the thickness of oil zone 103m, if the density of water is 1060 kg/m3
Determine the oil and gas density. what is the pressure at the depth of 2200m ?
what is the depth at which the pressure is 1900 MPa? Determine the gas-oil and
oil- water contact depth.
A car with tires pressurized to 270 kPa leaves
Los Angeles with the tire temperature at 30°C
Estimate the tire pressure
(gage) when the
car arrives in New York with a tire temperature of 65°C .
I think Charles law should work here
Solved this question??????????????????
I do not know!!!!!!!!
In order to properly construct this road you have determined that you need to bring in 135,000 CCY of fill material to construct the sub-base. The fill material that is available is dry earth, with a weight of 2800 lbs/BCY and a moisture content of 7%. The percent swell for this soil is 25%. The final compaction level will be 118 PCF dry density and the optimum moisture content is 6%. How many BCY of earth must be excavated to meet this requirement
Answer:
The volume of water needed to be removed = 4301100 lbs/ [ 62.43 × 27] = 2551.66.
Explanation:
The first thing to do right now is to determine the weight of soil solids. This can be gotten below as:
The weight of soil solids = 1 × 118 × 27 = 3186 lbs.
Also, there is the need to determine the value of the water that we have and the one that is needed. This is calculated below as:
The weight of water needed = 3186 × 6% = 191.16 lbs.
The water available = 3186 × 7% = 223.02%.
The next thing to determine is the value for the total BCY of available of soil.
Thus, the total BCY of available of soil = [ 3186 + 223.02 ]/ 2800 × 135000 = 3409.02/ 2800 = 1.217551 BCY × 135000 = 164,363.46 BCY.
The last thing to do here is to determine the value for the weight and volume of water to be removed.
The weight of water to be removed = [ 223.02 lbs - 191.16 lbs] × 135,000 = 4301100 lbs
The volume of water needed to be removed = 4301100 lbs/ [ 62.43 × 27] = 2551.66.
3. WHAT IS THE DIFFERENCE BETWEEN LAYOUT AND MEASUREMENT?
difference between theory and practice?
Answer:
There is a huge difference between theory vs. practice. Theory assumes an outcome, while practice allows you to test the theory and see if it is accurate.
Theory and Practice Explained
Practice is the observation of disparate concepts (or a phenomenon) that needs explanation. A theory is a proposed explanation of the relationship between two or more concepts, or an explanation for how/why a phenomenon occurs.
HELPP!!! Calculating voltage drop
Answer:
srry
Explanation:
While recharging a refrigerant system, the charging stops before the required amount of refrigerant has been inserted. What should a technician do at this time?
A) start the engine and charge through the high side
B) open the low side valve to allow charging through both sides
c) start the engine and charge through the low side
d) reevacuate and recharge the system again
Answer:
Answer C
Explanation:
That is the correct way.
A wastewater treatment plant discharges 2.0 m^3/s of effluent having an ultimate BOD of 40.0 mg/L into a stream flowing at 15.0 m^3/s. Just upstream from the discharge point, the stream has an ultimate BOD of 3.5 mg/L. The deoxygenation constant kd is estimated at 0.22/day.
a. Assuming complete and instantaneous mixing, find the ultimate BOD of the mixture of wastewater and river just downstream from the outfall.
b. Assuming a constant cross sectional area for the stream equal to 55 m^2 , what ultimate BOD would you expect to find at a point 10,000 m downstream?
Answer:
What grade is this is for??
At steady state, the power input of a refrigeration cycle is 500 kW. The cycle operates between hot and cold reservoirs which are at 550 K and 300 K, respectively. a) If cycle's coefficient of performance is 1.6, determine the rate of energy removed from the cold reservoir, in kW. b) Determine the minimum theoretical power required, in kW, for any such cycle operating between 550 K and 300 K
Answer:
The answer is below
Explanation:
Given that:
Hot reservoir temperature ([tex]T_H[/tex]) = 550 K, Cold reservoir temperature ([tex]T_C[/tex]) = 300 K, power input ([tex]W_{cycle}=500 \ kW[/tex]), cycle's coefficient of performance([tex]\beta_{actual}[/tex]) = 1.6
a) The rate of energy removal in the cold reservoir ([tex]Q_C[/tex]) is given by the formula:
[tex]Q_C=\beta_{actual}* W_{cycle}\\\\Q_C=1.6*500\\\\Q_C=800\ kW[/tex]
b) The maximum cycle's coefficient of performance([tex]\beta_{max}[/tex]) is:
[tex]\beta_{max}=\frac{T_C}{T_H-T_C}=\frac{300}{550-300}=1.5\\\\For\ minimum\ theoretical\ power\ \beta_{max}=\beta_{actual}=1.5\\\\W_{cycle}=\frac{Q_C}{\beta_{actual}} =\frac{800}{1.5} \\\\W_{cycle}=533.3\ kW[/tex]
Question # 3
Multiple Choice
Which events significant to the United States transportation industry occurred in the 1970s and 1980s?
A.The FCC was created.
B.Laws to begin regulating pipeline transportation were enacted.
C.Many transportation industries were deregulated.
D.The regulation of transportation industries began.
Answer:
c Many transportation industries were deregulated.
Explanation:
correct on edge 2022