Answer:
where is the picture?
Explanation:
Three-fourths of the illuminated side of the moon be visible from Earth.
What are the phases of the moon?There are 12 phases of the moon which are visible and different from each other.
These are new moon, waxing crescent, first quarter, waxing gibbous, full moon, waning gibbous, last quarter, waning crescent.
Thus, three-fourths of the illuminated side of the moon be visible from Earth.
Learn more about phases of the moon
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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
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
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.
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]
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 .
Solved this question??????????????????
I do not know!!!!!!!!
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
A garden hose attached with a nozzle is used to fill a 20-gallon bucket. The inner diameter of the hose is 1 inch and it reduces to 0.5 inch at the nozzle exit. If the average velocity in the hose is 8ft/s, determine
a) the volume and mass flow rates of water through the hose.
b) how long it will take to fill the bucket with water.
c) the average velocity of water at the nozzle exit.
Answer:
a). [tex]$0.0436 \ ft^3/s$[/tex] , [tex]$2.72 \ lb \ m/s$[/tex]
b). [tex]$61.32 \ s$[/tex]
c). 32. ft/s
Explanation:
a). The volume flow rate of the water is given by :
[tex]$\dot V = uA$[/tex]
[tex]$=u \pi \left( \frac{d}{2}\right)^2$[/tex]
[tex]$=\frac{u \pi d^2}{4}$[/tex]
[tex]$=\frac{8\ ft/s \ \pi \left(\frac{1}{12}\right)^2}{4}$[/tex]
[tex]$= 0.0436 \ ft^3/s$[/tex]
The mass flow rate of the water is given by :
[tex]$\dot m = \rho \dot V$[/tex]
[tex]$= 62.4 \times 0.0436$[/tex]
[tex]$=2.72 \ lb \ m/s$[/tex]
b). The time taken to fill the container is
[tex]$\Delta t = \frac{V}{\dot V}$[/tex]
[tex]$=\frac{20 \ gal}{0.0436 \ ft^3/s}\left( \frac{1 \ ft^3}{7.4804 \ gal}\right)$[/tex]
[tex]$=61.32 \ s$[/tex]
c). The average velocity at the nozzle is :
[tex]$u=\frac{\dot V}{A}$[/tex]
[tex]$=\frac{\dot V}{\frac{\pi d^2}{4}}$[/tex]
[tex]$=\frac{0.0436}{\frac{\pi \left(\frac{0.5}{12}\right)^2}{4}}$[/tex]
= 32. ft/s
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
At steady state, a well-insulated compressor takes in air at 60F, 14.2 lbf/in.2 , with a volumetric flow rate of 1200 ft3 /min, and compresses it to 500F, 120 lbf/in.2 Kinetic and potential energy changes from inlet to exit can be neglected. Determine the compressor power, in hp, and the volumetric flow rate at the exit, in ft3 /min.
Answer:
the compressor power is -223.12 hP { -ve indicate work done }
Volumetric flow rate at exit = 262.74 ft³/s
Explanation:
Given the data in the question;
p₁ = 14.2 psi = 0.978 bar
p₂ = 120 psi = 8.268 bar
T₁ = 60°F = 288.706 K
T₂ = 500°F = 533.15 K
Q₁ = 1200 ft³/min = 20ft³/s = 0.566 m³/s
δ₁ = p₁/RT₁ = 0.978 × 10⁵ / ( 287 × 288.706) = 1.18 kg/m³
δ₂ = P₂/RT₂ = 8.268 × 10⁵ / ( 287 × 533.15 ) = 5.4 kg/m³
so
ω₁ = δ₁Q₁ = 1.18 × 0.566 = 0.668 kg/s
we know that; ω₁ = ω₂ { in a steady flow }
ω₂ = δ₂Q₂
Q₂ = ω₂/δ₂
Q₂ = 0.668 / 5.4 = 0.1237 m³/s
Hence Volumetric flow rate at exit = 262.74 ft³/s
from the steady state energy equation;
ω( h₁ - h₂ ) = dW/dt
{ where h = CpT)
( Cp = 1.005 )
dW/dt = ωCp( T₁ - T₂ )
we substitute
dW/dt = 0.668 × 1.005( 288.706 - 533.15 )
dW/dt = 0.67134 × -244.444
dW/dt = -164.105 kW = -223.12 hP
Hence, the compressor power is -223.12 hP { -ve indicate work done }
The entire population of a given community is examined, and all who are judged to be free from bowel cancer are questioned extensively about their diets. These people then are followed for several years to see whether their eating habits will predict their risk of developing bowel cancer. Which of the following study designs most appropriately characterizes this situation?
A. Cross-sectional study.
B. Case-control study.
C. Prospective cohort study.
D. Historical prospective cohort study.
E. Clinical trial.
F. Community trial.
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.
What is the real name of the town that Bud & Bugs stayed in while they were waiting for the train?
Group of answer choices
Hooverville
Chicago
Hooperville
Memphis
Answer: Hooverville
Explanation:
Bud, not Buddy is a book about a ten years old boy who ran away from home where he stayed with his foster parents. He was treated unfairly and beaten.
He left home to seek a better living and hope that he will find his father as well. He met another orphan named Bugs and they decided to go to Hooverville so that they can get a train going to California.
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 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 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:
11. Which of these is NOT true when dealing with refrigerants?
OA. An increase in temperature will cause an increase in pressure
OB. An increase in pressure will cause an increase in the boiling point
OC. An increase in pressure will cause an decrease in the boiling point
OD. An increase in pressure will cause an increase in temperature
(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.
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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.
Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.05 bar. The isentropic turbine efficiency is 90%. Assume ideal gas behavior for the air and ignore kinetic and potential energy effects. Determine the temperature at the exit, in K, and the work developed, in kJ per kg of air flowing.
Solution :
The isentropic efficiency of the turbine is given as :
[tex]$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$[/tex]
[tex]$=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$[/tex]
[tex]$=\frac{h_1-h_2}{h_1-h_{2s}}$[/tex]
The entropy relation for the isentropic process is given by :
[tex]$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$[/tex]
[tex]$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$[/tex]
[tex]$ \frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$[/tex]
[tex]$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$[/tex]
Now obtaining the properties from the ideal gas properties of air table :
At [tex]$T_1 = 1600 \ K,$[/tex]
[tex]$P_{r1}=791.2$[/tex]
[tex]$h_1=1757.57 \ kJ/kg$[/tex]
Calculating the relative pressure at state 2s :
[tex]$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$[/tex]
[tex]$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$[/tex]
[tex]$P_{r2}=63.296$[/tex]
Obtaining the properties from Ideal gas properties of air table :
At [tex]$P_{r2}=63.296$[/tex], [tex]$T_{2s}\approx 860 \ K$[/tex]
Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:
[tex]$\eta=\frac{h_1-h_2}{h_1-h_{2s}}$[/tex]
[tex]$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$[/tex]
[tex]$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$[/tex]
[tex]$0.9=\frac{1600-T_2}{1600-860}$[/tex]
[tex]$T_2= 938 \ K$[/tex]
So, at [tex]$T_2= 938 \ K$[/tex], [tex]$h_2=975.66 \ kJ/kg$[/tex]
Now calculating the work developed per kg of air is :
[tex]$w=h_1-h_2$[/tex]
= 1757.57 - 975.66
= 781 kJ/kg
Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.
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.)
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10 and its isentropic efficiency is 85%. At the inlet to the turbine, the pressure is 950 kPa, and the temperature is 1400 K. The turbine has an isentropic efficiency of 88% and the exit pressure is 100 kPa. On the basis of an air-standard analysis,
a. develop a full accounting of the net exergy increase of the air passing through the gas turbine combustor, in kW.
b. devise and evaluate an exergetic efficiency for the gas turbine cycle.
Answer:
a) 3581.15067 kw
b) 95.4%
Explanation:
Given data:
compressor efficiency = 85%
compressor pressure ratio = 10
Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K
At turbine inlet : pressure = 950 kPa, temperature = 1400k
Turbine efficiency = 88% , exit pressure of turbine = 100 kPa
A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW
attached below is a detailed solution to the given question
How can you throw a ball as hard as you can and have it come back to you, even if it doesn't
bounce off anything? There is nothing attached to it, and no one else catches or throws it back to
you.
Answer:
Magic
Explanation:
magic is the answer to everything unexplainable
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?
3. WHAT IS THE DIFFERENCE BETWEEN LAYOUT AND MEASUREMENT?
HELPP!!! Calculating voltage drop
Answer:
srry
Explanation:
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 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??