Answer:
The building will cost $$137.34 per square foot
Explanation:
Let capacity of building funded in Chicago in 1998 be Ca
Cb = Capacity of building library = Ca + 0.4Ca = 1.4Ca
e = Capacity factor = 0.8
($b/$a) = (Cb/Ca)^e = 1(.4Ca/Ca)^0.8 = 1.308
$B = $a * 1.308
$B = $105 * 1.308
$B = $137.34
The building will cost $$137.34 per square foot. Note: Inflation rate is not applied as it was not specified in the question.
A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50 mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?
Answer:Decay rate constant,k = 0.00376/hr
Explanation:
IsT Order Rate of reaction is given as
In At/ Ao = -Kt
where [A]t is the final concentration at time t and [A]o is the inital concentration at time 0, and k is the first-order rate constant.
Initial concentration = 80 mg/L
Final concentration = 50 mg/L
Velocity = 40 m/hr
Distance= 5000 m
Time taken = Distance / Time
5000m / 40m/hr = 125 hr
In At/ Ao = -Kt
In 50/80 = -Kt
-0.47 = -kt
- K= -0.47 / 125
k = 0.00376
Decay rate constant,k = 0.00376/hr
If the specific surface energy for soda–lime glass is 0.30 J/m2 and its modulus of elasticity is (69 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.
Answer: the critical stress for the sodalime glass = 5.7MPa
Explanation:
Ist Step
we Calculate half length of internal crack as
2a =0.8 mm
a = 0.8/2 = 0.4 mm
Changing to meters becomes = 0.4 / 1000 =0.0004m
2nd Step
Now the critical stress required for the propagation of the internal crack can be calculated using the formulae
Critical Stress (σc) = (2 E γs/ πa) 1/2
where E= modulus of elasticity
γs= specific surface energy for soda–lime glass
a= Length
= (2 x 69 x 10 ^9 x 0.30/ π x 0.0004)1/2
=[tex]\sqrt{ 32,940,802,036,919}[/tex]
= 5,739,407.8= 5.7 x 10^6 N/m^2
= 5.7MPa
My computer has a mass of 0.031080997078386 slug the Earth's surface.
a. What is its mass in pounds mass (lbm) on Mars where the acceleration of gravity is 5.35 ft/sec^2?
b. What is its weight in pounds force (lbf) on the Mars surface where the acceleration of gravity is 5.35 ft/sec^2 ?
Answer:
The answer is "0.187 lbm and 1 lbf".
Explanation:
The mass = [tex]0.031080997078386\ slug[/tex]
Calculating mass on Mars:
[tex]\to m=m_g\frac{g}{g_e}[/tex]
[tex]=0.031080997078386 \times \frac{32.2}{5.35}\\\\=0.187 \ lbm[/tex]
[tex]\to W=mg_e[/tex]
[tex]=0.187 \times 5.35\\\\=1 \ lbf[/tex]
At 800K, a plot of ln[cyclobutane] vs t gives a straight line with a slope of -1.6 s-1. Calculate the time needed for the concentration of cyclobutane to fall to 1/16 of its initial value.
Answer:
hmmm.........
Explanation:
What pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water?
According to the scenario, the pounds per square inch required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water is found to be approximately 2.0 Psig.
What do you mean by the Bubbler system?The bubbler system may be defined as a type of system that significantly measures water level based on the amount of pressure it takes to push an air bubble out of an orifice line and into the water body. This pressure often referred to as the “line pressure”, requires changes in the elevation of the water.
According to the context of this question,
The depth of bubbles produced by the bubbler system = 4 ft 7 inches.
The pounds per square inch = 2.31 Psig.
∴ The pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water = 4 ft 7 inches/2.31 = 2.03 Psig ≅ 2Psig.
Therefore, according to the scenario, the pounds per square inch required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water is found to be approximately 2.0 Psig.
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A settling tank has an influent rate of 0.6 mgd. It is 12 ft deep and has a surface area of 8000 ft². What is the hydraulic retention time?
Answer: hydraulic retention time,τ=28.67 hours
Explanation:
The hydraulic retention time τ (tau), is given as The volume of the settling tank(V) divided by the influent flowrate(Q)
τ =V/Q
But Volume is not known and is given as
Volume = surface area x depth of the tank
= 8000 ft² X 12 ft
= 96,000 ft³
Also, the influent flow rate is in mgd ( million gallons per day), we change it to ft³/sec so as to be in same unit with the volume in ft³
1 million gallons/day = 1.5472286365101 cubic feet/second
0.6mgd = 1.5472286365101 cubic feet/second x 0.6
=0.93cubic feet/second
τ =V/Q
96,000 ft³/0.93 ft³/sec
τ=103,225.8 secs
changing to hours
103,225.8 /3600 =28.67 hours
The hydraulic retention time =28.67 hours
Consider a circuit element, with terminals a and b, that has vab= -12V and iab= 3A. Over a period of 2 seconds, how much charge moves through the element? If electrons carry the charge, which terminal do they enter? How much energy is transferred? Is it delivered to the element or taken from it?
Answer:
a) 6 coulombs
b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current
c) = -72 joules
Energy is taken from element
Explanation:
Given data:
V ab = -12 v
I ab = 3A
period ( t ) = 2 seconds
a) determine how much charge moves through the element
q = I * t
= 3 * 2 = 6 coulombs
b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current
c) determine how much energy is transferred
= Vab * Iab * t
= -12 * 3 * 2
= -72 joules
Energy is taken from element
A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as shown. Determine the velocity v of the motorcycle when s=200 m. At this point, also determine the value of the derivative dv/ds.
Answer:
Follows are the solution to this question:
Explanation:
Calculating the area under the curve:
A = as
[tex]=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}[/tex]
Calculating the kinematics equation:
[tex]\to v^2 = v^2_{o} + 2as\\\\[/tex]
[tex]=0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}[/tex]
Calculating the value of acceleration:
[tex]\to a= \frac{dv}{dt}[/tex]
[tex]=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}[/tex]
[tex]\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\[/tex]
[tex]=\frac{0.092}{s}[/tex]
A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and the products are cooled to 25.00 degree C. Suppose the reactor is continuous. Take a basis of calculation of 1.000 mol/s of the fuel gas, assume some value for the percent excess oxygen fed to the reactor (the value you choose will not affect the results), and calculate - Q(kW), the rate at which heat must be transferred from the reactor if the water vapor condenses before leaving the reactor and if the water remains as a vapor. Now suppose the combustion takes place in a constant-volume batch reactor. Take a basis of calculation 1.000 mol of the fuel gas charged into the reactor, assume any percent excess oxygen, and calculate -Q(kJ) for the cases of liquid water and water vapor as products.
Answer:
A)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
Explanation:
Given data :
45.00 % mole of methane
55.00 % of ethane
attached below is a detailed solution
A) calculate - Q(kw)
- Q ( kw ) for vapor = -1258.05 kw
- Q ( kw ) for liquid = -1146.3 kw
B ) calculate - Q ( KJ )
- Q ( kj ) for vapor = -1258.05 kJ
- Q ( KJ ) for liquid = - 1146.3 KJ
since combustion takes place in a constant-volume batch reactor
A MBR treatment plant has a flow of 0.3 mgd with 220 mg/l BOD (the soluble portion is 120 mg/l) and 230 mg/l suspended solids (184 mg/l is volatile). Effluent soluble BOD is 2 mg/l. The design calls for 5000 mg/l MLSS and sludge age of 20 days with a Y = 0.8, kd = 0.4. Calculate the aeration basin volume, detention time, BOD loading, ratio, and waste solids, both VSS and TSS.
Answer:
attached below is the detailed solution
A) 8288.77 cu.ft
B) 4.96 hours
C) Vss = 131.21 IbVss/day
Tss = 164 IbTss/day
D) attached below
E ) 0.2
F) 287.23 Ib/day
Explanation:
A) Determine the aeration basin volume
Given
∅c = 20 days
Y = 0.8Ib VSS/Ib BOD
Q = 0.3 mgd
So = 120 mg/l
Se = 2 mg/l
X = 5000 mg/l
Kd = 0.04 per day
attached below is the detailed solution
B) Determine the detention time using this relation
t = ( V / Q )* 24
= ( 0.062 / 0.3 ) * 24 = 4.96 hours
C ) Determine Vss and Tss
we first calculate the excess biomass Px then assuming Vss ratio to be 80% for sludge age of 20 days
Vss = 131.21 IbVss/day
Tss = 164 IbTss/day
D ) determine BOD loading
Q = 0.3 mgd , BOD = 220mg/l , V = 8288.77 cu.ft
solution attached below
e) food to microorganism ratio
F/M = 0.2
solution attached below
f) determine the waste solid
waste solid = Q * SS * % removal of suspended solids
where : Q = 0.3 , SS = 220mgl , % = 50 %
waste solids = 0.3 * 230 * 0.5 * 8.34 = 287.23 Ib/day
An electric circuit is made up of a 100 m long manganin wire with a section of I mm^2; this wire constitutes 4/5 of the total resistance of the circuit itself and the intensity of the current circulating there is 2.5 A. Calculate the voltage applied to the terminals of the manganin wire, the energy dissipated on this wire in 30 minutes and the voltage applied by the generator across the circuit.
Answer:
a.dont know e
Explanation:
because d q tlga ammu
the differences between building technology vs architectural technology
4. The instant the ignition switch is turned to the start position,
A. The starter motor starts to rotate before energizing the starter p
O B. Only the pull-in winding is energized.
C. Only the hold-in winding is energized.
D. Both pull-in and hold-in windings are energized.
Answer:
D. Both pull-in and hold-in windings are energized.
Explanation:
The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.
The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.
What is the best way to collaborate with your team when publishing Instagram Stories from Hootsuite?
A generator has a voltage constant, KE, of 0.01 volts per rpm. Find the voltage when it is driven at 2400 rpm
a. 60 V
b. 24 V
c. 72 V
d. 54 V
Answer:
Total voltage = 24 V
Explanation:
Given:
Volts per rpm = 0.01
Total rpm = 2400
Find:
Total voltage
Computation:
Total voltage = Volts per rpm x Total rpm
Total voltage = 0.01 x 2400
Total voltage = 24 V
What is the amount of pearlite formed during the equilibrium cooling of a 1055 steel from 1000°C to room temperature?
Answer: 98.5% of pearlite was formed during the equilibrium cooling
Explanation:
First we calculate the fraction of pro-eutectoid phase which forms for equilibrium cooling of the 1085 steel from 1000°C at room temperature;
we know that in 1085 steel, last two digits denotes the carbon percentage
so 1085 steel contains 0.85% carbon.
Now from the diagram, carbon percentage is greater than the eutectoid com[psition
i.e 0.85 > 0.76
it is a hyper eutectoid steel
so
fraction of pro eutectoid phase W_Fe₃C = (0.85 - 0.76) / ( 6.7 - 0.76)
= 0.09 / 5.94 = 0.015 = 1.5%
Now, the amount of pearlite formed during the equilibrium cooling of the 1055 steel from 1000°C to room temperature will be;
pearlite (C') = (1 - W_Fe₃C)
= 1 - 0.015
= 0.985 = 98.5%
Therefore 98.5% of pearlite was formed during the equilibrium cooling
. Chemical manufacturers must present which Information on the product's label?
A) Product identifier
B) O Contact Information for the manufacturer
C) O Hazard pictograms
D) All of the above
Chemical manufacturers must present Product Identifier, Contact Information for the manufacturer and Hazard pictograms on the product's label.
What is a product label?A product label means a display of written, printed or graphic material that is printed on or affixed to a product or its immediate container.
Let's consider which of the following information must be presented on the product's label by chemical manufacturers.
A) Product identifier. Yes. The product identifier can also be found in Material Safety Data Sheet.B) Contact Information for the manufacturer. Yes. It should provide a phone number or mail to contact the manufacturer.C) Hazard pictograms. Yes. Hazard pictograms form part of the international Globally Harmonized System of Classification and Labelling of Chemicals and alert us to the presence of a hazardous chemical.D) All of the above. Yes.Chemical manufacturers must present Product Identifier, Contact Information for the manufacturer and Hazard pictograms on the product's label.
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In a CS amplifier, the resistance of the signal source Rsig = 100 kQ, amplifier input resistance (which is due to the biasing network) Rin = 100kQ, Cgs = 1 pF, Cgd = 0.2 pF, gm = 5 mA/V, ro = 25 kΩ, and RL = 20 kΩ. Determine the expected 3-dB cutoff frequency.
Answer:
406.140 KHz
Explanation:
Given data:
Rsig = 100 kΩ
Rin = 100kΩ
Cgs = 1 pF,
Cgd = 0.2 pF, and etc.
Determine the expected 3-dB cutoff frequency
first find the CM miller capacitance
CM = ( 1 + gm*ro || RL )( Cgd )
= ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )
= ( 11.311 ) pF
now we apply open time constant method to determine the cutoff frequency
Th = 1 / Fh
hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]
= [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] = 406.140 KHz
You are the project manager assigned to construct a new 10-story office building. You are trying to estimate the costs for this project. You start by assigning the costs associated with each of the project activity. Then you sum up all the individual costs into a final cost estimate. Which type of cost estimation technique did you use?
Answer:
Bottom-up Estimation
Explanation:
Bottom-up estimation is a type of project cost estimation that considers the cost of individual project activities and finally sums them up or finds the aggregates. The summation gives an idea of what the entire project will cost.
This is an effective way of estimating the cost of a project as it evaluates the costs on a wholistic basis. It also considers the tiniest details during the estimation process. The process moves from the simpler details to the more complicated details.
Rear defrosters generally have a relay with a timer. This allows ___.
Anyone help me please ?
Answer:
I can help but I need to know what it looking for
An oxygen–nitrogen mixture consists of 35 kg of oxygen and 40 kg of nitrogen. This mixture is cooled to 84 K at 0.1 MPa pressure. Determine the mass of the oxygen in the liquid and gaseous phase.
Answer:
The mass of oxygen in liquid phase = 14.703 kg
The mass of oxygen in the vapor phase = 20.302 kg
Explanation:
Given that:
The mass of the oxygen [tex]m_{O_2}[/tex] = 35 kg
The mass of the nitrogen [tex]m_{N_2}[/tex] = 40 kg
The cooling temperature of the mixture T = 84 K
The cooling pressure of the mixture P = 0.1 MPa
From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.
i.e.
Liquid percentage of [tex]O_2[/tex] = 70% = 0.70
Vapor percentage of [tex]O_2[/tex] = 34% = 0.34
The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively
Thus, the number of moles of each component is:
number of moles of oxygen = 35/32
number of moles of oxygen = 1.0938 kmol
number of moles of nitrogen = 40/28
number of moles of nitrogen = 1.4286 kmol
Hence, the total no. of moles in the mixture is:
[tex]N_{total} = 1.0938+1.4286[/tex]
[tex]N_{total} = 2.5224 \ kmol[/tex]
So, the total no of moles in the whole system is:
[tex]N_f + N_g = 2.5224 --- (1)[/tex]
The total number of moles for oxygen in the system is
[tex]0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)[/tex]
From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)
∴
0.7(2.5224 - N_g) + 0.34 N_g = 1.0938
1.76568 - 0.7 N_g + 0.34 N_g = 1.0938
1.76568 - 0.36 N_g = 1.0938
1.76568 - 1.0938 = 0.36 N_g
0.67188 = 0.36 N_g
N_g = 0.67188/0.36
N_g = 1.866
From equation (1)
[tex]N_f + N_g = 2.5224[/tex]
N_f + 1.866 = 2.5224
N_f = 2.5224 - 1.866
N_f = 0.6564
Thus, the mass of oxygen in the liquid and vapor phases is:
[tex]m_{fO_2} = 0.7 \times 0.6564 \times 32[/tex]
[tex]m_{fO_2} = 14.703 \ kg[/tex]
The mass of oxygen in liquid phase = 14.703 kg
[tex]m_{g_O_2} = 0.34 \times 1.866 \times 32[/tex]
[tex]m_{g_O_2} = 20.302 \ kg[/tex]
The mass of oxygen in the vapor phase = 20.302 kg
A laboratory furnace wall is constructed of 0.2 m thick fireclay brick having a thermal conductivity of 1.82 W/m-K. The wall is covered on the outer surface with insulation of thermal conductivity of 0.095 W/m-K. The furnace inner brick surface is at 950 K and the outer surface of the insulation material is at 300 K. The maximum allowable heat transfer rate through the wall of the furnace is 830 W/m^2. Determine how thick in cm the insulation material must be.
Answer:
The appropriate solution will be "6.4 cm".
Explanation:
The given values are:
Length,
l = 0.2 m
Thermal conductivity,
K₁ = 1.82 W/m-K
K₂ = 0.095 W/m-K
Temperature,
T = 950 K
T = 300 K
Heat transfer rate,
Q = 830 W/m²
Now,
⇒ [tex]Q = \frac{\Delta T}{\frac{L_1}{K_1 A} +\frac{L_2}{K_2 A} }=\frac{A \Delta T}{\frac{L_1}{K_1 } +\frac{L_2}{K_2 } }[/tex]
⇒ [tex]\frac{Q}{A} =\frac{\Delta T}{\frac{L_1}{K_1} +\frac{L_2}{K_2} }[/tex]
On substituting the above given values in the equation, we get
⇒ [tex]830=\frac{(980-300)}{\frac{0.2}{1.82} +\frac{x}{0.095} }[/tex]
On applying cross-multiplication, we get
⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{950-300}{830}[/tex]
⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{650}{830}[/tex]
⇒ [tex]x =0.639 \ m[/tex]
⇒ [tex]x=6.345 \ i.e., 6.4 \ m[/tex]
draw afd,sfd and bmd of frame
Answer:
uh, i dont understand?
Explanation:
) A Car is moving with a non-uniform velocity towards East.
Its velocity changes at different time intervals. Calculate the instantaneous
velocity at time 3 sec. The distance is given by equation 2t2 – 4t
Answer:
Instantaneous velocity = 8m/s
Explanation:
Given the following data;
Distance = 2t² - 4t
Time, t = 3 secs.
To find the instantaneous velocity;
[tex] Velocity = \frac {distance}{time} [/tex]
[tex] V(t) = \frac {dd}{dt} [/tex]
We would differentiate the equation for the distance with respect to time, t.
[tex] \frac {dd}{dt} = \frac {d(2t^{2} - 4t)}{dt}[/tex]
[tex] \frac {dd}{dt} = 4t - 4[/tex]
Substituting the value of "t" into the above equation, we have;
[tex] V(3) = 4(3) - 4[/tex]
[tex] V(3) = 12 - 4[/tex]
Instantaneous velocity = 8m/s
A sinusoidal voltage source has a peak voltage of 12 V and a frequency of 50 Hz. What is the voltage at 10 ms?
Answer:
0
Explanation:
Given that:
The peak of the voltage [tex]V_{peak}[/tex] = 12 V
The frequency f = 50 Hz
At the time t = 10 ms
[tex]V = V_p \ sin \ \omega t[/tex]
where;
[tex]\omega = 2 \pi f[/tex]
[tex]V = 1 2\ V \times sin \ ( 2 \pi \times 50 \times 10 \ )[/tex]
V = 12V sin (180)
V = 12 V × 0
V = 0
Estimating is important in construction industry because
Select one:
a. Contractors need to know the amount of mark-up
b. Projects are awarded on serious competition
c. Construction industry has tendency for complex constructions
d. The regulations of government for construction industry changes
Your answer is correct,
Answer:
a( contractors need to know the amount of markup)
because the contractor should have a long term vision for a proper satisfaction to the people.
It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Construction Carpenters? Check all that apply. Spend time sitting wear common protective or safety equipment face-to-face discussions exposed to hazardous equipment spend time using hands to handle, control, or feel objects, tools, or controls public speaking
Answer:
bcde!!
Explanation:
They also tend to be traditional, which means that they enjoy working in structured environments and are typically organized and detail-oriented. Thus option B,C,D, E is correct.
What are the characteristics of Construction Carpenters?Carpenters continue to have a bright future in their profession. According to Job Outlook, the carpentry industry is expanding quickly. The advantages of carpentry lead to employment security and a long-term career with this kind of industrial growth.
Carpentry is a physically demanding line of work that calls for endurance. You frequently spend the most of your shift standing, moving slowly, and crouching.
Therefore, As well as using hand tools to shape and cut wood, lifting big objects, moving heavy beams, furniture, or equipment are all possible.
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Technician A says vehicles with electronic throttle control do not need a separate cruise control module, stepper motor, or cable to control engine speed. Technician B says a faulty brake light switch may cause the cruise control to not operate. Who is correct?
Answer: its A
Explanation:
Air flow is measured in a Venturi meter that has a large diameter of 1.5 m and a small diameter of 0.9 m. A 1:12 scale model with water as the fluid is used to calibrate the meter. For the model, it is determined that when the volume flowrate is 0.07 m3 /s, the pressure drop from the large diameter portion (Section 1) to the small diameter portion (Section 2) is 172 kPa. Calculate:
This question is incomplete, the complete question is;
Air flow is measured in a Venturi meter that has a large diameter of 1.5 m and a small diameter of 0.9 m. A 1:12 scale model with water as the fluid is used to calibrate the meter. For the model, it is determined that when the volume flowrate is 0.07 m3 /s, the pressure drop from the large diameter portion (Section 1) to the small diameter portion (Section 2) is 172 kPa.
Calculate The corresponding flowrate in the prototype.
Assume a water temperature of 15°C and standard properties of air
Answer: The corresponding flowrate in the prototype is 10.21 m³/s
Explanation:
Given that;
Lm = Lp/12 and lp = 12Lm, Qm = 0.07 m³/s, ΔPm = 172 Kpa
properties of water at 15°C ---- Vm = 1.2015 × 10⁻⁶ m²/s, Sm = 1001.2 kg/m₂
Also for Air---- Vp = 1.46041 × 10⁻⁵, Sp = 1.225 kg/m³
Now by Using Reynold's model law; (Vm × Lm)/Vm = (Vp × Lp)/Vp
(Vm × Lm) / 1.2015 × 10⁻⁶ = (Vp ×12 × Lm) / 1.46041 × 10⁻⁵
Vm/Vp = 0.9872
we know that
Discharge = Area × Velocity
Qm/Qp = Lm²/Lpl × Vm/Vp
= (1/12)² × 0.9872
= 6.856 × 10⁻³
so Qp = (0.07 / 6.856 × 10⁻³) = 10.21 m³/s
Therefore The corresponding flowrate in the prototype is 10.21 m³/s