List five areas that increased energy prices impact.
Answer:
Supply, demand, global markets, imports and exports, and government Regulation.
Explanation:
When determining risk, it is necessary to estimate all routes of exposure in order to determine a total dose (or CDI). Recognizing this, estimate the total chronic daily intake of toluene from exposure to a city water supply that contains toluene at a concentration equal to the drinking water standard of 1 mg/L over a period of 10 years. Assume the exposed individual is an adult female that is exposed to the chemical via drinking water and inhaling gaseous toluene released while she showers. Use the given parameters to calculate the CDI for water consumption. For inhalation, assume the woman takes a 15-minute shower every day. Assume the average air concentration of toluene during the shower is 1 μg/m3 and that she breathes at the adult rate of 20 m3/day.
Answer:
The following are the solution to this question:
Explanation:
The Formula for calculating CDI:
[tex]\bold{CDI = \frac{C \times CR \times EF \times ED}{BW \times AT}}[/tex]
[tex]_{where} \\ CDI = \text{Chronic daily Intake rate} (\frac{mg}{kg-day})} \\\\\text{C = concentration of Toluene}\\\\\text{CR = contact rate} \frac{L}{day}\\\\\text{EF = Exposure frequency} \frac{days}{year}\\\\\text{ED = Exposure duration (in years)} = 10 \ \ years\\\\\text{BW = Body weight (kg) = 70 kg for adult}\\\\ \text{AT = average period of exposure (days) }[/tex]
calculating the value of AT:
[tex]= 365 \frac{days}{year} \times 70 \ year \\\\ = 25550 \ days[/tex]
calculating the value of Intake based drinking:
[tex]C = 1 \ \frac{mg}{L}[/tex]
[tex]CR = 2 \frac{L}{day}[/tex] Considering that adult females eat 2 L of water a day,
EF = 350 [tex]\frac{days}{year}[/tex] for drink
calculating the CDI value:
[tex]\to CDI = \frac{(1 \times 2 \times 350 \times 10)}{(70 \times 25550)}\\\\[/tex]
[tex]= \frac{(2 \times 3500)}{(70 \times 25550)}\\\\ = \frac{(7000)}{(70 \times 25550)}\\\\ = \frac{(100)}{(25550)}\\\\=0.00391 \frac{mg}{ kg-day}[/tex]
Centered on inhalation, intake:
[tex]C = \frac{1 \mu g} { m^3} \ \ \ or \ \ \ \ 0.001 \ \ \frac{mg}{m^3}\\\\CR = 20 \frac{m^3}{day}\\\\EF = 15 \frac{min}{day} \ \ or\ \ 5475 \frac{min}{yr} \ \ \ or \ \ 3.80 \frac{days}{year}\\[/tex]
calculating the value of CDI:
[tex]\to CDI = \frac{(0.001 \times 20 \times 3.80 \times 10)}{(70 \times 25550)}[/tex]
[tex]= \frac{(0.76)}{(1788500)}\\\\= 4.24 \times 10^{-7} \ \ \frac{mg}{kg-day}[/tex]
A 75-kg piano is hosted on a crane and delivered throughout the window of a 6th story apartment (20 meters above ground). What is the potential energy of the piano?
Answer:
14,700 J
Explanation:
PE = Mgh = (75 kg)(9.8 m/s²)(20 m) = 14,700 J
Seawater containing 3.50 wt% salt passes through a series of 11 evaporators. Roughly equal quantities of water are vaporized in each of the 11 units and then condensed and combined to obtain a product stream of fresh water. The brine leaving each evaporator but the 11this fed to the next evaporator. The brine leaving the 11th evaporator contains 5.00 wt% salt. It is desired to produce 2 x 104 L/h of fresh water. What is the mass flow rate of concentrated brine out of the process?
______kg/h Save for Later
Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Explanation:
F, W and B are the fresh feed, brine and total water obtained
w = 2 x 10^4 L/h
we know that
F = W + B
we substitute
F = 2 x 10^4 + B
F = 20000 + B .................EQUA 1
solute
0.035F = 0.05B
B = 0.035F/0.05
B = 0.7F
now we substitute value of B in equation 1
F = 20000 + 0.7F
0.3F = 20000
F = 20000/0.3
F = 66666.67 kg/hr
B = 0.7F
B = 0.7 * F
B = 0.7 * 66666.67
B = 46,666.669 kg/hr
the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Complete the given statement with the most appropriate word.
Robin is very
because she always ensures she is available if a teammate needs her assistance.
Answer:
A. approachable
Explanation:
Answer:
- approachable
Explanation:
Gear friction reduces power and engineers never use more gears than are need it.
A) True
B) False
Answer:
i personally think it is false
Explanation:
i think this because gear friction reduces next to no power
A lake with a surface area of 525 acres was monitored over a period of time. During onemonth period the inflow was 30 cfs (ie. ft3 /sec), the outflow was 27 cfs, and a 1.5 in seepage loss was measured. During the same month, the total precipitation was 4.25 inches. Evaporation loss was estimated as 6 inches. Estimate the storage change for this lake during the month.
Answer:
The storage of the lake has increased in [tex]4.58\times 10^{6}[/tex] cubic feet during the month.
Explanation:
We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:
[tex]\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}[/tex]
Where [tex]\Delta V_{storage}[/tex] is the monthly storage change of the lake, measured in cubic feet.
Monthly inflow
[tex]V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)[/tex]
[tex]V_{inflow} = 77.76\times 10^{6}\,ft^{3}[/tex]
Monthly outflow
[tex]V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)[/tex]
[tex]V_{outflow} = 66.98\times 10^{6}\,ft^{3}[/tex]
Seepage losses
[tex]V_{seepage} = s_{seepage}\cdot A_{lake}[/tex]
Where:
[tex]s_{seepage}[/tex] - Seepage length loss, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{seepage} = 1.5\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex]
[tex]V_{seepage} = 2.86\times 10^{6}\,ft^{3}[/tex]
Evaporation losses
[tex]V_{evaporation} = s_{evaporation}\cdot A_{lake}[/tex]
Where:
[tex]s_{evaporation}[/tex] - Evaporation length loss, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{evaporation} = 6\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex]
[tex]V_{evaporation} = 11.44\times 10^{6}\,ft^{3}[/tex]
Precipitation
[tex]V_{precipitation} = s_{precipitation}\cdot A_{lake}[/tex]
Where:
[tex]s_{precipitation}[/tex] - Precipitation length gain, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{precipitation} = 4.25\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex]
[tex]V_{precipitation} = 8.10\times 10^{6}\,ft^{3}[/tex]
Finally, we estimate the storage change of the lake during the month:
[tex]\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}[/tex]
[tex]\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}[/tex]
The storage of the lake has increased in [tex]4.58\times 10^{6}[/tex] cubic feet during the month.
The volume of water gained and the loss of water through flow,
seepage, precipitation and evaporation gives the storage change.
Response:
The storage change for the lake in a month is 1,582,823.123 ft.³How can the given information be used to calculate the storage change?Given parameters:
Area of the lake = 525 acres
Inflow = 30 ft.³/s
Outflow = 27 ft.³/s
Seepage loss = 1.5 in. = 0.125 ft.
Total precipitation = 4.25 inches
Evaporator loss = 6 inches
Number of seconds in a month is found as follows;
[tex]30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds[/tex]
Number of seconds in a month = 2592000 s.
Volume change due to flow, [tex]V_{fl}[/tex] = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³
1 acre = 43560 ft.²
Therefore;
525 acres = 525 × 43560 ft.² = 2.2869 × 10⁷ ft.²
Volume of water in seepage loss, [tex]V_s[/tex] = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³
Volume gained due to precipitation, [tex]V_p[/tex] = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³
Volume evaporation loss, [tex]V_e[/tex] = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³
[tex]Storage \, change \, \Delta V = \mathbf{V_{fl} - V_s + V_p - V_e}[/tex]Which gives;
ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123
The storage change, ΔV = 1,582,823.123 ft.³Learn more about water resources and hydrology here:
https://brainly.com/question/10555667
When starting your vehicle, what does it mean when the ABS indicator light 1 point
on instrument panel turns on for a few seconds before turning off? *
There is a problem with your braking system, do not drive
The ABS bulb has burned out, replace the bulb
O The vehicle safety check indicates the ABS is functioning normally
O The ABS system is not working properly but it's safe to drive
Answer:
C. The vehicle safety check indicates the ABS is functioning normally.
Explanation:
ABS, an antilock braking system, is a safe and secure slip-resistant braking system in cars and air-crafts. An ABS is there to prevent wheel-locking while using brakes in vehicles.
When one ignites a car, the ABS indicator will light up briefly as a part of safety check. The ABS indicator light comes on for a few seconds before it turns off again. This indicates that the ABS system is functioning normally. But, if the ABS indicator light remains after turning on the ignition, this indicates that there is a problem in the system.
In the given scenario, the ABS indicator is functioning properly, thus the correct answer is the third option (C).
An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.
Answer:
The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
Explanation:
Given that,
Thickness of A = 8.0 m
Conductivity = 25.0 m/d
Thickness of B = 2.0 m
Conductivity = 142 m/d
Thickness of C = 34 m
Conductivity = 40 m/d
We need to calculate the horizontal conductivity
Using formula of horizontal conductivity
[tex]K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}[/tex]
Put the value into the formula
[tex]K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}[/tex]
[tex]K_{H}=41.9\ m/d[/tex]
We need to calculate the vertical conductivity
Using formula of vertical conductivity
[tex]K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}[/tex]
Put the value into the formula
[tex]K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}[/tex]
[tex]K_{V}=37.2\ m/d[/tex]
Hence, The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
An engine manufacturer wants to develop a gasoline engine that produces 300 hP at 6000 rpm. They can achieve 10 bar BMEP at peak power with a naturally aspirated engine, and 20bar BMEP with a turbocharged engine. Most modern light-duty engines have displacements of ~0.5L per cylinder. Determine the displacement and minimum number of cylinders (rounded to nearest whole number) required to meet these specs for: a. A naturally aspirated engine. b. A turbocharged engine.
Answer:
a). 5 cylinders
b). 3 cylinders
Explanation:
To develop a gasoline engine that produces
Power, P = 300 hp
= 300 x 746 = 223710 watt
at speed, N = 600 rpm = 100 rps
and stroke volume [tex]$V_s=0.5$[/tex] L
= [tex]$0.5 \times 10^{-3}\ m^3$[/tex]
a). A naturally aspirated engine
BMEP at peak power = 10 bar
= [tex]$10 \times 10^5\ N/m^2$[/tex]
Suction volume, V = [tex]$V_s \times N$[/tex]
= [tex]$0.5 \times 10^{-3} \times 100 = 0.05\ m^3/s$[/tex]
Power produced in one engine cylinder , p = BMEP x V
= [tex]$10 \times 10^5 \times 0.05$[/tex]
= 50000 watts
No. of cylinders required, n = [tex]$\frac{P}{p}$[/tex]
= [tex]$\frac{223710}{50000}$[/tex]
= 4.47
So number of cylinders ≈ 5 nos.
b). A turbo charged engine
BMEP = 20 bar = [tex]$20 \times 10^5\ N/m^2$[/tex]
and Volume V = [tex]$0.05\ m^3 /s$[/tex]
Power produced in one engine cylinder, p = BMEP x V
= [tex]$20\times 10^5 \times 0.05$[/tex]
= 100000 watts
Therefore, number of cylinders, n = [tex]$\frac{P}{p}$[/tex]
= [tex]$\frac{223710}{100000}$[/tex]
= 2.23
So number of cylinders ≈ 3 nos.
Air enters a counterflow heat exchanger operating at steady state at 27 C, 0.3 MPa and exits at 12 C. Refrigerant 134a enters at 0.4 MPa, a quality of 0.3, and a mass flow rate of 35 kg/h. Refrigerant exits at 10 C. Stray heat transfer is negligible and there is no significant change in pressure for either stream.(a) For the Refrigerant 134a stream, determine the rate of heat transfer, in kJ/h.(b) For each of the streams, evaluate the change in flow exergy rate, in kJ/h, and interpret its value and sign.Let T0 = 22 C, p0 = 0.1 MPa, and ignore the effects of motion and gravity.
Answer:
A) 337.21 kj/h
b) -224.823 kj/h
Explanation:
Steady state temperature = 27⁰C
Pressure = 0.3 MPa
exit temperature = 12⁰c
Refrigerant 134a :
entering pressure = 0.4 MPa
mass flow rate = 35 kg/h
quality = 0.3
inlet temp = 8.93⁰c
A) Determine the rate of heat transfer in kJ/h for Refrigerant 134a stream
calculate the specific enthalpy of refrigerant at the inlet
hm = hf + x ( hg - hf )
at 0.4 MPa : hf = 64 kj/kg , hg = 256 kj/kg
x = 0.3 ( quality )
hence : hm = 64 + 0.3 ( 256 - 64 ) = 64.3 + 57.6 = 121.9 kj/kg
next calculate the specific enthalpy at outlet
Tout = 10⁰c , T sat = 8.93⁰c
since the Tsat < Tout the refrigerant is super heated and from the super heated refrigerant table at P = 0.4 MPa, hout = 257 kj/kg
Heat gained by refrigerant ( Qin)
Qin = mref ( hout - hin )
= 35 ( 257 - 112) = 5075 kj/h
To determine the rate of heat transfer we have to apply the equation below
heat gained by refrigerant = heat gained by air
Qin = Qout
5075 kj/h = mair ( hout - hin ) ------------ 1
considering ideal conditions
at T = 300 k , hout = 300.19 kj/kg ( specific enthalpy )
at T = 285 k , hin = 285.14 kj/kg ( specific enthalpy )
back to equation 1
5075 kj/h = mair ( 300.19 - 285.14 )
mair ( rate of energy transfer ) = 5075 / 15.05 = 337.21 kj/h
B) evaluating the change in flow energy rate in kJ/h
attached below is the detailed solution
a team of engineer's is designing a new Rover to explore the surface of Mars which statement describes the clearest constraint that applies to the solution
Answer:it must operate at temperatures below 0’C
Explanation:
Answer:
it must operate at temperatures below 0 degrees Celsius
Explanation:
1. How many board feet in piece of Oak that is 1" thick 6" wide and 8' long?
Answer:
4
Explanation:
A "board foot" is 1/12 of a cubic foot. It is the volume of a board 1" thick and 1 foot square.
Here, the board is 1" thick, so the number of board feet is numerically equal to the number of square feet of area it has.
(1/2 ft)(8 ft) = 4 ft²
The number of board feet is 4.
Technician A says you should place the air ratchet setting to
clockwise to loosen a fastener. Technician B says a fastener
must be set to the proper torque after using an air ratchet. Who
is correct?
A group of scientists studied the environmental impact of internal combustion engines burning hydrocarbon fuels. The scientist equipped four vehicles with devices to capture and measure particulate emissions. One vehicle burned diesel fuel, one burned ordinary gasoline, one burned a gasoline/ethanol mixture and one burned natural gas. The four vehicles have equal masses and carried identical cargo. The scientists drove each vehicle 400 km, recording the volume of fuel burns in the quantity of particulate emissions generated. What is the independent variable in this experiment?
Answer: Combustion of Hydrocarbons
Explanation:
The Independent variable in an experiment is the one whose effect on the dependent variable is being measured. The independent variable therefore is controlled to see the effect it will have in the experiment.
In this experiment, the scientists combusted different types of hydrocarbons (diesel, gasoline, natural gas and a gasoline/ethanol mixture) as they aimed to find out the effect that this burning would have on the environment thereby making the combustion of hydrocarbons the independent variable.
Answer:
A. Type of Fuel
Explanation: The quantity of particulate Matter (PM) primarily depends upon the type of fuel used. Fine carbonaceous particles are mainly responsible for PM emissions. Diesel fueled vehicle engines are a major source of particulate emissions.
PLEASE HELP!!!! You are going to create a transcript for a persuasive presentation to convince your client that your design improvement is the best possible solution for their needs. Take your client through the different design options you considered for the product design improvement. Explain why you discarded the designs that were not the best possible solution. Finally, explain why your final product design covers each of the requirements and specifications you used to create your needs statement.
A test of a driver’s perception/reaction time was conducted on a special test track in a rural area. The friction factor f = 0.6. The driver’s initial speed is 55 MPH. The track is on level ground.
a) When a driver is not texting on his smart phone, a stop can be made just in time to avoid hitting an unexpected object that is first visible 520 feet ahead.
b) When a driver is texting on his smart phone, but with all other conditions exactly the same, the driver is distracted and fails to stop and hits the object at a speed of 35 MPH. Determine the driver’s perception/reaction time for both situations, for (a) and for (b).
Answer:
a) tD = 4.36 sec
b) tD = 5.2 sec
Explanation:
a)
V = 0
work-done = change in kinetic energy
f.m.g(d) = 1/2m(55^2 - v^2)
divided both sides by m
1/2(55^2 - v^2) = f.g(d)
1/2(55 * 1.467)^2 ft^2/sec^2 = 0.6 * (9.81 * 3.28) * d
3255.03 = 19.306d
d = 3255.03 / 19.306
d = 168.6 ft
now D = 520 - 168.6 = 351.4
therefore reaction/perception time = tD
tD = 351.4 / ( 55 * 1.467)
tD = 4.36 sec
b)
Also here, V = 35 mph
so
1/2(55^2 - 35^2)*(1.467)^2 ft^2/sec^2) = 0.6 * (9.81 * 3.28) * d
1936.88 = 19.306d
d = 1936.88 / 19.306
d = 100.325ft
also D = 520ft - 100.325ft = 419.675
so tD = 419.675 / ( 55 * 1.467)
tD = 5.2 sec
5. What are the 3 basic types of electrical circuits?
OA. Simple, AC, and AC direct
OB. Series, Parallel, and DC indirect
OC. AC, DC, and DC direct
OD. Series, Parallel, and Series/Parallel
Answer:
D. Series, Parallel, and Series/Parallel
Explanation:
Categorized according to topology, circuits are ...
Series, Parallel, and Series/Parallel
__
Circuits can also be categorized according to the nature of the voltages and currents found in them. These categories would include AC, DC, and some mixture of AC and DC.
The terms "simple" and "direct" and "indirect" don't seem to have any defined meaning in this context.
Answer:
D. Series, Parallel, and Series/Parallel
Explanation:
on todays class
Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of the window, calculate the heat loss rate through L = 12-mm-thick polycarbonate, soda lime glass, and aerogel windows, respectively. The thermal conductivities of the aerogel and polycarbonate are kag = 0.014 W/m ⋅ K and kpc = 0.21 W/m ⋅ K, respectively.
Evaluate the thermal conductivity of the soda lime glass at 300 K. If the aircraft has 130 windows and the cost to heat the cabin air is $1/kW ⋅ h, compare the costs associated with the heat loss through the windows for an 8-hour intercontinental fight.
Answer:
HEAT LOST
polycarbonate = 252 W
soda lime glass = 1680 W
aerogel = 16.8 W
COST associated with heat loss
polycarbonate = $ 262.08
soda lime glass = $ 1,747.2
aerogel = $ 17.472
The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel
Explanation:
Given that;
surface area for each window = 0.4m * 0.4m = 0.16m^2
DeltaT = 90°C, L = 12mm = 0.012m
thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER
so at 300K
KsL = 1.4 W/mK
Kag = 0.014 W/mK
Kpc = 0.21 W/mK
Now HEAT LOSS
for polycarbonate;
Qpc = -KA dt/dx
NOTE ( heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)
so Qag = (0.21 * 0.16 * 90) / 0.012
= 252 W
for soda lime glass;
Qsl = (1.4 * 0.16 * 90) / 0.012
= 1680 W
for aerogel
Qaq = (0.014 * 0.16 * 90) / 0.012
= 16.8 W
Now for COST associated with heat lost
for polycarbonate;
cost = Qpc * 130 * 8 * 1/1000
= 252 * 130 * 8 * 1/1000
= $ 262.08
for soda lime glass;
cost = 1680 * 130 * 8 * 1/1000
= $ 1,747.2
for aerogel
cost = 16.8 * 130 * 8 * 1/1000
= $ 17.472
Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel
Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft.
A) Determine the pressure at this elevation, based on the equation
p = pa(1 - βz/T0)g/Rβ
β = .00650 K/m = absolute pressure at z =0 R= 286.9 J/kg*k.
B) If the air is assumed to have a constant specific weight of 0.07647 lb/ft3 what would the pressure be at this altitude?
C) If the air is assumed to have a constant temperature of 59˚F what would the pressure be at this elevation? For all three cases assume standard atmospheric conditions at sea level.
Properties of U.S Standard Atmosphere at Sea Level
Property SI Units BG Units
Temperature, T 288.15 K(15 oC) 518.67 oR(59.00 oF)
Pressure, P 101.33 kPa(abs) 2116.2 Ib/ft2(abs)
(14.696 Ib/in.2(abs))
Density 1.225 kg/m3 0.00237 slugs/ft3
Specific weight 12.014 N/m3 0.07647 Ib/ft3
Viscosity, mu 1.789 * 10-5 N.s/m2 3.737 * 10-7 Ib.s/ft2
Answer:
A. 1236.49 lb/ft²
B. 1037.21 lb/ft²
C 2269.7 lb/ft²
Explanation:
P = pa(1-βz/Ta)^g/Rβ
a.)
Pa [1-(0.00357)(14110)/518.67]^32.2/1716*0.00357
= 2116.2(1-0.09712)^5.26
= 2116.2(0.5843)
= 1236.49566 lb/ft²
b.)
Pressure at higher elevation
P = Pa - yz
= 2116.2-(0.07647)(14110)
= 1037.2083 lb/ft³
C.
Pressure at higher elevation if air is isothermal
= (Pa)*exp[-(32.2)(14110)/(1716)(519)]
= 2116.2*exp[-454342/890604]
= 1269.7 lb/ft²
You often experiment with everyday materials around you, trying to create new and unique items. You enjoy designing and creating things that improve your daily tasks. What characteristic of a STEM student are you displaying?
A. technologically-literate
B. inventor
C. innovator
D. problem-solver
answer:D. problem-solver
The answer is:
B) Inventor
ion 2 23
Ohm's law tells us that it takes 1 volt to push 1 amp through how many ohms?
A. 422
B. 292
O c. 10
D. 32
The evaporator section of a refrigeration unit consists of thin-walled, 10-mm-diameter tubes through which refrigerant passes at a temperature of −18°C. Air is cooled as it fows over the tubes, maintaining a surface convection coeffcient of 100 W/m2 ⋅ K, and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of −3°C, what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator’s defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range 0 ≤δ ≤ 4 mm. Frost may be assumed to have a thermal conductivity of 0.4 W/m ⋅ K. (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which T[infinity] = 20°C and natural convection maintains a convection coeffcient of 2 W/m2 ⋅ K, how long will it take for the frost to melt? The frost may be assumed to have a mass density of 700 kg/m3 and a latent heat of fusion of 334 kJ/kg
Remy noticed that after oiling his skateboard wheels, it was easier to reach the speeds he needed to perform tricks. How did the oil help? A. The oil reduced friction between the moving parts of the skateboard. B. The oil increased friction between the moving parts of the skateboard. C. The oil reduced friction between the skateboard wheels and the ground. D. The oil increased friction between the skateboard wheels and the ground.
Answer:
The oil reduced friction between the moving parts of the skateboard. ( A )
Explanation:
The oil reduced Friction between the moving parts of the skateboard and this is because the reduction in friction between moving parts causes an increase in speed.
Remy will oil the moving parts that connects the wheels of the skateboard to the Board, because this is where the most friction is found. the friction between the wheels and the ground cannot be affected by oiling the wheels
1. Current in a series circuit is:
OA. Greatest at the point of highest resistance
OB. Least at the point of highest resistance
OC. The same throughout
OD. Equal to the Voltage minus the Resistance
Answer:
C. The same throughout
Explanation:
There is only one current path in a series circuit. The current is the same everywhere on that path.
what are two types of drag that act on an aircraft in flight
Answer:
Parasite drag and induced drag.
You can use this to listen to light from the big bang.
A. a microwave
C.
B. a cell phone
C. a radio
D. a tv
6. Dr. Li boils water using a kettle with a 1.5 kW Nichrome (80% Ni and 20% Cr) heating element (resister heater). The diameter and length of this heating element is 8 mm and 20 cm, respectively. When exposed to liquid water, the convection heat transfer between water and heating element is 800 W m2K . Please find: (25 Points) (1) The surface temperature and maximum temperature of the heating element when water is boiling in the kettle in Dr. Li’s office, ESB 751. (2) The surface temperature and maximum temperature when all liquid water has been evaporated and the heating element is exposed completely to the superheated vapor. You can assume the convection heat transfer between superheated water vapor and heating element is 24 W m2K . You can assume the temperature of the superheated water vapor is 100 °C. (3) Please justify if the heating element can survive in case (2) without melting.
Assuming that:
The heat generation is uniform throughout the heating element, Nichrome wire.
The cross-sections are insulated and heat transfer is taking place only in the radial direction.
All the heat generated is conducted, there is energy storage.
Now, from the properties of Nichrome:
The melting temperature of Nichrome, [tex]T_m=1400 ^{\circ}C[/tex]
Thermal conductivity, [tex]K=11.3 W/m^{\circ}C[/tex]
Given that:
The power generated by the heating element, [tex]Q_g=1.5kW=1500W[/tex]
Diameter, [tex]D=8mm=8\times10^{-3}m[/tex]
So, radius, [tex]R=4\times10^{-3}m[/tex]
Length, [tex]L=20cm=0.2m[/tex]
The volume of the Nichrome wire,
[tex]V=\pi R^2L=\pi\times (4\times10^{-3})^2\times0.2=1.005\times10^{-5}m^3[/tex]
Heat generation rate per unit volume,
[tex]q_g=\frac{Q_g}{V}=\frac{1500}{1.005\times10^{-5}}=149.2\times10^{6}W/m^3[/tex]
With the assumptions made above, this is the case of heat transfer in one direction.
Let [tex]T[/tex] be the temperature at the radius [tex]r[/tex].
Now, the heat generated within the cylinder of radius [tex]r[/tex] is conducted in a radially outward direction. i.e
[tex]q_g(\pi r^2)L=-K(2\pi r L)\frac{dT}{dr}[/tex] [where K is the thermal conductivity oof Nichrome]
[tex]\Rightarrow -\frac{q_g}{2K}rdr=dT[/tex]
[tex]\Rightarrow T=-\frac{q_g}{4K}r^2+ C_0[/tex] , where [tex]C_0[/tex] is constant.
Let the surface temperature is [tex]T_s[/tex], i.e at [tex]r=R, T=T_s[/tex].
Putting this boundary condition to get [tex]C_0[/tex], we have
[tex]T=T_s+\frac{q_g}{4K}(R^2-r^2)[/tex]
This is the temperature profile within the Nichrome wire, which is maximum at [tex]r=0[/tex].
So, the maximum temperature,
[tex]T_{max}=T_s+\frac{q_g}{4K}R^2\;\cdots(i)[/tex]
(1) The water is boiling, to the temperature of the water is, [tex]T_b=100^{\circ}C[/tex].
The total heat generated within the heating element is convected to the water from the surface. i.e
[tex]Q_g=h_w(2\pi RL)(T_s-T_b)[/tex]
where, [tex]h_w=800W/m^2K=800W/m^2^{\circ}C[/tex] is the convective heat transfer constant (Given).
[tex]\Rightarrow 1500=800\times (2\pi\times4\times10^{-3}\times 0.2(T_s-100)[/tex]
[tex]\Rightarrow T_s=100+373=473^{\circ}C[/tex]
So, the surface temperature is [tex]473^{\circ}C[/tex].
From equation (i), the maximum temperature is at the center of the wire which is
[tex]T_{max}=473+\frac{149.2\times10^{6}}{4\times11.3}(4\times10^{-3})^2[/tex]
[tex]\Rightarrow T_{max}=473+53=526^{\circ}C[/tex]
(2) In this case, the temperature of the superheated water vapor, [tex]T_b = 100^{\circ}C[/tex] (Given)
The heat transfer coefficient between the superheated water vapor and the heating surface is, [tex]h_v=24W/mK=24W/m^{\circ}C[/tex].
Similarly, [tex]Q_g=h_v(2\pi RL)(T_s-T_b)[/tex]
[tex]\Rightarrow 1500=24\times (2\pi\times4\times10^{-3}\times 0.2(T_s-100)[/tex]
[tex]\Rightarrow T_s=100+12434=12534^{\circ}C[/tex]
So, the surface temperature is [tex]12534^{\circ}C[/tex].
From equation (i), the maximum temperature is at the center of the wire which is
[tex]T_{max}=12534+\frac{149.2\times10^{6}}{4\times11.3}(4\times10^{-3})^2[/tex]
[tex]\Rightarrow T_{max}=12534+53=12587^{\circ}C[/tex]
(3) The melting temperature of Chromium is [tex]1400 ^{\circ}C[/tex].
So, the 1st case when the heating element is surrounded by water is safe as the maximum temperature within the element is below the melting temperature.
But, it the 2nd case, the heating element will melt out as the maximum temperature is much higher than the melting temperature of the element.
5.5 A scraper with a 275 hp diesel engine will be used to excavate and haul earth for a highway project. An evaluation of the job-site conditions indicates the scraper will operate 40 min./hr. For this project it is anticipated that the total cycle time will be 20 min. for a round trip. Previous job records show the scraper operated at full power for the 1.5 min. required to fill the bowl of the scraper and at 80% of the rated hp for the balance of the cycle time. Calculate the gallons per hour for fuel consumption of the scraper.
Answer: Fuel consumption per hour of the scrapper is 6 gal/hr
Explanation:
Given that;
Rated power = 275 hP
The scraper will work = 40min per hr
anticipated total cycle time = 20min
Previous Scrapper operated full power for 1.5min at 80% hP
First we find the Engine factor;
filling the bucket = 1.5/20 * 100% = 0.075
rest of cycle = (20-1.5)/20 * 80% = 0.74
so total engine factor = 0.075 + 0.74 = 0.815
now Time factor will be 40/60 = 0.6666 =
therefore operating factor = 0.6666 * 0.815 = 0.543
Now we know that for a diesel engine, a range fuel consumption = 0.04 gal/hP
so
Fuel consumption per hour of the scrapper is;
= 0.543 * 275 * 0.04
= 5.97 = 6 gal/hr
Singularity is an important property of a square matrix. This is also known as degenerate. What is the value of the determinant of a singular matrix?
Answer:
For a Singular matrix, the determinant must be equivalent to 0.
Explanation:
A matrix is a rectangular array in which elements are arranged in rows and columns.
Each square matrix has a determinant. The determinant is a numerical idea that has a fundamental function in finding the arrangement just as investigation of direct conditions. For a Singular matrix, the determinant must be equivalent to 0.
