Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T* t )
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = [tex]\frac{Q}{4\pi T} * W (u)[/tex]
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
what do you expect the future trends of an operating system in terms of (a) cost (b) size (c) multitasking (d) portability (e) simplicity
Answer:
plz follow in titkok
Explanation:
When do you need to apply for program completion and review?
Answer:
1/2 semesters before completion
Explanation:
Program completion and review are both a necessary part and parcel of a successful completion of any program. The recommended period for applying for program completion and review is one to two semesters before the program is to be completed. This is mandatory, so that there is different time span for review and any other issues that are needed to be carried out. And then afterwards, finally the certificates are then issued to the qualified person.
Igneous rocks are formed due to the
a cooling of magma
b weathering of rocks
c changes in pressure
d deposition of sediment
Answer:
a. cooling of magma
Explanation:
they form when magma cools
PLLLLLSSSSSSS HELPPPPPPPP!
Answer:
i'll help you but there is no question to answer??
Oxygen enters an insulated 12-cm-diameter pipe with a velocity of 70 m/s. At the pipe entrance, the oxygen is at 240 kPa and 20°C, and at the exit it is at 200 kPa and 18°C. Calculate the rate at which entropy is generated in the pipe.
Answer:
S = 0.10253 kW/k
Explanation:
Given data:
Velocity of oxygen ( V ) = 70 m/s
Diameter of pipe = 12-cm = 0.12 m
At entrance
pressure of oxygen ( p1 ) = 240 kPa
Temperature of oxygen ( T1 ) = 20°c = 293 k
At exit
pressure of oxygen ( p2 ) = 200 kPa
temperature of oxygen ( T2 ) = 18°c = 291 k
First calculate specific volume of oxygen at inlet
V1 = [tex]\frac{RT1}{P1}[/tex] -------- ( 1 )
R = 0.2598 KJ/kgk ( property of oxygen )
T1 = 293 k
P1 = 240 kpa
substitute values into equation 1
V1 = 0.3172 m^3/kg
next we calculate the mass flow rate of Oxygen
m = [tex]\frac{A1V}{v1}[/tex] ----- ( 2 )
A1 ( area of pipe ) = 0.0113 m^2 ( calculated )
V = 70 m/s
V1 = 0.3172 m^3/kg
substitute value into equation 2
m ( mass flow rate of oxygen ) = 2.4936 kg/s
Finally calculate the rate at which entropy is generated in the pipe
S = [tex]m( C_{p} In\frac{T2}{T1} - RIn\frac{P2}{P1} )[/tex] --------- ( 3 )
[tex]C_{p}[/tex] = 0.918 kj/kgk ( property of oxygen )
T2 = 291 k
T1 = 293 k
P2 = 200 kPa
P1 = 240 kPa
substitute values into equation 3 above
S = 0.10253 kW/k
A rigid, insulated vessel is divided into two equal-volume compartments connected by a valve. Initially, one compartment con tains 1 m3 of water at 20°C, x = 50%, and the other is evacuated. The valve is opened and the water is allowed to fill the entire volume. For the water, determine the final temperature, in °C, and the amount of entropy produced, in kJ/K.
Solution :
Given
Volume, [tex]$V_1 = 1 \ m^3$[/tex]
Temperature, [tex]$T_1=20 \ ^\circ C[/tex]
[tex]$x_1=0.5$[/tex]
From the saturated water table, corresponding to [tex]$T_1=20 \ ^\circ C[/tex], we get the saturated liquid, vapor specific and the entropy.
[tex]$v_f=1.0010 \ m^3/kg$[/tex]
[tex]$v_g=57.791 \ m^3/kg$[/tex]
[tex]$s_f=0.2966 \ kJ/kg-K$[/tex]
[tex]$s_g=8.6672 \ kJ/kg-K$[/tex]
Now calculating the initial specific volume
[tex]$v_1=v_f+x_1 \cdot(v_g-v_f)$[/tex]
[tex]$=1.0018+05 \cdot(57.791-1.0018)$[/tex]
[tex]$= 29.8973 \ m^3/kg$[/tex]
Calculating the initial specific entropy:
[tex]$s_1=s_f+x+1 \cdot (s_g-s_f)$[/tex]
[tex]$=0.2966+0.5 \cdot (8.6673 - 0.2966)$[/tex]
[tex]$= 4.48 \ kJ/kg-K$[/tex]
So final volume of the vessel is two times bigger as the initial volume
[tex]$V_2=2 .V_1$[/tex]
[tex]$= 2 \times 29.8973 = 59.8 \ m^3/kg$[/tex]
If we interpolate the values from tables between [tex]$v_g=57.791 \ m^3/kg$[/tex] and [tex]$v_g=61.293 \ m^3/kg$[/tex], we can get final temperature and specific entropy corresponding to value of [tex]$v_2$[/tex] :
Final temperature, [tex]$T_2= 19.6 ^\circ C$[/tex]
and [tex]$s_2 = 8.68 \ kJ/kg-K$[/tex]
Calculating change in entropy
[tex]$\Delta s = s_2-s_1$[/tex]
[tex]$=8.68-4.48 = 4.2 \ kJ/kg-K$[/tex]
A typical printed page of text contains 50 lines of 80 characters each. Imagine that a certain printer can print 6 pages per minute and that the time to write a character to the printer’s output register is so short it can be ignored. Does it make sense to run this printer using interrupt‐driven I/O if each printer requires an interrupt that takes 50 µsec all‐in to service?
Solution :
Given
The number of line that can be written by printer per page = 50 lines
The number of characters that can be written by printer per page = 80 characters.
Number of pages that can be written per minute = 6 pages
The speed for writing characters to the printer can be found by
= 50 x 80 x 6
= 400 characters per sec
In one second the printer has the writing speed of 400 character.
Now each of the character uses = 50 µsec of the CPU time for interrupt.
So in each second, the interrupt overhead = 20 msec
Now using the interrupt driven input/output, 980 msec time can be available for other work.
Thus, the interrupt overhead charges only 2 percent of the CPU, that will hardly affect the program to run.
Discuss the aging that occurs in asphalt cement during mixing with aggregates and in service. How can the different types of aging of asphalt cement be simulated in the laboratory?
Answer:
It can be simulated with RTFO and PAV method
Explanation:
The aging which occurs for the binders is always tested using rolling thin film oven and also pressure aging vessels. The rolling thin film oven is a method used to calculate the short term aging that occurs in the asphalt cement when it mixes with the aggregates as well as in service. While the long term process of aging in asphalt cement is carried out by pressure aging vessel.
The short term process of aging can be simulated with the help of RTFO and the long term is simulated by the PAV method.
The cross section of a heat exchanger consists of three circular pipes inside a larger pipe. The internal diameter of the three smaller pipes is 2.5 cm, and the pipe wall thickness is 3 mm. The inside diameter of the larger pipe is 8 cm. If the velocity of the fluid in the region between the smaller pipes and larger pipe is 10 m/s, what is the discharge in m^3/s?
Answer:
0.0432 m^3/s
Explanation:
Internal diameter of smaller pipes = 2.5 cm = 0.025 m
pipa wall thickness = 3 mm = 0.003 m
internal diameter of larger pipes = 8 cm = 0.8 m
velocity of region between smaller and larger pipe = 10 m/s
Calculate discharge in m^3/s
First we calculate the area of the smaller pipe
A = [tex]\pi Dt[/tex] = [tex]\pi ( 0.025 ) ( 0.003 )[/tex] = 0.00023571 m^2
next we calculate area of fluid between the smaller pipes and larger pipe
A = [tex][\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})[/tex]
= [tex][ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )][/tex]
= [ 0.00502857 - 0.00070713 ]
= 0.00432144 m^2
hence the discharge in m^3/s
Q = AV
= 0.00432144 * 10
= 0.0432 m^3/s
The set of line styles and line thickness used on drawings used on drawings are referred to as the ____.
Answer:
I think it's referred as the outline
Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the t's are in minutes and the x's are in thousands of vehicles per hour. The travel times on the routes are known to be in user equilibrium. If an observation for route 1 finds that the gaps between 30% of the vehicles are less than 6 seconds. Estimate the volume and average travel times for the two routes
Solution :
Given
[tex]$t_1=2+x_1$[/tex]
[tex]$t_2=1+x_2$[/tex]
Now,
[tex]$P(h<5)=1-P(h \geq5)$[/tex]
[tex]$0.4=1-P(h \geq5)$[/tex]
[tex]$0.6=P(h \geq5)$[/tex]
[tex]$0.6= e^{\frac{-x_1 5}{3600}}$[/tex]
Therefore, [tex]$x_1=368 \ veh/h$[/tex]
[tex]$=\frac{368}{1000} = 0.368$[/tex]
Given, [tex]$t_1=2+x_1$[/tex]
= 2 + 0.368
= 2.368 min
At user equilibrium, [tex]$t_2=t_1$[/tex]
∴ [tex]$t_2$[/tex] = 2.368 min
[tex]$t_2=1+x_2$[/tex]
[tex]$2.368=1+x_2$[/tex]
[tex]$x_2 = 1.368$[/tex]
[tex]$x_2 = 1.368 \times 1000$[/tex]
= 1368 veh/h
what are the benefits of NSTP CWTS?
Explanation:
The purpose of this program is to recognize the Youth's vital role in nation- building, promote consciousness among youth and develop their physical, moral, spiritual, intellectual and social well-being. It shall inculcate in the youth patriotism, nationalism, and advance their involvement in public and civic affairs.
Discoloration on walls, work surfaces, ceilings, walls, and pipes may indicate a leak that is causing you to waste raw materials.
Answer:
True :)
Explanation:
If this is a true or false question.
The discoloration is when the color of a particular thing comes off or changes, fading from its original color. In other words, when the original color of anything changes or wanes off, then that is called discoloration. The statement is true.
What is Discoloration?This chemical shift occurs for a number of reasons, including its interaction with other elements nearby or the unfavorable effects of other elements it comes into touch with.
Water leaks may create cases of discoloration on walls, surfaces, even ceilings, and pipes. As a result, the surface progressively changes and becomes moist.
Discoloration puts the surrounding region in danger in addition to altering the part(s) or area's look and durability.
Since a change in that region or component is required, the raw material(s) used are wasted. The claim that discoloration results in the waste of raw materials are thus accurate.
Learn more about discoloration here:
brainly.com/question/6378613
#SPJ6
How are you today First to awnser is brainlyest.
Answer:
Happy
Explanation:
1. Every employer shall keep the records of all accidents, dangerous occurrences, occupation
diseases and occupational poisoning at the workplace for
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A 360 kg/min stream of steam enters a turbine at 40 bar pressure and 100 degrees of superheat. The steam exits the turbine as a 100% saturated vapor at a pressure of 5 bar. Write and simplify the appropriate energy balance and then determine the energy generated by the steam as it passes through the turbine in kW.
How are glasses a form of technology and how do they involve engineering aspects?
Please help!!!!
Draw a circuit diagram of one lamp controlled by one switch and show how insulation resistance test is carried out
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What is the process of a Diesel engine uses to convert fuel to mechanical energy
Answer:
A diesel engine is a type of heat engine that uses the internal combustion process to convert the energy stored in the chemical bonds of the fuel into useful mechanical energy. ... First, the fuel reacts chemically (burns) and releases energy in the form of heat.
Which company produces comprehensive patch management software?
Quest
Motorola
Apple
Sony
What should be a concern as a weldment becomes larger as more parts are added?
will mark brainliest if correct
When a tractor is driving on a road, it must have a SMV sign prominently displayed.
True
False
Answer: true
Explanation:
For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of alloy that is liquid.
Answer:
Hello your question is incomplete attached below is the complete question
answer: wt.% of alloy that is solid = 61.5%
wt.% of allot that is liquid = 38.5%
Explanation:
To determine the wt.% of the alloy that is solid
= [tex]\frac{R}{R +S } * 100[/tex]
= [tex]\frac{53-45}{58-45} * 100[/tex] = 61.5%
To determine the wt.% of the alloy that is liquid
= [tex]\frac{S}{S+R} * 100\\[/tex]
= [tex]\frac{58-53}{58-45} *100[/tex] = 38.5%
attached below is a free hand sketch as well
To purchase a new car, you borrow $20,000. The bank offers a 6-year loan at an interest rate of 3.25% compounded annually. If you make only one payment at the end of the loan period, repaying the principal and interest: Which time value factor should be used to solve this problem?
Answer:
SPCA factor
Single payment compound amount factor.
Total amount pay A = $24,230.95 (Approx)
Interest paid = $4,230.95 (Approx)
Explanation:
Given:
P = $20,000
n = 6 year
r = 3.25%
Find:
Total amount pay A
Computation:
A=p(1+r)ⁿ
A=20,000[1+3.5%]⁶
A=20,000[(1.0325)⁶]
Total amount pay A = $24,230.95 (Approx)
Interest paid = $24,230.95 - 20,000
Interest paid = $4,230.95 (Approx)
An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 2.01cm and the young's modulus of aluminum is 69 GPa
The question is incomplete. The complete question is --
An aluminum rod is press fitted onto an aluminum collar. The collar has an inner radius of 1 cm and an outer radius of 2 cm. Given the rod has a diameter of 1.01 cm and the young's modulus of aluminum is 69 GPa, determine the following :
1. the interference value, i
2. the radial pressure at the interference of the collar and the rod
3. the maximum effective stress in the collar
4. if the yield strength of aluminium is 200 MPa and assume a safety factor of 1.5, will the aluminium collar break
Solution:
Given :
Inner radius of the collar = 1 cm
So, inner diameter, [tex]$d_1$[/tex] = 2 cm
Outer radius of the collar = 12 cm
So, outer diameter, [tex]$d_2$[/tex] = 4 cm
The aluminium rod diameter, d = 1.01 cm
Now, from the figure, we can see that there will be no interference and so the rod will easily insert inside the collar.
1. So, the interference , i =0
2. The radial pressure is also 0.
3. There will be no stress developed. So the maximum effective stress is 0
4. The collar will not break
A soil sample, taken from a borrow pit has a specific gravity of soil solids of 2.66. The sample was taken to a materials laboratory and tested. The results of a standard Proctor test are tabulated below.
Weight of Soil (lb) Moisture Content (%)
3.20 12.8
3.78 13.9
4.40 15.0
4.10 15.7
3.70 16.6
3.30 18.1
The maximum dry density in lb/ft3 is most nearly:_______
Answer:
115 Ib/ft^3
Explanation:
To determine the maximum dry density in Ib/ft3 we have to calculate :
Bulk unit weight ( yb ) ; W / v
Dry unit weight ( yd. ) : yb / ( 1 + w )
For every set of data given
assuming v = 1/30 ft^3
calculating for the 3 data set ( maximum dry density )
weight of soil (W) = 4.40
moisture content (%) (w) = 15.0 = 0.15
Bulk unit weight (yb) = 4.40 / (1/30) = 132 Ib/ft^3
Dry unit weight ( yd. ) = 132 / ( 1 + 0.15 ) = 114.702 Ib/ft^3
therefore after calculations the maximum dry density in Ib/ft^3 ≈ 115 Ib/ft^3
2) The switch in the circuit below has been closed a long time. At t=0, it is opened.
Find the inductor current for il(t) for t> 0.
Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes
Find the minimum sum of products expression using Quine-McCluskey method of the function. F(A,B,C,D)= Σ m(1,5,7,8,9,13,15)+ Σ d(4,6,11)
Answer:
Digital electronics deals with the discrete-valued digital signals. In general, any electronic system based on the digital logic uses binary notation (zeros and ones) to represent the states of the variables involved in it. Thus, Boolean algebraic simplification is an integral part of the design and analysis of a digital electronic system.
Explanation:
Hi please follow me also I you can and thanks for the points. Have a good day.
10. Which one of the following items would you expect to find on a floor framing plan?
ОООО
A. Sizes of joists
O B. Height of objects
C. Location of plumbing features
D. Arrangement of heating features
Answer:
c
Explanation:
the pluming usually is on the bottom floor plan
Lower headlight beams must be used when approaching within __________ of an oncoming vehicle or when following within __________ of the rear of another vehicle.
Answer:
500 feet, 300 feet
Explanation: