Which of the following router queuing policies might result in a situation where it is possible for a datagram to get stuck in the queue indefinitely (without being dropped)?
O Process the datagram with the shortest payload first
First-in-first-out (FIFO)
Random selection of a datagram
Round Robin across multiple queues
Consider the subnet 123.45.24.0/21, which can support up to 2048 hosts. Which of the following sets of 4 subnets represent a partitioning of this subnet into 4 equally sized subset subnets of size 512 hosts each?
123.45.24.0/22 123.45.24.1/22 123.45.24.2/22 123.45.24.3/22
123.45.24.0/23 123.45.26.0/23 123.45.28.0/23 123.45.30.0/23
123.45.24.0/23 123.45.25.0/23 123.45.26.0/23 123.45.27.0/23
123.45.24.0/23 123.45.24.1/23 123.45.24.2/23 123.45.24.3/23
123.45.24.0/22 123.45.24.2/22 123.45.24.4/22 123.45.24.6/22

Answer:

True

Explanation:

Because the Electric Magnetic Field is the work done per unit charge where other forms of energy is tranferred to electrical energy

Based on the given data, (A) The given rung of logic will check if the value in Source A is less than or equal to 99. If it is, then the output will be turned on. ; (B) The given rung of logic will check if the value in Source A is less than 75, greater than 100, or equal to 85. If it is, then the output will be turned on.

Here is the rung of logic to check if a value is less than or equal to 99:

015 LES LESS THAN

Source A

99

-EQU EQUAL

Source A

99

23. Turn on an output if the statement is true.

This rung of logic will check if the value in Source A is less than or equal to 99. If it is, then the output will be turned on.

Here is the rung of logic to check if a value is less than 75 or greater than 100 or equal to 85:

0:5 LES LESS THAN

Source A

75

01

Source B

GRT GREATER THAN

Source A

100

01

Source C

EQU EQUAL

Source A

85

23. Turn on an output if the statement is true.

This rung of logic will check if the value in Source A is less than 75, greater than 100, or equal to 85. If it is, then the output will be turned on.

Thus, based on the given data, (A) The given rung of logic will check if the value in Source A is less than or equal to 99. If it is, then the output will be turned on. ; (B) The given rung of logic will check if the value in Source A is less than 75, greater than 100, or equal to 85. If it is, then the output will be turned on.

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Answer:

are there any options, but if no I think the answer might be INTEREST

Fault location techniques are used in power systems for accurate pinpointing of the fault position.

This paper presents a comparative study between two fault location methods in distribution network with Distributed Generation (DG). Both methods are based on computing the impedance using fundamental voltage and current signals. The first method uses one-end information and the second uses both ends

Answer:

0.00695 A

Explanation:

µ represents \(10^{-6}\). Multiply this by 6,950.

Answer:

10.007

Explanation:

Assuming we have to find out the compression ratio of the engine

Given information

Cubic capacity of the engine, V = 245 cc

Clearance volume, V_c = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

Volume of the engine V =\(\frac{\pi}{4} D^2L\)

plugging values we get

245 = \(\frac{\pi}{4} D^3/1.1\)

Solving we get D =7 cm

therefore,  L= 7/1.1 =6.36 cm

Now,

the compression ratio is given as:

r =(V+V_c)/V_c

on substituting the values, we get

r = (245+27.2)/27.2 =10.007

Hence, Compression ratio = 10.007

Option C is correct. Swage joints must be used to connect two pieces of copper tubing by soldering or brazing without the use of any fittings.

Swaging is the process of fitting the end of one copper tube over the other so that the junction may be brazed or soldered. It is used to unite two copper tubes of the same diameter.

In order to permanently unite fittings to pipes or cables (also known as wire ropes), swaging is most frequently used. After the parts are loosely assembled, the fitting is compressed and deformed by a mechanical or hydraulic tool.

A swage joints is a wire rope component by definition. By squeezing the fitting on the wire, it is fastened to the wire.

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In order to find the temperature required to obtain 0.5% C at a distance of 0.5mm beneath the surface of a 0.2% C steel in 2 hours, given 1.1% C at the surface, we can make use of the Jominy end-quench test.

In the Jominy end-quench test, a standard size and shape of steel sample is heated to a high temperature and then quickly quenched at one end by spraying water on it. The water quenches the steel, causing it to cool rapidly from the high temperature. As the steel cools, it undergoes a transformation from austenite to a mixture of ferrite and pearlite.

As the steel cools, it undergoes a transformation from austenite to ferrite and pearlite. The distance from the quenched end to the point where this transformation is complete is a measure of the steel's hardenability. The hardenability of the steel depends on its composition and the cooling rate. A higher carbon content and a faster cooling rate will result in a higher hardenability.

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Answer:

STICK SKILL DRILL...

Explanation:

The skill drill uses a Whip and Pause technique.The reason for whip and pause is that 6010 electrodes are fast freezing electrodes and the momentary whips allow the puddle to solidify a bit ..Length of Art..Too long an arc causes problems so try to keep the arc equal to or shorter than diameter of electrode.

Answer:

A) F = 0.011 N

B) ΔP = 5.6 N/m²

Explanation:

We are given;

surface tension coefficient; S = 0.07 N/m

Radius; r = 2.5 mm = 0.025 m

A) Formula to find the surface tension force(F) is given by;

F = SL

Where L is effective length = 2πr

F = 0.07 × 2π × 0.025

F = 0.011 N

B) Formula for difference in pressure droplet is;

ΔP = 2S/r

Thus;

ΔP = (2 × 0.07)/0.025

ΔP = 5.6 N/m²

Answer: None of the above

Explanation:

Business process modeling refers to the graphical representation of the business processes of a company, which is vital in the identification of potential improvements.

Business pticess modelling can be done through graphing methods, like data-flow diagram, flowchart etc. It is vital as business managers can effectively and quickly communicate their ideas.

It also enhances the customization of business processes, enhances the competitive advantage and enhances the process communication as well.

Therefore, the answer to the question will be "None of the above".

Pressure altitude is the vertical distance above the standard datum plane, whereas Density altitude is the height in the International Standard Atmosphere at which the air density is equal to the actual air density at the place of observation.

Pressure Altitude Pressure Altitude is a term used to describe the altitude of an aircraft above a given datum plane. It is measured by an altimeter that has been calibrated to read pressure rather than altitude. This is because pressure is directly proportional to the altitude, and so changes in pressure can be used to determine changes in altitude. Density Altitude Density Altitude is the altitude in the International Standard Atmosphere (ISA) at which the air density is equal to the actual air density at the place of observation.

It is affected by the air temperature, atmospheric pressure, and humidity, and is usually higher than the pressure altitude.Q2: The effect of pressure, humidity, and temperature on air density is described below:Air Pressure: When air pressure increases, air density also increases.Humidity: Humidity decreases air density because water molecules are lighter than air molecules and displace some of the air molecules in a given space.Temperature: When air temperature increases, air density decreases. Conversely, when air temperature decreases, air density increases.Q3: The primary factors that affect the performance of an aircraft are the following:Thrust: The forward force that propels the aircraft forward. Weight: The downward force exerted on the aircraft due to gravity.

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Answer: A, B, C or D

Explanation: its either those

Answer:

\(\mathbf{K_p =2.857}\)

\(\mathbf{\tau_1= 400 \ s}\)

\(\mathbf{ \tau_d = 100 \ s}\)

Explanation:

We can determine each parameter by using the first method of Zeigler Nichols

\(K_p = 1.2 \times \dfrac{\tau}{KL}\)

\(\tau_1= 2L\)

\(\tau_d = 0.5 L\)

For this process;

the initial state of the output = \(A_o\)

\(\tau = constant\)

For calculating the gradient, we use the equation:

\(y-y_o = G(t-t_o) \\ \\ y-A_o = 0.01 \% A_o (t-L)\)

where.

Time constant relates to the time \(y = K = 1.05 A_o\)

∴

\(1.05A_o -A_o = 0.01\% A_o (t-200)\)

\(0.05 A_o = 0.0001 A_o (t-200)\)

\(t = 500A_o + 200\)

To find time t

\(\tau = (500A_o +200) -200\)

\(\tau = 500 \ A_o\)

Recall that:

Using the first method of Zeigler Nichols

\(K_p = 1.2 \times \dfrac{\tau}{KL}\)

\(\tau_1= 2L\)

\(\tau_d = 0.5 L\)

\(K_p = 1.2 \times \dfrac{500 \ A_o}{1.05\ A_o \times 200}\)

\(K_p = 1.2 \times 2.38095\)

\(\mathbf{K_p =2.857}\)

\(\tau_1= 2L\)

\(\tau_1= 2(200)\)

\(\mathbf{\tau_1= 400 \ s}\)

\(\tau_d = 0.5 \ L\)

\(\tau_d = 0.5 \times 200\)

\(\mathbf{ \tau_d = 100 \ s}\)

For a counter-flow heat exchanger, the rate of heat transfer is given by the formula: Q = m_dot_water * cp_water * (T4 - T3) = m_dot_steam * h1 - m_dot_steam * h2

where Q is the rate of heat transfer, m_dot_water is the mass flow rate of water, cp_water is the specific heat of water, T3 and T4 are the inlet and outlet temperatures of water, m_dot_steam is the mass flow rate of steam, h1, and h2 are the specific enthalpies of steam at inlet and outlet conditions respectively.

i) To find the required mass flow rate of the steam if the steam is at P1= 1000 kPa, we need to determine the values of h1 and h2 at the given conditions. Using a steam table, we can find that h1 = 2706.9 kJ/kg and h2 = 191.6 kJ/kg. Substituting these values into the above formula, we get:

1 * 4.18 * (50 - 20) = m_dot_steam * (2706.9 - 191.6)
m_dot_steam = 0.267 kg/s

ii) The entropy generated by the heat exchanger can be calculated using the formula:

delta_S = m_dot_water * cp_water * ln(T4/T3) - m_dot_steam * (s2 - s1)

where s1 and s2 are the specific entropies of steam at inlet and outlet conditions respectively. Using a steam table, we can find that s1 = 6.9918 kJ/kg-K and s2 = 0.7602 kJ/kg-K. Substituting the given values into the above formula, we get:
delta_S = 1 * 4.18 * ln(50/20) - 0.267 * (0.7602 - 6.9918)
delta_S = 1.399 kJ/K

iii) To find the required mass flow rate of the steam if the steam is at P1= 500 kPa, we need to repeat the above calculation using the specific enthalpies of steam at the new conditions. Using a steam table, we can find that h1 = 2676.5 kJ/kg and h2 = 165.8 kJ/kg. Substituting these values into the formula, we get:

1 * 4.18 * (50 - 20) = m_dot_steam * (2676.5 - 165.8)
m_dot_steam = 0.337 kg/s

iv) Similarly, we can calculate the entropy generated by the heat exchanger when the steam is at P1= 500 kPa using the same formula as in (ii), but with s1 and s2 values obtained from the steam table for the new conditions. Substituting the given values, we get:

delta_S = 1 * 4.18 * ln(50/20) - 0.337 * (0.6769 - 6.5836)
delta_S = 1.530 kJ/K

v) Comparing the results from (i) and (iii), we see that the required mass flow rate of steam is higher for P1 = 500 kPa than for P1 = 1000 kPa. This is because, at lower pressure, the steam has lower specific enthalpy, which means that more steam is required to transfer the same amount of heat. However, the entropy generated by the heat exchanger is also higher for P1 = 500 kPa, which means that more irreversibility occurs during the process. Therefore, P1 = 1000 kPa is the preferable case because it requires less steam and generates less entropy.

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Answer:

Explanation: 2 is thy answer

The process that a ceramist uses to knead and remove pockets of air from wet clay is called wedging.

Wedging involves a technique of working the clay by hand to remove any air bubbles or inconsistencies in the clay's texture. The ceramist takes a lump of wet clay and repeatedly kneads, compresses, and folds it to ensure uniformity and eliminate air pockets. This process helps to improve the clay's plasticity, remove excess moisture, and create a more workable and consistent material for shaping and forming. Wedging is an essential step in the preparation of clay before it can be used for various ceramic techniques such as throwing, slabbing, or coiling. It ensures that the clay is free from air bubbles that could cause structural weaknesses or uneven drying during the firing process.

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Therefore, roughly 12 drain holes with a diameter of 0.4 inches would be required to prevent the sink from overflowing with water.

Two gal of water per minute enter the sink. The water will eventually go through the overflow drain holes if the drain is closed as opposed to over the edge of the sink. number of 0.4-in.

Equation: The rate of flow through a circular hole can be determined.

\(Q = C \times A \times \sqrt{(2gH) (2gH)}\)

Calculate the following to determine the cross-sectional area of a hole with dimension d:

\(A = \frac{\pi}{4} \times d^2\)

To prevent the water from overflowing the sink, we can set the flow rate through each hole to be equal to the incoming flow rate and solve for the necessary number of holes.

\(n = \frac{2}{(0.61 \times \frac{\pi}{4} \times 0.16 \times \sqrt{(64.4 \times 6)})}\)

  \(= \frac{2}{(0.61 \times \frac{\pi}{4} \times 0.16 \times 2.51)}\)

  ≈ 12

where,

2 gal/min

\(= n \times C \times \frac{\pi}{4} \times d^{2} \times \sqrt{(2gH)}\)

Now,

\(n = \frac{2}{C \times \frac{\pi}{4} \times d^{2} \times \sqrt{(2gH)}}\)

  \(= \frac{2}{(0.61 \times \frac{\pi}{4} \times (0.4^{2}\ i\ n^2) \times \sqrt{(2 \times 32.2 \times H)})}\)

  \(= \frac{2}{(0.61 \times \frac{\pi}{4} \times 0.16 \times \sqrt{(64.4 \times H)})}\)

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The frequency (in Hz) of the radiation released by the transition of an electron in a hydrogen atom from the n = 5 level to the n = 3 level is 7.68 x 10¹⁴ Hz

The frequency (in Hz) of the radiation released by the transition of an electron in a hydrogen atom can be calculated using the Rydberg formula:

1/λ = R (1/n1² - 1/n2²)

where λ is the wavelength of the radiation,

R is the Rydberg constant (3.29 x 10¹⁵ Hz), and

n1 and n2 are the initial and final energy levels of the electron in the hydrogen atom, respectively.

For the transition from n=5 to n=3, we have:

1/λ = R (1/3² - 1/5²)

1/λ = R (1/9 - 1/25)

1/λ = R (16/225)

λ = 225/16R

λ = 225/(16 x 3.29 x 10¹⁵)

λ ≈ 390.6 nm

The frequency of the radiation can be obtained by using the speed of light (c) and the wavelength (λ):

c = fλ

f = c/λ

f = (2.998 x 10⁸ m/s)/(390.6 x 10⁻⁹ m)

f ≈ 7.68 x 10¹⁴ Hz

Therefore, the frequency is approximately 7.68 x 10¹⁴ Hz.

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The equipment aspect of the corporate environment is exemplified by tools like phones, fax machines, and computers.

Equipment includes items like computers, vehicles, and processing equipment. They are tangible because they have a physical form, as opposed to intangible assets, which don't (such patents, trademarks, or copyrights), which are the things required for a specific activity or purpose. When we say "the set of goods you require for a certain function," we mean "the uncountable noun equipment," such as tools or clothing: New sleeping bags and a small refrigerator are among the camping supplies we've purchased. They are the most fundamental methods for employing leverage (or mechanical advantage) to increase force that are currently understood. Simple machines include the wheel and axle, inclined plane, lever, wedge, pulley, and screw basic machines

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Answer:

For most uses you'll want your water heated to 120 F(49 C) In this example you'd need a demand water heater that produces a temperature rise and it will take about 2 hours

Polymorphism is a concept in object-oriented programming that allows objects of different classes to be treated as if they are objects of the same class. It provides a flexible way of working with objects that share a common behavior, but have different implementations.

When it comes to arrays in Java, traditionally, they can only contain objects of the same type. However, with polymorphism, this limitation can be overcome. Because polymorphism allows objects of different classes to be treated as if they are objects of the same class, it's possible to create an array of the parent class and add objects of the child classes to it.

For example, let's say we have a parent class called Shape, and two child classes called Circle and Square. We can create an array of Shape objects and add both Circle and Square objects to it. This is possible because both Circle and Square classes inherit from the Shape class and therefore share a common behavior.

This flexibility allows for more efficient and organized code, as we can now work with objects of different classes as if they are of the same class, simplifying our code.

In conclusion, polymorphism can benefit the usage of arrays in Java by allowing for arrays of parent classes to contain objects of child classes. This provides flexibility and allows for more efficient code.

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Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L\(\frac{di(t)}{dt}\)             -----------(i)

Where;

L = inductance of the inductor

i(t) = current through the inductor at a given time

t = time for the flow of current

From the question:

i(t) = \(10e^{-t/2}\)A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = \((10*10^{-3})\frac{d(10e^{-t/2})}{dt}\)

Solve the differential

v(t) = \((10*10^{-3})\frac{-1*10}{2} (e^{-t/2})\)

v(t) = -0.05 \(e^{-t/2}\)

At t = 8s

v(t) = v(8) = -0.05 \(e^{-8/2}\)

v(t) = v(8) = -0.05 \(e^{-4}\)

v(t) = -0.05 x 0.223

v(t) = -0.01116V

(b) To get the power, we use the following relation:

p(t) = i(t) x v(t)

Power at t = 8

p(8) = i(8) x v(8)

i(8) = i(t = 8) = \(10e^{-8/2}\)

i(8) = \(10e^{-4}\)

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

Answer:

The voltage is - 0.9158 mV

The power is - 0.1677 mW

Explanation:

Given;

current through the inductor, i(t)  = \(10e^{-t/2}\) -----equation (1)

inductance, L = 10 mH

given time, t  = 8 s

The voltage across the inductor is given by;

\(V_L = L\frac{di}{dt} \\\\V_L = (10 *10^{-3})\frac{d}{dt} (10e^{-t/2})\\\\V_L = (10 *10^{-3})\frac{10}{-2}(e^{-t/2})\\\\ V_L = -0.05e^{-t/2} \ ----equation (2)\)

when t = 8 s, the voltage will be ;

\(V_L = -0.05 e^{-t/2}\\\\V_L = -0.05 e^{-8/2}\\\\V_L = -0.05 e^{-4}\\\\V_L = -9.158 *10^{-4} \ V\\\\V_L = -0.9158 \ mV\)

Power is given by;

P = I V

When t = 8, the current "I" is given by;

\(i(t) = 10e^{-t/2}\\\\i(8) = 10e^{-8/2}\\\\I = 10e^{-4}\\\\I = 0.1832 \ A\)

P = 0.1832 x (-9.158 x 10⁻⁴)

P = -1.677 x 10⁻⁴ W

P = -0.1677 mW

Answer:

rain

Explanation:

evaoration causes clouds

clouds condense and rain

Answer:

rain

Explanation:

The strain in member AB is measured to be 8.9 x 10^-4.Axial strain,

ε = ΔL/LLet the total length of the member AB be ‘L’. The change in length in inches can be given as,

ΔL = εL

= (8.9 x 10^-4) x LInches

= 8.9 x 10^-4 x L x 12 inches

= 0.01068 L inches Axial force in the member AB,

F = σAWhere σ is the stress developed in the member and ‘A’ is the cross-sectional area of the member.σ can be calculated by,σ = E εWhere E is the modulus of elasticity of the material. From AISC manual, the modulus of elasticity of A36 steel is given as 29,000 ksi. Cross-sectional area of L3 x 2 1/2 x 1/4

= 3.52 in^2Let us assume that the force in the member AB is in tension. Direction of the force is from joint A to joint B.

F = σA= 29,000 × 8.9 x 10^-4 × 3.52

= 91.03 kips Thus, Therefore, the total force applied to the truss is 2 x 10 + 91.03 = 111.03 kips.The vertical reactions can be calculated using the force balance equation in the vertical direction,RAV + RDV = 111.03 kipsSince the truss is symmetric about the y-axis,

= 111.03/2

= 55.52 kipsThe horizontal reactions can be calculated using the force balance equation in the horizontal direction,RAH + RDH

= 0Therefore, RAV

= - RDHLet us take the direction of RAH to be from A to D, and RDH to be from D to A. Then, RAH

= - RDH

= - RAHRAH

= RDH

= 0.5 x 111.03

= 55.515 kips The maximum tension force one needs to consider for the design of this truss is the force in member AB, i.e., 91.03 kips.

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Answer:

Biochemistry

Explanation:

Hope this helps :)

Answer:

Biochemistry

Have an amazing day!

Answer:

All nonmetallic elements are generally poor conductors of heat and electricity. There are only 17 nonmetallic elements, while more than 75 percent of the known elements are either pure metals or metalloids, which are better conductors of heat and electricity to a varying degree.

Explanation:

Answer:

final temperature = 26.5°

Explanation:

Initial volume of water is 1 x 1 x 1 = 1 \(m^{3}\)

Initial temperature of water = 20° C

Density of water = 1000 kg/\(m^{3}\)

volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 \(m^{3}\)

Initial temperature of copper block = 100° C

Density of copper = 8960 kg/\(m^{3}\)

Final volume of water = 1 - 0.097 = 0.903 \(m^{3}\)

Assumptions:

mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg

specific heat capacity of water c = 4186 J/K-kg

heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules

mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg

specific heat capacity of copper is 385 J/K-kg

heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules

total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules

this heat will be distributed in the entire system

heat energy of water within the system = mcT

where T is the final temperature

= 903 x 4186 x T = 3779958T

for copper, heat will be

mcT = 869.12 x 385 = 334611.2T

these component heats will sum up to the final heat of the system, i.e

3779958T + 334611.2T = 109.05 x \(10^{6}\)

4114569.2T = 109.05 x \(10^{6}\)

final temperature T = (109.05 x \(10^{6}\))/4114569.2 = 26.5°

To determine the total system head for flow rates from 0 to 20l/s in steps of 5 l/s, we need to calculate the head loss due to friction and fittings on both the suction and delivery sides.

First, we need to calculate the Reynolds number to determine the flow regime in the pipe. The Reynolds number is given by:
Re = (ρVD)/μ
Where:
ρ = density of water = 1000 kg/m3
V = flow velocity
D = diameter of pipe = 100 mm = 0.1 m
μ = dynamic viscosity of water at 27°C = 0.9x10-3 Pa.s
At a flow rate of 10 l/s, the flow velocity is given by:
V = Q/A = (10x10-3 m3/s)/(π(0.1/2)2) = 2.54 m/s
Therefore, the Reynolds number is given by:
Re = (1000x2.54x0.1)/0.9x10-3 = 282222
Since the Reynolds number is greater than 4000, the flow regime is turbulent.
Next, we need to calculate the friction factor using the Colebrook equation:
1/sqrt(f) = -2.0log((ε/D)/3.7 + 2.51/(Re*sqrt(f)))
Where:
ε/D = relative roughness = 0.0015/0.1 = 0.015
Solving this equation iteratively using a spreadsheet or calculator, we get a friction factor of 0.018.
Using this friction factor, we can now calculate the head loss due to friction on both the suction and delivery sides using the Darcy-Weisbach equation:
hf = f(L/D)(V^2/2g)
Where:
L = length of pipe
D = diameter of pipe
V = flow velocity
g = acceleration due to gravity = 9.81 m/s2
For the suction side, the length of the pipe is 2.5 m and the fittings have a total equivalent length (TEL) of:
TEL = 2(0.05) + 10.0 + 0.9 = 11.95
Therefore, the total length of the suction side is 2.5 + 11.95 = 14.45 m.
The head loss due to friction on the suction side is given by:
hfs = 0.018(14.45/0.1)(2.54^2/2x9.81) = 1.14 m
For the delivery side, the length of the pipe is 80 m and the fittings have a TEL of:
TEL = 0.4 + 12.5 + 16(0.05) + 2(0.9) + 0.2 + 1 = 15.15
Therefore, the total length of the delivery side is 80 + 15.15 = 95.15 m.
The head loss due to friction on the delivery side is given by:
hfd = 0.018(95.15/0.1)(2.54^2/2x9.81) = 5.96 m
The total system head is given by the sum of the head losses due to friction and fittings on both the suction and delivery sides, plus the static head difference between the sump and tank:
hsystem = hfs + hfd + (60 - 1) = 64.1 m
To plot the system head curve, we repeat these calculations for flow rates of 5, 10, 15, and 20 l/s, and plot the results on a graph. The x-axis represents the flow rate and the y-axis represents the system head. The resulting curve shows the relationship between the flow rate and system head for this water supply system.

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The symbol usually used to designate a current source is:

B) An arrow in a circle.

Current, in the context of electrical circuits, refers to the flow of electric charge. It is the rate at which electric charges, typically electrons, move through a conductor. Current is measured in amperes (A) and is denoted by the symbol "I."

In simple terms, current is the movement of electric charges from a region of higher electric potential (voltage) to a region of lower electric potential. It is analogous to the flow of water in a pipe: the current represents the amount of water flowing through the pipe, while the voltage corresponds to the pressure driving the water flow.

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