Answer:
power input = -66.2798 kW
Explanation:
Steady state pressure ( Cp ) = 1.05 bar,
Temperature ( T1 ) = 300 K
Volumetric flow rate = 30 m^3/min
exit pressure = 12 bar
exit temperature ( T2 ) = 400 K
Heat transfer rate ( Q ) = 5 kW
Calculate the power input in kW
p1v1 = m RT
m = p1v1 / RT1
= (1.05 * 10^2 * ( 30/60 )) / ( 0.287 * 300 )
= 0.60975 m^3/sec
also
h1 + Q = h2 + w
∴ w = m ( h1 - h2 ) + Q
= mCp ( t1 - t2 ) + Q ----- ( 1 )
where : ( Q ) = 5 kW , Cp = 1.05 bar, t1 = 300 K, t2 = 400 k ( input values into equation 1 )
w = -66.2798 kW
Diseña un mecanismo multiplicador con un engranaje motriz cuya relación de transmisión sea de 0.5 y que transmita el movimiento entre ejes distantes. Inserta una captura de pantalla indicando la relación entre los diámetros y la velocidad de giro del engranaje motriz.
Explain the process of energy conversion by describing how energy was converted from the windmill design brief. Discuss the different forms of energy and what technology was used to convert the energy from one form to another.
Answer:
Wind energy is converted to Mechanical energy which is then converted in to electrical energy
Explanation:
In a wind mill the following energy conversions take place
a) Wind energy is converted into Mechanical energy (rotation of rotor blades)
b) Mechanical energy is converted into electrical energy (by using electric motor)
This electrical energy is then used for transmission through electric lines.
A pump draws water from a reservoir through a 150 mm diameter suction pipe and delivers it to a 75 mm diameter discharge pipe.
a. The end of the pipe is 2 m below the free surface of the reservoir.
b. The pressure gage on the discharge pipe (located 2 m above the reservoir surface) reads 170 kPa.
c. The average velocity in the discharge pipe is 3 m/s.
Assume no head loss in the suction or discharge pipe. Determine the power added to the fluid by the pump.
Answer:
3.42 kW
Explanation:
calculate the mass flow rate of the water = density * velocity * area
= 13.3 kg/s
where Area of pipe = π/4 * d^2 = π/4 * 0.075^2 m = 0.0044 m^2
given that : 1 Kg = 1 N s^2 / m, 1 Watt = 1 N m /s
calculate : power produced by discharge pipe
Discharge Pressure x Volume flow =
Pressure * Area * velocity
P * 0.0044 * 3 = 2253 N m /s
calculate: power due the mass of water
mass of water * 2 = 258 N m/s
where: mass of water = density * volume
Calculate power produced due the velocity of exiting water
= m * V2/2 = 59.4 N m/s
hence The power added to fluid = 2253 watts
power added to the fluid by pump = 2253 / 0.75 = 3.42 kW
a) The initial moisture content of a food product is 77% (wet basis), and the critical moisture content is 30% (wet basis). If the constant drying rate in a fluidized bed dryer is 0.1 kg water removed/m2-s, determine the time required for the product to begin the falling-rate drying period. The product has a cube shape with 5-cm sides; the initial product density is 950 kg/m3.
Answer:
≈ 53 seconds
Explanation:
calculate Time required for the product to begin the falling-rate drying period
Initial moisture content = 0.77 kg water /kg of product
= 3.35 kg water /kg solids
Critical moisture content = 0.3 kg water / kg product
= 0.43 kg water / kg solids
∴ amount of water to be removed = 3.35 - 0.43 = 2.95kg water /kg solids
next: calculate surface are a of product during drying
= (0.05 * 0.05 ) * 6
= 0.015 m^3
Drying rate = 0.1 kg water m^2.s^-1 * 0.015 m^3 = 1.5 * 10^-3 kg water s^-1
applying product density
initial product mass = 0.11875 * 0.23 = 0.0273kg solid
hence total amount of water to removed = 2.92 * 0.0273 = 0.07972 kg
therefore : Time required for the product to begin the falling-rate drying period
= 0.07972 / 1.5 * 10^-3
= 53 seconds
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is 40 kh at 520 rev/min. The application factor is 1.4. The radial load is 2600 lbf. The reliability goal is 0.90.
Required:
Determine the C10 value in kN for this application and design factor.
Answer:
[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]
Explanation:
From the information given:
Life requirement = 40 kh = 40 [tex]40 \times 10^{3} \ h[/tex]
Speed (N) = 520 rev/min
Reliability goal [tex](R_D)[/tex] = 0.9
Radial load [tex](F_D)[/tex] = 2600 lbf
To find C10 value by using the formula:
[tex]C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}[/tex]
where;
[tex]x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}[/tex]
[tex]\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}[/tex]
The Weibull parameters include:
[tex]x_o = 0.02[/tex]
[tex](\theta - x_o) = 4.439[/tex]
[tex]b= 1.483[/tex]
∴
Using the above formula:
[tex]C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}[/tex]
[tex]C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}[/tex]
[tex]C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}[/tex]
[tex]C_{10} = 30962.449 \ lbf[/tex]
Recall that:
1 kN = 225 lbf
∴
[tex]C_{10} = \dfrac{30962.449}{225}[/tex]
[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]
Two technicians are explaining what exhaust gas emissions tell you about engine operation. Technician A says that the higher the level of CO2 in the exhaust stream, the more efficiently the engine is operating. Technician B says that CO2 levels of 20 to 25 percent are considered acceptable. Who is correct?
A. Both Technicians A and B
B. Neither Technicians A and B
C. Technician A
D. Technician B
Technicians A is correct in the given scenario. The correct option is C.
What is exhaust gas?Exhaust gas is a byproduct of combustion that exits the tailpipe of an internal combustion engine.
It consists of a gas mixture that includes carbon dioxide (CO2), carbon monoxide (CO), nitrogen oxides (NOx), hydrocarbons (HC), and particulate matter (PM).
Technician B is mistaken. CO2 levels in the exhaust should be less than 15%, preferably between 13% and 14.5% for petrol engines and 11% to 13% for diesel engines.
High CO2 levels can actually indicate inefficient engine operation, as it means that not all of the fuel in the engine is being burned and is being wasted as exhaust.
Thus, C is the correct answer. A technician is correct.
For more details regarding exhaust gas, visit:
https://brainly.com/question/11779787
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An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE
Answer:
true
Explanation:
true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE
the complexity of bfs and dfs
Answer:
BFS uses Queue to find the shortest path. DFS uses Stack to find the shortest path. ... Time Complexity of BFS = O(V+E) where V is vertices and E is edges. Time Complexity of DFS is also O(V+E) where V is vertices and E is edges.
Explanation:
brainly and points if you want
Answer:
thank you
Explanation:
have a nice day
Answer:
thankd
...... ..............
[tex] \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ [/tex]
A chemical process stream enters a shell-and-tube exchanger at a temperature of 200.0°Fand does two passes on the shell side, exiting the exchange at 170.0°F. The Heat exchanger Spring 2021 has 200 stainless steel tubes that are 2-in.ODand 10.0ft long. Indicate whether the temperature of the process stream will increase, decrease, or remain the sameunder the following scenarios. You must justify your answerto receive full credit.a)The flow rate of the cooling fluid is increased.b)There are 200 tubes that are 1.0-in. OD and 20.0ft long.c)The number of shell passes is doubled.d)The tube material is changed to copper.
Answer:
a) Decrease
b) Decrease
c) Decrease
d) Decrease
Explanation:
Ti= 200°F ,
Te = 170°F
Area of heat exchanger = [tex]\pi *(2 )* 10[/tex] = 20π
A) when The flow rate of the cooling fluid is increased
Temperature of process stream will decrease and this is because the tube side heat transfer coefficient will increase and this will increase the rate of heat transfer thereby decreasing the temperature of the process stream.
B) when There are 200 tubes that are 1.0-in. OD and 20.0ft long
The temperature of the process stream will decrease and this is because the heat transfer coefficient will increase likewise the heat transfer rate
C) When The number of shell passes is doubled
This will cause an increase in the overall length of the shell, an increase in velocity of constant volumetric flowrate, hence the Temperature of the process steam will decrease as well
D) When The tube material is changed to copper.
Due to the high thermal conductivity of copper when compared to steel , switching to copper will cause a decrease in the temperature of the process steam
which statement best describes the velocity of a bus traveling along its route
Answer:
Option A is the correct answer. The bus traveled at 50 mph for 20 minutes
Explanation:
The complete question is
Which of the following choices best describes velocity of a bus traveling along its route? A. The bus traveled at 50 mph for 20 minutes. B. The airplane traveled southwest at 280 mph. C. The car went from 35 mph to 45 mph. D. The train made several stops, with an average rate of 57 mph.
Solution
In option A the bus is travelling at a speed of 50 miles per hour. This describes the velocity of bus along its route.
The other options are about the velocity of airplane, car and train
Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 20oC and exits at 3.0 bar. The refrigerant undergoes a throttling process. Determine the temperature, in oC, and the quality of the refrigerant at the exit of the expansion valve. Step 1 Determine the temperature of the refrigerant at the exit, in oC.
Answer:
[tex]T_{2}[/tex] = -9.24 °C
x = 0.1057
Explanation:
The tables used in this answer and explanation come from Fundamentals of Engineering Thermodynamics 9th Edition.
Using Table A-14: Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table and the given [tex]P_{2}[/tex], [tex]T_{2}[/tex] can be determined by finding the temperature that corresponds with [tex]P_{2}[/tex] on the table. In this case, [tex]T_{2}[/tex] = -9.24 °C.
The quality of the refrigerant can be determined by using data from the same table and [tex]h_{2} =274.26[/tex] kJ/kg.
Necessary data (P=3bar):
[tex]h_{f}=137.42[/tex] kJ/kg
[tex]h_{g}=1431.47[/tex] kJ/kg
The formula to calculate quality is [tex]h_{2} =h_{f}+x(h_{g}-h_{f})[/tex].
Rearranging for x:
[tex]x=\frac{h_{2}-h_{f} }{h_{g}-h_{f} }= \frac{274.26-137.42}{1431.47-137.42}=0.1057[/tex]
Question 5
Not yet answered
Marked out of 1.00
P Flag question
Which one of the following torque is produced by the spring in PMMC instrument?
O a. Damping
O b. Forcing
OC. Deflection
O d. Controlling
Answer:
A
Explanation:
Actually I don't know anything about American history, I chose it because South Africa is not in the least
An interest rate of norminal 12% per year , compounded weekly is
Answer: It is a nominal rate per year
me that both a triaxial shear test and a direct shear test were performed on a sample of dry sand. When the triaxial test is performed, the specimen was observed to fail when the major and minor principal stresses were 100 lb/in2 and 20 lb/in2, respectively. When the direct shear test is performed, what shear strength can be expected if the normal stress is 3000 lb/ft2
Answer:
shear strength = 2682.31 Ib/ft^2
Explanation:
major principal stress = 100 Ib / in2
minor principal stress = 20 Ib/in2
Normal stress = 3000 Ib/ft2
Determine the shear strength when direct shear test is performed
To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a
for direct shear test
use Mohr Coulomb criteria relation between normal stress and shear stress
Shear strength when normal strength is 3000 Ib/ft = 2682.31 Ib/ft^2
attached below is the detailed solution
other examples of joption
Answer:
Profit from stock price gains with limited risk and lower cost than buying the stock outright.
Profit from stock price drops with limited risk and lower cost than shorting the stock.
Profit from sideways markets by selling options and generating income.
Get paid to buy stock.
Explanation:
I don’t know if you meant option but I hope this helps.
A wind turbine designer is considering installing a horizontal axis wind turbine at a location in Michigan. To increase the power extracted from wind, the designer is debating between increasing the blade radius by 10% versus increasing the height from 40 m to 60 m, which would increase the average wind speed from 6 mph to 6.5 mph. Considering the only the impact on power extraction, the designer should go with the height increase.
a. True
b. False
Answer:
False ( B )
Explanation:
considering that the wind turbine is a horizontal axis turbine
Power generated/extracted by the turbine can be calculated as
P = n * 1/2 * p *Av^3
where: n = turbine efficiency
p = air density
A = πd^2 / 4
v = speed
From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine
wadadadadwadawdawd;ougaow7dtgwo8qa7gtydao8gdailydgw8dfaoufgsuytd
Answer:
what?????
Explanation:
What is being shown in the above Figure?
A. Camshaft is being aligned
B. Engine is being timed
C. Camshaft is being removed
D. Camshaft gear backlash is being checked
During peak systole, the heart delivers to the aorta a blood flow that has a velocity of 100cm/sec at a pressure of 120mmHg. The aortic root has a mean diameter of 25mm. Determine the force (Rz)acting on the aortic arch if the conditions at the outlet are a pressure of 110mmHg and a diameter of 21mm. The density of blood is 1050 kg/m3. Assume that aorta is rigid non-deformable and blood is incompressible and steady state. Ignore the weight of blood vessel and the weight of blood inside the blood vessel (i.e. body force is zero).
Solution :
Given :
Velocity, [tex]$V_1 =100$[/tex] cm/sec
Pressure, [tex]$P_1 = 120 $[/tex] mm Hg
Then, [tex]$P_1 = \rho_1 g h$[/tex]
[tex]$P_1 = 0.120 \times 13.6 \times 1000 \times 9.81$[/tex]
= 16.0092 kPa
[tex]$P_2 = 110 $[/tex] mm Hg
[tex]$P_2 = \rho_2 g h$[/tex]
[tex]$= 0.110 \times 13.6 \times 1000 \times 9.81$[/tex]
= 14.675 kPa
Then blood is incompressible,
[tex]$A_1v_1=A_2v_2$[/tex]
[tex]$\frac{\pi}{4}(25)^2\times 100=\frac{\pi}{4}(21)^2\times v_2$[/tex]
[tex]$v_2=141.72 \ cm/s$[/tex]
Then the linear momentum conservation fluid :
(Blood ) in y - direction
[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]
[tex]$m_1=m_2=P_1A_1v_1$[/tex]
[tex]$=1.50 \times \frac{\pi}{4}\times (0.025)^2 \times 1.00$[/tex]
= 0.515 kg/ sec
Then the linear conservation of momentum of blood in y direction.
[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]
[tex]$16.0092 \times 1000 \times \frac{\pi}{4} \times (0.025)^2+14.675 \times 1000\times \frac{\pi}{4}\times (0.021)^2$[/tex]
[tex]$-F_y=0.515(-1.4172-1)$[/tex]
7.858+5.0828- Fy = 0.515(-2.4172)
Fy = 14.1856 N
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be deformed using a tensile load of 18,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.5 x 10-2 mm. Would the 1040 steel be a possible candidate for this application? Justify your choice(s) using calculations. For the 1040steel: Elastic modulus is 205 GPa, Yield strength is 450 MPa, and Poisson’s ratio is 0.27. (b) If you are asked to perform a strain hardening process to increase the yield strength so the steel can be used in another application with larger force load. How much cold work would be required to reduce the diameter of the steel to 6.0 mm?
Answer:
A) 1040 steel is not a possible candidate for this application
B) 35.94%
Explanation:
Initial length = 100 mm = 0.1 m
Initial diameter ( d ) = 7.5 mm = 0.0075 m
Tensile load ( p ) = 18,000 N
Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m
A) would the 1040 steel be a possible candidate for this application
Yield strength of 1040 steel < stress ( in order to be a possible candidate )
stress = p / A0 = ( 18000 ) / ( [tex]\frac{\pi }{4}[/tex] ) * 0.0075^2
= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa
Yield strength of 1040 steel = 450 MPa
stress = 407.424 MPa
∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )
Therefore 1040 steel is not a possible candidate for this application
B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm
Area1 = ( [tex]\frac{\pi }{4}[/tex] ) ( 0.006 )^2 = 2.83 * 10^-5 m^2
therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%
Let S = { p q |p, q are prime numbers greater than 0} and E = {0, −2, 2, −4, 4, −6, 6, · · · } be the set of even integers. . Prove that |S| = |E| by constructing a bijection from S to E.
Answer:
prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S
Explanation:
Given that S = { p q |p, q are prime numbers greater than 0}
E = {0, −2, 2, −4, 4, −6, 6, · · · }
To prove by constructing a bijection from S to E
detailed solution attached below
After the bijection :
prove that | S | = | E | : every element of S there is an Image on E , while not every element on E has an image on S
∴ we can say sets E and S are infinite sets
3
Select the correct answer
Which statement is true about a corporation?
A
B.
The shareholders hold no liability for the corporation's debts.
The shareholders hold limited liability for the corporation's debts.
The shareholders hold complete liability for the corporation's debts.
C.
D.
There are no shareholders in a corporation.
OE.
A single individual owns the corporation,
Which industry does a shoe manufacturer belong?
20 points!!!! In this unit, we learned that, ideally, the reductionist and holistic approaches to engineering work together to ensure an optimal manufacturing process. We also discussed how problems may arise without a systems engineer to oversee the process. What if the tables were turned, however? What might happen if a systems engineer overlooked a detail that may have been discovered by a manufacturing engineer? Explain using an example.
Explanation:
When a problem is discovered in the system design or manufacturing process, it generally gets reported to all concerned (if the company has an appropriate culture). Depending on the nature of the problem, a "quick fix" may be found by the discoverer, or by someone to whom the discovery is reported. In some situations, what seems a simple problem requires a rethinking of the entire manufacturing process, possible product recalls, possible plant retrofits, and even larger ramifications. This can happen regardless of where along the line the problem is discovered.
__
Hypothetical example:
A test technician determines that a lithium battery charger sometimes gets confused and doesn't shut down properly--continuing to charge the battery when it should not. If the proximate cause is a wire harness routing or a component improperly installed, a manufacturing engineer may be called in to address the issue. The manufacturing engineer's investigation may determine that a number of battery chargers have been delivered with the problem. Further investigation may reveal a potential for fire in situations where injury or loss of life are possible outcomes.
Assessment of the design by manufacturing, process, and design engineers may reveal more than one potential cause of the shutdown/fire-hazard issue, and that the location and nature of any fire may release toxins or cause damage to systems and equipment beyond those in the immediate vicinity of the defective battery and/or charger.
Such ramifications may require the attention of one or more systems engineers and/or a rethinking of system failure modes and effects, including fire detection and suppression, throughout the product. The product may be effectively "grounded" (taken out of service), with possible customer revenue implications, until such time as the issues can be satisfactorily resolved.
(Note: 787 Dreamliner battery problems were caused by the physics of the battery construction, not the charger. The rest of the scenario above is not a bad match.)
What disadvantages can a resort come across
Answer:
the disadvantages can be people's thoughts about them and also a rival resort nearby which looks more posh pls mark brainliest i need 5 more for next rank thank you
Explanation:
A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at X kPa, and exits at 800 kPa and 60oC. The refrigerant leaves the condenser as saturated liquid at 800 kPa. The number of letters in your first name multiplied by 10 plus 50
Answer:
Hello your question has some missing information below are the missing information
The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump
answer : 2.49
Explanation:
For vapor-compression refrigeration cycle
P1 = P4 ; P1 = 140 kPa
P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa
From pressure table of R 134a refrigerant
h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg
h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )
= 296.8kJ/kg
h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg
also h4 = 95.47 kJ/kg
To determine the coefficient of performance
Cop = ( h1 - h4 ) / ( h2 - h1 )
∴ Cop = 2.49
We have a credit charge that is trying to process but we do not remember signing up and email login is not working? Is there a way to check?
Answer:
Yes
Explanation:
In such a case, one way to check the credit charge is to contact your bank, doing so would allow the bank to check your account properly to determine where the transaction was originated from.
Another way you could check is to contact the online merchant where such a transaction was initiated.
Q5) Write C++ program to find the summation of sines of the even values that can be divided by 7 between -170 and -137.
Answer:
In C++:
#include <iostream>
#include <cmath>
using namespace std;
int main(){
double total = 0;
for(int i = -170; i<=-130;i++){
if(i%2 == 0 && i%7==0){
double angle = i*3.14159/180;
total+=sin(angle); } }
cout<<"Total: "<<total;
return 0; }
Explanation:
This initializes the total to 0
double total = 0;
This iterates from -170 to - 130
for(int i = -170; i<=-130;i++){
This checks if the current number is even and is divided by 7
if(i%2 == 0 && i%7==0){
This converts the number from degrees to radians
double angle = i*3.14159/180;
This adds the sine of the number
total+=sin(angle); } }
This prints the calculated total
cout<<"Total: "<<total;
The irreversible losses in the penstock of a hydroelectric dam are estimated to be 7 m. The elevation difference between the reservoir surface upstream of the dam and the surface of the water exiting the dam is 140 m. If the flow rate through the turbine is 4000 L/min, determine (a) the power loss due to irreversible head loss, (b) the efficiency of the piping, and (c) the electric power output if the turbine-generator efficiency is 84 percent.
Answer:
a) the power loss due to irreversible head loss is 4.57 kW
b) the efficiency of the piping is 95%
c) the electric power output is 72.9918 kW
Explanation:
Given the data in the question below;
Irreversible loses [tex]h_L[/tex] = 7m
Total head, H = 140 m
flow rate Q = 4000 L/min = 0.0666 m³/s
Generator efficiency n₀ = 84% = 0.84
we know that density of water is 1000 kg/m³
g = 9.81 m/s²
a) power loss due to irreversible head loss [tex]P_L[/tex] is;
[tex]P_L[/tex] = p × Q × g × [tex]h_L[/tex]
we substitute
[tex]P_L[/tex] = 1000 × 0.0666 × 9.81 × 7
[tex]P_L[/tex] = 4573.422 W
[tex]P_L[/tex] = 4573.422 / 1000
[tex]P_L[/tex] = 4.57 kW
Therefore, the power loss due to irreversible head loss is 4.57 kW
b) the efficiency of the piping n is;
n = (Actual head / maximum head) × 100
n = (( H - [tex]h_L[/tex] ) / H) × 100
so we substitute
n = (( 140 - 7 ) / 140) × 100
n = (133/140) × 100
n = 0.95 × 100
n = 95%
Therefore, the efficiency of the piping is 95%
c) the electric power output if the turbine-generator efficiency is 84 percent;
n₀ = [tex]Power_{outpu[/tex] / [tex]power_{inpu[/tex]
[tex]Power_{outpu[/tex] = n₀ × [tex]power_{inpu[/tex]
[tex]Power_{outpu[/tex] = n₀ × ( pQg( H - [tex]h_L[/tex] ))
so we substitute
[tex]Power_{outpu[/tex] = 0.84 × ( 1000 × 0.0666 × 9.81( 140 - 7 ))
[tex]Power_{outpu[/tex] = 0.84 × 653.346( 133)
[tex]Power_{outpu[/tex] = 0.84 × 86895.018
[tex]Power_{outpu[/tex] = 72991.815 W
[tex]Power_{outpu[/tex] = 72991.815 / 1000
[tex]Power_{outpu[/tex] = 72.9918 kW
Therefore, the electric power output is 72.9918 kW
what technology has been used for building super structures
Answer: Advanced technologixal machines
Explanation: such as big cranes, multiple workers helping creat said structure, and big bull dozers
