Answer:
17060700 btu/h
Explanation:
Power output from power plant = 150 kW.
This power output is at 3% efficiency, this means power available in pond is 0.30P
Now, 150 kW = 0.03P
P = 150/0.03 = 5000 kW of power in the pond.
From basic conversion,
1 kW = 3412.14 btu/h
5000 kW = 5000 x 3412.14
average value of the required solar energy collection rate equal
17060700 btu/h
Solar pond power plant with 3% efficiency and 150 kW output needs ~17,060,000 Btu/h solar energy collection rate.
To find the average value of the required solar energy collection rate in Btu/h, we need to calculate the total solar energy collected per unit time and then convert it into Btu/h.
First, let's find the total solar energy collected per unit time:
Given:
- Net power output = 150 kW
- Efficiency = 3%
We know that efficiency is the ratio of output power to input power, so:
Efficiency = (Net Power Output / Solar Energy Input) * 100
Rearranging the equation to find the solar energy input:
Solar Energy Input = (Net Power Output / Efficiency) * 100
Now, let's substitute the given values:
Solar Energy Input = (150 kW / 3%) * 100
Solar Energy Input = (150,000 W / 0.03) * 100
Solar Energy Input = 5,000,000 W
Now, we convert this into Btu/h. To do this, we'll use the conversion factor:
1 watt = 3.412 Btu/h
So, the solar energy input in Btu/h will be:
Solar Energy Input (Btu/h) = (Solar Energy Input in watts) * (3.412 Btu/h per watt)
Solar Energy Input (Btu/h) = 5,000,000 W * 3.412 Btu/h per watt
Solar Energy Input (Btu/h) ≈ 17,060,000 Btu/h
Therefore, the average value of the required solar energy collection rate is approximately 17,060,000 Btu/h.
You need to display output for all of the values between the starting and ending values. First two values are temperatures in Fahrenheit. You need to display all of the values from the first temperature to the last temperature. You increment from one temperature to the next by the increment value (the third value you read in). You need to convert these temperatures to Celsius and Kelvin. You need to output the temperatures as Fahrenheit, Celsius, and Kelvin. The numbers should be 18 characters wide with 4 digits of precision and need to be in fixed format. Do not use tab characters (\t) to output the values.
Answer:
Check the explanation
Explanation:
#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
double temp1,temp3,inc,cel;
int i=1;
while(i==1)
{
i=0;
cin>>temp1>>temp3>>inc;
if(temp3<temp1||inc<=0)
{
i=1;
cout<<"Starting temperature must be <= ending temperature and increment must be >0.0\n";
}
}
cout<<endl;
cout<<setw(18)<<"Fahrenheit"<<setw(18)<<"Celsius";
while(temp1<=temp3)
{
cel=(temp1-32)/1.8;
cout<<endl;
cout<<fixed<<setprecision(4)<<setw(18)<<temp1<<setw(18)<<cel;
temp1+=inc;
}
}
A certain process requires 3.0 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded 90o elbows, determine the pipe diameter. Elevation differences are negligible.
Answer:
diameter of the pipe = 0.4932ft.
Explanation:
assuming d = 0.4932
Re = 3.16 x 10∧5/0.4932
= 6.4 x 10 ∧5
E/d = 0.0005/0.4932
= 0.0010 from moody chat t = 0.02
if 0.02 is beign substituted in equation 2 we will get the same required diameter of the pipe which is 0.4932ft.
check the attachment for better explanation.thanks
A four‐lane freeway (two lanes in each direction) operates at flow rate of 1700 during the peak hour. It has 11‐ft lanes, 4‐ft shoulders, and there are three ramps within three miles upstream of the segment midpoint and four ramps within three miles downstream of the segment midpoint. The freeway has only regular users. There are 8% heavy trucks, and it is on rolling terrain with a peak‐hour factor of 0.85. It is known that 12% of the AADT occurs in the peak hour and that the directional factor is 0.6.
What is the freeway’s AADT?
Answer:
Please see the attached file for the complete answer.
Explanation:
A photovoltaic panel of dimensions 6 m x 5 m is located on top of the roof of a house. Solar irradiation Gs = 900 W/m2 is incident on the panel. The panel has an absorptivity to solar irradiation, αs of 0.92. The freestream air temperature is Tinf and the surrounding temperature for radiation exchange with the sky is Tsurr. For this particular problem, it is given that Tinf= Tsurr. The convective heat transfer coefficient of air blowing over the panel is h W/m2-K.
Answer:
2.7 W/m^2K
Explanation:
Area of pane = 5 m x 6 m = 30 m^2
Solar irradiation Gs = 900 W/m2
Heat rate on panel = Gs x area = 900 x 30 = 27000 W
absorptivity to solar irradiation αs = 0.92
Therefore, absorbed heat is
0.92 x 27000 = 24840 W
For heat gain,
From E = §AT^4
Where § = stefan's constant = 5.7x10^-8 Wm^-2K^-1
T = temperature of panel
24840 = 5.7x10^-8 x 30 x T^4
24840 = 1.71x10^-6 x T^4
1.453x10^10 = T^4
T = 347.167 K
For net heat gain,
From E = §A(T^4 - T^4sur)
24840 = 5.7x10^-8 x 30 x (T^4 - T^4sur)
24840 = 1.71x10^-6 x (1.453x10^10 - T^4sur)
24840 = 24846.3 - 1.71x10^-6(T^4sur)
-6.3 = -1.71x10^-6(T^4sur)
3684210.526 = T^4sur
Tsur = 43.81 K
Also for convective heat,
E = Ah(T - Tsur)
24840 = 30h(347.167 - 43.81)
24840 = 30h x 303.357
81 = 30h
h = 2.7 W/m^2K
Given that the size of Elmer's property is 180 acres, and given that there are 20 volunteers, what would be a reasonable approach to surveying the land for houndstongue?
Answer:
Weed Mapping Application is the most reasonable approach for surveying.
Answer:
The most reasonable approach to survey the property would be a GPS survey using a dual-frequency GPS system.
Explanation:
Surveying makes use of coordinates to determine positions of objects, it is the first step towards developing a property. There are different types of survey which are specific to various needs and sizes of a property.
GPS SURVEY
GPS survey makes use of the global positioning system to determine positions of a property or object. It is quite accurate and also an efficient type of survey based on its flexibility.
In the illustration, the property size is 180 acres and 20 volunteers are expected to work on it. To carry out this survey, a station whose coordinate is known normally referred to the master station is used to get the coordinates of the remote stations. The volunteers are stationed at 20 remote stations and the surveyor at the master station makes use of dual-frequency GPS system to pick the coordinates of the remote stations. The dual-frequency GPS system would be used because it covers more distance and for a survey as massive as 180 acres this is reasonable.
For the survey to be carried out seamlessly by the 20 volunteers, a dual-frequency GPS system would be used for the GPS survey.
As a project manager of Permagam Construction you want to plan renting a fleet of 25 cu yd tractor-scrapers and have them hauling between the pit and a road construction job. The haul road is a rutted dirt road that deflects slightly under the load of the scraper. There is a slight grade of 5% from the pit at the fill location. The return road is level. The haul distance to the dump location is 0.90 miles and the return distance is 0.75 miles. The scrapers are push loaded in the pit. The cycle time for the pusher is 1.5 minutes and the cycle time for the scrapers is 8 minutes. Assume that the weight on the wheels is 75 tons (full) and 50 tons
(empty). (Use Tables 14.1 and 14.2)
What are the rolling resistances and grade resistance?
What are the effective grades?
How many scrapers you recommend to be rented? Explain.
What is the production of the system in case 4 scrapers would be rented?
Answer:
See explaination
Explanation:
Rolling resistance which in some occassions can be called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface.
In order to calculate our rolling resistance, there should be a force.
Please kindly check attachment for the step by step solution of the given problem.
Calculate the headloss through a filter bed consisting of 30.0 in. of stratified sand with the gradation given below. Assume a filtration rate of 5.0 gpm/ft2, a clean bed porosity of 0.42 and a temperature of 40i F.
U.S. Sieve Number 50
40 30 20 18 16
Sieve Opening (mm) 0.297
0.420 0.590 0.840 1.000 1.190
Percent Passing (by wt.) 0.10
6.50 22.00 76.00 90.00 98.00
Answer:
1.5258m
Explanation:
Please see attachment
Determine the average power, complex power and power factor (including whether it is leading or lagging) for a load circuit whose voltage and current at its input terminals are given by: v(t) = 100cos(377t-30)v, i(t) = 2.5cos(377t-60)A
Answer:
Average power = 108.25W
Complex power = 125W
PF = 0.866
Explanation:
Check attachment for step by step instructions.
Answer:
Average power: 108.25 W
Complex power: 125 VA
Power factor: 0.866
Refer below for the explanation.
Explanation:
Refer to the picture for brief explanation.
A cylinder fitted with a movable piston contains water at 3 MPa with 50% quality, at which point the volume is 20 L. The water now expands to 1.2 MPa as a result of receiving 600 kJ of heat from a large source at 300◦C. It is claimed that the water does 124 kJ of work during this process. Is this possible?
Answer:
The process is possible:
Explanation:
We are going to find out if the entropy generated is greater than 0, if it is greater than 0, then the process is feasible. If it is not, the process is not feasible.
[tex]P_{1} = 3 MPa[/tex]
[tex]x_{1} = 50 % = 0.5[/tex]
[tex]V_{1} = 20 L = 0.02 m^{3}[/tex]
[tex]P_{2} = 1.2 MPa[/tex]
[tex]T_{H} = 300^{0} C = 573 K[/tex]
Received heat energy, [tex]Q_{12} = 600 kJ[/tex]
Work done, [tex]W_{12} = 124 kJ[/tex]
At state 1, using the steam table:
[tex]T_{1} = T_{s} = 233.9^{0} C\\v_{f1} = 0.001216 m^{3} /kg\\v_{fg1} = 0.06546m^{3} /kg\\u_{f1} = 1004.76 kJ/kg\\u_{fg1} = 1599.34 kJ/kg\\s_{f1} = 2.6456 kJ/kg-K\\s_{fg1} = 3.5412kJ/kg-K[/tex]
[tex]v_{1} = v_{f1} + x_{1} * v_{fg1}[/tex]
[tex]v_{1} = 0.001216 + 0.5*(0.06546)\\v_{1} = 0.03395 m^{3} /kg[/tex]
[tex]M = \frac{V_{1} }{v_{1} } \\M = 0.02/0.03395\\M = 0.5892 kg[/tex]
[tex]u_{1} = u_{f1} + x_{1} * u_{fg1}\\u_{1} = 1004.76 + 0.5*1599.34\\u_{1} = 1804.43 kJ/kg[/tex]
[tex]s_{1} = s_{f1} + x_{1} * s_{fg1}\\s_{1} = 2.6456 + 0.5*3.5412\\s_{1} = 4.4162 kJ/kg[/tex]
[tex]Q_{12} = m(u_{2} - u_{1} ) + W_{12} \\600 = 0.5892(u_{2} -1804.43) + 124\\[/tex]
Solving for u₂
[tex]u_{2} = 2612.3 kJ/kg[/tex]
Since P₂ = 1.2 MPa, u₂ = 2612.2 kJ/kg,
then from steam table, T₂ = 200°C, S₂ = 6.5898 kJ/kg-K
The entropy generated will be:
[tex]\triangle S = m(S_{2} -S_{1} ) - \frac{Q_{12} }{T_{H} }\\ \triangle S= 0.5892(6.5898 - 4.4162) - \frac{600 }{573 }\\ \triangle S =0.233 kJ/K[/tex]
Since ΔS > 0, this process is possible
Answer:
Yes it is possible
Explanation:
Attached is the solution
Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow. Who is right?
Answer:
A
Explanation:
Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow Person A is right.
Who is technician?A technician is a worker in the technology industry who possesses the necessary knowledge, abilities, and skills as well as a practical comprehension of the theoretical underpinnings.
Expert technicians in a certain tool domain often have expert competency in technique and a moderate comprehension of theory. As a result, technicians in that field of technology are typically much more knowledgeable about technique than the average layperson and even general professionals.
For instance, although not as knowledgeable in acoustics as acoustical engineers, audio technicians are more adept at using sound equipment and are likely to know more about acoustics than other studio staff members, such as performers.
Therefore, Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow Person A is right.
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A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mechanics is applicable. Each particle has a mass m and is free to perform one-dimensional oscillations about its equilibrium position. Calculate the heat capacity of this system of particles at this temperature in each of the following cases:
Answer:
the restoring force is = 3/4NKT
Explanation:
check the attached files for answer.
A technician connects a voltmeter in parallel across two points in a circuit. Technician A says this will provide a reading of the potential difference in volts. Technician B says that this will also show the amperage flowing in the circuit. Who is correct?
A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. Technician A is correct while Technician B is wrong.
What is a voltmeter?A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. It is linked in parallel. It generally has a high resistance and draws very little current from the circuit.
Given the technician connects the voltmeter in parallel across two points in a circuit. Therefore, the technician A is correct the meter will provide a reading of the potential difference in volts.
Hence, Technician A is correct while Technician B is wrong.
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// global variables int a = 5, b = 6, c = 7, d = 8; void sub2() { int a = 0, b = 3; //local variables { int b = 5; //local variable System.out.println("a=" + a); System.out.println("b=" + b); System.out.println("c=" + c); System.out.println("d=" + d); } } void sub1() { int a = 2, b = 4, c = 1; //local variables sub2(); } void main() { int a = 1, b = 2, c = 3, d = 4; //local variables sub1(); }\
Answer:
1)
Static scoping
A=0
B=5
C=3
D=4
2) For dynamic scoping
A=2
B=5
C=1
D=8
Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.28 mm and that has a tip radius of curvature of 0.002 mm when a stress of 1430 MPa is applied.
Answer:
theoretical fracture strength = 16919.98 MPa
Explanation:
given data
Length (L) = 0.28 mm = 0.28 × 10⁻³ m
radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m
Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa
solution
we get here theoretical fracture strength s that is express as
theoretical fracture strength = [tex]s_{0} \times \sqrt{\frac{L}{r} }[/tex] .............................1
put here value and we get
theoretical fracture strength = [tex]1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }[/tex]
theoretical fracture strength = [tex]16919.98 \times 10^6[/tex]
theoretical fracture strength = 16919.98 MPa
Water flows inside a smooth circular thin-walled tube of diameter D = 25 mm at a mass flow rate of 50 g/s. Outside of the tube, air moves in cross flow over the tube at a velocity of V = 20 m/s and a temperature of T[infinity] = 10°C. If the mean temperature of the water is Tm = 50°C, determine (a) The Darcy friction factor for the water flow inside the tube
Answer:
See explaination
Explanation:
We can say that that the The Darcy Friction factor or Equation is a theoretical equation that predicts the frictional energy loss in a pipe based on the velocity of the fluid and the resistance due to friction. It is used almost exclusively to calculate head loss due to friction in turbulent flow.
Please kindly check attachment for the step by step solution of the given problem.
4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and plot it. Determine the runoff hydrograph for a 30-minute storm where there is 4 cm of runoff in the first 10 min, 2.5 cm of runoff in the second 10 min, and 2 cm of runoff in the third 10 min. Plot the runoff hydrograph for each 10-minute rainfall excess along with the aggregated (total) runoff hydrograph on the same axes. Also report the peak flow of the aggregated runoff hydrograph. Also report the total volume of runoff. You do not need to report any tables of data.
Answer:
The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s
Peak flow of the aggregated runoff hydrograph is 420.58 m³/s
The total volume of runoff is 2125000 m³/s
Explanation:
We have
A = 25 km²
tr = 10 min = 1/6 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr
Qp = 2.08×25/1.043 = 49.84 m³/s
Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr
Since the area is
Time (min) Runoff (cm) Volume of runoff m³
0 0 0
10 4 1000000 m³
20 2.5 625000 m³
30 2 500000 m³
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
For the 1st 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×4/1.043 = 197.92 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 2nd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 123.7 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 3rd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 98.96 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
Peak flow of aggregate runoff is given by
Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s
Total volume of runoff is given by
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the Mach number at the exit is 0.3, determine the direction and the rate of heat transfer. For the same conditions at the inlet, determine the amount of heat that must be transferred to the system if the flow is to be sonic at the exit of the duct.
Answer:
a. 318.2k
b. 45.2kj
Explanation:
Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.
See attachment for detailed analysis
investigation, determine when the Elastic Potential Energy is zero. Make sure you test your idea with several masses, all three springs and vary the stiffness of spring three. Write down how you determined the zero location(s) and explain why the position for zero makes sense.
Elastic Potential Energy is zero detailed description is given below.
Explanation:
It is the energy stored in stretched or compressed elastic materials. This also means that elastic potential energy is zero in objects that have not been stretched or compressed.To determine the gravitational potential energy of an object, a zero height position must first be arbitrarily assigned. Typically, the ground is considered to be a position of zero height. But this is merely an arbitrarily assigned position that most people agree upon. Since many of our labs are done on tabletops, it is often customary to assign the tabletop to be the zero height position. Again this is merely arbitrary. If the tabletop is the zero position, then the potential energy of an object is based upon its height relative to the tabletop. For example, a pendulum bob swinging to and from above the tabletop has a potential energy that can be measured based on its height above the tabletop. By measuring the mass of the bob and the height of the bob above the tabletop, the potential energy of the bob can be determined. Potential energy is the energy that is stored in an object due to its position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height. An object possesses elastic potential energy if it is at a position on an elastic medium other than the equilibrium position.
Since the gravitational potential energy of an object is directly proportional to its height above the zero position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling of the height will result in a tripling of the gravitational potential energy.
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This is a multi-part question. Once an answer is submitted, you will be unable to return to this part As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.
Determine the angular velocity at t20 s. (You must provide an answer before moving on to the next part)
The angular velocity is [ ] rpm.
Answer:
60 rpm
Explanation:
At t = 0,
Angular speed = 0
At t = 10 sec
Angular speed = 20/10 = 2 rev/s
Average speed = (2 - 0)/2 = 1 rev/s
= 1 x 60 = 60 rpm
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain. Assume elastic modulus of Cu to be 110GPa. (Points: 5).
Answer:
The resulting strain is [tex]1.39\times 10^{-3}[/tex].
Explanation:
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm
Force, F = 44,500 N
Th elastic modulus of Cu to be 110 GPa
The resulting strain is given by the formula as follows :
[tex]\epsilon=\dfrac{F}{AE}[/tex]
E is elastic modulus of Cu is are of cross section
[tex]\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}[/tex]
So, the resulting strain is [tex]1.39\times 10^{-3}[/tex].
Select all of the true statements.
A. Diodes are used in voltage regulators and limiters.
B. Different diode models are used to replace the real diode with a simpler version that approximates the i-v characteristics of the real diode.
C. In the ideal diode plus voltage source model, the forward bias region is characterized by VD
Answer:
Diodes consists of two-terminal electronic parts which conducts current mainly in one direction. It has low resistance in one direction and high resistance in the other direction. These are correct statements about a Diode.
A. Diodes are used in voltage regulators and limiters.
C. In the ideal diode plus voltage source model, the forward bias region is characterized by VD
Explanation:
Diodes
Steam enters an insulated pipe at 200 kPa and 200°C and leaves at 150 kPa and 150°C. The inlet-to-outlet diameter ratio for the pipe is D1/D2 = 1.80. Determine the inlet and exit velocities of the steam.
Answer:
Inlet and exit velocities are 143.71 m/sec and 465.697 m/sec
Explanation:
At inlet of the pipe
[tex]P_1=150kPa[/tex] and [tex]T_1=150^{\circ}C[/tex]
At this pressure and temperature from steam table.
[tex]h_1=2870kj/kg[/tex] and [tex]s_1=7.508kj/kgK[/tex]
At pressure [tex]P_2=200kPa[/tex] and [tex]T_2=200^{\circ}C[/tex]
By steam table from interpolation method.
[tex]h_2=2776.38+(2768.80-2776.38)(\frac{150-100}{200-100})[/tex]
[tex]h_2=2772.59kj/kg[/tex]
[tex]Q=A_1V_1=A_2V_2[/tex]
[tex]D_1^2V_1=D_2^2V_2[/tex]
We have given [tex]\frac{D_1}{D_2}=1.80[/tex]
[tex]\frac{V_1}{V_2}=\frac{D_2^2}{D_1^2}[/tex]
[tex]\frac{V_1}{V_2}=(\frac{1}{1.80})^2[/tex]
[tex]V_1=0.3086V_2[/tex]
Now energy equation in the pipe
[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}[/tex]
[tex]2870.7\times 10^3+\frac{0.3086V_2^2}{2}=2772.59\times 10^3+\frac{V^2}{2}[/tex]
[tex]V_2=465.697m/sec[/tex]
[tex]V_1=0.3086\times 465.697=143.714m/sec[/tex]
(a) Design a half-subtractor circuit with inputs x and y and outputs Diff and B out . The circuit subtracts the bits x – y and places the difference in D and the borrow in B out .
(b) Design a full-subtractor circuit with three inputs x , y , B in and two outputs Diff and B out . The circuit subtracts x – y – B in , where B in is the input borrow, B out is the output borrow, and Diff is the difference.
Answer:
See the attachment
Explanation:
A half subtractor circuit is made up of on NOT gate, one XOR and one AND gate.
A full subtractor is made up of two half subractor with their outputs in an OR gate for Bout as shown in attachment
The purification of hydrogen gas by diffusion through a palladium sheet. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m2 at 500°C. Assume a diffusion coefficient of 1.0 × 10-8 m2/s and that the concentrations at the high and low pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per m3 of palladium. Assume steady state conditions.
Answer:
The mass of the hydrogen for one hour would be 0.0039168 kg/hr
Explanation:
The concentration and the distance at concentration point is calculated before calculating the mass of hydrogen. The attached images show a clear explanation;
The mass of the hydrogen for one hour would be 0.0039168 kg/hr. The derivation of the diffusion coefficient has been attached in the image below:
According to Fick's law, the mass of a substance dM is proportional to the concentration gradient grad c of this substance as it diffuses in time dt over a surface dF normal to the diffusion direction: dM = D grad c dF dt.
Physically, the diffusion coefficient, therefore, suggests that for a concentration gradient of unity, the mass of the substance diffuses through a unit surface in a unit of time. A square meter per second corresponds to D in the SI system. Physical constants such as temperature, pressure, and the size of the dispersing substance's molecules all affect the diffusion coefficient.
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Your program this week will have the same output as lab 10, except that instead of redirecting the filename for the input file, you will get the filename from the command-line, and use a file pointer to open the file. You will also dynamically allocate space in memory after reading in the first value from the file indicating the number of lean proteins that are in the file.
Answer:
Check the explanation
Explanation:
defs.h
#ifndef DEFS_H
#define DEFS_H
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
char item[20];
char quantity[10];
int calories;
float protein;
float carbs;
float fats;
} food;
int size;
void printArray(int size, food arr1[]);
#endif
arrayProcessing.c
#include "defs.h"
void printArray(int size, food arr1[])
{
int i = 0;
printf("\nFOOD ITEM\t\tQUANTITY\tCALS\tPRO\tCARBS\tFAT");
for (i = 0; i < size; i++)
{
printf("\n%i.%s", i + 1, arr1[i].item);
printf("\t\t%s", arr1[i].quantity);
printf("\t\t%i", arr1[i].calories);
printf("\t%.2f", arr1[i].protein);
printf("\t%.2f", arr1[i].carbs);
printf("\t%.2f\n", arr1[i].fats);
}
}
lab12.c
#include "defs.h"
int main(int argc, char *argv[])
{
int i = 0;
FILE *inFile;
inFile = fopen(argv[1], "r");
if(inFile==NULL){
fprintf(stderr, "File open error. Exiting program\n");
exit(1);
}
fscanf(inFile, "%i", &size);
food *arr = (food *)malloc(sizeof(food)*size);
for (i = 0; i < size; i++)
{
fscanf(inFile, "\n%[^\n]s", arr[i].item);
fscanf(inFile, "%s", arr[i].quantity);
fscanf(inFile, "%i", &arr[i].calories);
fscanf(inFile, "%f", &arr[i].protein);
fscanf(inFile, "%f", &arr[i].carbs);
fscanf(inFile, "%f", &arr[i].fats);
}
printArray(size, arr);
return 0;
}
Kindly check the Output below,
A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to be 35 mm. The screw is to be used to lift and lower a load of 7 kN. A coefficient of friction of 0.05 is to be used for friction at the thread and at the collar. Determine the following: (i) Torque required to raise the load, TR, (Equation 8.1) (ii) Torque required to lower the load, TL, (Equation 8.2) (iii) A conservative estimate of self locking condition is to set TL in equation 8.2 to zero and calculate the minimum coefficient of friction to ensure self locking. What is the minimum coefficient of friction to ensure self locking
Answer:
i) The torque required to raise the load is 15.85 N*m
ii) The torque required to lower the load is 6.91 N*m
iii) The minimum coefficient of friction is -0.016
Explanation:
Given:
dm = mean diameter = 0.03 m
p = pitch = 0.004 m
n = number of starts = 1
The lead is:
L = n * p = 1 * 0.004 = 0.004 m
F = load = 7000 N
dc = collar diameter = 0.035 m
u = 0.05
i) The helix angle is:
[tex]tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43[/tex]
The torque is:
[tex]T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm[/tex]
ii) The torque to lowering the load is:
[tex]T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm[/tex]
iii)
[tex]T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\ 0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}[/tex]
Clearing u:
u = -0.016
Let f(t) be an arbitrary signal with bandwidth Ω. Determine the minimum sampling frequencies ωs needed to sample the following analog signals without causing aliasing error. (a) f1(t) = f(t) sin(4000πt) (b) f2(t) = f(t) ∗ sin(4000πt) (c) f3(t) = f(t) ∗ f(sample the following analog signals without causing aliasing error
Answer:
See explaination
Explanation:
We can describr Aliasing as a false frequency which one get when ones sampling rate is less than twice the frequency of your measured signal.
please check attachment for the step by step solution of the given problem.
5. A straight round shaft is subjected to a torque of 5000 lb - in. Determine the required diameter, using steel with a tensile yield strength of 60 ksi and a safety factor of 2 based on initial yielding: (a) According to the maximum-normal-stress theory. (b) According to the maximum-shear-stress theory. (c) According to the maximum-distortion-energy theory. Discuss briefly the relative validity of the three predictions.
Answer:
a. 0.95 in
b. 1.19 in
c. 1.137 in
Explanation:
Express the factor of safety equation for maximum-normal-stress theory as:
S SF = Eau
Here, the factor of safety is SF, the yield strength is S„ and the maximum stress :I
Modify the above equation for shear stress acting on the solid rod as:
S. SF = To
Here, the combined shear stress on solid rod s
Calculate the combined shear stress for solid rod.
16T r2:1 = trd3
(1)
Here, the torque is T, and the diameter of the solid rod is d.
Substitute 5,000 'bin. for T.
— 16(5,000 lb • in ) v ird 80 000lb - in. _
rd3
60ksi 2 — 80,000lb -in. trd
Solve the above equation for d.
60 x 1031bAn? 80,00016 -in. ird3 — 2(80, 000 lb in.) d3 rrt60x103Ibfln?)
v3 d —[ 2(80,000lb-in.) rz-(60 x103112,An?) = 0.8488 in.3r3 =0.95 in.
check the attached files for clear cut details
When will stemuless checks come I was told not to file my taxes because I get a pension check every month direct deposit money into my bank account is this true I don't have to do my taxes because of this
The IRS and the U.S Department of the treasury declared that social security recipients are not required to file a simple tax return to receive stimulus payments under the CARES Act.
Explanation:
Due to the impact of corona virus problem, the CARES Act calls for stimulus payment to be sent to Americans based on their gross income.
The Social security recipients are not required to file a tax return and do not take action and they will receive the payments directly to their bank accounts.
The Automatic payments will begin by next week. The eligible taxpayers who filed tax returns for 2019 or 2018 and chose direct deposit for their refund will automatically receive a stimulus payment of $1,200 for individuals or $2,400 for married couples and $500 for each qualifying child.
A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are: The year must be divisible by 4 If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400 Some example leap years are 1600, 1712, and 2016. Write a program that takes in a year and determines whether that year is a leap year. Ex: If the input is 1712, the output is: 1712 is a leap year. Ex: If the input is 1913, the output is: 1913 is not a leap year.
Answer:
Explanation:
def is_leap_year(year):
if(year % 400 == 0):
return True
elif year % 100 == 0:
return False
elif year%4 == 0:
return True
else:
return False
if __name__ == '__main__':
n = int(input())
if(is_leap_year(n)):
print(n,"- leap year")
else:
print(n, "- not a leap year")
check the attachment for the output
Answer:
def is_leap_year(year):
if(year % 400 == 0):
return True
elif year % 100 == 0:
return False
elif year%4 == 0:
return True
else:
return False
if __name__ == '__main__':
n = int(input())
if(is_leap_year(n)):
print(n,"is a leap year.")
else:
print(n, "- not a leap year")
Explanation: