Answer:
See the explanation for the answer
Explanation:
Check to determine the redesigned beam is satisfactory for cracking:
crack width is controlled by establishing a minimum spacing.
Steps followed to check for cracking in the beams
i) According to aci code 10.6.7 if depth of beam isgreater than 36 in then skin reinforcement has to provide.the skin reinforcement to be provided should be such that it should not be greater than the actual main tension reinforcement.
ii) In second step steel stress is determined:
steel stressfs=Ms/(As*(d-hf/2)) or fs= 0.60fy
where Ms=service load moment
As*(d-hf/2)=area of reinforcement *moment area
iii) S=540/fs-2.5Cc is less than or equal to 12*(36/fs)
here Cc = clear spacing
if S=center to center spacing is with in the limit as specified above then the cracking is with in the control if not then redesign has to done.
In the given problem the data given is :
grade of steel in desigened beam is 75 and in redesigned beam is 60 so the stress in steel is 75*0.6=45ksi and 0.6*60=36ksi respectively
now the spacing is calculated for the two design and redesigned beams
the center to center spacing is given by S=540/fs-2.5Cc
For designed beam
S=540/45-(2.5*2.25)=6.375in which is less than 12*36/fs=12*36/45=9.6in hence it is safe
For redesigned beam
S=540/36-(2.5*2.25)=9.375in and it is less than the maximum spacing which is given by 12*36/fs=12*36/36=12in
Hence, the beam is within the limits and the beam is safe against the cracking.
Modifications to reduce the number of reinforcing bars
The addition of steel does not prevent cracking due to restrained shrinkage but it limits the width of crack by causing the formation of the number of narrow cracks rather than single wide crack.
Larger size bars leads to fewer cracks but wider cracks while smaller size bars leads to number of narrow cracks hence it is advisable to provide number of smaller diameter bars of equal strength of designed bars rather than larger bars.
The brakes on a vehicle work OK for a while, then the vehicle slows because the brakes self- applied. Technician A says that an overfilled master cylinder could be the cause. Technician B says that a blocked vent port (compensating port) could be the cause. Which technician is correct?
A. Technician A only
B. Technician B only
C. Both technician A and B
D/ Neither technician A nor B
Answer:
C. Both technician A and B
Explanation:
If the master cylinder is overfilled it will not allow enough room for the brake fluid to expand due to heat expansion. This blocks the vent port. If a vent port is not open, brake fluid pressure will increase as brakes heat up. This will cause the brakes to self apply, cause more heat in the brake fluid and the vehicle will slow down.
There, we can conclude that Both technician A and B are correct.
Answer:
C. Both technician A and B
Explanation:
The event that made both cylinders to be over filled especially the master cylinder and the blocking of the vent port, this will cause the vehicle brake to apply itself after just a little motion of the vehicle.
Therefore both technicians are correct from the information given above.
Hence, we can boldly say the correct answer is C. ie Both technician A and B
Which of these actions can be taken to minimize number of victims or prevent injury? (check all that apply) A. Hire good lawyers to defend those responsible for the disater. B. Regular inspections of man-made structures by licensed inspectors. C. Alarm systems and plans for evacuation and shelter of all communities neighboring a site of a potential disaster. D. Include fail-safe mechanisms in the design of life-critical systems
Answer:
The options that apply are:
B, C and D.
Explanation:
There have been a number of accidents all over the world resulting from Acts of God, professional negligence amongst other things.
These may not be avoided completely but the actions above speak to how they can be mitigated or reduced.
Cheers!
The mathematical form of the stream-function for 2-D SSSF satisfies the Laplace Equation inside a 2-D region in space (or inside a control volume) with specific boundary conditions on the entire control surface enclosing the region, The values for the velocity components at any given point in the flow field are determined from:
A. The velocity components at arty point are found by solving the clinical Armour, equation
B. The velocity components at any pant are found by integrating the stream function around the boundary (control surface)
C. The velocity components at any pant are found by integrating the stream functran over the entire flow field region (control volume)
D The velocity components at any point are found from the first partial derivatives of the stream function at the given point
Answer:
D. The velocity components at any point are found from the first partial derivatives of the stream function at the given point
Explanation:
The stream function can be used to plot the streamlines of the flow and find the velocity. For two-dimensional flow the velocity components can be calculated in Cartesian coordinates by u = −∂ψ/∂y and v = ∂ψ/∂x,
where u and v are the velocity vectors (components) in the x and y directions, respectively, ψ is the stream function.
From the above, the velocity components at any point are found from the first partial derivatives of the stream function at the given point.
Utilizing the charge neutrality equation and the mass-action law derive an expression for the concentration of electrons and holes present inside a p-type compensated semiconductor material. How can the resulting expression be modified for practical cases where (????????AA − ????????DD) ≫ 2nn�
Answer:
The expression for the concentration of electrons is P = NA - ND
Explanation:
Please look at the solution in the attached Word file
A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.5 mm long and that the value of Y is 1.1, which of the following can be said about this plate when a tensile stress of 1000 MPa is applied?
A. It is not possible to determine whether or not the plate will fracture.
B. The plate will definitely not fracture
C. The plate will definitely fracture.
Answer:
option B is correct. Fracture will definitely not occur
Explanation:
The formula for fracture toughness is given by;
K_ic = σY√πa
Where,
σ is the applied stress
Y is the dimensionless parameter
a is the crack length.
Let's make σ the subject
So,
σ = [K_ic/Y√πa]
Plugging in the relevant values;
σ = [50/(1.1√π*(0.5 x 10^(-3))]
σ = 1147 MPa
Thus, the material can withstand a stress of 1147 MPa
So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.
You are asked by your college crew to estimate the skin friction drag in their eight-seat racing shell. The hull of the shell may be approximated as half a circular cylinder with 450 mm diameter and 7.32 m length. The speed of the shell through the water is 6.71 m/s. Estimate the location of the transition from laminar to turbulent flow in the boundary layer on the hull of the shell. Calculate the thickness of the turbulent boundary layer at the rear of the hull. Determine the total skin friction drag on the hull under the given conditions
Answer:
The total skin friction drag on the hull under these conditions is 276N
Explanation:
In this question, we are asked to determine the total skin friction drag on the hull under the given conditions.
Please check attachment for complete solution and step by step explanation
You’re engineering an energy-efficient house that will require an average of 6.85 kW to heat on cold winter days. You’ve designed a photovoltaic system for electric power, which will supply on average 2.32 kW. You propose to heat the house with an electri-cally operated groundwater-based heat pump. What should you specify as the minimum acceptable COP for the pump if the pho-tovoltaic system supplies its energy?
Answer:
2.95 approximately 3
Explanation:
For a heat pump,
COP = Q/W
Where Q = power needed for heating process
W = power input into heat pump.
Power for heating Q = 6.85 kW
Proposed power input to heat pump W = 2.32 kW
Minimum COP = 6.85/2.32 = 2.95
Approximately 3
Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.
Part A
Determine the specific energy.
Express your answer to three significant figures and include the appropriate units.
E =
SubmitRequest Answer
Part B
Determine the alternate depth that provides the same specific energy for the same volumetric flow.
Express your answer to three significant figures and include the appropriate units.
Answer:
specific energy = 2.65 ft
y2 = 1.48 ft
Explanation:
given data
average speed v = 6.5 ft/s
width = 5 ft
depth of the water y = 2 ft
solution
we get here specific energy that is express as
specific energy = y + [tex]\frac{v^2}{2g}[/tex] ...............1
put here value and we get
specific energy = [tex]2 + \frac{6.5^2}{2\times 9.8\times 3.281}[/tex]
specific energy = 2.65 ft
and
alternate depth is
y2 = [tex]\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})[/tex]
and
here Fr² = [tex]\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}[/tex]
Fr² = 0.8025
put here value and we get
y2 = [tex]\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})[/tex]
y2 = 1.48 ft
Water at 20◦C is pumped through 1000 ft of 0.425 ft diameter pipe at a volumetric flowrate of 1 ft3/s through a cast iron pipe that connects to connects two reservoirs. The elevation difference between the two reserviors is 120 ft. Find the pumping power delivered to the water. The minor losses only include a wide-open globe valve with KL = 10. Use a density of 1.94 slug/ft3 and a dynamic viscosity of 2.32 × 10−5 lbf·s/ft2 .
Answer:
7582.9 ft.Ibf/s
Explanation:
Given
L=1000ft,d=0.425ft,Q=1ft^3/s,z2-z1=120ft,Kl=10,d=1.94slug/ft^3, vicosity u= 2.32*10-5ibf.s/ft2
Reynold Re= Density*diameter*velocity/ viscosity
But Q=AV
V= 4/3.142*0.425=2.99ft/s
Re= 1.94*0.425*2.99/2.32*10-5)=106455.3
Friction factor=1/√f=-1.8log[((e/d)/3.7)^1.11+6.9/Re] is very neglible hence equals 0
Pump head Hp= z2-z1+v^2/2g[FL/f+KL]
Hp=120+2.99^2/2*32.2(0+10)=121.4ft
Pump power = density*g*Q*hp
1.94*32.2*121.4=7582.9 ft.Ibf/s
The high electrical conductivity of copper is an important design factor that helps improve the energy efficiency of electric motors. This is important because motors and motor-driven systems are significant consumers of electricity, accounting for 43% - 46% of all global electricity consumption and 69% of all electricity used by industry. Inefficient motors waste electrical energy and are indirect contributors to greenhouse gas emissions. ElectroSpark, Inc. has been developing a new copper die-cast rotor technology specifically for premium efficiency motors, replacing the standard aluminum rotor. There are multiple reasons for doing so, including the possibility that the motor will consume less energy. They designed an experiment to test their idea in a common ¾ Horse power (HP) motor that is normally manufactured with an aluminum rotor. They designed a copper rotor that fit in their ¾ HP motor housing and ran a production line for a day producing the motors. They randomly selected 20 copper-rotor motors from that output and 20 aluminum-rotor motors produced from the same line the day before. These 40 motors were all run for 8 hours a day for 30 days and the energy consumed was measured in total Kilowatt Hours (example data below, using alpha=.05):
Copper: 560.145 539.673 556.834 559.873
Aluminium: 564.674 573.912 553.385 574.078
What is the correct hypothesis to test the problem described in this scenario?
A. H0: μD (copper-aluminum) ≥ 0; H1: μD (copper-aluminum) < 0
B. H0: μ_copper – μ_aluminum ≥ 0; H1: μ_copper – μ_aluminum < 0
C. H0: μD (copper-aluminum) ≥ 0; H1: μD (copper-aluminum) > 0
D. H0: μ_copper – μ_Aluminum ≤ 0; H1: μ_copper – μ_aluminum > 0
Answer:
B
Explanation:
This is a two sample t-test and not a matched pair t-test
null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors
alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.
So, option D is rejected
The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.
Option A and C are also rejected
A minor road intersects a major 4-lane divided road with a design speed of 55mph and a median width of 8 feet. The intersection is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 3%.
Answer:
stopping sight diatnce is =158 meters
Explanation:
The Design Speed =55 MPH(Miles per Hour)
1 miles per hour =0.447 meter pe second
then 55MPH=55*0.447=24.58 meter per second
Consider recation time =2.5 second
Consider Co efficinet friction is 0.35
As per the Stopping sight distance based on the Breaking distance+ Lag distance
SSD=Vt+(V^2/(2gf))
SSD=24.58+(24.58^2/(2*9.81*0.35))
The total stopping sight distance is =150 meters
if we consider the Approaching garde on the minor road is 3%
SSD=Vt+(V^2/(2g(f-n/100))
SSD=24.58+(24.58^2/(2*9.81*(0.35-3/100))
then stopping sight diatnce is =158 meters
See attachment for workings
The rainfall rate in a certain city is 20 inches per year over an infiltration area that covers 33000 acres. Twenty percent of the rainfall percolates into the groundwater, with the remaining 80% running off into the river. The city uses 83000 acre-ft per year, some of which comes from the river and the remainder from groundwater. The groundwater volume of fresh water is currently at 1.1 × 105 acre-ft and is expected to last for 30 years before being completely depleted of fresh water (assume uniform withdrawal each). Assuming the groundwater lasts exactly as expected, determine the rate at which water is being withdrawn from the river by the city.
Answer:
The rate at which water is being withdrawn from the river by the city is 57353 acre-ft/y
Explanation:
Please look at the solution in the attached Word file
Diesel-fueled generators are frequently used as backup electrical power sources for homes and hospitals. Consider a Diesel powered generator with an efficiency of 39 percent for an engine speed of idle to about 1,800 rpm (revolutions per minute). If Diesel fuel has a chemical formula of C12H23.
a. Determine the chemical reaction for 1 kmol of Diesel fuel burning with the stoichiometric amount of air.
b. For each kg of Diesel fuel burned, how much CO2 is generated, in kg?
c. Find the higher heating value (HHV) for C12H23 at 25°C, 1 atm.
d. Calculate the amount of Diesel fuel, in kg and in gallons, required to produce a power output of 18 kW to a home for a period of 8 h.
e. Comment on your results.
Answer:
a. C12H23 + 84.5 moles of air —-> 12CO2(g)+ 11.5H2O(g)
b. 3.2kg of CO2 per 1kg of C12H23
c. HVV of C12H23 is -1724.5 KCal/mol
d. Total weight required is 30.742kg
e. The amount of CO2 produced per kg of C12H23 is too much. CO2 is harmful to the environment and should be produced in weights as low as possible
Explanation:
Please check attachment for complete solution and step by step explanation
thermodynamics A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power of 3 MW. Steam exits the reactor core at 100 bar, 620°C and expands through the turbine to the condenser pressure of 1 bar. Saturated liquid exits the condenser and is pumped to the reactor pressure of 100 bar. Isentropic efficiencies of the turbine and pump are 87% and 78%, respectively. Cooling water enters the condenser at 15°C with a mass flow rate of 114.79 kg/s. Determine: (a) the percent thermal efficiency. (b) the temperature of the cooling water exiting the condenser, in °C.
Answer:
(a) the percent thermal efficiency is 27.94%
(b) the temperature of the cooling water exiting the condenser is 31.118°C
Explanation:
A piston-cylinder assembly contains 2 lb of air at a temperature of 540 °R and a pressure of 1 atm. The air compressed to a state where the temperature is 840 °R and the pressure is 6 atm. During the compression, there is a heat transfer from the air to the surroundings equal to 20 Btu.
1. Using the ideal gas model for air, determine the work during the process in Btu.
Answer:
123.9 Btu
Explanation:
The energy balance on the air is:
∆E = E2 − E1 = ∆KE + ∆PE + ∆U = Q + W
ignore ∆KE and ∆PE,
W = ∆U − Q = m(u2 − u1) − Q; (u2 − u1 = 51.94 Btu/lb)
ideal gas properties is attached
W = (2 lb)(143.98 − 92.04) Btu/lb − (− 20 Btu) = 123.9 Btu
u2 − u1 ≈ cv(T2 − T1) = (0.173 Btu/lb°R)(840 − 540) °R = 51.9 Btu/lb
The net work done in compressing the air as given is; W = -123.8 Btu
What is the Energy Balance equation?The equation for Energy Balance in thermodynamics is;
Q - W = ΔU
where;
Q is
ΔU is change in the internal energy of the system
Q is the net heat transfer
W is Net work done
Now, ΔU can also be written as;
ΔU = mC_v(T₂ - T₁)
C_v for air is 0.173 Btu/bm.R
We are given;
m = 2 lb
T₁ = 540 °R
T₂ = 840 °R
Q = -20 Btu (negative because heat is transferred to the surrounding)
Thus;
ΔU = 2 * 0.173 * (840 - 540)
ΔU = 103.8 Btu
Work done during the process is;
W = Q - ΔU
W = -20 - 103.8
W = -123.8 Btu
Read more about Energy Balance at; https://brainly.com/question/25329636
An insulated rigid tank is divided into two compartments by a partition. One compartment contains 7 kg of oxygen gas at 40°C and 100 kPa, and the other compartment contains 4 kg of nitrogen gas at 20°C and 150 kPa. Now the partition is removed, and the two gases are allowed to mix. Determine
(a) the mixture temperature and
(b) the mixture pressure after equilibrium has been established.
Answer:
(a) The mixture temperature, T₃ is 305.31 K
(b) The mixture pressure, P₃ after establishing equilibrium is 114.5 kPa
Explanation:
Here we have the initial conditions as
Oxygen compartment
Mass of oxygen = 7 kg
Molar mass of oxygen = 32.00 g/mol
Pressure in compartment, P₁ = 100 kPa
Temperature of oxygen, T₁ = 40 °C = 313.15 K
Number of moles of oxygen, n₁ is given by
[tex]Number \ of\ moles \ of \ oxygen, n_1 = \frac{Mass \ of \ oxygen}{Molar \ mass \ of \ oxygen} = \frac{7000}{32} = 218.75 \ moles[/tex]
From the universal gas equation, we have;
P·V = n·R·T
[tex]V_1 = \frac{n_1RT_1}{P_1} =\frac{218.75 \times 8.3145 \times 313.15 }{100000 } = 5.696 m^3[/tex]
For the Nitrogen compartment, we have
Mass of nitrogen = 4 kg
Molar mass of oxygen = 28.0134 g/mol
Pressure in compartment, P₂ = 150 kPa
Temperature of oxygen, T₂ = 20 °C = 293.15 K
Number of moles of nitrogen, n₂ is given by
[tex]Number \ of\ moles \ of \ nitrogen, n_2 = \frac{Mass \ of \ nitrogen}{Molar \ mass \ of \ nitrogen} = \frac{4000}{28.0134} = 142.79 \ moles[/tex]
From the universal gas equation, we have;
P·V = n·R·T
[tex]V_2 = \frac{n_2RT_2}{P_2} =\frac{142.79\times 8.3145 \times 293.15 }{150000 } = 2.32 m^3[/tex]
Therefore
We have for nitrogen
[tex]\frac{c_p}{c_v} = 1.4[/tex]
[tex]c_p - c_v = 296.8 J/KgK[/tex]
Therefore;
[tex]Nitrogen, \ c_p = 1036 \ J/(kg \cdot K)[/tex]
[tex]Nitrogen, \ c_v = 740\ J/(kg \cdot K)[/tex]
The molar heat capacities of Nitrogen are therefore as follows;
[tex]Nitrogen, \ \tilde{c_p} = 29.134 \ kJ/(kmol \cdot K)[/tex]
[tex]Nitrogen, \ \tilde{c_v} = 20.819 \ kJ/(kmol \cdot K)[/tex]
For oxygen we have
[tex]Oxygen, \ \tilde{c_p} = 29.382 \ kJ/(kmol \cdot K)[/tex]
[tex]Nitrogen, \ \tilde{c_v} = 21.068 \ kJ/(kmol \cdot K)[/tex]
The final volume, V₃ then becomes
V₃ = V₁ + V₂ = 5.696 m³ + 2.32 m³ = 8.016 m³
(a) For adiabatic mixing of gases the final temperature of the mixture is then found as follows
Therefore before mixing
U₁ = [tex]\sum \left (n_i\tilde{c_v}_i T_i \right )[/tex] = 0.21875 × 21.068 × 313.15 + 0.14279×20.819×293.15 = 2,314.65 kJ
After mixing, we have
U₂ = [tex]T_3 \sum \left (n_i\tilde{c_v}_i \right )[/tex] = T (0.21875 × 21.068 + 0.14279×20.819) = T×7.58137001
Therefore the final temperature, T is then
[tex]T_3 = \frac{\sum \left (n_i\tilde{c_v}_i T_i \right )}{\sum \left (n_i\tilde{c_v}_i \right )} =\frac{2,314.65 }{7.58137001} =305.30761550 \ K[/tex]
The mixture temperature, T₃ = 305.31 K
(b) The mixture pressure, P₃ after equilibrium has been established is given as
[tex]P_3 = \frac{n_3 \tilde{R}T_3}{V_3}[/tex]
Where:
n₃ = n₁ + n₂ = 0.21875 + 0.14279 = 0.36154 kmol = 361.54 moles
[tex]\tilde{R}[/tex] = 8.3145 J/(gmol·K)
Therefore ,
[tex]P_3 = \frac{361.54 \times 8.3145 \times 305.31 }{8.016 } = 114,492.1766706961 Pa[/tex]
P₃ ≈ 114.5 kPa.
Water flows steadily through a fire hose and nozzle. The hose is 75 mm inside diameter, and the nozzle tip is 25 mm inside diameter; water gage pressure in the hose is 510 kPa, and the stream leaving the nozzle is uniform. The exit speed 32 m/s and pressure is atmospheric. Determine the force transmitted by the coupling between the nozzle and hose. (25 points)
Answer:
R = 1804 N
Explanation:
Given:-
- The density of water, ρ = 997 kg/m^3
- The inside diameter of the hose, dh = 75 mm
- The gauge pressure of water in the hose, P1 = 510 KPa
- The exit speed of the water, V2 = 32 m/s
- The inside diameter of the nozzle tip, dn = 25 mm
- The atmospheric pressure (gauge), P2 = 0 KPa ... P = 1 atm (Absolute).
Find:-
Determine the force transmitted by the coupling between the nozzle and hose.
Solution:-
- We will first develop a control surface at the hose-nozzle interface.
- Assuming steady and one dimensional flow - (x-direction).
- Since there are no fictitious unbalanced forces acting on the fluid flow due to roughness of hose any any losses of energy from the fluid are negligible.
- The use of conservation of momentum of fluid flow is valid for an isolated system, where the flow of fluid into the control volume is denoted by (-) and the flow of fluid going out of the control volume is denoted by (+):
- The principle of conservation of momentum, the pair of equal force (Newton's third law) act on the control volume at (nozzle-hose) interface:
R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)
Where, Q: Flow rate
V1: The velocity of fluid in hose
A1: Cross sectional area of the hose
A2: Cross sectional area of the nozzle exit
- We see that the reaction force (R) that acts on nozzle-hose interface is due to changes in dynamic and hydrostatic pressures.
- Compute the required quantities Q, A1 and A2 and V1 using the given data:
- The flow rate Q for any flow in the hose can be given, where the cross sectional area of hose (A1)is:
[tex]A1 =\pi\frac{d_h^2}{4} = \pi\frac{0.075^2}{4} \\\\A1 = 0.00441 m^2\\\\\\[/tex]
- The cross sectional area of the nozzle tip with diameter dn = 25 mm is:
[tex]A2 =\pi\frac{d_n^2}{4} = \pi\frac{0.025^2}{4} \\\\A2 = 0.00049 m^2\\\\\\[/tex]
- The flow rate (Q) can now be calculated:
[tex]Q = A2*V2\\\\Q = (0.00049)*(32)\\\\Q = 0.01570 \frac{m^3}{s}[/tex]
- Since, the density of the water does not vary along the direction of flow, the flow rate (Q) remains constant throughout. So from continuity equation we have:
[tex]Q = A2*V2 = A1*V1\\\\V1 = \frac{Q}{A1} = \frac{0.0157}{0.00441} \\\\V1 = 3.56189 \frac{m}{s}[/tex]
- Now use the calculated quantities and compute the pair of reaction force at the nozzle-hose interface:
R = ρ*Q*(V2 - V1) + (P2*A2 - P1*A1)
R = (997)*(0.01570)*(32-3.56189) + (0 - 510*0.00441)*1000
R = 445.13889 - 2,249.1
R = - 1803.961 ≈ -1,804 N
- Here the negative sign denotes the direction of in which the force (R) is exerted. Since, (-) denotes into the control volume it acts opposite to the flow of water.
The coupling between the nozzle and hose is -1.81N
This question relates to flow rate of a liquid
Data given:
The density of water = 997kg/m^3
The inside diameter of the hose = 75mm = 0.0075m
The gauge pressure of water in the hose = 510kPa
The exit speed of the water = 32m/s
The inside diameter of the nozzle tip = 25mm = 0.0025m
The atmospheric pressure = 0kPa or 1atm
Let's calculate the inlet velocity
[tex]v_1=v_2=A_2/A_1\\v_1=V_2(\frac{d_2}{d_1})^2\\v_1=32(\frac{25}{75})^2\\v_1=3.50m/s[/tex]
Calculating the force transmitted by coupling between the nozzle and hose
[tex]R_x+p_1gA_1=v_1[-|pv_1A_1|]+v_2[|pv_2A_2|]\\[/tex]
μ[tex]_1[/tex]=[tex]v_1[/tex] and μ[tex]_2[/tex] =[tex]v_2[/tex]
[tex]R_x=-p_1gA_1-v_1pv_1A_1+v_2pv_2A_2\\R_x=-p_1gA+pv_2A_2(v_2-v_1)\\R_x=-510*10^3N/m^3*\frac{\pi }{4}(0.075m)^2+997kg/m^3*32m/s*\frac{\pi }{4} (0.025m)^2(32-3.50)=-1805=-1.81kN[/tex]
The force between the nozzle and hose is -1.81
Learn more about flow rate;
https://brainly.com/question/17151453
Design a Mealy machine for a 20 cent candy dispensing machine which accepts nickel (n) , dime (d) and quarter(q) . It gives candy as well change in the form of nickels only. To help you, here are the some of the elements of the machine: States: 0, 5, 10 and 15 represent the amount of money already inserted in the machine. n, d and q representing coins inserted in the machine Output: c0 (candy and no change), c1 (one nickel as change), c2(two nickels as change ), c3 (three nickels as change), and c4 (four nickels as change)
Find the answer in the attachment
A shaft is loaded in bending and torsion such that Ma = 70 N ? m, Ta = 45 N ? m, Mm = 55 N ? m, and Tm = 35 N ? m. For the shaft, Su = 700 MPa and Sy = 560 MPa, and a fully corrected endurance limit of Se = 210 MPa is assumed. Let Kf = 2.2 and Kfs = 1.8. With a design factor of 2.0 determine the minimum acceptable diameter of the shaft using
a. ) DE-Gerber criterion.
b) DE-ASME Elliptic criterion.
c) DE-Soderberg criterion.
d) DE-Goodman criterion.
Answer:
Please see the attached picture for the compete answer.
Explanation:
Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a cold reservoir at 250 K and rejects 950 KJ to a hot reservoir at 300 K while receiving 70 kJ of work to operate. Draw a schematic of the cycle clearly indicating the hot and cold reservoir and the direction of heat and work transfers. (10 pts.)
Answer:
Impossible.
Explanation:
The ideal Coefficient of Performance is:
[tex]COP_{i} = \frac{250\,K}{300\,K-250\,K}[/tex]
[tex]COP_{i} = 5[/tex]
The real Coefficient of Performance is:
[tex]COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}[/tex]
[tex]COP_{r} = 12.571[/tex]
Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.
The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter dd. Express your answer in terms of some or all of the variables rrr, RRR, ddd, and TTT. nn
Answer:
n = 2r³/Rd²
Explanation:
See the attached file for the derivation.
The function below takes a single string parameter called sentence. Your function should return True if the sentence contains at least one copy of each of the following vowels: a, e, i, o, and u. Otherwise, return False. The vowels can be either upper case or lower case.
student.py
Bef contains_all_values (sentence): 1 w
The question asks for a function to check if a sentence contains all five vowels at least once, regardless of case sensitivity. A solution involves creating a set of vowels and comparing it to a set of found vowels in the sentence, returning true if all vowels are present.
Explanation:The question relates to determining whether a given sentence contains all five vowels (a, e, i, o, u) at least once, ignoring case sensitivity. This problem is typically solved using a function that iterates through each character in the sentence, checks if it is a vowel, and then keeps track of whether all vowels have been encountered. The essential steps involve converting the sentence to lowercase (to ignore case sensitivity), then checking for the presence of each vowel. A simple approach is to use a set to keep track of the vowels found, and once the set contains all five vowels, the function can return True. Otherwise, it returns False after checking the entire sentence.
An example implementation could be:
def contains_all_vowels(sentence):This code creates a set of vowels and then iterates over the sentence, adding each encountered vowel to another set. If, by the end of the sentence, the second set is equal to the set of all vowels, the function returns True; otherwise, it returns False.
A commercial enclosed gear drive consists of a 20o spur pinion having 16 teeth driving a 48-tooth gear. The pinion speed 350 rev/min, the face width 2 in, and the diametral pitch 6 teeth/in. The gears are grade 1 steel, through-hardened at 240 Brinell, made to No. 6 quality standards, uncrowned, and are to be accurately and rigidly mounted. Assume a pinion life of 108 cycles and a reliability of 0.90. Determine the AGMA bending and contact stresses and the corresponding factors of safety if 5 hp is to be transmitted.
Answer:
Check the explanation
Explanation:
Generally, when there is a pair of gears meshing, the minor gear is referred to as the pinion gear. in addition, it involves the meshing of cylindrical gear with a rack in a rack-and-pinion system which transforms to linear motion from a rotational motion.
Rack and pinion gears are mostly utilized in converting rotation into linear motion.
Kindly check the attached images below for the full step by step explanation to the question above.
Software, such as a word processor, search engine, or mobile interface, typically includes plug-in support specific to a language to aid with spelling. In this assignment, you will implement a class that provides general language support; such a class could presumably be (re)used in these broader software applications. For the purpose of spell checking, a simple language model is a set of valid words. By convention, a language specification may include both capitalized and uncapitalized words. A word that is is entirely lowercased in the language specification can be used in either capitalized or uncapitalized from (e.g., if 'dog'is in the language specification, then both 'dog' and 'Dog' are legitimate usages). However, any word that includes one or more uppercased letters in the original language reflects a form that cannot be modified (e.g., 'Missouri' is acceptable but 'missouri' is not; 'NATO' is acceptable, but neither 'Nato', 'nato', nor 'nAto' would be acceptable). The goals of the new class will be to answer the following types of queries: • Is a given string a legitimate word in the language? (based on the above conventions regarding capitalization) • Given a string, which may or may not be in the language, produce a list of suggestions that are valid words in the language and reasonably "close" to the given string in terms of spelling. (We will say more below, about the notion of distance between words.) Formally, you are to provide a file named language_tools.py that defines aLanguageHelper class with the following three methods. _init__(self, words) The words parameter can be any iterable sequence of strings that define the words in the language. For example, the parameter may be a list of strings, or a file object that has one word per line. All you should assume about this parameter is that you are able to do a loop, for w in words: to access its entries. The class is responsible for recording all words from the language into an internal data representation, and stripping any extraneous whitespace from each entry (such as newline characters that will appear in a file). For the sake of efficiency, we recommend that you store the language words in a Python set instance. (We discuss sets in a later section.) _contains_(self, query) The query parameter is a string. This method should determine whether the string is considered a legitimate word, returning True if the word is contained in the language and False otherwise. This method should adhere to the aforementioned conventions regarding capitalized and uncapitalized words. For example, dog, Dog and Missouri are contained in the English language, yet missouri and Missourri are not. The _contains_ special method is used by Python to support the in operator. It allows the standard syntax "Missouri' in language which is implicitly translated by Python to the internal call language. _contains_('Missouri') presuming that language is an instance of our LanguageHelper class. getSuggestions (self, query) Given a query string, this method should return an alphabetical list of "nearby" words in the language. Doing a good job at offering suggestions is the most difficult part of writing a good language helper. We discuss this aspect of the project in a later section. S = set() create a new set instance (which is initially an empty set). s.add(value) adds the given value to the set (value will be a string in our application). value in s returns True if the given value is currently in the set, and False otherwise.
Answer:
Check the explanation
Explanation:
class LanguageHelper:
language=set()
#Constructor
def __init__(self, words):
for w in words:
self.language.add(w)
def __contains__(self,query):
return query in self.language
def getSuggestionns(self,query):
matches = []
for string in self.language:
if string.lower().startswith(query) or query.lower().startswith(string) or query.lower() in string.lower():
matches.append(string)
return matches
lh = LanguageHelper(["how","Hi","What","Hisa"])
print('how' in lh)
print(lh.getSuggestionns('hi'))
===========================================
OUTPUT:-
==================
True
['Hisa', 'Hi']
====
Boron fibers (of total mass 5kg, rho B = 2.3g/cc, σy B = 55MPa) are uniaxially introduced into 8kg of an Al matrix (rho Al = 2.7g/cc, σy Al = 10MPa). There is no porosity before and after composite fabrication. Calculate the density of the composite, yield strengths parallel and perpendicular to fiber orientations.
Answer:
Explanation:
Check attachment for step by step solution
An incandescent lightbulb is an inexpensive but highly inefficient device that converts electrical energy into light. It converts about 10 percent of the electrical energy it consumes into light while converting the remaining 90 percent into heat. The glass bulb of the lamp heats up very quickly as a result of absorbing all that heat and dissipating it to the surroundings by convection and radiation.
Consider an 8-cm-diameter 60-W lightbulb in a room at 258C. The emissivity of the glass is 0.9. Assuming that 10 percent of the energy passes through the glass bulb as light with negligible absorption and the rest of the energy is absorbed and dissipated by the bulb itself by natural convection and radiation, determine the equilibrium temperature of the glass bulb. Assume the interior surfaces of the room to be at room temperature.
An incandescent light bulb's equilibrium temperature is determined by considering energy conversion, dissipation processes, and emissivity of the glass.
Explanation:An incandescent light bulb converts electrical energy into light and heat. In the given scenario, the glass bulb absorbs and dissipates heat through convection and radiation. To determine the equilibrium temperature of the glass bulb, we need to consider energy conversion and emissivity.
We know that 10% of the energy passes through the glass bulb as light while the remaining 90% is absorbed and dissipated. By calculating the energy balance and accounting for the emissivity of the glass, the equilibrium temperature of the glass bulb can be found.
Factors like the wattage of the bulb, its size, and the room temperature play a role in determining the final equilibrium temperature of the glass bulb based on the energy conversion and dissipation processes involved.
Implement this C program by defining a structure for each payment. The structure should have at least three members for the interest, principle and balance separately. And store all the payments in a structure array (the max size of which could be 100). Name this C program as loanCalcStruct.c
Answer:
Explanation:
check the attached files for the solution and output result.
Each of the two drums and connected hubs of 13-in. radius weighs 210 lb and has a radius of gyration about its center of 30 in. Calculate the magnitude of the angular acceleration of each drum. Friction in each bearing is negligible.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
Explanation:
From the question we are told that
Their radius of the drum is [tex]r = 13 in = \frac{13}{12} ft = 1.083ft[/tex] each
The weight is [tex]W = 210 lb[/tex]
The mass is [tex]M = \frac{210 lb}{32.2 ft /s^2} = 6.563\ lb s^2 ft^{-1}[/tex]
Their radius of gyration is [tex]z=30 in= \frac{30 }{12} = 2.5 ft[/tex]
The free body diagram of a drum and its hub and 30lb and in the case the weight is connect to the hub separately is shown on the second uploaded image
The T in the diagram is the tension of the string
Now taking moment about the center of the the drum P we have
[tex]\sum M_p = I_p \alpha[/tex]
=> [tex]T * r = Mz^2 * \alpha[/tex]
Where r is the radius ,z is the radius of gyration about the center O , M is the mass of the drum including the hub, and [tex]\alpha[/tex] is the angular acceleration
Inputting
[tex]T * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
=> [tex]T = 37.87\alpha[/tex]
Considering the force equilibrium in the vertical direction (Looking at the second free body diagram now )
The first on is
[tex]\sum F_y = ma[/tex]
=> [tex]30lb - T = m(r \alpha )[/tex]
Where m is the mass of the hanging block which has a value of
[tex]m = \frac{30lb}{32.2 ft/s^2} = 0.9317 \ lb ft^{-1} s^2[/tex]
a is the acceleration of the hanging block
inputting values we have
[tex]30- 37.87 \alpha = 0.9317* 1.083 \alpha[/tex]
[tex]30 = 37.87\alpha + \alpha[/tex]
[tex]\alpha = \frac{30}{38.87 }[/tex]
[tex]\alpha = 0.792 rad/s^2[/tex]
So the angular acceleration for first drum [tex]\alpha = 0.792 rad/s^2[/tex]
The free body diagram of a drum and its hub when the only on the string is 30lb is shown on the third uploaded image
So here we would take the moment about O
[tex]\sum M_o = I_O \alpha[/tex]
So [tex]\sum M_o = 30* 1.083[/tex]
and [tex]I = M z^2[/tex]
Therefore we will have
[tex]30 * 1.083 = (Mz^2 )\alpha[/tex]
inputting values
[tex]30 * 1.083 = 6.563 * 2.5^2 \alpha[/tex]
[tex]32.49=41.0\alpha[/tex]
[tex]\alpha =\frac{41}{32.49}[/tex]
[tex]\alpha =1.262[/tex]
So the angular acceleration for the second drum is [tex]\alpha =1.262[/tex]
Evaluate the performance of the proposed heat pump for three locations Using R134a. Discuss the effect of outdoor temperature on the performance of the heat pump. What happens to the COP if the heat exchangers are only 80% effective, Philadelphia only. Discuss improvements to the design for Philadelphia that will increase the COP closer to the theoretical max (you must show supporting numbers). Your report should include tables that list the information at each state, the evaluated work, heat transfer, and calculated COP values. Sample calculations of your work should be included in an appendix.
Answer:Table 2.2: Differences in runstitching times (standard − ergonomic).
1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42
.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43
-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82
A paired t-test is the standard procedure for testing this null hypothesis.
We use a paired t-test because each worker was measured twice, once for Paired t-test for
each workplace, so the observations on the two workplaces are dependent. paired data
Fast workers are probably fast for both workplaces, and slow workers are
slow for both. Thus what we do is compute the difference (standard − er-
gonomic) for each worker, and test the null hypothesis that the average of
these differences is zero using a one sample t-test on the differences.
Table 2.2 gives the differences between standard and ergonomic times.
Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-
ences in the sample. We assume that these differences are independent sam-
ples from a normal distribution with mean µ and variance σ
2
, both unknown.
Our null hypothesis is that the mean µ equals prespecified value µ0 = 0
(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the
workers to be faster in the ergonomic workplace.
The formula for a one sample t-test is
t =
¯d − µ0
s/√
n
,
where ¯d is the mean of the data (here the differences d1, d2, . . ., dn), n is the The paired t-test
sample size, and s is the sample standard deviation (of the differences)
s =
vuut
1
n − 1
Xn
i=1
(di − ¯d )
2 .
If our null hypothesis is correct and our assumptions are true, then the t-
statistic follows a t-distribution with n − 1 degrees of freedom.
The p-value for a test is the probability, assuming that the null hypothesis
is true, of observing a test statistic as extreme or more extreme than the one The p-value
we did observe. “Extreme” means away from the the null hypothesis towards
the alternative hypothesis. Our alternative here is that the true average is
larger than the null hypothesis value, so larger values of the test statistic are
extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of
freedom from the observed t-value to the right. (If the alternative had been
µ < µ0, then the p-value is the area under the curve to the left of our test
Explanation: The curve represents the sum total of the evaluation
A small vehicle is powered by a pulsejet. The available net thrust is 6000 N and the traveling speed is 200 km/hr. The gases leave the engine with an average velocity (Ve) of 360 m/s Assume pressure equilibrium exists at the outlet plane and the fuel to air ratio is 0.06.
a. Compute the mass flow rate required
b. Calculate the inlet area (assume To is 16 °C and Po is one atmosphere)
C. Calculate the thrust power
d. Calculate the propulsive efficiency
Answer:
a) The mass flow rate is 19.71 kg/s
b) The inlet area is 0.41 m²
c) The thrust power is 333.31 kW
d) The propulsive efficiency is 26.7%
Explanation:
Please look at the solution in the attached Word file.