Answer:
a) t = √(2*h/g)
b) v₀ = 67*√(g/(2*h))
c) ∅ = tan⁻¹(2*h/67)
Explanation:
Suppose that the height of the building is known (h), then we have
a) y = h = g*t²/2
then the time of flight is
t = √(2*h/g)
b) The initial speed (v₀) can be obtained as follows
x = v₀*t ⇒ v₀ = x / t
where
x = xmax = 67 m and t = √(2*h/g)
So, we have
v₀ = 67 / √(2*h/g) = 67*√(g/(2*h))
c) In order to get the speed and the angle with respect to the horizontal of the velocity vector at impact, we obtain vx and vy as follows
vx = v₀ = 67*√(g/(2*h))
and
vy = gt = g*√(2*h/g) ⇒ vy = √(2*h*g)
then we apply
v = √(vx² + vy²) = √((67*√(g/(2*h)))² + (√(2*h*g))²)
⇒ v = √(g*(4489 + 4*h²)/(2*h))
The angle is obtained as follows
tan ∅ = vy / vx ⇒ tan ∅ = √(2*h*g) / 67*√(g/(2*h))
⇒ tan ∅ = 2*h / 67 ⇒ ∅ = tan⁻¹(2*h/67)
A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the air in the tank is 9.4 psig and the atmospheric pressure is 12.5 psia, the fluid-level difference between the two columns, h, in feet is
Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
A horizontal rigid bar ABC is pinned at end A and supported by two cables at points B and C. A vertical load P 5 10 kN acts at end C of the bar. The two cables are made of steel with a modulus elasticity E 5 200 GPa and have the same cross-sectional area. Calculate the minimum cross-sectional area of each cable if the yield stress of the cable is 400 MPa and the factor of safety is 2.0. Consider load P only; ignore the weight of bar ABC and the cables.
To determine the minimum cross-sectional area of the steel cables, calculate the allowable stress and use it along with the provided load. The result is a minimum area of 50 mm² for each cable.
Explanation:The question involves calculating the minimum cross-sectional area of each steel cable designed to support a load with a safety factor, given the yield stress of the material. First, determine the allowable stress by dividing the yield stress by the factor of safety. In this case, the allowable stress is 200 MPa (400 MPa / 2.0). To find the minimum cross-sectional area (A), use the formula A = P / σ, where P = 10 kN = 10,000 N (the load) and σ (sigma) is the allowable stress in N/m². Convert 200 MPa to N/m² to get 200,000 N/m². Therefore, the minimum cross-sectional area required for each cable is 50 mm² (10,000 N / 200,000 N/m²).
It is usually easy to minimize errors due to the input bias current of an opamp by adding a resistor in the input terminal, but this still leaves a small error due to the input offset current. Select one: True False
Answer:True
Explanation:
Answer:
True
Explanation:
Input bias current:
It is a small current that flows in parallel with the input terminals of op-amp to bias the input transistors. This current gets converted into voltage and amplified which results in incorrect output results. This bias current Ib+ and Ib- flows in the positive and negative input terminals of the op-amp.
Ib+ and Ib- create errors of opposite polarity. Therefore, bias current can be minimized by carefully adding a resistor in the positive input terminal.
Input offset current:
Unfortunately, a small error still remains due to the mismatch between input currents Ib+ and Ib-.
This input offset current error can be adjusted by adding a potentiometer and resistor in the negative input terminal.
5. A driver is traveling at 90 km/h on a wet road. An object is spotted on the road 140m ahead and the driver is able to come to a stop just before hitting the object. Assuming standard reaction time and using the practical-stopping distance equation, determine the grade of the road.
Answer: Check the attached
Explanation:
An engineering student claims that a country road can be safely negotiated at 65 mi/h in rainy weather. Because of the winding nature of the road, one stretch of level pavement has a sight distance of only 510 ft. Assuming practical stopping distance, comment on the student
Answer:
Negotiated speed should be lower. Perception/reaction time is too less than design values.
Explanation:
Given:
- The claimed safe speed V_1 = 65 mi/h
- Sight distance D = 510 ft
- The practical deceleration a = 11.2 ft/s ... according to standards
Find:
Assuming practical stopping distance, comment on the student whether the claim is correct or not
Solution:
- Calculate the practical stopping distance:
d = V_1^2 / 2*a
d = ( 65 * 1.46 )^2 / 2*11.2 = 402.054 ft
- Solve for reaction distance d_r is as follows:
d_r = D - d = 510 - 402.054 = 107.945 ft
- The perception/time reaction is:
t_r = d_r/V_1 = 107.945 / 94.9
t_r = 1.17 sec
Answer: The perception/reaction time t_r = 1.17 s is well below the t = 2.3 s.
Hence, the safe speed should be lower.
The terms batten seam, standing seam, and flat seam all describe types of:
(A) architectural sheet metal roofing.
(B) methods for glazing large windows.
(C) types of EPDM roofing membranes.
(D) framing methods for hollow steel doors.
(E) built-up roofing systems.
Answer:
(A) architectural sheet metal roofing
Explanation:
By the name itself we can judge that the 'Architectural sheet metal roofing' is a kind of metal roofing.
And these type of metal roofing is primarily used for small and big houses, small buildings and as well as in a building that is for commercial use they can be totally flat as well as little bit sloped.
And the words similarly like batten and standing seam, and flat seam all tells us that these are the types of architectural sheet metal roofing.
Water flows downward through a vertical 10-mm-diameter galvanized iron pipe with an average velocity for 5.0 m/s and exits as a free jet. There is a small hole in the pipe 4 meters above the outlet. Will water leak out of the pipe through this hole, or will air enter into the pipe through the hole? Repeat the problem if the average velocity is 0.5 m/s.
Water is likely to leak out of the hole in the pipe when the average velocity is 5.0 m/s due to high dynamic pressure. With a slower velocity of 0.5 m/s, air might enter the pipe if the static pressure at the hole is less than atmospheric, but this requires additional details to confirm.
Explanation:The question addresses the behavior of water within a pipe system under different flow conditions, involving principles of fluid dynamics. Specifically, it asks whether water will leak out of a small hole in a vertical pipe or if air will enter into the pipe through the hole given two different average velocities of water flow.
Case 1: Average velocity of 5.0 m/s
With an average velocity of 5.0 m/s, the dynamic pressure of the flowing water is considerable, and thus, water is likely to leak out of the hole due to the higher pressure inside the pipe compared to atmospheric pressure.
Case 2: Average velocity of 0.5 m/s
With a decreased velocity of 0.5 m/s, the dynamic pressure is significantly lower. If the static pressure at the hole's location is less than the atmospheric pressure, air might enter the pipe; however, if it is still higher than atmospheric, water would continue to leak out. The determination requires additional information, such as the height of the water column above the hole and any applied pressures at the water's source.
Generally, the behavior can be predicted using Bernoulli's principle and the continuity equation for incompressible flow, which together relate the velocities, pressures, and cross-sectional areas in different sections of a pipe.
The calorie is a unit of energy defined as the amount of energy needed to raise 1 g of water by 1°C. a. How many calories are required to bring a pot of water at 1°C to a boil? The pot is full to the brim, with diameter 20 cm and depth 20 cm. The density of water is 1000 kg/m3. b. If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?
Answer:
a. Calories required = 622710 calories
b. Energy = 1000 times much energy
Calories = 622710000 calories
Explanation:
Given:
h = Depth of pot = 20cm = 0.2m
Diameter of pot = 20cm = 0.2m
r = ½ *diameter = ½ * 0.2
r = 0.1m
Density = 1000kg/m³
Water temperature = 1°C
a.
First, we calculate the volume of the water(pot)
V = Volume = πr²h
V = 22/7 * 0.1² * 0.2
V = 0.044/7
V = 0.00629m³
M = Mass of water = Volume * Density
M = 0.00629m³ * 1000kg/m³
M = 6.29kg
M = 6.29 * 1000 grams
M = 6290g
The water is at 1°C, so it needs to gain 99°C to reach boiling point
So, Calories = 99 * 6290
Calories required = 622710 calories
b.
If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?
Depth of pot = 20cm
Depth of pot = 0.2m
Depth of hot tube = 2m
Energy is directly proportional to D³
Since the depth of hot the is 10 times greater than that of the pot
It'll require 10³ much more energy
Energy = 10³
Energy = 1000 times much energy
Calories required = 622710 * 1000
Calories = 622710000 calories
A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at one end or not. Assume the speed of sound to be 340 m/s.
Answer:
The pipe is open ended and the length of pipe is 3.4 m.
Explanation:
For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below
Equation for nth frequency for open end pipe is given as
[tex]f_n=\frac{nv}{2L}[/tex]
For (n+1)th value the frequency is
[tex]f_{n+1}=\frac{(n+1)v}{2L}[/tex]
Taking a ratio of both equation and solving for n such that the value of n is a whole number
[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\[/tex]
So n is a whole number this means that the pipe is open ended.
For confirmation the nth frequency for a closed ended pipe is given as
[tex]f_n=\frac{(2n+1)v}{4L}[/tex]
For (n+1)th value the frequency is
[tex]f_{n+1}=\frac{(2n+3)v}{4L}[/tex]
Taking a ratio of both equation and solving for n such that the value of n is a whole number
[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\[/tex]
As n is not a whole number so this is further confirmed that the pipe is open ended.
Now from the equation of, with n=6, v=340 m/s and f=300 Hz
[tex]f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m[/tex]
The value of length is 3.4m.
Assume that the flow of air through a given duct is isentropic. At one point in the duct, the pressure and temperature are pl = 1800 lb/ft2 and TI = 500°R, Problems respectively. At a second point, the temperature is 400"R. Calculate the pressure and density at this second point.
Answer:
pressure is 825 lb/ft²
density is 1.20 × [tex]10^{-3}[/tex] slug/ft²
Explanation:
given data
p1 = 1800 lb/ft²
T1 = 500°
T2 = 400°
solution
we use here isentropic flow relation that is
[tex]\frac{P2}{P1} = (\frac{T2}{T1})^{\gamma / \gamma - 1 }[/tex]
put here value we get pressure P2
P2 = 1800 × [tex](\frac{400}{500})^{3.5}[/tex]
P2 = 825 lb/ft²
and we know pressure is
pressure = [tex]\rho RT[/tex]
so for pressure 825 we get here [tex]\rho[/tex]
825 = [tex]\rho[/tex] × 1716 × 400
[tex]\rho[/tex] = 1.20 × [tex]10^{-3}[/tex] slug/ft²
Consider a constant volume process involving heat addition to a closed system consisting of an ideal gas with no changes in kinetic or potential energy. Is the required heat transfer for raising the temperature from 295 to 305 K the same as the heat transfer required from 345 to 355 K?
Answer:
Yes and no
Explanation:
The thermodynamic equation for the heat transfer in a constant volume process is the following:
[tex]Q=\Delta U=mC_V\Delta T[/tex]
where Q is the required heat, U is the internal energy, m the mass of the gas, C_V the heat capacity assuming consant volume and [tex]\Delta T[/tex] is the change in temperature.
If you assume the heat capacity doesn't change with temperature at which the gas is currently at then the heat transfer depends solely on the change in temperature. With this assumption the transfered heat would be the same in both cases.
In reality the heat capacity does change with respect to temperature. Depending on the type of gas. In reality there would be difference in heat transfered between 295/205 K and 245/255 . Only then you wouldn't use the [tex]\Delta T[/tex] expression since the integral would be different depending on the heat capacity in relation to temperature.
Given the following materials and their corresponding thermal conductivity values, list them in order from most conductive to least conductive.Sheet Rock: k = 0.43 W/(m*K)Masonite: k = 0.047 W/(m*K)Glass: k = 0.72 W/(m*K)Lexan: k = 0.19 W/(m*K)b) Given the following information, calculate the thermal conductivity using Fourier's Equation.q = 100 WA = 8 m^2ATΔT= 10L = 7 m
Answer:
1) Glass
2) Rock sheet
3) Lexan
4) Masonite
b) k = 8.75 W/m.K
Explanation:
Given:
The thermal conductivity of certain materials as follows:
-Sheet Rock: k = 0.43 W/(m*K)
-Masonite: k = 0.047 W/(m*K)
-Glass: k = 0.72 W/(m*K)
-Lexan: k = 0.19 W/(m*K)
Data Given:
- Q = 100 W
- A = 8 m^2
- dT = 10 C
- L = 7 m
Find:
a) list the materials in order from most conductive to least conductive
b) calculate the thermal conductivity using Fourier's Equation
Solution:
- We know from Fourier's Law the relation between Heat transfer and thermal conductivity as follows:
Q = k*A*dT / L
- From the relation above we can see that rate of heat transfer is directly proportional to thermal conductivity k.
- Hence, the list in order of decreasing conductivity is as follows:
- The list of materials in the decreasing order of thermal conductivity k is:
1) Glass k = 0.72 W/m.K
2) Rock sheet k = 0.43 W/m.K
3) Lexan k = 0.19 W/m.K
4) Masonite k = 0.047 W/m.K
- Use the relation given above we can compute the thermal conductivity k with the given data:
k = Q*L / (A*dT)
k = (100 W * 7 m) / (8 m^2*10 C)
k = 8.75 W/m.K
the correlation between a car's engine size and its fuel economy is r = -0.774. what fraction of the variability in fuel economy is accounted for by the engine size?
Answer:
59.9%
Explanation:
R^2 =(-0.774)^2 = 0.599
59.9% of fuel is accounted for
Air is compressed slowly in a piston–cylinder assembly from an initial state where p1 = 1.4 bar, V1 = 4.25 m3 , to a final state where p2 = 6.8 bar. During the process, the relation between pressure and volume follows pV = constan
The work done by the gas is -940 kJ
Explanation:
In this process, we are told that the product of pressure and volume remains constant:
[tex]pV=const.[/tex]
so we can write
[tex]p_1 V_1 = p_2 V_2[/tex]
where
[tex]p_1 = 1.4 bar[/tex] is the initial pressure
[tex]p_2 = 6.8 bar[/tex] is the final pressure
[tex]V_1=4.25 m^2[/tex] is the initial volume
Solving for [tex]V_2[/tex], we find the final volume:
[tex]V_2=\frac{p_1V_1}{p_2}=\frac{(1.4)(4.25)}{6.8}=0.875 m^3[/tex]
Now by looking at the equation of state of an ideal gas:
[tex]pV=nRT[/tex] (1)
we notice that since [tex]pV=const.[/tex], this means that also the absolute temperature of the gas T remains constant (because the number of moles n does not change). Therefore this is an isothermal process: the work done in an isothermal process is given by
[tex]W=nRTln(\frac{V_2}{V_1})[/tex]
And by looking again at (1), we can substitute (nRT) with (pV), so we get
[tex]W=p_1 V_1 ln (\frac{V_2}{V_1})[/tex]
Converting the pressure into SI units,
[tex]p_1 = 1.4 bar = 1.4\cdot 10^5 Pa[/tex]
So the work done is
[tex]W=(1.4\cdot 10^5)(4.25)ln(\frac{0.875}{4.25})=-9.4\cdot 10^5 J[/tex]
Which means -940 kJ. This value is negative since the work is done by the surroundings on the gas (because the gas is compressed).
Learn more about ideal gases:
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A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that the R-134a is kept at constant pressure until a final state is reached with a quality of 25%. Calculate the heat transfer in the process.
Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = [tex]\int\limits^a_b P \, dV[/tex] = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
A liquid phase reaction, A → B, is to be carried out in an isothermal, well mixed batch reactor with a volume of 1L. Initially there are 6 moles of A. The rate of destruction of A is given by –rA =k1CA/ (1+k2CA), where k1=4, k2 =5. The unit of time in the rate constants is hours. Calculate the time, in hours, that the reaction must proceed in the reactor in order to result in 3 moles of A remaining in the reactor.
Answer:
the time, in hours = 4.07hrs
Explanation:
The detailed step by step derivation and appropriate integration is as shown in the attached files.
You are traveling along an interstate highway at 32.0 m/s (about 72 mph) when a truck stops suddenly in front of you. You immediately apply your brakes and cut your speed in half after 6.0 s.(a) What was your acceleration, assuming it was constant?
Answer:
a= - 2.6 m/s².
Explanation:
u = 32 m/s
The speed after 6 s is half of u
[tex]v= \dfrac{32}{2}=16\ m/s[/tex]
t= 6 s
The average acceleration = a
We know v = u +at
v=final velocity
u=initial velocity
Now by putting the values in the above equation
16= 32 + a x 6
[tex]a=\dfrac{16-32}{6}\ m/s^2[/tex]
[tex]a=-2.6\ m/s^2[/tex]
Therefore the acceleration will be - 2.6 m/s².
a= - 2.6 m/s².
Negative indicates that velocity and acceleration is is opposite direction.
Consider a large plane wall of thickness L = 0.3 m, thermal conductivity k = 2.5 W/m · °C, and surface area A =12 m2. The left side of the wall at x = 0 is subjected to a net heat flux of q0 = 700 W/m2 while the temperature at that surface is measured to be T1 =80°C. Assuming constant thermal conductivity and no heat generation in the wall, a. Express the differential equation and the boundary conditions for steady onedimensional heat conduction through the wall. b. Obtain a relation for the variation of temperature in the wall by solving the differential equation. c. Evaluate the temperature of the right surface of the wall at x = L.
Answer:
a) -k* dT / dx = q_o
b) T(x) = -280*x + 80
c) T(L) = -4 C
Explanation:
Given:
- large plane wall of thickness L = 0.3 m
- thermal conductivity k = 2.5 W/m · °C
- surface area A =12 m2.
- left side of the wall at net heat flux q_o = 700 W/m2 @ x = 0
- temperature at that surface is measured to be T1 =80°C.
Find:
- Express the differential equation and the boundary conditions for steady one dimensional heat conduction through the wall.
- Obtain a relation for the variation of temperature in the wall by solving the differential equation
- Evaluate the temperature of the right surface of the wall at x = L.
Solution:
- The mathematical formulation of Rate of change of temperature is as follows:
d^2T / dx^2 = 0
- Using energy balance:
E_out = E_in
-k* dT / dx = q_o
- Integrate the ODE with respect to x:
T(x) = - (q_o / k)*x + C
- Use the boundary conditions, T(0) = T_1 = 80C
80 = - (q_o / k)*0 + C
C = 80 C
-Hence the Temperature distribution in the wall along the thickness is:
T(x) = - (q_o / k)*x + 80
T(x) = -(700/2.5)*x + 80
T(x) = -280*x + 80
- Use the above relation and compute T(L):
T(L) = -280*0.3 + 80
T(L) = -84 + 80 = -4 C
Differential equation: [tex]\(\frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0\).[/tex]Temperature variation: [tex]\(T(x) = T_1\).[/tex]Temperature at x = L is [tex]\(T(L) = T_1\)[/tex].
a. The differential equation for steady one-dimensional heat conduction through the wall is given by Fourier's law:
[tex]\[ \frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0 \][/tex]
This equation states that the rate of change of heat flux with respect to distance ( x ) is constant and equal to zero in steady-state conditions.
The boundary conditions are:
1. At x = 0 : [tex]\( q = q_0 \)[/tex], [tex]\( T = T_1 \)[/tex]
2. At x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex], as there is no heat flux across the right surface of the wall.
b. To solve the differential equation, integrate it twice:
[tex]\[ k \frac{{dT}}{{dx}} = C_1 \][/tex]
[tex]\[ \frac{{dT}}{{dx}} = \frac{{C_1}}{{k}} \][/tex]
[tex]\[ T = \frac{{C_1}}{{k}} x + C_2 \][/tex]
Apply the boundary conditions:
At x = 0 : [tex]\( T = T_1 \)[/tex]
[tex]\[ C_2 = T_1 \][/tex]
At x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex]
[tex]\[ \frac{{C_1}}{{k}} = 0 \][/tex]
[tex]\[ C_1 = 0 \][/tex]
Therefore, the temperature variation in the wall is given by:
[tex]\[ T(x) = T_1 \][/tex]
c. The temperature of the right surface of the wall at x = L is equal to [tex]( T(L) = T_1 \)[/tex], as there is no variation in temperature along the wall according to the solution obtained in part b.
In a conduit with a diameter of 4.5 ft, the depth of flow is 4.0 ft. (a) Determine the hydraulic radius, hydraulic depth, and section factors for critical and normal flows. (b) Determine the alternate depth of flow that will carry the same discharge.
Answer:
(a) 1.125 ft, Section factor = 22.78
(b) 42.75 ft
Explanation:
Hydraulic radius is given by [tex]R_{H} = \frac{A}{P}[/tex] Where
A = Cross sectional area of flow and
P = Perimeter h
Since the cross section is a circle then at depth 4 of 4.5 the perimeter
[tex]=2 \pi r-\frac{\theta }{360} *2 \pi r[/tex] where r = 2.25 and θ = 102.1 °
perimeter = 10.1 ft and the area = [tex]=\pi r^2-\frac{\theta }{360} * \pi r^2[/tex] = 11.39 ft²
Therefore [tex]R_{H} = \frac{11.39}{10.1} = 1.125 ft[/tex]
Section factor is given by for critical flow = Z = A×√D
= 11.39 ft² × √(4 ft) = 22.78
for normal flow Z =[tex]Z_{} ^{2} = \frac{A^{3}}{T}[/tex] = 22.78
(b) The alternate depth of flow is given by
for a given flow rate, we have from chart for flow in circular pipes
Alternative depth = 0.9×45 = 42.75 ft
A water jet that leaves a nozzle at 95 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets located on the perimeter of a wheel. Determine the power generation potential of this water jet. The power generation potential of the water jet is kW.
Answer:
P= 541.5 kW.
Explanation:
Given that
velocity of water after leaving the nozzle ,v= 95 m/s
The mass flow rate of the water , m= 120 kg/s
The power generated P is given as
[tex]P=\dfrac{1}{2}mv^2[/tex]
Now by putting the values in the above equation we get
[tex]P=\dfrac{1}{2}\times 120\times 95^2\ W[/tex]
P=541500 W
The power in kW will be 541.5 kW.
Therefore the answer will be 541.5 kW
P= 541.5 kW.
The power generation potential of the water jet is 542.25 kW.
Explanation:To determine the power generation potential of the water jet, we need to calculate the kinetic energy of the water jet and then convert it to power. The kinetic energy of the water jet can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass flow rate of the water and v is the velocity of the water jet. Given that the flow rate is 120 kg/s and the velocity is 95 m/s, we can calculate the kinetic energy to be KE = 0.5 * 120 * 95^2 = 0.5 * 120 * 9025 = 542,250 J/s.
To convert the kinetic energy to power, we divide by the time taken to deliver the energy. Since the flow rate is given in kg/s, we can assume the time taken is 1 second. Therefore, the power generation potential of the water jet is 542,250 J/s, or 542.25 kW.
Code a Boolean expression that tests if a decimal variable named currentSales is greater than or equal to 1000 or a Boolean variable named newCustomer is equal to true. Code this statement so both conditions will always be tested, and code it in the shortest way possible]
Answer:
Given
Decimal variable: currentSales
Decimal test value: 1000
Boolean variable: newCustomer
Boolean default value: true
The following code segment is written in Java
if(currentSales == 1000 || newCustomer)
{
//Some statements
}
The Program above tests two conditions using one of statement
The first condition is to check if currentSales is 1000
== Sign is s a relational operator used for comparison (it's different from=)
|| represents OR
The second condition is newCustomer, which is a Boolean variable
If one or both of the conditions is true, the statements within the {} will be executed
Meaning that, both conditions doesn't have to be true;
At least 1 condition must be true for the statement within the curly braces to be executed
A Scotch-yoke mechanism is used to convert rotary motion into reciprocating motion. As the disk rotates at the constant angular rate , a pin A slides in a vertical slot causing the slotted member to displace horizontally according to x = r sin(t) relative to the fi xed disk center O. Determine the expressions for the velocity and acceleration of a point P
Answer:
The question continues ; Determine the expressions for the velocity and acceleration of a point P as a function of time t, and determine the maximum velocity of point P during one cycle. Use the values r = 75mm and w = pie-rads/s
Explanation:
The diagram and the detailed step by step explanation is as shown in the attachment
In the Scotch-yoke mechanism, if the displacement is given by x = r sin(t), the velocity v = r cos(t) and acceleration a = -r sin(t). The negative sign in acceleration indicates that it is in the opposite direction to displacement.
Explanation:The Scotch-yoke mechanism which converts rotary motion into reciprocating motion can be analyzed using principles of kinematics. If we have the displacement given by x = r sin(t), the velocity and acceleration can be derived from this displacement equation.
The velocity (v) is the time derivative of the displacement function, i.e., v = dx/dt = r cos(t).
The acceleration (a) is the time derivative of the velocity function, so a = dv/dt = -r sin(t).
The negative sign signifies that acceleration is in the opposite direction to displacement.
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A cylindrical tank has a thin barrier and it carries two fluids, one of the fluids has specific gravity of 2.0 and the other fluid has a specific weight of 100 lbf/ft3. The mass of the tank is 20lb-mass. Determine the magnitude of the vertical force required to give the tank a downward acceleration of 10 ft/s2.
Answer:
attached below
Explanation:
A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz. Draw the circuit
The question is incomplete! Complete question along with answers and explanation is provided below.
Question:
A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.
a) Draw the circuit
b) Calculate the power drawn by the lamp
c) Calculate the apparent power
d) Calculate the power factor
e) Calculate the reactive power
f) Calculate the size of the capacitor necessary to provide unity power factor correction
Explanation:
a) draw the circuit
Refer to the attached image.
As you can see in the attached drawing, it is a series circuit containing two resistors and one inductor.
In a series circuit, current remains same throughout the circuit
The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.
The current flowing in the circuit can be found by ohm's law
I = V/Z
where V is the voltage and Z is the total impedance of the circuit
Z = R + XL
where XL is the inductive reactance
XL = j2 π f L
XL = j2*π*60*0.9
XL = j339.29Ω
Total resistance is
R =200 + 80 = 280 Ω
Total impedance is
Z = 280 + j339.29 Ω
b) Calculate the power drawn by the lamp
First calculate the current
I = V/Z
I = 120/(280 + j339.29)
I = 0.272<-50.46° A (complex notation)
P = I²R
P = (0.272)²200
P ≈ 15 W
Power drawn by the circuit
P=V*I*cos(50.46°)
P=20.77 W
c) Calculate the apparent power
A = VI*
A = 120*0.272<50.46°
A = 32.64<50.46° VA
d) Calculate the power factor
PF = cos(50.46)
PF = 0.63
e) Calculate the reactive power
Q = VIsin(50.46)
Q = 120*0.272<-50.46*sin(50.46)
Q = 25.13<-50.46 VAR
f) Calculate the size of the capacitor necessary to provide unity power factor correction
The required reactive compensation power is
Qc = P (tan(old) - tan(new))
Qc = 20.77 (tan(50.46) - tan(0))
Qc = 25.16 VAR
C = Qc/2πfV²
C = 25.16/2*π*60*120²
C = 4.63 uF
Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.
The space shuttle fleet was designed with two booster stages. If the first stage provides a thrust of 686.68 Mega-newtons(MN) and the space shuttle has a mass of 5,470,0005, 470,000 pound-mass, what is the acceleration of the space craft in miles per hour squared?
Answer:
6.30 miles/hour
Explanation:
Newton's second law applies here. In simple terms:
[tex]F = ma[/tex]
where F = Force (Thrust) in N
a = acceleration (m/s²)
The acceleration can be given by rearranging the formula to give:
[tex]a = \frac{F}{N}[/tex]
= [tex]\frac{(686.68*10^{6} )}{24811505120150.2656} \\= 0.0000277 m/s\\= 6.03 miles/hr[/tex]
For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream under steady-state conditions with a first-order reaction: reactor volume = 280 m3, flow rate = 14 m3 · day−1, and reaction rate coefficient = 0.05 day−1.
Answer:
The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.
Xₚբᵣ = 0.632
X꜀ₘբᵣ = 0.5
Xₚբᵣ > X꜀ₘբᵣ
Explanation:
From the reaction rate coefficient, it is evident the reaction is a first order reaction
Performance equation for a CMFR for a first order reaction is
kτ = (X)/(1 - X)
k = reaction rate constant = 0.05 /day
τ = Time constant or holding time = V/F₀
V = volume of reactor = 280 m³
F₀ = Flowrate into the reactor = 14 m³/day
X = conversion
k(V/F₀) = (X)/(1 - X)
0.05 × (280/14) = X/(1 - X)
1 = X/(1 - X)
X = 1 - X
2X = 1
X = 1/2 = 0.5
For the PFR
Performance equation for a first order reaction is given by
kτ = In [1/(1 - X)]
The parameters are the same as above,
0.05 × (280/14) = In (1/(1-X)
1 = In (1/(1-X))
e = 1/(1 - X)
2.718 = 1/(1 - X)
1 - X = 1/2.718
1 - X = 0.3679
X = 1 - 0.3679
X = 0.632
The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.
To determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream, we compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation r = kC. In a CMFR, the volume is constant, while in a PFR, the volume varies. Therefore, a PFR may be more efficient depending on the reactor design.
Explanation:To determine whether a CMFR (Continuous Mixed Flow Reactor) or a PFR (Plug Flow Reactor) is more efficient in removing a reactive compound from the waste stream, we need to compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation: r = kC, where r is the reaction rate, k is the reaction rate coefficient, and C is the concentration of the reactive compound.
In a CMFR, the volume is constant, so the reactor volume (280 m3) is equal to the product of the flow rate (14 m3·day−1) and the residence time (t), which is the time it takes for the fluid to pass through the reactor. Therefore, t = V/Q = 280/14 = 20 days.
In a PFR, the volume varies along the length of the reactor, and the residence time is defined as the integral of the volume divided by the flow rate. Using the equation t = ∫V/Q, we can calculate the residence time for a PFR.
Since the residence time for a CMFR is fixed at 20 days, and the residence time for a PFR can be longer or shorter depending on the reactor design, a PFR may be more efficient in removing the reactive compound from the waste stream under steady-state conditions with a first-order reaction.
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A supersonic nozzle has an exit area 2.5 time the throat area. For a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, find the Mach number at the throat and at the exit plane for:
a. Pa/Pt = 0.06
b. Pa/Pt = 0.9725
Answer:
The Mach number of the throat for supersonic flow = M* = 1
and the Mach number at exit = 2.44
For
a. Pa/Pt = 0.06, Me = 2.484 Supersonic flow
b. when Pa/Pt = 0.9725 Me = 0.1999 ≅ 0.2 or subsonic flow The mach number at the throat could also be determined given the temperature parameter
Explanation:
To solve the question we note that for a supersonic nozzle, the mach number at the throat = 1
Therefore M* = 1
[tex]\frac{A_{e} }{A^{*} } = 2.5[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(\gamma -1)M_{e} ^{2} }{\gamma +1} )^{\frac{\gamma +1}{2(\gamma -1)} }[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(0.4)M_{e} ^{2} }{2.4} )^{3 }[/tex] = [tex]\frac{1}{M_{e} } ({2+(0.4)M_{e} ^{2} })^{3 } = 34.56[/tex]
34.56Me = (2+(0.4)M²)³ expanding and collecting like terms we have
Possible solutions of Me = 0.2395, 2.44, 0.90
Since flow is supersonic, Me = 2.44
a)
Solving for [tex]M_{e}[/tex] we have [tex]\frac{P_{a} }{P_{t} } =(1+\frac{\gamma -1}{2} M^{2} _{e} )^{\frac{-\gamma}{\gamma -1} }[/tex]
When Pa/Pt = 0.06 =[tex](1+\frac{1.4 -1}{2} M^{2} _{e} )^{\frac{-1.4}{1.4 -1} }[/tex] = [tex](1+0.2M^{2} _{e} )^{-3.5 }[/tex]
Solving, we get Me = 2.484 Supersonic flow
b)
When Pa/Pt = 0.9725, Me = 0.1999 ≅ 0.2 or subsonic flow
The Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.
To find the Mach number at the throat and at the exit plane of a supersonic nozzle with an exit area 2.5 times the throat area for a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, we can use the following steps:
Calculate the critical pressure ratio:
pr_crit = (2 / (gamma + 1)) ** (gamma / (gamma - 1))
pr_crit = 0.5283
Calculate the Mach number at the throat:
Mt = sqrt((1 - pr_crit) / (gamma - 1))
Mt = 1.0859
Calculate the Mach number at the exit plane:
Me = sqrt((1 - (Pa / Pt)) / (gamma - 1))
Part a:
Pa/Pt = 0.06
Me = sqrt((1 - 0.06) / (1.4 - 1))
Me = 1.5329
Part b:
Pa/Pt = 0.9725
Me = sqrt((1 - 0.9725) / (1.4 - 1))
Me = 0.2622
Therefore, the Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.
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Your organization spans multiple geographical locations. The name resolution is happening with a single DNS zone for the entire organization. Which of the following is likely to happen if you continue with the single DNS zone? [Choose all that apply.]
Name resolution traffic goes to the single zone
Granular application of policies
Centralized Management
Higher security
Administrative burden
Submit
Answer:Name resolution traffic goes to the single zone
Administrative burden
Submit
Centralized Management
Explanation:DNS(Domain name system) is a term used in the internet which Describes the conversion of alphabetical names into Numerical representations,he a large Organisation as stated which spans through different Geographical areas continue with a single Domain name system it will lead to the following.
Name resolution traffic will increase which might delay the execution of tasks
Administrative burden will be Increased as it is carrying out a wide range of activities.
Centralized management which may affect the flow of work.
Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room constructor requires parameters for length, width, and height fields (all of type int); use a variety of values when constructing the objects. The Room class also contains the following fields: Area - The wall area of the Room (as an int) Gallons - The number of gallons of paint needed to paint the room (as an int)
Answer:
Explanation:
Code used will be like
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace PaintingWall
{
class Room
{
public int length, width, height,Area,Gallons;
public Room(int l,int w,int h)
{
length = l;
width = w;
height = h;
}
private int getLength()
{
return length;
}
private int getWidth()
{
return width;
}
private int getHeight()
{
return height;
}
public void WallAreaAndNumberGallons()
{
Area = getLength() * getHeight() * getWidth();
if (Area < 350)
{
Gallons = 1;
}
else if (Area > 350)
{
Gallons = 2;
}
Console.WriteLine ("The area of the Room is " + Area);
Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);
}
}
class PaintingDemo
{
static void Main(string[] args)
{
int l, w, h;
Room[] r = new Room[8];
for (int i = 0; i <= 7; i++)
{
Console.WriteLine("Room "+(i+1));
Console.Write("Enter Length : ");
l = Convert.ToInt32(Console.ReadLine() );
Console.Write("Enter Width : ");
w = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter Height : ");
h= Convert.ToInt32(Console.ReadLine());
r[i] = new Room(l,w,h);
Console.WriteLine();
}
for (int i = 0; i <= 7; i++)
{
Console.WriteLine("Room " + (i + 1));
r[i].WallAreaAndNumberGallons();
}
Console.ReadKey();
}
}
}
An airplane starts from rest, travels 5000ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi/h. it then climbs in a straight line with a uniform acceleration of 3 ft/s^s until it reaches a constant speed of 220 mi/h. draw the st, vt, and at graphs that describe the motion.
Answer:
Explanation:
Given
Take off speed [tex]v=162\ mph\approx 237.6\ ft/s[/tex]
distance traveled in runway [tex]d=5000 ft[/tex]
using motion of equation
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](237.6)^2=2\times a\times 5000[/tex]
[tex]a=5.64\ ft/s^2[/tex]
Acceleration after take off [tex]a_2=3\ ft/s^2[/tex]
time taken to reach [tex]237.6 ft/s[/tex]
[tex]v=u+at[/tex]
[tex]237.6=0+5.64\times t[/tex]
[tex]t=42.127\ s [/tex]
after take off it take [tex]t_2[/tex] time to reach [tex]220 mph\approx 322.67[/tex]
[tex]322.67=237.6+3\times t_2[/tex]
[tex]t_2=28.35\ s[/tex]
total time taken [tex]t_0=t+t_1[/tex]
[tex]t_0=70.48\ s[/tex]