Answer:
Shape factor
Explanation:
Shape factor is the ratio of maximum plastic moment to maximum elastic moment.Shape factor is denoted by K.
Shape factor can be given as
[tex]K=\dfrac{M_p}{M_y}[/tex]
[tex]K=\dfrac{\sigma _yZ_p}{\sigma _y Z}[/tex]
[tex]K=\dfrac{Z_p}{ Z}[/tex]
For a solid rectangular section made from ductile material shape factor is 1.5 .
S1.1 The Acre-Foot. Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumped 2.0 in. of rain in 30 minutes on a town of area 26 km2. What volume of water, in acre-feet, fell on the town? S1.2 The Thunderstorm. What mass of water fell on the town in S1.1The Acre-Foot during the thunderstorm? One cubic meter of water has a mass of 103kg.
Answer:
1072 acre foot
1331424000 kg
Explanation:
1 feet has 12 inches, so 2 in is 0.167 feet.
1 km^2 has 1 million m^2.
1 acre is 4074 m^2.
So, 1 km is 247 acres.
Then 26 km^2 is 6422 acres.
So, the volume of water is
6422 * 0.167 = 1072 acre-foot
Since one cubic meter of water has 1000 kg
One inch is 25.4 mm = 0.0254 m
One feet is 12 * 0.0254 = 0.3048 m
An acre-feet has a volume of
4074*0.3048 = 1242 m^3
And that is a mass of water of
1242 * 1000 = 1242000 kg/acre-feet
Therefore the mass of rainwater in the town is of
1072 * 1242000 = 1331424000 kg = 1331424 tons
One horsepower is the ability to lift 550 pounds a height of one foot in one second or one horsepower is equal to 550ft-lb/s. How many ft-lb/s are there in 32 horsepower?
Answer:
32 horsepower will be equal to 17600 ft-lb/sec
Explanation:
We have given 1 horsepower = 550 ft-lb/sec
We have to convert 32 horsepower into ft-lb/sec
As we know that 1 horsepower is equal to 550 ft-lb/sec
So for converting horsepower to ft-lb/sec we have multiply with 550
We have given 32 horse power
So [tex]32\ horsepower =32\times 550=17600ft-lb/sec[/tex]
So 32 horsepower will be equal to 17600 ft-lb/sec
The time factor for a doubly drained clay layer
undergoingconsolidation is 0.2
a. What is the degree of consolidation (Uz) at z/H=0.25,
0.5,and 0.75
b. If the final consolidation settlement is expected to be
1.0m, how much settlement has occurred when the time factor is 0.2
andwhen it is 0.7?
Answer with Explanation:
Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is
[tex]T_v=\frac{\pi }{4}(\frac{U}{100})^2[/tex]
Solving for 'U' we get
[tex]\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%[/tex]
Since our assumption is correct thus we conclude that degree of consolidation is 50.46%
The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2
i)[tex]\frac{z}{H}=0.25=U=0.71[/tex] = 71% consolidation
ii)[tex]\frac{z}{H}=0.5=U=0.45[/tex] = 45% consolidation
iii)[tex]\frac{z}{H}=0.75U=0.3[/tex] = 30% consolidation
Part b)
The degree of consolidation is given by
[tex]\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm[/tex]
Thus a settlement of 50.46 centimeters has occurred
For time factor 0.7, U is given by
[tex]T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59[/tex]
thus consolidation of 85.59 % has occured if time factor is 0.7
The degree of consolidation is given by
[tex]\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm[/tex]
Describe a physical meaning of velocity and acceleration.
Answer and Explanation:
Velocity : Velocity gives us information about how much we can travel in a particular interval of time.
Velocity is a rate of change of distance if we have information about velocity and time we can easily calculate how much distance we travel in that particular time.
And if we have information about distance and time then we can find how much velocity we should have to travel that distance in given time
Acceleration : Acceleration gives us information about how the velocity is changes with respect to time
If the velocity is increases with time then there is positive acceleration and if velocity is decreases with time then it is negative acceleration.
An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50 km/h? If the vehicle accelerates to 100 km/h in 15 seconds, what is the change in power required?
Answer:
a)Ek=115759.26J
b)23.15kW
Explanation:
kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation
[tex]Ek=0.5mv^{2}[/tex]
for the first part
m=1200Kg
V=50km/h=13.89m/s
solving for Ek
[tex]Ek=0.5(1200)(13.89)^2[/tex]
Ek=115759.26J
For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.
m=1200kg
V=100km/h=27.78m/s
[tex]Ek=0.5(1200)(27.78)^2=462963J\\[/tex]
taking into account all of the above the following equation is inferred
ΔE=[tex]\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15} =23146.916W=23.15kW[/tex]
An ideal Otto Cycle with air as the working fluid has a compression ratio of10. The minimum temperature and pressure in the cycle are 25 degrees celsius and 100 kPa respectively. The highest temperature in the cycle is 1000 degrees celsius. Using the PG model for air with properties: k =1.4, c_p = 1 kJ.kg.K, c_v = 0.717 kJ/kg.K, determine the exhaust temperature (temperature at the end of the expansion stroke) in Kelvins.
Answer:
[tex]T_4[/tex]= 506.79 K
Explanation:
Given that
Minimum temperature [tex]T_1[/tex]= 25 C
Pressure = 100 KPa
Highest temperature [tex]T_3[/tex]= 1000 C
Compression ratio r= 10
We know that for ideal Otto cycle
[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]
Where [tex]T_3[/tex] is the exhaust temperature.
Now by putting the values
[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]
[tex]\dfrac{1000+273}{T_4}=10^{1.4-1}[/tex]
[tex]T_4[/tex]= 506.79 K
Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Consider a drag of 12 kN on the shuttle. Calculate the work done by the shuttle engine.
Answer:
work done = 48.88 × [tex]10^{9}[/tex] J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × [tex]10^{9}[/tex] J
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws 12.5 hp for 16 h/day, five days per week. Compute the energy used by the mo- tor for one year. Express the result in ft.lb and W.h
The energy used by the motor for one year is 102,313,596,240 ft.lb/year, expressed in both ft.lb and W.h.
1 horsepower (hp) is approximately equal to 745.7 watts (W).
So,[tex]12.5 hp * 745.7 W/hp[/tex]
= 9312.5 W.
The motor draws 9312.5 watts for 16 hours per day.
So, [tex]9312.5 W * 16 hours/day[/tex]
= 148,200 W.h/day.
Multiply the daily energy by the number of days per year:
The motor operates for 5 days per week, so in one year (considering 52 weeks), it operates for 5 days/week × 52 weeks/year = 260 days/year.
[tex]148,200 W.h/day *260 days/year[/tex]
= 38,532,000 W.h/year.
To express the result in ft.lb, we need to convert the energy from watt-hours to foot-pounds:
1 watt-hour (Wh) is approximately equal to 2655.22 foot-pounds (ft.lb).
So, [tex]38,532,000 W.h/year * 2655.22 ft.lb/W.h[/tex]
≈ 102,313,596,240 ft.lb/year.
To 3 significant digits, what is the temperature of water in degrees C, if its pressure is 350 kPa and the quality is 0.01
Answer:
138.9 °C
Explanation:
The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C
A large fraction of the thermal energy generated in the engine of a car is rejected to the air by the radiator through the circulating water. Should the radiator be analyzed as an open system?
Answer:
Yes
Explanation:
Type of control system
1.Open system
When mass and energy is allowed to pass through the system ,then this type of system is called open system.
2.Close system
When only energy is allowed to pass through the system ,then this type of system is called close system.
3.Isolated system
When mass and energy did not allowed to pass through the system ,then this type of system is called isolated system.
In the engine radiator ,mass of hot fluid enters and interact with the cold air then after heat transfer take place between these two fluids,that is why we can say that radiator is an open system.
It become necessity to use the radiator in all type of engine because it produce cooling effect to the engine cylinder .If temperature of engine cylinder will become too high then it will leads to the high thermal stresses and may be it will break the engine cylinder.
Yes, the radiator should be analyzed as an open system.
In thermodynamics, a system is classified as open if both mass and energy can cross its boundaries. The radiator in a car's engine cooling system is designed to dissipate heat from the engine to the air. This dissipation occurs through the circulation of water (or coolant) which absorbs heat from the engine and releases it to the air flowing through the radiator.
In the radiator:
1. Mass Flow: The coolant continuously enters and exits the radiator, carrying thermal energy with it.
2. Energy Transfer: Heat is transferred from the hot coolant to the cooler ambient air through the radiator fins, driven by forced convection due to airflow.
3. Open System Characteristics: Both mass (coolant) and energy (heat) cross the boundaries of the radiator.
Considering these aspects, the radiator facilitates the flow of coolant (mass) and the transfer of thermal energy (heat), making it an open system.
The radiator in a car should be analyzed as an open system due to the continuous flow of coolant and heat transfer involved in its operation. This perspective is essential for accurately understanding and optimizing the cooling process.
A bar of uniform cross section is 86.9 in longand weighs 89.1 lb. A weight of 79.0 lb is suspended from one end. The bar and weight combination is to be suspended from a cable attached at the balance point. How far (in) from the weight should the cable be attached, and what is the tension (lb) in the cable?
Answer:
y = 20.41 in
T= 168.1 lb
Explanation:
From diagram
Total force balance in vertical direction
T= 89.1 + 79 lb
T= 168.1 lb
Now taking moment about point m
Mm= 0 Because system is in equilibrium position
79 x 43.45 = T x y
Now by putting the value of T
79 x 43.45 = 168.1 x y
y = 20.41 in
So the cable attached at distance of 20.41 in from the mid point of bar.
Tension in the cable = 168.1 lb
Water enters a tank from two pipes, one with a flow rate of 0.3 kg/s and the other with a flow rate of 0.1 kg / s. The tank has a small hole through which water leaks out at a rate of 0.03 kg / s. If the tank initially contained 40 kg of water, how much will it have after 2 minutes?
Answer:
total amount of water after 2 min will be 84.4 kg/s
Explanation:
Given data:
one tank inflow = 0.1 kg/s
2nd tank inflow = 0.3 kg/s
3rd tank outflow = 0.03 kg/s
Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s
From third point, outflow is 0.03 kg/s
Therefore, resultant in- flow = 0.4 - 0.03
Resultant inflow is = 0.37 kg/s
Tank has initially 40 kg water
In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg
So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg
What does stall mean?
Answer:
Explanation:
In aeronautics stall is the separation of the boundary layer from the wings of plane. This causes loss of lift because the otherwise low pressure area above the wing become filled with turbulent whirls.
Stall occurs when the critical angle of attack of the wing is exceeded.
A disk is rotating around an axis located at its center. The angular velocity is 0.6 rad/s and the angular acceleration is 0.3 rad/s^2. The radius of the disk is 0.2 m. What is the magnitude of the acceleration at a point located on the outer edge of the disk, in units of m/s?
Answer:
[tex]a=0.0937 \ m/s^2[/tex]
Explanation:
Given that
Angular velocity (ω)= 0.6 rad/s
Angular acceleration (α)= 0.3 [tex]rad/s^2[/tex]
Radius (r)= 0.2 m
We know that is disc is rotating and having angular acceleration then it will have two acceleration .one is radial acceleration and other one is tangential acceleration.
So
[tex]Radial\ acceleration(a_r)=\omega ^2r\ m/s^2[/tex]
[tex]Radial\ acceleration(a_r)=0.6^2 \times0.2 \ m/s^2[/tex]
[tex](a_r)=0.072 \ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=\alpha r\ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=0.3\times 0.2\ m/s^2[/tex]
[tex](a_t)=0.06 \ m/s^2[/tex]
So the total acceleration ,a
[tex]a=\sqrt{a_t^2+a_r^2}\ m/s^2[/tex]
[tex]a=\sqrt{0.06^2+0.072^2}\ m/s^2[/tex]
[tex]a=0.0937 \ m/s^2[/tex]
A mass of 2 kg is hanging to a spring of 200 n/m spring constant vertically. Calculate the period of the period if the spring is set in simple harmonic motion.
Answer:
The period is 0.628s
Explanation:
As the system mass-spring is in simple harmonic motion, the period is given by the equation:
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where m is the mass of 2kg and k is the spring constant [tex]200\frac{N}{m}[/tex]
Replacing the values in the equation, we have:
[tex]T=2\pi \sqrt{\frac{2Kg}{200\frac{N}{m}}}[/tex]
And finally we find the value of the period, that is:
[tex]T=0.628s[/tex]
Given 10 parts per billion. What is the concentration
inmg/L?
Answer:
The concentration in mg/L equals 0.01 mg/L
Explanation:
Since the concentration is given as 10 parts per billion
We know that
1 part per billion equals 0.001 mg/L
Thus 10 parts per billion concentration equals
[tex]10\times 0.001mg/L\\\\=0.01mg/L[/tex]
The high-pressure air system at OSU's Aerospace Research Center is fed by a set of two cylindrical tanks. Each tank has an outer height of 50 ft and an outer diameter of 4.6 ft. The tanks are made of 0.1 ft thick steel (steel = 499 lbm/ft?) and store air at a maximum pressure of 2500 psi at -10 °F. How much load must the support structure at the base of the tanks carry?
Answer:
179000 lb
Explanation:
The supports must be able to hold the weight of the tank and the contents. Since tanks are pressure tested with water, and the supports cannot fail during testing, we disregard the air and will consider the weight of water.
The specific weight of water is ρw = 62.4 lbf/ft^3
These tanks are thin walled because
D / t = 4.6 / 0.1 = 46 > 10
To calculate the volume of steel we can approximate it by multiplying the total surface area by the thickness:
A = 2 * π/4 * D^2 + π * D * h
The steel volume is:
V = A * t
The specific weight is
ρ = δ * g
ρs = 499 lbm/ft^3 * 1 lbf/lbm = 499 lbf/ft^3
The weight of the steel tank is:
Ws = ρs * V
Ws = ρs * A * t
Ws = ρs * (2 * π/4 * D^2 + π * D * h) * t
Ws = 499 * (π/2 * 4.6^2 + π * 4.6 * 50) * 0.1 = 37700 lb
The weight of water can be approximated with the volume of the tank:
Vw = π/4 * D^2 * h
Ww = ρw * π/4 * D^2 * h
Ww = 62.4 * π/4 * 4.6^2 * 50 = 51800 lb
Wt = Ws + Ww = 37700 + 51800 = 89500 lb
Assuming the support holds both tanks
2 * 89500 = 179000 lb
The support must be able to carry 179000 lb
A closed, rigid, 0.50 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the pressure is P2 = 4 bar. Determine the initial quality, X1, and the heat transfer, in kJ.
Answer:
X1= 41%
heat transfer = -3450.676 KJ
Explanation:
To get the properties for pure substance in a system we need to know at least to properties. These are usually pressure and temperature because they’re easy to measure. In this case we know the initial pressure (20 bar) which is not enough to get all the properties, but they ask to determine quality, this a property that just have meaning in the two-phase region (equilibrium) so with this information we can get the temperature of the system and all its properties.
There is another property that we can calculate from the data. This is the specific volume. This is defined as [tex]\frac{volume}{mass}[/tex]. We know the mass (12 Kg) and we can assume the volume is the volume of the tank (0.5 [tex]m^{3}[/tex]) because they say that the tank was filled.
With this we get a specific volume of
Specific volume = [tex]\frac{0,5 m^{3}}{ 12 kg}= 0.04166667 \frac{m^{3}}{Kg}[/tex]
From the thermodynamic tables we can get the data for the saturated region with a pressure of 20 bar.
Temperature of saturation = 212.385 °C
Specific volume for the saturated steam (vg) = 0.0995805 [tex]\frac{m^{3}}/{Kg}[\tex]
Specific volume for the saturated liquid (vf)= 0.00117675 [tex]\frac{m^{3}}/{Kg}[\tex]
The specific volume that we calculate before 0.04166667 m^3/Kg is between 0.00117675 m^3/Kg and 0.0995805 m^3/Kg so we can be sure that we are in two-phase region (equilibrium).
The quality (X) is defined as the percentage in mass of saturated steam in a mix (Two-phase region)
The relation between specific volume and quality is
[tex]v = (1-x)*v_{f} + x*v_{g}[\tex]
where
v in the specific volume in the condition (0.04166667 m^3/Kg)
vf = Specific volume for the saturated liquid (0.00117675 m^3/Kg)
vg = Specific volume for the saturated steam (0.0995805 m^3/Kg)
x = quality
clearing the equation we get:
[tex]X = \frac{(v-v_{f})}{(v_{g}-v_{f})}[/tex]
[tex]X =\frac{(0.04166667- 0.00117675)}{ (0.0995805 – 0.00117675)} = 0.411[/tex]
The quality is 41%
To calculate the heat transfer we use the next equation.
Q = m * Cp * delta T
Where
Q = heat transfer (Joules, J)
m= mass of the substance (g)
Cp = specific heat (J/g*K) from tables
Delta T = change in temperature in K for this equation.
The mass of the substance is 12 kg or 12000 g for this equation
Cp from tables is 4,1813 J/g*K. You can find this value for water in different states. Here we are using the value for liquid water.
For delta T, we know the initial temperature 212.385 °C.
We also know that the system was cooled. Since we don’t have more information, we can assume that the system was cooled until a condition where all the steam condensates so now we have a saturated liquid. Since we know the pressure (4 bar), we can get the temperature of saturation for this condition from the thermodynamics tables. This is 143.613 °C, so this is the final temperature for the system.
T(K) = T°C +273
T1(K) = 212.385 + 273.15 = 485.535 K
T2 (K) = 143.613 +273.15= 416.763 K
Delta T (K) = (T2-T1) =416.763 K - 485.535 K = -68.772 K
Now we can calculate Q
Q = 12000g * 4,1813 J/g*K* (-68.772 K) = -3450676.36 J or -3450.676 KJ
Is negative because the heat is transfer from the water to the surroundings
Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 mm, determine the minimum flow rate that the pump must provide.
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
[tex]Q = A \times V[/tex]
where A is Area
[tex]A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2[/tex]
[tex]Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s[/tex]
minimum flow rate provided by pump is 0.02513 m^3/s
What is the significance of Saint Venant's principle?
Answer:
While calculating the stresses in a body since we we assume a constant distribution of stress across a cross section if the body is loaded along the centroid of the cross section , this assumption of uniformity is assumed only on the basis of Saint Venant's Principle.
Saint venant principle states that the non uniformity in the stress at the point of application of load is only significant at small distances below the load and depths greater than the width of the loaded material this non uniformity is negligible and hence a uniform stress distribution is a reasonable and correct assumption while solving the body for stresses thus greatly simplifying the analysis.
The torque required to drive a 40 mm diameter power screw having double square threads with a pitch of 6 mm is 1.0 kN-m. The friction for the thread is 0.10. Is the condition self-locking? Show your calculation.
Answer:
This is self locking.
Explanation:
Given that
Diameter d= 40 mm
Pitch (P)= 6 mm
Torque T= 1 KN.m
Friction(μ) = 0.1
Thread is double square.
Condition for self locking
[tex]\mu >tan\theta[/tex]
[tex]tan\theta=\dfrac{L} {\pi d}[/tex]
L= n x P
N= number of start
Here given that double start so n=2
L=2 x 6 = 12 mm
[tex]tan\theta=\dfrac{L} {\pi d}[/tex]
[tex]tan\theta=\dfrac{12} {\pi \times 40}[/tex]
[tex]tan\theta=0.095[/tex] (μ = 0.1)
[tex]\mu >tan\theta[/tex]
So from we can that this is self locking.
Derive the following conversion factors: (a) Convert a pressure of 1 psi to kPa (b) Convert a vol ume of 1 liter to gallons (c) Convert a viscosity of 1 lbf .s/ft^2 to N s/m^2
Answer:
a)6.8 KPa
b)0.264 gallon
c)47.84 Pa.s
Explanation:
We know that
1 lbf= 4.48 N
1 ft =0.30 m
a)
Given that
P= 1 psi
psi is called pound force per square inch.
We know that 1 psi = 6.8 KPa.
b)
Given that
Volume = 1 liter
We know that 1000 liter = 1 cubic meter.
1 liter =0.264 gallon.
c)
[tex]1\ \frac{lb.s}{ft^2}=47.84\ \frac{Pa.s}{ft^2}[/tex]
An airplane flies horizontally at 80 m/s. Its propeller delivers 1300 N of thrust (forward force) to overcome aerodynamic drag (backward force). Using dimensional reasoning and unity conversion ratios, calculate the useful power delivered by the propeller in units of kW and horsepower.
Answer:
Power in kW is 104 kW
Power in horsepower is 139.41 hp
Solution:
As per the question:
Velocity of the airplane, [tex]v_{a} = 80 m/s[/tex]
Force exerted by the propeller, [tex]F_{p} = 1300 N[/tex]
Now,
The useful power that the propeller delivered, [tex]P_{p}[/tex]:
[tex]P_{p} = \frac{Energy}{time, t}[/tex]
Here, work done provides the useful energy
Also, Work done is the product of the displacement, 'x' of an object when acted upon by some external force.
Thus
[tex]P_{p} = \frac{F_{p}\times x}{time, t}[/tex]
[tex]P_{p} = \frac{F_{p}\times x}{time, t}[/tex]
[tex]P_{p} = F_{p}\times v_{a}[/tex]
Now, putting given values in it:
[tex]P_{p} = 1300\times 80 = 104000 W = 104 kW[/tex]
In horsepower:
1 hp = 746 W
Thus
[tex]P_{p} = \frac{104000}{746} = 139.41 hp[/tex]
Calculate the efficiency of a Carnot Engine working between temperature of 1200°C and 200°C.
Answer:
efficiency = 0.678
Explanation:
First we have to change temperatures from °C to K (always in thermodynamics absolute temperature is used).
[tex]\text{hot body temperature,}T_H = 1200 \circC + 273.15 = 1473.15 K[/tex]
[tex] \text{cold body temperature,} T_C = 200 \circC + 273.15 = 473.15 K[/tex]
Efficiency [tex] \eta [/tex] of a carnot engine can be calculated only with hot body and cold body temperature by
[tex] \eta = 1 - \frac{T_C}{T_H}[/tex]
[tex] \eta = 1 - \frac{473.15 K}{1473.15 K}[/tex]
[tex] \eta = 0.678[/tex]
A particle is moving along a straight line and has an
acceleration of kV^2 m/s^2, where V is the velocity of the
particle. At time t=0, the velocity of the particle = 4m/s, and the
time t=30s the velocity = 26m/s and displacement at time t = Dt
metres. Derive expressions for both velocity and displacement as a
function of time t.
Answer with Explanation:
By definition of acceleration we have
[tex]a=\frac{dv}{dt}[/tex]
Given [tex]a=kv^{2}[/tex], using this value in the above equation we get
[tex]kv^2=\frac{dv}{dt}\\\\\frac{dv}{v^{1}}=kdt[/tex]
Upon integrating on both sides we get
[tex]\int \frac{dv}{v^2}=\int kdt\\\\-\frac{1}{v}=kt+c[/tex]
'c' is the constant of integration whose value can be found out by putting value of 't' = 0 and noting V =4 m/s
Thus [tex]c=-\frac{1}{4}[/tex]
the value of 'k' can be found by using the fact that at t= 30 seconds velocity = 26 m/s
[tex]\frac{-1}{26}=k\times 30-\frac{1}{4}\\\\\therefore k=\frac{\frac{-1}{26}+\frac{1}{4}}{30}\\\\k=7.05\times 10^{-3}[/tex]
Hence the velocity as a function of time is given by
[tex]v(t)=\frac{-1}{7.05\times 10^{-3}t-\frac{1}{4}}[/tex]
By definition of velocity we have
[tex]v=\frac{dx}{dt}[/tex]
Making use of the obtained velocity function we get
[tex]\frac{dx}{dt}=\frac{-1}{kt-\frac{1}{4}}\\\\\int dx=\int \frac{-dt}{kt-\frac{1}{4}}\\\\x(t)=\frac{-1}{7.05\times 10^{-3}}\cdot ln(7.05\times 10^{-3}t-\frac{1}{4})+x_o[/tex]
here [tex]x_o[/tex] is the constant of integration
Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158 mm, what are the temperature and velocity at the exit?
Answer:
[tex]v_2 = 160.23 m/s[/tex]
[tex]T_2 = 475.797 k[/tex]
Explanation:
given data:
Diameter =[tex] d_1 = 200mm[/tex]
[tex]t_1 =195 degree[/tex]
[tex]p_1 =500 kPa[/tex]
[tex]v_1 = 100m/s[/tex]
[tex]p_2 = 85kPa[/tex]
[tex]d_2 = 158mm[/tex]
from continuity equation
[tex]A_1v_1 = A_2v_2[/tex]
[tex]v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}[/tex]
[tex]v_2 = \frac{d_2v_1}{d_2^2}[/tex]
[tex]v_2 = [\frac{d_1}{d_2}]^2 v_1[/tex]
[tex]= [\frac{0.200}{0.158}]^2 \times 100[/tex]
[tex]v_2 = 160.23 m/s[/tex]
by energy flow equation
[tex]h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w[/tex]
[tex]z_1 =z_2[/tex] and q =0, w =0 for nozzle
therefore we have
[tex]h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2} [/tex]
[tex]dh = \frac{1}{2} (v_1^2 -v_2^2)[/tex]
but we know dh = Cp dt
hence our equation become
[tex]Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)[/tex]
[tex]Cp (T_2 -T_1) = 7836.94[/tex]
[tex](T_2 -T_1) = \frac{7836.94}{1.005*10^3}[/tex]
[tex](T_2 -T_1) = 7.797 [/tex]
[tex]T_2 = 7.797 +468 = 475.797 k[/tex]
For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 316 mm2 without plastic deformation? (b) If the original specimen length is 128 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Answer:
a) 93.852 kN
b) 128.043 mm
Explanation:
Stress is load over section:
σ = P / A
If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:
P = σ * A
316 mm^2 = 3.16*10^-4
P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN
The stiffness of the specimen is:
k = E * A / l
k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m
Hooke's law:
x' = x0 * (1 + P/k)
x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm
Starting from position s = 0 m at time t = 0 s, a particle travels on a straight-line path. The particle's velocity is given by the function v = 5 sin (4s2) m/s, where s is in meters and the argument of the sine function is unitless. Find the particle's acceleration when s = 0.25 m and s = 1.25 m.
Answer:
acceleration is 0.2181 m/s²
acceleration is 27.05 m/s²
Explanation:
given data
s = 0
time t = 0
velocity V = 5 sin (4s²) m/s
to find out
particle's acceleration
solution
we have given velocity = 5 sin(4s²)
and we know here that velocity = [tex]\frac{ds}{dt}[/tex]
so acceleration will be a = [tex]\frac{dv}{dt}[/tex]
put here velocity v
acceleration = [tex]\frac{dv}{dt}[/tex]
acceleration = [tex]\frac{d(5sin(4s^4))}{dt}[/tex]
acceleration = 5 cos4s² × 8s × [tex]\frac{ds}{dt}[/tex]
acceleration = 5 cos4s² × 8s × 5 sin4s²
acceleration = 200 s ×cos4s² × sin4s²
put here s = 0.25 and s = 1.25
so
acceleration = 200× 0.25 ×cos(4×0.25²) × sin(4×0.25²)
acceleration = 0.2181 m/s²
acceleration = 200× 1.25 ×cos(4×1.25²) × sin(4×1.25²)
acceleration = 27.05 m/s²
At the grocery store you place a pumpkin with a mass of 12.5 lb on the produce spring scale. The spring in the scale operates such that for each 5.7 lbf applied, the spring elongates one inch. If local acceleration of gravity is 32.2 ft/s2, what distance, in inches, did the spring elongate?
Answer:
x=2.19in
Explanation:
This is the equation that relates the force and displacement of a spring
F=Kx
m=mass=12.5lbx1slug/32.14lb=0.39slug
F=mg=0.39*32.2=12.52Lbf
then we calculate the spring count in lbf / ft
K=F/x
K=5.7lbf/1in=5.7lbf/in=68.4lbf/ft
Finally we calculate the displacement with the initial equation
X=F/k
x=12.52/68.4=0.18ft=2.19in
In this exercise we want to calculate how much the spring was elogate, for this we have to be:
the spring stayed x=2.19in elogante
organizing the information given in the statement we have that:
mass of 12.5 lb force each 5.7 lbf appliedacceleration of gravity is 32.2 ft/sRecalling the basic equation of the spring we find that:
[tex]F=Kx[/tex]
Where:
F is the applied forceK is the spring constantX is how much the spring has been elongatedSo calculating the force we have:
[tex]F=mg\\=0.39*32.2\\=12.52[/tex]
Putting the value of the force in the given formula:
[tex]K=F/x\\K=5.7/1\\=5.7\\X=F/k=12.52/68.4\\=0.18ft\\=2.19in[/tex]
See more about spring at brainly.com/question/4433395
Thermosets burn upon heating. a)-True b)- false?
Answer:
true
Explanation:
True, there are several types of polymers, thermoplastics, thermosets and elastomers.
Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.
Because of the above, thermosetting polymers burn when heated.