Answer:
[tex]V = 0.208 m^3[/tex]
saturated pressure = 0.91 bar 0r 91 kPa
Explanation:
Given data:
Mass of liquid water 200 kg
Steel tank temperature is 97 degree celcius = 97°C
At T = 97 degree celcius, saturated pressure = 0.91 bar 0r 91 kPa
And also for saturated liquid
Specific volume [tex]\nu[/tex] is 0.00104 m^3/kg
Volume of tank is given as
[tex]V = \nu * m[/tex]
[tex]V = 0.00104 * 200[/tex]
[tex]V = 0.208 m^3[/tex]
You are to define and illustrate the following in regards to a 4 bar linkage mechanism: a) Illustrate and briefly define an open-loop 4 bar linkage system. b) illustrate and briefly define a closed-loop 4 bar linkage system.
Answer:
Answered
Explanation:
Open- loop 4 bar linkage system:
Open- loop 4 bar linkages are not mechanically constrained, meaning in this type of linkages the degree of freedom is are more than 1.
DOF> 1 example = industrial robots, epicyclic gear trains etc.
Closed loop four bar linkage system:
These types are linkages are mechanically constrained and have degree of freedom as one. examples, four bar chains, slider crank mechanism.
DOF=1
Draw and label a typical true stress-strain curve for a ductile material.
Answer:
this is a typical stress-strain curve for a ductile material
Explanation:
A: proportional limit
B:Elastic limit
C:upper yield point
D:lower yield point
E:ultimate strength
F:rupture strength
In a reversing 2-high mill, a series of cold rolling process is used to reduce the thickness of a plate from 45mm down to 20mm. Roll diameter is 600mm and coefficient of friction between rolls and strip 0.15. The specification is that the draft is to be equal on each pass. Determine a) Minimum number of passes required? b) Draft for each pass?
Answer:
Explanation:
Given
Initial Thickness=45 mm
Final thickness=20 mm
Roll diameter=600 mm
Radius(R)=300 mm
coefficient of friction between rolls and strip ([tex]\nu [/tex])=0.15
maximum draft[tex](d_{max})=\nu ^2R[/tex]
[tex]=0.15^2\times 300=6.75 mm[/tex]
Minimum no of passes[tex]=\frac{45-20}{6.75}=3.70\approx 4[/tex]
(b)draft per each pass
[tex]d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}[/tex]
[tex]d=\frac{45-20}{4}=6.25 mm[/tex]
The flow of a real fluid has (more —less - same ) complexity of an ideal fluid, owing to the phenomena caused by the existence of (viscosity-pressure drop- friction
Answer:
The flow of a real fluid has more complexity as compared to an ideal fluid owing to the phenomena caused by existence of viscosity
Explanation:
For a ideal fluid we know that there is no viscosity of the fluid hence the boundary condition need's not to be satisfied and the flow occur's without any head loss due to viscous nature of the fluid. The friction of the pipe has no effect on the flow of an ideal fluid. But for a real fluid the viscosity of the fluid has a non zero value, the viscosity causes boundary layer effects, causes head loss and also frictional losses due to pipe friction hugely make the analysis of the flow complex. The losses in the energy of the flow becomes complex to calculate as frictional losses depend on the roughness of the pipe and Reynolds number of the flow thus increasing the complexity of the analysis of flow.
What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?
The resultant force on one side of a 25cm diameter circular plate at the bottom of 3m of pool water is approximately 1447 Newtons, calculated using the principles of fluid pressure and hydrostatic force.
Explanation:The question asks for the resultant force on one side of a 25cm diameter circular plate at the bottom of 3m of pool water. To calculate this, we first need to understand that the pressure exerted by a fluid in a static situation is given by the equation P = ρgh, where ρ is the density of the fluid (water in this case, which is approximately 1000 kg/m³), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the depth of the fluid above the point of interest (3m in this case).
To find the force exerted by the water on the plate, we also need the area of the plate, which can be derived from its diameter (D = 0.25m). The area (A) of a circle is given by πD²/4, which in this case gives us an area of approximately 0.0491 square meters. The force (F) exerted by the water can then be calculated using the equation F = P*A.
Substituting the values into the equations gives us a pressure (P) of 29430 Pa and, subsequently, a resultant force of approximately 1447 Newtons. This represents the sum of all the hydrostatic forces acting perpendicularly to the surface area of the circular plate due to the water pressure at the depth of 3 meters.
The velocity of a point mass that moves along the s-axis is given by s' = 40 - 3t^2 m/s, where t is in seconds. Find displacement of the particle from t = 2 s to t 6 s.
Answer:
The displacement is -48m.
Explanation:
Velocity is the rate of change of displacement. So, the instantaneous displacement is calculated by the integration of velocity function with respect to time. Displacement in range of time is calculated by integrating the velocity function with respect to time with in time range.
Given:
Velocity along the s-axis is
[tex]s{}'=40-3t^{2}[/tex]
time range is t=2s to t=6s.
Calculation:
Step1
Displacement in the time range t=2s to t=6s is calculated as follows:
[tex]\frac{\mathrm{d}s}{\mathrm{d}t}=40-3t^{2}[/tex]
[tex]ds=(40-3t^{2})dt[/tex]
Step2
Integrate the above equation with respect to time with the lower limit as 2 and upper limit as 6 as follows:
[tex]\int ds=\int_{2}^{6}(40-3t^{2})dt[/tex]
[tex]s=(40\times6-40\times2)-3(\frac{1}{3})(6^{3}-2^{3})[/tex]
s=160-208
s=-48m
Thus, the displacement is -48m.
The displacement of a point mass moving along the s-axis from t = 2 s to t = 6 s, with a given velocity function s' = 40 - 3t^2 m/s, is calculated by integrating the velocity to get the position function and then evaluating it between the two time limits, resulting in a displacement of 152 meters.
Explanation:The question requires finding the displacement of a point mass moving along the s-axis between t = 2 s and t = 6 s. The velocity is given as s' = 40 - 3t^2 m/s. To find the displacement, we will integrate the velocity function with respect to time from 2 to 6 seconds.
To integrate s' = 40 - 3t^2, we get:
Integral of 40 dt is 40tIntegral of -3t^2 dt is -t^3The displacement (s) is thus
s = 40t - t^3 evaluated from t = 2 to t = 6. Inserting the limits, we subtract the value at t = 2 from the value at t = 6:
s(6) - s(2) = (40(6) - 6^3) - (40(2) - 2^3) = 240 - 216 - 80 + 8 = 152 m
So, the displacement of the particle from t = 2 s to t = 6 s is 152 meters.
A cylindrical specimen of some metal alloy having an elastic modulus of 123 GPa and an original cross-sectional diameter of 3.3 mm will experience only elastic deformation when a tensile load of 2340 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
maximum length of the specimen before deformation = 200 mm
Explanation:
Hi!
If we have a cylinder with length L₀ , and it is elasticaly deformed ΔL (so the final length is L₀ + ΔL), the strain is defined as:
[tex]\epsilon =\frac{\Delta L}{L_0}[/tex]
And the tensile stress is:
[tex]\sigma = \frac{F}{A}\\F = \text{tensile load}\\A = \text{ cross section area}[/tex]
Elastic modulus E is defined as:
[tex]E = \frac{\sigma}{\epsilon }[/tex]
In this case ΔL = 0.45 mm and we must find maximum L₀. We know that A=π*r², r=(3.3/2) mm. Then:
[tex]\sigma=\frac{2340N}{\pi (1.65 \;mm)^2}=273.68 MPa = [/tex]
[tex]E=123\;GPa=\frac{L_0 \;(273.68MPa)}{0.45\;mm } \\ L_0 = 200\; mm[/tex]
Evaluate each of the following and express with an appropriate prefix: (a) (430 kg)^2 (b) (0.002 mg)^2, and (c) (230 m)^3
The expressions squared or cubed with appropriate metric prefixes are: (a) 184.9 Mg^2, (b) 4 x 10^-18 kg^2, and (c) 12.167 km^3. Metric prefixes are used for clarity in representing large quantities.
Explanation:The question deals with the operation of squaring and cubing given quantities and expressing them with the appropriate metric prefixes. Let's evaluate each expression step by step:
(a)It is important to use the proper metric prefixes when
expressing large numbers to ensure clarity and avoid confusion.
The primary heat transfer mechanism that warms me while I stand next to a campfire is: a)- Conduction b)- Impeadance c)- Convection d)- Radiation
Answer:
convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.
Explanation:
convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.
Convection is the transition of heat between distinct temperature fields by moving fluid (liquid or gas). Dry air is less thick than moist air and in the presence of a temperature difference, convection currents can form.
A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure methane. If the tank is to contain 90% liquid and 10% vapor, by volume, at 150 K, what mass of LNG (kg) will the tank hold? What is the quality in the tank?
Answer:
mass of LNG: 129501.3388 kg
quality: 0.005048662
Explanation:
Volume occupied by liquid:
400 m^3*0.9 = 360 m^3
Volume occupied by vapor
400 m^3*0.1 = 40 m^3
A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present
For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg
From specific volume definition:
vf = liquid volume/liquid mass
liquid mass = liquid volume/vf
liquid mass = 360 m^3/0.002794 m^3/kg
liquid mass = 128847.5304 kg
vg = vapor volume/vapor mass
vapor mass = liquid volume/vg
vapor mass = 40 m^3/0.06118 m^3/kg
vapor mass = 653.8084341 kg
total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg
Quality is defined as the ratio between vapor mass and total mass
quality = 653.8084341 kg/129501.3388 kg = 0.005048662
The velocity of a particle along the s-axis is given by v = 14s^7/6 where s is in millimeters and v is in millimeters per second. Determine the acceleration when s is 5.5 millimeters.
Answer:
The acceleration is [tex]2220.00m/s^{2}[/tex]
Explanation:
We know that the acceleration is given by
[tex]a=v\frac{dv}{ds}........................(i)[/tex]
The velocity as a function of position is given by [tex]v=14\cdot s^{7/6}[/tex]
Thus the acceleration as obtained from equation 'i' becomes
[tex]a=14\cdot s^{7/6}\times \frac{d}{ds}(14\cdot s^{7/6})\\\\a=14\cdot s^{7/6}\times \frac{7}{6}\times 14\cdot s^{1/6}\\\\a=\frac{686}{3}\cdot s^{8/6}[/tex]
Hence acceleration at s = 5.5 equals
[tex]a(5.5)=\frac{686}{3}\times (5.5)^{8/6}\\\\\therefore a=2220.00mm/s^{2}[/tex]
The pressure and temperature at the beginning of compression of a cold air-standard Diesel cycle are 100 kPa and 300K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2250 K. Assume constant specific heats evaluated at 300 K. Determine the cut-off ratio. There is a +/- 5% tolerance.
Answer:
Cut off ratio=2.38
Explanation:
Given that
[tex]T_1=300K[/tex]
[tex]P_1=100KPa[/tex]
[tex]P_2=P_3=7200KPa[/tex]
[tex]T_3=2250K[/tex]
Lets take [tex]T_1[/tex] is the temperature at the end of compression process
For air γ=1.4
[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}[/tex]
[tex]\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}[/tex]
[tex]T_2=1070K[/tex]
At constant pressure
[tex]\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}[/tex]
[tex]\dfrac{2550}{1070}=\dfrac{V_3}{V_2}[/tex]
[tex]\dfrac{V_3}{V_2}=2.83[/tex]
So cut off ratio
[tex]cut\ off\ ratio =\dfrac{V_3}{V_2}[/tex]
Cut off ratio=2.38
A closed rigid tank contains water initially at 10,000 kPa and 520ºC and is cooled to a final temperature of 270° C. Determine the final pressure of the water.
Answer:
final pressure is 6847.41 kPa
Explanation:
given data:
[tex]P_1 = 10,000 kPa[/tex]
[tex]T_1 =520\ degree\ celcius = 793 K[/tex]
[tex]T_2 = 270 degree celcius = 543 K[/tex]
as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream
for ischoric process we know that
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]
[tex]\frac{10*10^6}{793} = \frac{P_2}{543}[/tex]
[tex]P_2 = 6.84741*10^6 Pa[/tex]
final pressure is 6847.41 kPa
A 25 lb sacrificial Mg anode is attached to the steel hull of a container ship. If the anode completely corrodes within 3 months, what is the average current produced by the anode?
Answer:
The average current will be of 6.36 A.
Explanation:
The anode capacity of magnesium is C = 550 A*h/lb
A month has 30 days.
A day has 24 hours.
Therefore 3 months have:
t = 3 * 30 * 24 = 2160 hours.
The average current is then:
I = C * m / t
I = 550 * 25 / 2160 = 6.36 A
The average current will be of 6.36 A.
An air-conditioned room at sea level has an indoor design temperature of 80°F and a relative humidity of 60%. Determine a) Humidity ratio, b) Enthalpy, c) Density, d) Dew point, and e) Thermodynamic wet bulb temperature (Use Psychrometric charts)
Answer:
a)Humidity ratio =0.013 kg/kg
b) Enthalpy=60.34 KJ/kg
c) Density = 1.16 [tex]Kg/m^3[/tex]
d) Dew point temperature = 64.9°F
e) Wet bulb temperature = 69.53°F
Explanation:
Given that
Dry bulb temperature = 80°F
relative humidity = 60%
Given air is at sea level it means that total pressure will 1 atm.
So P= 1 atm.
As we know that psychrometric charts are always drawn at constant pressure.
Now from charts
We know that dry bulb temperature line is vertical and relative humidity line is curve and at that point these two line will meet ,will us property all property.
a)Humidity ratio =0.013 kg/kg
b) Enthalpy=60.34 KJ/kg
c) Density = 1.16 [tex]Kg/m^3[/tex]
d) Dew point temperature = 64.9°F
e) Wet bulb temperature = 69.53°F
How does fouling affects the performance of a heat exchanger?
Answer:
Fouling :
When rust or undesired material deposit in the surface of heat exchanger,is called fouling of heat exchanger.
Effect of heat exchanger are as follows"
1.It decreases the heat transfer.
2.It increases the thermal resistance.
3.It decreases the overall heat transfer coefficient.
4.It leads to increase in pressure drop.
5.It increases the possibility of corrosion.
A cylindrical tank with a radius of 2-m is filled with oil and water. The water has a density of rho = 1000 kg/m3 while the oil has a density of rho = 800 kg/m^3 . If the depth of the water is 2.5 m, and the pressure difference between the top of the oil and the bottom of the water is 80 kPa, determine the depth of the oil, in m. Assume that gravity is 9.81 m/s^2 .
Answer:
Height of oil is 7.06 meters.
Explanation:
The situation is shown in the attached figure
The pressure at the bottom of the tank as calculated by equation of static pressure distribution is given by
[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(\rho _{water}h_{water}+\rho_{oil}h_{oil})=P_{bottom}-P_{top}[/tex]
Applying the given values we get
[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(1000\times 2.5+800\times h_{oil})=80\times 10^{3}\\\\\therefore 800\times h_{oil}=\frac{80\times 10^{3}}{9.81}-2500\\\\\therefore 800\times h_{oil}=5654.94\\\\\therefore h_{oil}=7.06m[/tex]
Given two resistors R1=40 ohm and R2=30 ohm connected in series, what is the total resistance of this configuration? Enter the value in the box below without the unit. Round the result to two decimal places if necessary. For example if the answer is 10.333 ohm, put 10.33 in the box.
Answer:
Rt=70.00 ohm
Explanation:
ohm's theory tells us that the connected resistors in series add up directly
Rt=R1+R2+R3+R4.......
Therefore, for this case, the only thing we should do is add the resistance directly
Rt=R1+R2
Rt=40.00 ohm+30.00 Ohm.
Rt=70.00 ohm
the total resistance of this configuration is 70.00 ohm
Discuss the difference between the observed and calculated values. Is this error? If yes, what is the source?
Measurement error is the difference between observed and calculated values, including both random and systematic errors. Adjusting parameters such as K can help fit experimental data to models, with potential errors addressed through improved methods and investigation of discrepancies.
Explanation:The difference between the observed and calculated values is known as measurement error or observational error. This error can be classified into two types: random error, which occurs naturally and varies in an unpredictable manner, and systematic error, which is consistent and typically results from a flaw or limitation in the equipment or the experimental design. To identify the sources of these errors, one can compare the experimental data with calculated models and adjust parameters, such as the constant K, to find the best fit.
If the experimental values do not match the accepted values, sources of error should be investigated and procedures modified to improve accuracy. For example, in a physics experiment, one could compare the experimental acceleration to the standard acceleration due to gravity (9.8 m/s²) and identify factors contributing to any discrepancy. Common sources of error might include environmental factors, instrument calibration issues, or procedural mistakes.
A rigid tank with a volume of 0.5 m3 contains air at 120 kPa and 300 K. Find the final temperature after 20 kJ of heat is added to the air using (a) constant specific heats and (b) ideal gas tables.
Answer:
T=340. 47 K
Explanation:
Given that
Volume of tank =0.5 [tex]m^3[/tex]
Pressure P=120 KPa
Temperature T=300 K
Added heat ,Q= 20 KJ
Given that air is treated as ideal gas and specific heat is constant.
Here tank is rigid so we can say that it is constant volume system.
We know that specific heat at constant volume for air
[tex]C_v=0.71\ \frac{KJ}{kg.K}[/tex]
We know that for ideal gas
P V = m R T
For air R=0.287 KJ/kg.K
P V = m R T
120 x 0.5 = m x 0.287 x 300
m=0.696 kg
[tex]Q=mC_v\Delta T[/tex]
Lets take final temperature of air is T
Now by putting the values
[tex]Q=mC_v\Delta T[/tex]
[tex]20=0.696\times 0.71\times (T-300)[/tex]
T=340. 47 K
So the final temperature of air will be 340.47 K.
Describe the physics associated with the concept of thermal resistance.
Answer:
Thermal resistance:
Thermal resistance is the property which oppose the the heat transfer.It is the property for measurement of heat flow.
As we know that heat transfer take place from the high temperature to low temperature.Heat transfer Q given as
[tex]Q=\dfrac{\Delta T}{R_{th}}[/tex]
Where Δt is the temperature difference
[tex]{R_{th}}[/tex] is the thermal resistance.
Connection:
1. Series connection
[tex]R_{total}={R_1}+{R_2}+{R_1}----{R_n}[/tex]
2. Parallel connection
[tex]\dfrac{1}{R_{total}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}-----\dfrac{1}{R_n}[/tex]
The 10mm diameter rod is made of Kevlar 49. Determine the change in
length and the change in diameter.
Lenght of the rod = 100mm
Therefore there are two 80KN forces are pulling the rod from both
sides.
Answer:
0.815 mm
Explanation:
The rod in made of Kevlar 49, so it has an Young's modulus of
E = 125 GPa
The stiffness of a rod is given by:
k = E * A / L
k = E * π/4 * d^2 / L
So:
k = 125*10^9 * π/4 * 0.01^2 / 0.1 = 98.17 MN/m
Of the pulling forces only one is considered because when you pull on something there is always another force on the other side of equal magnitude and opposite direction to maintain equilibrium.
Hooke's law:
Δl = P/k
Δl = 80*10^3 / 98.17*10^6 = 0.000815 m = 0.815 mm
Is CO, an air pollutant? How does it differ from other emissions resulting from the combustion of fossil fuels?
Answer:
Explanation:
CO, carbon monoxide is a toxic gas. It casues asphixiation on people and animals by interfering with hemoglobin, not allowing blood to transport oxygen to the cells in the body.
The normal emissions resulting from the combustion of fussil fuels are CO2 (carbon dioxide) and H2O (water). Carbon monoxide is formed by an incomplete combustion of fossil fuels or carbon containing fuels in general, this not only produces toxic gas, but also is an inefficient combustion that wastes energy.
What is the definition of a fluid?
Answer: Basically it's a substance with no shape and it's one of the different states of water.
Explanation:
Determine the factor of safety for a 9 foot long hollow steel
column 3.5 inches on a side that has a wall thickness of 0.225
inches and is loaded with a 22 kip load. Use the steel E of 29 *10
^6 psi and assume the column is pin connected at each end.
Answer:
factor of safety for A36 structural steel is 0.82
Explanation:
given data:
side of column = 3.5 inches
wall thickness = 0.225 inches
load P = 22 kip
Length od column = 9 ft
we know that critical stress is given as
[tex]\sigma_{cr} = \frac{\pi^2 E}{(l/r)^2}[/tex]
where
r is radius of gyration[tex] = \sqrt{\fra{I}{A}}[/tex]
Here I is moment od inertia [tex]= \frac{b_1^2}{12} - \frac{b_2^2}{12}[/tex]
[tex] I == \frac{3.5^2}{12} - \frac{3.05^2}{12} = 5.294 in^4[/tex]
For hollow steel area is given as [tex]A = b_1^2 -b_2^2[/tex]
[tex]A = 3.5^2 -3.05^2 = 2.948 in^2[/tex]
critical stress [tex]\sigma_{cr} = = \frac{\pi^2\times 29\times 10^6}{((9\times12)/(1.34))^2}[/tex]
[tex]\sigma_{cr} = 44061.56 lbs/inc^2[/tex]
considering Structural steel A36
so A36[tex] \sigma_y = 36ksi[/tex]
factor of safety [tex]= \frac{yield\ stress}{critical\ stress}[/tex]
factor of safety =[tex]\frac{36\times10^3}{44061.56} = 0.82[/tex]
factor of safety for A36 structural steel is 0.82
The student's query involves calculating the factor of safety for a hollow steel column, but without complete material strength data, the calculation cannot be precisely done.
Explanation:The student's question pertains to determining the factor of safety for a hollow steel column with specified dimensions, material properties (steel E), and loading conditions. To calculate the factor of safety for the column, we would need to compare the column's actual stress under load to its maximum allowable stress. However, due to insufficient data about material yield strength or ultimate strength and considering the provided information does not match the examples given, this calculation cannot be accurately completed without further specifics on the steel's properties.
What is the ideal cooling system for low horsepower motor? For example1hp motor
Answer:
Air cooling.
Explanation:
Low power motors are supposed to be low cost, and they dissipate little heat. Therefore a low cost solution is ideal.
Air cooling can be achieved with very little cost. Fins can be added to a cast motor casing and a fan can be places on the shaft to use a small amount of the motor power to move air to cool it.
An element has two naturally occurring isotopes, isotope 1 with an atomic weight of 78.918 amu and isotope 2 with an atomic weight of 80.916 amu. If the average atomic weight for the element is 79.903 amu, calculate the fraction of occurrences of these two isotopes.
Answer:
The fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.
Explanation:
Given that
Weight of isotope 1 = 78.918 amu
Weight of isotope 2 = 80.916 amu
Average atomic weight= 79.903 amu
Lets take fraction of isotope 1 is x then fraction of isotope 2 will be 1-x.
The total weight will be summation of these two isotopes
79.903 = 78.918 x + 80.916(1-x)
By solving above equation
80.916 - 79.903 = (80.916-78.918) x
x=0.507
So the fraction of isotope 1 is 50.7 % and fraction fraction of isotope 1 is 49.2 %.
A petrol engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency, and how much power is rejected to the ambient surroundings?
Answer:
efficiency =42.62%
AMOUNT OF POWER REJECTED IS 20.080 kW
Explanation:
given data:
power 20 hp
heat energy = 35kW
power production = 20 hp = 20* 746 W = 14920 Watt [1 hp =746 watt]
[tex]efficiency = \frac{power}{heat\ required}[/tex]
[tex]efficiency = \frac{14920}{35*10^3}[/tex]
[tex]= 0.4262*10^100[/tex]
=42.62%
b) [tex]heat\ rejected = heat\ required - amount\ of\ power\ generated[/tex]
[tex]= 35*10^3 - 14920[/tex]
= 20.080 kW
AMOUNT OF POWER REJECTED IS 20.080 kW
A disk with radius of 0.4 m is rotating about a centrally located axis with an angular acceleration of 0.3 times the angular position theta. The disk starts with an angular velocity of 1 rad/s when theta = 0. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
Answer:
a₁= 1.98 m/s² : magnitud of the normal acceleration
a₂=0.75 m/s² : magnitud of the tangential acceleration
Explanation:
Formulas for uniformly accelerated circular motion
a₁=ω²*r : normal acceleration Formula (1)
a₂=α*r: normal acceleration Formula (2)
ωf²=ω₀²+2*α*θ Formula (3)
ω : angular velocity
α : angular acceleration
r : radius
ωf= final angular velocity
ω₀ : initial angular velocity
θ : angular position theta
r : radius
Data
r =0.4 m
ω₀= 1 rad/s
α=0.3 *θ , θ= 2π
α=0.3 *2π= 0,6π rad/s²
Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
We calculate ωf with formula 3:
ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687
ωf=[tex]\sqrt{24.687}[/tex] =4.97 rad/s
a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²
a₂=α*r = 0,6π * 0.4 = 0.75 m/s²
For a rod of annealed AISI 1018 steel with a cross sectional area of 0.65 in^2?; what is the maximum tensile load Pmax that should be placed on it given a design factor of 3 to avoid yielding?
Answer:
maximum tensile load Pmax is 11.91 ksi
Explanation:
given data
area = 0.65 in²
design factor of safety = 3
to find out
what is the maximum tensile load Pmax
solution
we know here area is 0.65 in² and FOS = 3
so by steel table for rod of annealed AISI 1018 steel table we know σy = 55 ksi
so
we use here design factor formula that is
[tex]\frac{ \sigma y}{FOS} = \frac{Pmax}{area}[/tex] .............1
put here all these value we get Pmax in equation 1
[tex]\frac{55}{3} = \frac{Pmax}{0.65}[/tex]
Pmax = 11.91 ksi
so maximum tensile load Pmax is 11.91 ksi