Answer:
Power output, [tex]P_{out} = 178.56 kW[/tex]
Given:
Pressure of steam, P = 1400 kPa
Temperature of steam, [tex]T = 350^{\circ}C[/tex]
Diameter of pipe, d = 8 cm = 0.08 m
Mass flow rate, [tex]\dot{m} = 0.1 kg.s^{- 1}[/tex]
Diameter of exhaust pipe, [tex]d_{h} = 15 cm = 0.15 m[/tex]
Pressure at exhaust, P' = 50 kPa
temperature, T' = [tex]100^{\circ}C[/tex]
Solution:
Now, calculation of the velocity of fluid at state 1 inlet:
[tex]\dot{m} = \frac{Av_{i}}{V_{1}}[/tex]
[tex]0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}[/tex]
[tex]0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}[/tex]
[tex]v_{i} = 3.986 m/s[/tex]
Now, eqn for compressible fluid:
[tex]\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}[/tex]
Now,
[tex]\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}[/tex]
[tex]\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}[/tex]
[tex]\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}[/tex]
[tex]v_{e} = 19.33 m/s[/tex]
Now, the power output can be calculated from the energy balance eqn:
[tex]P_{out} = -\dot{m}W_{s}[/tex]
[tex]P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}[/tex]
[tex]P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW[/tex]
The power output of the Turbine is; 225.69 kW
What is the Power Output?We are given
Pressure of steam; P = 1400 kPa
Temperature of steam at state 1; T = 350°C
Diameter of pipe; d₁ = 8 cm = 0.08 m
Mass flow rate; m' = 0.1 kg/s
Diameter of exhaust pipe; d₂ = 15 cm = 0.15 m
Pressure at exhaust; P' = 50 kPa
Temperature at state 2; T' = 100°C
Area; A = πd²/4
A = π * 0.08²/4
A = 0.0016π m²
We can find the find initial velocity from the formula;
v₁ = m' * V₁/A
v₁ = (0.1 * 0.2004/(0.0016π))
v₁ = 3.986 m/s
From equation of compressible fluid, we know that;
(A₁ * v₁)/V₁ = (A₂ * v₂)/V₂
A₂ = πd₂²/4
A₂ = π * 0.15²/4
A₂ = 0.005625π m²
(0.0016π * 3.986)/0.2004 = (0.005625π * v₂)/3.4181
Solving for v₂ gives;
v₂ = 19.33 m/s
Finally power output is gotten from the expression;
P_out = -m'(H₂ - H₁) + (v₂² - v₁²)/2
P_out = -0.1(2682.6 - 3150.7) + (19.33² - 3.986²)/2
P_out = 225.69 kW
Read more about Power Output at; https://brainly.com/question/25573309
A material’s chemical property could be described as it’s reaction with other chemicals (gases, liquids and solid materials). a) True b) False
Answer:
The given statement is true.
Explanation:
A chemical property of any material be it solid liquid or gas is defined as how it interacts chemically with other substances after an interaction takes place between them. The interaction in language of chemistry is known as chemical reaction. Different materials in nature show different interactions with various other substances.The 2 substances can react with each other and form a different compound or may not react with each other and are termed as inert chemicals. Infact it is this interaction between the various chemicals that we can group different into classes based on their behavior with different chemicals. The interaction of different materials act as their signatures that help us in identifying them.
A 4,000-km^2 watershed receives 102cm of precipitation in one
year.The avg. flow of the river draining the watershed is 34.2
m^3/s.Infiltration is est. to be 5.5 x 10^(-7) cm/s
andevapotranspiration is est. to be 40 cm/y. Determine the change
instorage in the watershed over one year. The ratio of runoff
toprecipitation (both in cm) is termed the runoff
coefficient.Compute the runoff coefficient for this
watershed.
Answer:
1) The change in storage of the catchment is 707676800 cubic meters.
2) The runoff coefficient of the catchment is 0.83.
Explanation:
The water budget equation of the catchment can be written as
[tex]P+Q_{in}=ET+\Delta Storage+Q_{out}+I[/tex]
where
'P' is volume of precipitation in the catchment =[tex]Area\times Precipitation[/tex]
[tex]Q_{in}[/tex] Is the water inflow
ET is loss of water due to evapo-transpiration
[tex]\Delta Storage[/tex] is the change in storage of the catchment
[tex]Q_{out}[/tex] is the outflow from the catchment
I is losses due to infiltration
Applying the values in the above equation and using the values on yearly basis (Time scale is taken as 1 year) we get
[tex]4000\times 10^{6}\times 1.02+0=0.40\times 4000\times 10^{6}+\Delta Storage+34.2\times 3600\times 24\times 365\times 5.5\times 10^{-9}\times 4000\times 10^{6}\times 3600\times 24\times 365[/tex]
[tex]\therefore \Delta Storage=707676800m^3[/tex]
Part b)
The runoff coefficient C is determined as
[tex]C=\frac{P-I}{P}[/tex]
where symbols have the usual meaning as explained earlier
[tex]\therefore C=\frac{102-5.5\times 10^{-7}\times 3600\times 24\times 365}{102}=0.83[/tex]
The head difference between the inlet and outlet of a 1km long pipe discharging 0.1 m^3/s of water is 0.53 m. If the diameter is 0.6m, what is the friction factor? Is a) 0.01 b) 0.02 c) 0.03 d) 0.04 e) 0.05
Answer:
The correct option is 'e': f = 0.05.
Explanation:
The head loss as given by Darcy Weisbach Equation is as
[tex]h_{l}=\frac{flv^{2}}{2gD}[/tex]
where
[tex]h_{l}[/tex] is head loss in the pipe
'f' is the friction factor
'l' is the length of pile
'v' is the velocity of flow in pipe
'D' is diameter of pipe
From equation of contuinity we have [tex]v=\frac{Q}{A}[/tex]
Thus using this in darcy's equation we get
[tex]h_{l}=\frac{flQ^{2}}{2gDA}[/tex]
where
'Q' is discharge in the pipe
'A' is area of the pipe [tex]A=\frac{\piD^2}{4}[/tex]
Applying the given values we get
[tex]h_{l}=\frac{8flQ^{2}}{\pi ^{2}gD^{5}}[/tex]
Solving for 'f' we get
[tex]f=\frac{0.53\times \pi ^{2}\times 9.81\times 0.6^{5}}{1000\times 0.1^{2}\times 8}\\\\f=0.05[/tex]
A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.
Answer:
[tex]\sigma _1=10.33MPa[/tex]
[tex]\sigma _2=5.16MPa[/tex]
Explanation:
Given that
Diameter(d)=62 mm
Thickness(t)= 300 μm=0.3 mm
Internal pressure(P)=100 KPa
Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .
The hoop stress
[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]
Longitudinal stress
[tex]\sigma _l=\dfrac{Pd}{4t}[/tex]
Now by putting the values
[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]
[tex]\sigma _h=\dfrac{100\times 62}{2\times 0.3}[/tex]
[tex]\sigma _h=10.33MPa[/tex]
[tex]\sigma _l=5.16MPa[/tex]
So the principle stress are
[tex]\sigma _1=10.33MPa[/tex]
[tex]\sigma _2=5.16MPa[/tex]
A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The system is heated until the volume is 0.2 m3 and the pressure is 180 kPa. What is the work done by the air? External pressure is 100 kPa.
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = [tex]\frac{60+180}{2}[/tex] = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
calculate the density of air(as a perfect gas) when the pressure
is1.01325*105 N/m2 and the temperature
is288.15 K.
Answer:
1.2253 kg/m3
Explanation:
from ideal gas equation we know that
[tex]PV = n\timesRT[/tex]
[tex]\frac{n}{V} = \frac{P}{(RT)}[/tex]
where,
pressure P = 1.01325 x 10^5 N/m2 = 101325 Pascals
gas constant R = 8.314 J/mol/K
T = 288.15 K
[tex]\frac{n}{V} = \frac{101325}{(8.314*288.15K)}[/tex]
[tex]\frac{n}{V} = 42.2948967 Pa*mol/J[/tex]
we know that 1 Pa = 1 J/m3
so[tex]\frac{n}{V} = 42.2948967\ mol/m3[/tex]
as we know, Air consist of 21% oxygen and 79% nitrogen
therefore
Molar mass of air:
[tex]0.21 *(32 g/mol) + 0.79 *(28 g/mol) = 28.97 g/mol[/tex]
So [tex](42.2948967\ mol/m3) \times 28.97\ g/mol[/tex]
= 1 225.27 g/m^3
= 1.22528316 kg/m^3 ~ 1.2253 kg/m3
Yield and tensile strengths and modulus of elasticity with increasing temperature. (increase/decrease/independent)
Answer:
Decrease
Explanation:
Generally with increasing the temperature the mechanical properties of material decreases.But some materials have exceptions like tempered steel because when temperature increase then young modulus of elasticity of tempered steel increases.
So we can say that when with increasing temperature the properties of materials Yield and tensile strengths and modulus of elasticity decreases.
Decrease
Answer:
Decreases
Explanation:
Yield and tensile strength and modulus of elasticity decreases with increasing temperature. As the temperature is increased most materials decrease in their elasticity.
The modulus of elasticity is proportional to its tensile strength.
What are the two types of furnaces used in steel production?
Explanation:
The two types of furnaces used in steel production are:
Basic oxygen furnace
In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.
Electric arc furnace
Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.
Define drag and lift forces
Answer with Explanation:
Drag and lift are the forces that act on an object which moves in a fluid and are explained as under:
1)Drag: Drag is the force that opposes the motion of an object moving in a fluid and hence is analogous to fluid frictional force in a fluid. Power is needed to be spent by an object to overcome the frictional drag. The drag force is different from the classical friction as we know that the frictional force on a dry surface is independent of the velocity of the object but for drag the force is proportional to the square of the velocity of the object.
Mathematically
[tex]F_{Drag}=\frac{1}{2}C_{d}A\rho _{fluid}V^2[/tex]
where
[tex]C_d[/tex] is a constant known as coefficient of drag and A is the projected area of the object.
2) Lift: Lift force is a component of the force that a fluid exerts on an object moving in it that counters the weight of the object and hence has the tendency to lift the object and hence is known as lift force. The lift force is perpendicular to the incoming fluid. Lift force is useful as it allows the aeroplanes to fly as the lift force that is generated by the air on the wings of the plane is used to overcome the force of gravity on the plane.
Mathematically
[tex]F_{Lift}=\frac{1}{2}C_{L}A\rho _{fluid}V^2[/tex]
where
[tex]C_L[/tex] is a constant known as coefficient of lift and A is the projected area of the object moving with velocity 'v'.
A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine the force R which the elevator floor exerts on her feet and the lifting force L which she exerts on the package during the acceleration in-travel. If the elevator supports cables suddenly and completely fail, what values would R and L acquire
Answer:
force R = 846.11 N
lifting force L = 110.36 N
if cable fail complete both R and L will be zero
Explanation:
given data
mass woman mw = 60 kg
mass package mp = 9 kg
accelerates rate a = g/4
to find out
force R and lifting force L and if cable fail than what values would R and L acquire
solution
we calculate here first reaction R force
we know elevator which accelerates upward
so now by direction of motion , balance the force that is express as
R - ( mw + mp ) × g = ( mw + mp ) × a
here put all these value and a = g/4 and use g = 9.81 m/s²
R - ( 60 + 9 ) × 9.81 = ( 60 + 9 ) × g/4
R = ( 69 ) × 9.81/4 + ( 69 ) 9.81
R = 69 ( 9.81 + 2.4525 )
force R = 846.11 N
and
lifting force is express as here
lifting force = mp ( g + a)
put here value
lifting force = 9 ( 9.81 + 9.81/4)
lifting force L = 110.36 N
and
we know if cable completely fail than body move free fall and experience no force
so both R and L will be zero
Six kilograms of nitrogen at 30 °C are cooled so that the internal energy decreases by 60 kJ. Find the final temperature of the gas. Assume that the specific heat of nitrogen is 0.745 kJ / kg °C.
Answer:
Final temperature will be 16.57°C
Explanation:
We have given mass of nitrogen m = 6 kg
Initial temperature [tex]T_1=30^{\circ}[/tex]
Decrease in internal energy [tex]\Delta U=-60KJ[/tex]
Specific heat of nitrogen [tex]c_v[/tex]= 0.745 KJ/kg
Let final temperature is [tex]T_2[/tex]
Change in internal energy is given by [tex]\Delta U=mc_v\Delta T=mc_v(T_2-T_1)[/tex]
[tex]-60=6\times 0.745(T_2-30)[/tex]
[tex]T_2=16.57^{\circ}C[/tex]
So final temperature will be 16.57°C
A piston-cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m^3. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior for the air, determine the mass of the air, in kg, and the work and heat transfer, each in kJ.
Answer:
The mass of the air is 4.645 kg.
The work done or heat transfer is 277.25 kj.
Explanation:
In isothermal process PV=constant. Take air as an ideal gas. For air gas constant is 287 j/kgK.
Given:
Initial pressure of the gas is 2 bar.
Initial volume of the gas is 2 m³.
Initial temperature is 300 K.
Final pressure is 1 bar.
Calculation:
Step1
Apply ideal gas equation for air as follows:
PV=mRT
[tex]2\times10^{5}\times2=m\times 287\times300[/tex]
m = 4.645 kg.
Thus, the mass of the air is 4.645 kg.
Step2
For isothermal process work done is same as heat transfer.
Work done or the heat transfer is calculated as follows:
[tex]W=P_{i}V_{i}ln\frac{P_{i}}{P_{f}}[/tex]
[tex]W=2\times10^{5}\times2\times ln\frac{2}{1}[/tex]
[tex]W=2.7725\times10^{5}[/tex] j
Or,
W=277.25 kj.
Thus, the work done or heat transfer is 277.25 kj.
During an office party, an office worker claims that a can of cold beer on his table warmed up to 20oC by picking up energy from the surrounding air, which is 25oC. Is there any truth to his claim? Explain.
Answer:
Yes
Explanation:
As we know that heat transfer take place from high temperature body to low temperature body.
In the given problem ,the temperature of the air is high as compare to the temperature of can of bear ,so the heat transfer will take place from air to can of bear and at the last stage when temperature of can of bear will become to the temperature of air then heat transfer will be stop.Because temperature of the both body will become at the same and this stage is called thermal equilibrium.
So an office worker claim is correct.
A convenient and cost-effective way to store biogas is to use light-weight, rigid gas containers. The pressure and temperature of biogas stored in a 10 m^3 container are 1.5 bar and 5°C, respectively, as measured at an early morning time (state 1). The temperature of the biogas is expected to increase to 40°C at noon on the same day, without any significant change in the volume of the container or amount of methane in the container (state 2). The biogas can be approximated as methane (CH4) modelled as an ideal gas with constant specific heat ratio, k = 1.4. The effects of gravity and motion are negligible. The reference environment pressure and temperature are, respectively, po = 100 kPa and To = 0 °C. The molecular weight of methane is M = 16.04 gmol-1. The universal gas constant is R= 8.31 J mol-1K-1. (a) Calculate the mass, in kg, and amount of substance, in mol, for the methane in the container.
Answer:
10.41 kg
Explanation:
The gas state equation is:
p * V = n * R * T
For this equation we need every value to be in consistent units
1.5 bar = 150 kPa
5 C = 278 K
n = p * V / (R * T)
n = 150000 * 10 / (8.31 * 278) = 649 mol
Multiplying the amount of moles by the molecular weight of the gas we obtain the mass:
m = M * mol
m = 16.04 * 649 = 10410 g = 10.41 kg
What is the governing ratio for thin walled cylinders?
Answer:
D/t>20
Explanation:
Lets take
D =Diameter of thin cylinder
t =Thickness of thin cylinder
So a cylinder is called thin cylinder if the ratio of diameter to the thickness is greater than 20 (D/t>20 ).
But on the other hand a cylinder is called thick cylinder is ratio of thickness to the diameter is greater than 20 (t/D>20 ).
So the governing ratio of thin walled cylinder is 20.
How much would a 10.0 inch long, 0.25 inch diameter AISI 1020 Q&T bolt stretch when loaded with 2000 lbs?
Answer:
0.014 in
Explanation:
The Young's module of steel is of E = 210 GPa = 30*10^6 psi
The section of the bolt would be:
A = π/4 * D^2
The stiffness would be:
k = E * A / L
k = π * D^2 * E / (4 * L)
k = π * 0.25^2 * 30*10^6 / (4 * 10) = 147000 lb / in
If I apply a force of 2000 lb, I calculate the stretching with Hooke's law:
Δx = f / k
Δx = 2000 / 147000 = 0.014 in
a valueable preserved biological specimen is weighed by suspeding it from a spring scale. it weighs 0.45 N when it is suspendedin air and 0.081 N when it is suspended in a bottle of alchol what is its dencity?
Answer:
ρ=962.16kg/m^3
Explanation:
The first thing we must do to solve is to find the mass of the specimen using the weight equation
w = mg
m=w/g
m=0.45/9.81=0.04587kg
To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).
We must bear in mind that the principle of archimedes indicates that the buoyant force is given by
F = ρgV
where V is the specimen volume and ρ is the density of alcohol = 789kg / m ^ 3
considering the above we have the following equation
0.081=0.45-(789)(9.81m/s^2)V
solving for V
V=(0.081-0.45)/(-789x9.81)
V=4.7673x10^-5m^3
finally we found the density
ρ=m/v
ρ=0.04587kg/4.7673x10^-5m^3
ρ=962.16kg/m^3
What is the function of air preheater? How does air preheating save fuel?
Answer:
Air preheater:
Air preaheater is a heat exchanger which take heat from flue gases and increase the temperature of the air before entering into the combustion process.
As we know that ,flue gases contains high temperature and that temperature can be used by air preheater and it will reduce the amount of heat addition in the combustion process and that leads to increase the efficiency of plant.When heat addition reduces for the same out put of power it means that less amount of fuel is required for the same out put of power as without using air preheater. The main purpose of air preheater is to increase the thermal efficiency.
By increasing the temperature of air it will reduce the amount of fuel reduce for the combustion.
Specific cutting energy increases with increasing the cutting speed. a) True b) False
Answer:
b)false
Explanation:
Specific cutting energy:
Energy required to remove unit volume of material is called specific energy.In other words we can say that the ratio of energy to the volume removal rate is called specific cutting energy.
When cutting speed is increases then specific energy goes to decrease.As well as when depth of cut and feed of tool is increase then specific cutting energy will decrease.
What is the relation between Poisson's ration, young's modulus, shear modulus for an material?
Answer:
Explanation:
Poisson's ration= is the ratio between the deformation that occurs in the material in the direction perpendicular to the force applied with the deformation suffered by the material in the same direction of the force.
young's modulus= is the ratio between the stress applied to a material with respect to its unit deformation when this material complies with the hooke's law.
shear modulus=change in the way an elastic material experiences when subjected to shear stresses
What is the definition of diameter pitch?
Answer:
Diameter pitch is the parallel thread of the imaginary cylinder of the diameter that basically intersect in the surface of thread. It is also known as effective diameter. The diameter pitch is basically used to determine the threated parts.
The pitch diameter is the essential component for determining the basic compatibility in the externally and internally thread parts. The diameter pitch is the sensitive measuring tool for determine the specific measurements.
What are the parameters that affect life and drag forces on an aerofoil?
Answer:
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
Explanation:
Drag force:
Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.
Drag force given as
[tex]F_D=\dfrac{1}{2}\rho\ A\ V^2[/tex]
So we can say that drag force depends on following properties
1.The velocity of fluid
2.Fluid properties.
3.Projected area of object(geometry of the object).
What impact does modulus elasticity have on the structural behavior of a mechanical design?
Answer with Explanation:
The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:
1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.
2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.
3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity the resistance it offer's to shear forces and the torques is more.
While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.
A rocket developed by an amateur was traveling upwards at a velocity given by v = (11 + 0.2s) m/s, where s is in meters. Determine the time for the rocket to reach an altitude of s = 80 m. Initially, s = 0 when t = 0. [Hint: obtain initial velocity].
Answer:
Time taken to reach 80 meters equals 4.4897 seconds.
Explanation:
We know that velocity is related to position as
[tex]v=\frac{ds}{dt}[/tex]
Now it is given that [tex]v=(11+0.2v[/tex]
Using the given velocity function in the above relation we get
[tex]\frac{ds}{dt}=(11+0.2s)\\\\\frac{ds}{(11+0.2s)}=dt\\\\\int \frac{ds}{(11+0.2s)}=\int dt\\\\[/tex]
Now since the limits are given as
1) at t = 0 , s=0
Using the given limits we get
[tex]\int_{0}^{80} \frac{ds}{(11+0.2s)}=\int_{o}^{t} dt\\\\\frac{1}{0.2}[ln(11+0.2s)]_{0}^{80}=(t-0)\\\\5\times (ln(11+0.2\times 80)-ln(11))=t\\\\\therefore t=4.4897seconds[/tex]
Derive the following conversion factors: (a) Convert a viscosity of 1 m^2/s to ft^2/s. (b) Convert a power of 100 W to horsepower. (c) Convert a specific energy of 1 kJ/kg to Btu/lbm.
Answer:
(a) 10.76 [tex]ft^2/sec[/tex]
(b) 0.134 lbm
(c) 0.4308 Btu/lbm
Explanation:
We have to do conversion
(a) conversion of viscosity [tex]1m^2/sec\ to\ ft^2/sec[/tex]
We know that [tex]1m^2=10.76feet^2[/tex]
So [tex]1m^2/sec=10.76ft^2/sec[/tex]
(b) We have to convert power 100 W in hp
We know that 1 W = 0.00134 hp
So 100 W = 100×0.00134=0.134 hp
(c) We have to convert 1 KJ/kg to Btu/lbm
We know that 1 KJ = 0.94780 Btu
And 1 kg = 2.20 lbm
So [tex]1KJ/kg=\frac{0.9478Btu}{2.20lbm}=0.4308Btu/lbm[/tex]
A room with dimensions of 3 x 10 x 20 m is estimated to have outdoor air brought in at an infiltration rate of 1/4 volume changes per hour. Determine the infiltration rate in m^3/s.
Answer:
The infiltration rate is of 0.042 m^3/s.
Explanation:
The total volume of the room is:
V = 3 * 10 * 20 = 600 m^3
If the air infiltration rate is of 1/4 volume per hour:
v = V/4 * 1 hour /3600 seconds
Replacing:
v = 600/4 * 1/3600 = 0.042 m^3/s
The infiltration rate would then be of 0.042 m^3/s.
Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is [tex]1.557\times10^{-7}[/tex]m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:
[tex]Re=\frac{vd}{\nu}[/tex]
Here, v is velocity, [tex]\nu[/tex] is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:
[tex]Re=\frac{vd}{\nu}[/tex]
[tex]Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}[/tex]
Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.
A type 3 wind turbine has rated wind speed of 13 m/s. Coefficient of performance of this turbine is 0.3. Calculate the rated power density of the wind that is hitting the turbine. Calculate the mechanical power developed at the shaft connecting rotor and generator. Assume rotor diameter 100 m and air density 1.225 kg/m^3.
Answer:
Rated power = 1345.66 W/m²
Mechanical power developed = 3169035.1875 W
Explanation:
Wind speed, V = 13 m/s
Coefficient of performance of turbine, [tex]C_p[/tex] = 0.3
Rotor diameter, d = 100 m
or
Radius = 50 m
Air density, ρ = 1.225 kg/m³
Now,
Rated power = [tex]\frac{1}{2}\rho V^3[/tex]
or
Rated power = [tex]\frac{1}{2}\times1.225\times13^3[/tex]
or
Rated power = 1345.66 W/m²
b) Mechanical power developed = [tex]\frac{1}{2}\rho AV^3C_p[/tex]
Here, A is the area of the rotor
or
A = π × 50²
thus,
Mechanical power developed = [tex]\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3[/tex]
or
Mechanical power developed = 3169035.1875 W
A freight train and a passenger train share the same rail track. The freight train leaves station A at 8:00 A.M. The train travels at a speed of 45 km/h for the first 10 minutes and then continues to travel at a speed of 60 km/h. At 8:35 A.M., the passenger train leaves station A. The pas¬senger train travels first at a speed of 75 km/h for 5 minutes and then continues to travel at a speed of 105 km/h. Determine the location of the siding where the freight train will have to be parked to allow the faster passenger train to pass through. As a safety precaution, it is determined that the time headway between the two trains should not be allowed to fall below 5 minutes.
Answer:
74.2 km from station A.
Explanation:
We set a frame of reference with the station at the origin and the X axis pointing in the direction the trains run.
The freight train leaves at 8 AM and travels 10 minutes at 45 km/h.
For this problem it is better to convert the speeds to km/min
45 km/h = 0.75 km/min
The equation for position under constant speed is:
X(t - t0) = X0 + v0 * (t - t0)
Since we know the time it will stop moving at this speed:
X(10 - 0) = 0 + 0.75 * (10 - 0) = 7.5 km
After it ran those 7.5 km it will keep running at 60 km/h.
60 km/h = 1 km/min
The position equation for it is now:
X(t - 10) = 7.5 + 1 * (t - 10)
The passenger train leaves the station at 8:35 AM. It travels at 75 km/h for 5 minutes.
75 km/h = 1.25 km/min
After those 5 minutes it will have traveled:
X(40 - 35) = 0 + 1.25 * (40 - 35) = 6.25 km
Then it travels at 105 km/h
105 km/h = 1.75 km/min
Its position equation is now:
X(t - 40) = 6.25 + 1.75 * (t - 40)
Equating both positions we find the time at which they would meet:
7.5 + 1 * (t - 10) = 6.25 + 1.75 * (t - 40)
7.5 + t - 10 = 6.25 + 1.75*t - 70
t - 1.75*t = 6.25 - 70 +10 - 7.5
-0.75*t = -61.25
t = 61.25 / 0.75
t = 81.7 minutes
The freight train will have to be parked 5 minutes before this at t = 76.7 minutes.
At that moment the freight train will be at:
X(76.7 - 10) = 7.5 + 1 * (76.7 - 10) = 74.2 km
Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance +/- 0.000005 m^3/kg)?
Answer:
[tex]specific\ volume=0.00097\ m^3/kg[/tex]
Explanation:
Given that
Specific gravity of sea water = 1.025
So density of sea water = 1.025 x 1000 [tex]kg/m^3[/tex]
Density of sea water = 1025 [tex]kg/m^3[/tex]
We know that
[tex]Density=\dfrac{mass}{Volume}[/tex] ---1
Specific volume
[tex]specific\ volume=\dfrac{Volume}{mass}[/tex] ---2
From equation 1 and 2
We can say that
[tex]specific\ volume=\dfrac{1}{density}\ m^3/kg[/tex]
[tex]specific\ volume=\dfrac{1}{1025}\ m^3/kg[/tex]
[tex]specific\ volume=0.00097\ m^3/kg[/tex]