Answer:
5.9*10^-3 Pa*s
Explanation:
The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.
τ = μ * du/dr
u(r) = 0.4/r - 1000*r
The derivative is:
du/dr = -1/r^2 - 1000
On the radius of the inner cylinder this would be
u'(0.02) = -1/0.02^2 - 1000 = -3500
So:
τ = -3500 * μ
We don't care about the sign
τ = 3500 * μ
That is a tangential force per unit of area.
The area of the inner cylinder is:
A = h * π * D
And the torque is
T = F * r
T = τ * A * D/2
T = τ * h * π/2 * D^2
T = 3500 * μ * h * π/2 * D^2
Then:
μ = T / (3500 * h * π/2 * D^2)
μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s
If the original length of a specimen is L0 = 10"" and new length of the specimen after applied load is L = 12.5"". The value of true strain is: a) 0.5 b) 0.25 c) 0.223 d) 0.4
Answer:
The correct answer is option 'b':0.25
Explanation:
By definition of strain we have
[tex]\epsilon =\frac{L_f-L_o}{L_o}[/tex]
where
[tex]\epsilon [/tex] is the strain
[tex]L_o[/tex] is the original length of specimen
[tex]L_f[/tex] is the elongated length of specimen
Applying the given values we get
[tex]\epsilon =\frac{12.5''-10''}{10''}=0.25[/tex]
Find the error in the following preudocode.Constant Real GRAVITY = 9.81 Display "Rates of acceleration of an object in free fall:" Display "Earth: ", GRAVITY, " meters per second every second." Set GRAVITY = 1.63 Display "Moon: ", GRAVITY, " meters per second every second."
Answer:
It is attempting to change the value of a constant.
Explanation:
In this pseudocode program "GRAVITY" is declared as a constant value, therefore it cannot be changed during runtime. If you tried mo write this code in a real language and compile it you would get a compilation error because of that forbidden operation.
A rotating cup viscometer has an inner cylinder diameter of 2.00 in., and the gap between cups is 0.2 in. The inner cylinder length is 2.50 in. The viscometer is used to obtain viscosity data on a Newtonian liquid. When the inner cylinder rotates at 10 rev/min, the torque on the inner cylinder is measured to be 0.00011 in-lbf. Calculate the viscosity of the fluid. If the fluid density is 850 kg/m^3, calculate the kinematic viscosity
Answer:
The dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.
Explanation:
Step1
Given:
Inner diameter is 2.00 in.
Gap between cups is 0.2 in.
Length of the cylinder is 2.5 in.
Rotation of cylinder is 10 rev/min.
Torque is 0.00011 in-lbf.
Density of the fluid is 850 kg/m3 or 0.00095444 slog/in³.
Step2
Calculation:
Tangential force is calculated as follows:
T= Fr
[tex]0.00011 = F\times(\frac{2}{2})[/tex]
F = 0.00011 lb.
Step3
Tangential velocity is calculated as follows:
[tex]V=\omega r[/tex]
[tex]V=(\frac{2\pi N}{60})r[/tex]
[tex]V=(\frac{2\pi \times10}{60})\times1[/tex]
V=1.0472 in/s.
Step4
Apply Newton’s law of viscosity for dynamic viscosity as follows:
[tex]F=\mu A\frac{V}{y}[/tex]
[tex]F=\mu (\pi dl)\frac{V}{y}[/tex]
[tex]0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}[/tex]
[tex]\mu =1.3374\times 10^{-6}[/tex]lb-s/in².
Step5
Kinematic viscosity is calculated as follows:
[tex]\upsilon=\frac{\mu}{\rho}[/tex]
[tex]\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}[/tex]
[tex]\upsilon=1.4012\times 10^{-3}[/tex] in2/s.
Thus, the dynamic viscosity and kinematic viscosity are [tex]1.3374\times 10^{-6}[/tex] lb-s/in2 and [tex]1.4012\times 10^{-3}[/tex] in2/s.
A series R-L circuit is given. Circuit is connected to an AC voltage generator. a) Derive equations for magnitude and phase of current and voltages on resistor and inductor in the phasor domain. Assume that the resistance of the resistor is R, inductance of the inductor is L, magnitude of the source voltage is Vm and phase of the source voltage is θ. Note that you don’t have numbers in this step, so to find the magnitude and phase for current I and voltages VR and VL you must first derive both numerator and denominator in polar form using variables R, omega, L, Vm, Vphase (do not use numbers). The solutions should look like equations in slide 24/27! b) In this step, assume that R
Answer:
The equations for magnitude and phase of current and voltages on resistor and inductor are:
[tex]I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]
[tex]V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]
[tex]V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}[/tex]
Explanation:
The first step is to find the impedances of the resistance (R) and the inductor (L).
The impedance of the resistor is:
Rectangular form: [tex]Z_R=R[/tex]Polar form: [tex]Z_R=R\angle 0^{\circ}[/tex]The impedance of the inductor is:
Rectangular form: [tex]Z_L=j\omega L[/tex]Polar form: [tex]Z_L=\omega L \angle 90^{\circ}[/tex]Where [tex]\omega [/tex] is the angular frequency of the source, and the angle is [tex]90^{\circ} [/tex] because a pure imaginary number is on the imaginary axis (y-axis).
The next step is to find the current expression. It is the same for the resistor and inductor because they are in series. The total impedance equals the sum of each one.
[tex]I=\frac{V}{Z_R+Z_L}[/tex]
It is said that [tex]V=V_m\angle \theta[/tex], so, the current would be:
[tex]I=\frac{V_m\angle \theta }{R+j\omega L}[/tex]
The numerator must be converted to polar form by calculating the magnitude and the angle:
The magnitude is [tex]\sqrt{R^2+(\omega L)^2}[/tex]The angle is [tex]tan^{-1}(\omega L / R)[/tex]The current expression would be as follows:
[tex]I=\frac{V_m\angle \theta }{\sqrt{R^2+(\omega L)^2}\, \angle tan^{-1}(\omega L / R)}[/tex]
When dividing, the angles are subtracted from each other.
The final current expression is:
[tex]I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]
The last step is calculating the voltage on the resistor [tex]V_R[/tex] and the voltage on the inductor [tex]V_L[/tex]. In this step the polar form of the impedances could be used. Remember that [tex]V=I\cdot Z[/tex].
(Also remember that when multiplying, the angles are added from each other)
Voltage on the resistor [tex]V_R[/tex]
[tex]V_R=I\cdot Z_R=\bigg( \frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)\bigg) \cdot (R\angle 0^{\circ})[/tex]
The final resistor voltage expression is:
[tex]V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]
Voltage on the inductor [tex]V_L[/tex]
[tex]V_L=I\cdot Z_L=\bigg( \frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)\bigg) \cdot (\omega L \angle 90^{\circ})[/tex]
The final inductor voltage expression is:
[tex]V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}[/tex]
Summary: the final equations for magnitude and phase of current and voltages on resistor and inductor are:
[tex]I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]
[tex]V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)[/tex]
[tex]V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}[/tex]
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.
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]
Which is more detrimental to the tool life, a higher depth of cut or a higher cutting velocity? Why?
Answer:
Cutting velocity.
Explanation:
Velocity of cutting affects more as compare to the depth of cut to life of tool.
As we know that tool life equation
[tex]VT^n=C[/tex]
Where T is the tool life and V is the tool cutting speed and C is the constant.
So from we can say that when cutting velocity increase then tool life will decrease and vice versa.
The life of tool is more important during cutting action .The life of tool is less and then will reduce the production and leads to face difficulty .
A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m3m3. A constant pressure process gives 18 kJ of work out. What is the final temperature? You may assume ideal gas.
Answer:
1160 K.
Explanation:
Given that
Initial
Pressure P =600 KPa
Temperature T =290 K
Volume V =0.01 [tex]m^3[/tex]
If we assume that air is s ideal gas the
P V = mRT
R=0.287 KJ/kg.k
now by putting the values in above equation
600 x 0.01 = m x 0.287 x 290
m=0.07 kg
The work out at constant pressure given as
[tex]w=P(V_2-V_1)[/tex]
[tex]18=600(V_2-0.01)[/tex]
[tex]V_2=0.04\ m^3[/tex]
At constant pressure
[tex]\dfrac{T_2}{T_1}=\dfrac{V_2}{V_1}[/tex]
[tex]\dfrac{T_2}{290}=\dfrac{0.04}{0.01}[/tex]
[tex]T_2=1160\ K[/tex]
So the final temperature is 1160 K.
For turbulent now the friction factor is function of (Reynolds number —surface roughness -both the Reynolds number and the surface roughness) of the pipe.
Answer:
Both Reynolds and surface roughness
Explanation:
For turbulent flow friction factor is a function of both Reynolds and surface roughness of the pipe.But on the other hand for laminar flow friction factor is a function of only Reynolds number.
Friction factor for turbulent flow:
1. For smooth pipe
[tex]f=0.0032+\dfrac{0.221}{Re^{0.237}}[/tex]
[tex]5\times 10^4<Re<4\times 10^7[/tex]
2. For rough pipe
[tex]\dfrac{1}{\sqrt f}=2\ log_{10}\frac{R}{K}+1.74[/tex]
Where R/K is relative roughness
Friction factor for laminar flow:
[tex]f=\dfrac{64}{Re}[/tex]
If a pendulum is 10m long, (a) what is the natural frequency and the period of vibration on the earth, where the free-fall acceleration is 9.81 m/s^2 and (b) what is the natural frequency and the period of vibration on the moon, where the free-fall acceleration is 1.67 m/s^2?
Answer:
(a) Natural frequency = 0.99 rad/sec (b) 0.4086 rad/sec
Explanation:
We have given length of pendulum = 10 m
(a) Acceleration due to gravity [tex]=9.81m/sec^2[/tex]
Time period of pendulum is given by [tex]T=2\pi\sqrt{\frac{L}{g}}[/tex], L is length of pendulum and g is acceleration due to gravity
So [tex]T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{9.81}}=6.34sec[/tex]
Natural frequency is given by [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6.34}=0.99rad/sec[/tex]
(b) In this case acceleration due to gravity [tex]g=1.67m/sec^2[/tex]
So time period [tex]T=2\pi\sqrt{\frac{L}{g}}=2\times 3.14\times \sqrt{\frac{10}{1.67}}=15.3674sec\[/tex]
Natural frequency [tex]\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{15.36}=0.4086rad/sec[/tex]
A light bulb is switched on and within a few minutes its temperature becomes constant. Is it at equilibrium or steady state.
Answer:
The temperature attains equilibrium with the surroundings.
Explanation:
When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.
Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.
Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.
A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 constant. The process begins with p1 5 15 lbf/in.2, 1 5 1.25 ft3/lb and ends with p2 5 53 lbf/in.2, 2 5 0.5 ft3/lb. Determine (a) the volume, in ft3, occupied by the gas at states 1 and 2 and (b) the value of n. (c) Sketch Process 1–2 on pressure–volume coordinates.
Answer:
V1=5ft3
V2=2ft3
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5ft3
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= 2ft3
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Air enters an adiabatic turbine at 800 kPa and 870 K with a velocity of 60 m/s, and leaves at 120 kPa and 520 K with a velocity of 100 m / s. The inlet area of the turbine is 90 cm2. What is the power output?
Answer:
The power output of the turbine is 603 KW.
Explanation:
Turbine is the thermodynamic open system in which fluid looses thermal energy into kinetic energy. Kinetic energy then converted into electric energy.
Here, fluid is air which passes through turbine at 800 Kpa and 870 K with a velocity of 60 m/s.
The turbine is an adiabatic turbine that means there is no heat transfer from the surrounding. Finally the air leaves the turbine at 120 Kpa and 520 K with a velocity of 100 m/s. The turbine inlet area is 90 cm2
Given:
Inlet pressure is [tex]P_{1}=800[/tex]kpa.
Inlet temperature is [tex]T_{1}=870[/tex]K.
Inlet velocity is[tex]V_{1}=60[/tex] m/s.
Outlet pressure is [tex]P_{2}=120[/tex]Kpa.
Outlet temperature is [tex]T_{2}=520[/tex]K.
Outlet velocity is [tex]V_{2}=100[/tex] m/s.
Inlet area of turbine is A=90 cm2.
Step1
Convert the area into SI unit as follows:
[tex]A=90 cm^{2}(\frac{1 m^{2}}{10^{4}cm^{2}})[/tex]
[tex]A=0.009 m^{2}[/tex]
Step 2
Consider air as an ideal gas. So, ideal gas equation is applicable. For air, gas constant is 287 j/kgK.
Ideal gas equation is expressed as follows:
[tex]P=\rho RT[/tex]
Here, P is pressure, T is temperature and \rho is density.
Density of air is calculated by ideal gas equation as follows:
[tex]\rho =\frac{P}{RT}[/tex]
[tex]\rho =\frac{800\times 10^{3}}{287\times870}[/tex]
[tex]\rho =3.2039 kg/m^{3}[/tex]
Step 3
Mass flow rate is calculated as follows:
[tex]\dot{m}=\rho AV_{1}[/tex]
[tex]\dot{m}=3.2039\times 0.009\times60[/tex]
[tex]\dot{m}=1.73 Kg/s[/tex]
Step 4
Steady state equation is the equation of first law of thermodynamics for the open system
Steady state equation for the turbine as follows:
[tex]h_{1}+\frac{v^{2}_{1}}{2000}+Z_{1}+Q=h_{2}+\frac{v^{2}_{2}}{2000}+Z_{2}+W[/tex]
Heat transfer is zero as the process is adiabatic. So value of Q is zero.
Turbine is taken as at the same level. So the value of [tex]Z_{1}[/tex] is equal to [tex]Z_{2}[/tex].
Substitute the value of Q as zero and tex]Z_{1}[/tex] is equal to [tex]Z_{2}[/tex] in steady state equation as follows:
[tex]h_{1}+\frac{v^{2}_{1}}{2}+Z_{1}+0=h_{2}+\frac{v^{2}_{2}}{2}+Z_{1}+W[/tex]
[tex]h_{1}+\frac{v^{2}_{1}}{2}+0=h_{2}+\frac{v^{2}_{2}}{2}+W[/tex]
[tex]W=(h_{1}-h_{2})+\frac{v^{2}_{1}-v^{2}_{2}}{2000}[/tex]
[tex]W=c_{p}(T_{1}-T_{2})+\frac{60^{2}-100^{2}}{2000}[/tex]
Specific heat at constant pressure is 1.005 kj/kgK for air.
Substitute the values of temperature and specific heat at constant temperature in the above simplified steady state equation as follows:
W=1.005(870-520)-3.2
W=351.75-3.2
W=348.55 Kj/kg.
Step 5
Power of the turbine is calculated as follows:
[tex]P=\dot{m}W[/tex]
[tex]P=1.73\times348.55[/tex]
P=603 KW
Thus, the power output of the turbine is 603 KW.
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]
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
A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the density of air at this point?
Answer:
[tex]5.31\frac{kg}{m^3}[/tex]
Explanation:
Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to [tex]\frac{m}{M}[/tex], where m is the mass and M the molar mass of the gas, and the density is [tex]\frac{m}{V}[/tex].
For air [tex]M=28.66*10^{-3}\frac{kg}{mol}[/tex] and [tex]\frac{5}{9}R=K[/tex]
So, [tex]598.59 R*\frac{5}{9}=332.55K[/tex]
[tex]pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}[/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]
The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and reads 68.95 kPa gage;the discharge static gage is 0.61 m below the pump centre line and reads 344.75 kPagage. The gages are located close to the pump as much as possible. The area of the intake and discharge pipes are; 0.093 m2 and 0.069 m2 respectively. The pump efficiency is 74%. Take density of water equals 1000 kg/m3. What is the hydraulic power in kW
Answer:
Pump power is 23.09 kW
Explanation:
Data
gravitational constant, [tex] g = 9.81 m/s^2 [/tex]
mass flow, [tex] \dot{m} = 60.6 kg/s [/tex]
flow density, [tex] \rho = 1000 kg/m^3 [/tex]
pump efficiency, [tex] \eta = 0.74 [/tex]
output gage pressure, [tex] p_o = 344.75 kPa [/tex]
input gage pressure, [tex] p_i = 68.95 kPa [/tex]
output pipe area, [tex] A_o = 0.069 m^2 [/tex]
input pipe area, [tex] A_i = 0.093 m^2 [/tex]
output height, [tex] z_o = 1.22 m - 0.61 m = 0.61 m [/tex] (considering that pump is at the maximum height, i.e., 1.22 m)
input height, [tex] z_i = 0 m [/tex]
pump hydraulic power,[tex] P = ? kW [/tex]
First of all, volumetric flow (Q) must be computed
[tex] Q = \frac{\dot{m}}{\rho}[/tex]
[tex] Q = \frac{60.6 kg/s}{1000 kg/m^3} [/tex]
[tex] Q = 0.0606 m^3/s[/tex]
Then, velocity (v) must be computed for both input and output
[tex] v_o = \frac{Q}{A_o}[/tex]
[tex] v_o = \frac{0.0606 m^3/s}{0.069 m^2}[/tex]
[tex] v_o = 0.88 m/s [/tex]
[tex] v_i = \frac{Q}{A_i}[/tex]
[tex] v_i = \frac{0.0606 m^3/s}{0.093 m^2}[/tex]
[tex] v_i = 0.65 m/s [/tex]
Now, total head (H) can be calculated
[tex] H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g} [/tex]
[tex] H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2} [/tex]
[tex] H = 28.74m [/tex]
Finally, pump power is computed as
[tex] P = \frac{Q \, \rho \, g \, H}{\eta}[/tex]
[tex] P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}[/tex]
[tex] P = 23.09 kW [/tex]
The hydraulic power in kW is mathematically given as
P = 23.09 kW
What is the hydraulic power?Generally the equation for the volumetric flow (Q) is mathematically given as
Q=m/p
Therefore
Q=60/1000
Q=0.0606
Generally the equation for the velocity (v) is mathematically given as
v=Q/A
Hence for input
[tex]v_i = \frac{Q}{A_i}\\\\v_i = \frac{0.0606 }{0.093 }[/tex]
v_i = 0.65 m/s
For input
[tex]v_o = \frac{Q}{A_o}\\\\v_o = \frac{0.0606 }{0.069}[/tex]
v_o = 0.88 m/s
Therefore, Total head (H)
[tex]H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\mu g}[/tex]
[tex]H = (0.61 0 ) + \frac{{0.88 }^2 - {0.65 }^}{2 *9.81 } + \frac{(344.75 Pa-68.95 Pa)*10^3}{1000* 9.81}[/tex]
H = 28.74m
Generally the equation for the pump power P is mathematically given as
[tex]P = \frac{Q * \rho*g*H}{\eta}[/tex]
[tex]P = \frac{0.0606 * 100*9.81*28.74}{0.74}[/tex]
P = 23.09 kW
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What are two advantages of forging when compared to machining a part from a billet?
Answer:
Less material waste and time.
Explanation:
Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.
There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.
A battery is an electromechanical device. a)- True b)- False
Answer:
b)False
Explanation:
A battery is a device which store the energy in the form of chemical energy.And this stored energy is used according to the requirement.So battery is not a electromechanical device.Because it does have any mechanical component like gear ,shaft flywheel etc.
A flywheel is known as mechanical battery because it stored mechanical energy and supply that energy when more energy is required.Generally fly wheel is used during punching operation.
Explain what is the young's modulus?
Answer and Explanation:
Young's modulus is a mechanical property that estimates the solidness of a strong material. If we have information about stress and strain the we can easily found the young's modulus.
Young's modulus gives us information that how hard or how easy to bend a solid material
Young's modulus is given by [tex]young's\ modulus=\frac{stress}{strain}[/tex]
Stress = [tex]\frac{force}{area}[/tex] and strain is [tex]=\frac{chane\ in\ length}{actual\ length}=\frac{\Delta L}{L}[/tex]
A steel band blade, that was originally straight, passes
over8-in.-diameter pulleys when mounted on a band saw. Determine
themaximum stress in the blade, knowing that it is 0.018 in. thick
and0.625 in. wide. Use E = 29 x 106 psi.
Answer:
[tex]\sigma = 65.25\ ksi[/tex]
Explanation:
given data:
D = 8 inch
R =4 inch
thickness = 0.018 inch
width = 0.625 inch
[tex]E =29*10^6 psi[/tex]
from bending equation we know that
[tex]\frac{\sigma}{y} = \frac{M}{I} = \frac{E}{R}[/tex]
[tex]\sigma = \frac{Ey}{R}[/tex]
Where y represent distance from neutral axis
[tex]y = \frac{t}[2}[/tex]
[tex]y = \frac{0.018}{2}[/tex]
y = 0.009inch
[tex]\sigma = \frac{29*10^6*0.009}{4}[/tex]
[tex]\sigma = 65250\ psi[/tex]
[tex]\sigma = 65.25\ ksi[/tex]
Ductility (increases/decreases/does not change) with temperature.
Answer:
Increases
Explanation:
Ductility:
Ductility is the property of material to go permanent deformation due to tensile load.In other words the ability of material to deform in wire by the help of tensile load.
When temperature is increase then ductility will also increases.And when temperature decreases then the ductility will also decreases.As we know that at very low temperature material become brittle and this is know as ductile brittle transition.
Ammonia at 20 C with a quality of 50% and a total mass of 2 kg is in a rigid tank with an outlet valve at the bottom. How much saturated liquid can be removed from the tank in an isothermal process until there remains no more liquid?
Answer:
16.38L
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties.
Quality is defined as the ratio between the amount of steam and liquid when a fluid is in a state of saturation, this means that since the quality is 50%, 1kg is liquid and 1kg is steam.
then to solve this problem we find the specific volume for ammonia in a saturated liquid state at 20C, and multiply it by mass (1kg)
v(amonia at 20C)=0.001638m^3/kg
m=(0.01638)(1)=0.01638m^3=16.38L
Determine if the following errors are systematic or random. Justify your response. (a) Effect of temperature on the circuitry of an electronic measurement device. (b) Effect of parallax on the reading of a needle-type analog voltmeter. (c) Effect of using an incorrect value of emissivity in the readings of an infrared thermometer.
Answer:
a) temperature: random error
b) parallax: systematic error
c) using incorrect value: systematic error
Explanation:
Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.
(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of 0.1 mg/L. Calculate your dose for a 50 kg adult female who drinks 2 L lake water per day and consumes 30 g fish per day that is caught from the lake. Ans. 0.064 mg/kg-d (b) What percent of the total dose is from exposure to the water, and what percent is from exposure to the fish?
Answer:
0.064 mg/kg/day
6.25% from water, 93.75% from fish
Explanation:
Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.
The BCF = 10³, so the concentration of the chemical in the fish is:
10³ = x / (0.1 mg/kg)
x = 100 mg/kg
For 2 L of water and 30 g of fish:
2 kg × 0.1 mg/kg = 0.2 mg
0.030 kg × 100 mg/kg = 3 mg
The total daily intake is 3.2 mg. Divided by the woman's mass of 50 kg, the dosage is:
(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day
b) The percent from the water is:
0.2 mg / 3.2 mg = 6.25%
And the percent from the fish is:
3 mg / 3.2 mg = 93.75%
What is the difference between pound-mass and pound-force?
Answer:
Pound mass is the unit of mass and which is used in United states customary to describe the mass Slug is also a unit of mass .Pound mass is lbm. But on the other hand pound force is the unit of force.Pound force is lbf.
In SI unit ,the unit of mass is kilogram(kg) and the unit of force is Newton(N)
In CGS system the unit of mass is gram and unit of force is dyne.
The initial internal energy of a mug of coffee is known to be 168 Btu. The initial coffee temperature is approximately 200 F. The room temperature is 70 F. Time passes and the final amount of internal energy in the coffee is 68 Btu. How much heat flowed from the coffee to the room?
Answer:
100Btu
Explanation:
According to the First law of thermodynamics:
ΔQ = W + ΔU
Where,
ΔQ is the heat change
W is the amount of work done
ΔU is change in internal energy
Since, there is no work done on the system as heat is just passing from coffee mug to surroundings, So, W = 0
Thus,
Net heat change = change in internal energy.
Change in internal energy = [tex]U_f-U_i=(68-168)\ Btu=-100\ Btu[/tex]
Thus,
ΔQ = -100 Btu (negative sign indicates release of heat)
So,
Heat flown to the room = 100Btu
Engineering stress, strain vs true stress, strain.
Answer with Explanation:
Stress is defined as the force acting per unit area on a material.
Mathematically
[tex]\sigma =\frac{dF}{dA}[/tex]
where
[tex]\sigma [/tex] is the stress ,[tex]dF[/tex] is an infinitesimal force that acts on an infinitesimal area [tex]dA[/tex]
When a body is under stress it's dimensions change and this change in dimensions is known as strain.
Mathematically
[tex]\epsilon =\frac{\Delta x}{X}[/tex]
where
[tex]\epsilon=[/tex] strain in the object
[tex]\Delta x =[/tex] is the change in any dimension of the body
Now in the above relation of stress, the area involved also changes when the body is loaded as the load produces strain which changes the dimensions of the body.
Now while calculating the stress if we use the original area of the cross section of the body prior to loading the stress that we calculate is the engineering stress and the strain associated with it is the engineering strain.
On the other hand if we use the true cross section of the body when it is loaded the stress that we calculate is the true stress and the strain associated with it is the true strain.
Mathematically they are related as
[tex]\epsilon _{true}=ln(1+\epsilon _{engineering})}[/tex]
Thus the true stress is found to be larger than engineering stress.
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]