If the dry-bulb temperature is 95°F and the wet-bulb temperature is also 78°F, what is the relative humid- ity? What is the dew point? What is the humidity ratio? What is the enthalpy?

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]

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.

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]

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]

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

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:

The impedance of the inductor is:

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 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]

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]

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 .

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

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.

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

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.

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]

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.

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

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]

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:

Recalling the basic equation of the spring we find that:

[tex]F=Kx[/tex]

Where:

So 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]

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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.

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

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]

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]

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.

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]

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

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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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.

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]

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

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%

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.

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.