A Carnot heat engine receives heat at 900 K and rejects the waste heat to the environment at 300 K. The entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at –15°C at a rate of 295 kJ/min and rejects it to the same environment at 300 K. Determine the rate of heat supplied to the heat engine. (Round the final answer to one decimal place. You must provide an answer before moving to the next part.).The rate of heat supplied to the engine is ___ kJ/min.

Answer:

3.0772 m

Explanation:

Given:

Diameter of the aluminium rod, d = 20 mm = 0.02 m

Length of elongation, δL = 3.5 mm = 0.0035 m

Applied load, P = 25 KN = 25000 N

Modulus of elasticity, E = 70 GPa = 70 × 10⁹ N/m²

Now,

we have the relation

[tex]\delta L=\frac{\textup{PL}}{\textup{AE}}[/tex]

Now,

Where, A is the area of cross-section

A =  [tex]\frac{\pi}{4}d^2[/tex]

or

A = [tex]\frac{\pi}{4}\times0.02^2[/tex]

or

A = 0.000314 m²

L is the length of the member

on substituting the respective values, we get

[tex]0.0035=\frac{25000\times L}{0.000314\times70\times10^9}[/tex]

or

L = 3.0772 m

Answer:

1) s(3) = -32 feet.

2)v(5) = 3 feet/sec

3)a(4) = 12[tex]feet/s^{2}[/tex]

4) Velocity becomes zero at t = 5 seconds

Explanation:

Given that position as a function of time is

[tex]s(t)=t^{3}-6t^{2}-15t+40[/tex]

Now by definition of velocity we have

[tex]v=\frac{ds}{dt}\\\\v=\frac{d}{dt}(t^{3}-6t^{2}-15t+40)\\\\\therefore v(t)=3t^{2}-12t-15[/tex]

Now by definition of acceleration we have

[tex]a=\frac{dv}{dt}\\\\a=\frac{d}{dt}(3t^{2}-12t-15)\\\\\therefore a(t)=6t-12[/tex]

Applying values of time in corresponding equations we get

1) s(3)=[tex]3^{3}-6\times (3)^{2}-15\times 3+40=-32feet[/tex]

2)v(5)=[tex]3\times {5}^{2}-12\times 5-15=3feet/sec[/tex]

3)a(4)=[tex]6\times 4-12=12ft/s^{2}[/tex]

4)To obatin the time at which velocity is zero equate the velocity function with zero we get

[tex]3t^{2}-12t-15=0\\\\t^{2}-4t-5=0\\\\t^{2}-5t+t-5=0\\\\t(t-5)+1(t-5)=0\\\\(t-5)(t+1)=0\\\\\therefore t=5\\\\or\\\therefore t=-1[/tex]

Thus the correct time is 5 seconds at which velocity becomes zero.

Answer:

The micrometer with vernier graduation can measure reading to the nearest 0.0001 inches.  So, the statement is true.

Explanation:

Micrometer is the measuring device that used to measure length with more accuracy. Micrometer can measure the length in metric as well as in English unit. Micrometer is generally used to measure diameter and length of the mechanical component.

Working:

Micrometer is a screwed device. It contains spindle, anvil and thimble. Object is placed between spindle and anvil. Thimble is rotated that rotates the spindle till it touches the component completely. Two types of scales are used to measure the reading of micrometer, one is sleeve scale and other is thimble scale. Spindle moves toward component by 0.5 mm in or 0.025 in on every one rotation of spindle. There are three types of micrometer

Least count of micrometer:

Minimum measurement of any measuring device is the least count of that device. So, the least count for normal micrometer is 0.01 mm or 0.001 inches.

The micrometer is called vernier micrometer if the micrometer is provided with the vernier scale. The least count of vernier micrometer scale is 0.0001 inches.

Hence the micrometer with vernier graduation can measure reading to the nearest 0.0001 inches.

Thu, the statement is true.

Answer:

d) Is the thermal conductivity of the medium constant or variable.

Explanation:

As we know that

Heat equation with heat generation at unsteady state and with constant thermal conductivity given as

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}+\dfrac{\dot{q}_g}{K}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]

With out heat generation

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]

In 2 -D with out heat generation with constant thermal conductivity

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]

Given equation

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{a }\dfrac{dT}{dt}[/tex]

So we can say that this is the case of  with out heat generation ,unsteady state and with constant thermal conductivity.

So the option d is correct.

d) Is the thermal conductivity of the medium constant or variable.

Answer:

Maximum temperature of the cycle is 2231.3 K

Explanation:

See table (values there do not assume constant specific heat) and figure attached.

Assuming ideal gas behaviour, p1*v1 = p2*v2, rearranging p2/p1 = v1/v2

Data

[tex]p_1 = 1 bar [/tex]

[tex]T_1 = 300 K [/tex]

[tex] \frac{v_1}{v_2} = 8.5 [/tex] (compression ratio)

[tex] \frac{Q_{23}}{m} = 1400 kJ/kg [/tex]  (heat addition)

We can use the following relationship  for air

[tex] \frac{v_1}{v_2} = \frac{v_{r1}}{v_{r2}} [/tex]

[tex] v_{r1} [/tex] is only function of temperature and can be taken from table. In this case:

[tex] v_{r1} = 621.2 [/tex]  

Rearranging previous equation

[tex] v_{r2} = v_{r1} \times \frac{v_2}{v_1} [/tex]

[tex] v_{r2} = 621.2 \times \frac{1}{8}[/tex]

[tex] v_{r2} = 73.082 [/tex]

Interpolating from table

[tex] u_2 = 503.06 kJ/kg [/tex]

Energy balance in the process 2-3 gives

[tex] \frac{Q_{23}}{m} = u_3 - u_2 [/tex]  

[tex] u_3 = \frac{Q_{23}}{m} + u_2 [/tex]  

[tex] u_3 = 1400 kJ/kg + 503.06 kJ/kg [/tex]  

[tex] u_3 = 1903.06 kJ/kg [/tex]  

Interpolating from table

[tex] T_3 = 2231.3 K [/tex]

Answer:

a) 2.18 m/s^2

b) 9.83 m/s

Explanation:

The flywheel has a moment of inertia

J = m * k^2

Where

J: moment of inertia

k: radius of gyration

In this case:

J = 144 * 0.45^2 = 29.2 kg*m^2

The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.

T = p * r

r is the radius of the wheel.

T = m1 * g * r

T = 18 * 9.81 * 0.6 = 106 N*m

The torque will cause an acceleration on the flywheel:

T = J * γ

γ = T/J

γ = 106/29.2 = 3.63 rad/s^2

SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:

at = γ * r

at = 3.63 * 0.6 = 2.81 m/s^2

Now that we know the acceleration of the block we can forget about the flywheel.

The equation for uniformly accelerated movement is:

X(t) = X0 + V0*t + 1/2*a*t^2

We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:

X(t) = 1/2*a*t^2

Rearranging

t^2 = 2*X(t)/a

[tex]t = \sqrt{\frac{2*X(t)}{a}}[/tex]

[tex]t = \sqrt{\frac{2*18}{2.81}} = 3.6 s[/tex]

It will reach the 1.8 m in 3.6 s.

Now we use the equation for speed under constant acceleration:

V(t) = V0 + a*t

V(3.6) = 2.81 * 3.6 = 9.83 m/s

The cutting ratio is 0.027, and the shear angle is 88.46 degrees.

The Breakdown

- Initial diameter of the bar: 75 mm

- Final diameter of the bar after cutting: 73 mm

- Mean length of the cut chip: 73.5 mm

- Rake angle: 15 degrees

Calculate the cutting ratio.

Cutting ratio = (Initial diameter - Final diameter) / Mean length of the cut chip

Cutting ratio = (75 mm - 73 mm) / 73.5 mm

Cutting ratio = 0.027

Calculate the shear angle.

The shear angle (φ) can be calculated using the following formula:

tan(φ) = (1 - cutting ratio) / (cutting ratio × cos(α))

Where:

α = Rake angle (in radians)

Substituting the given values:

α = 15 degrees = 15 × π/180 = 0.2618 radians

Cutting ratio = 0.027

tan(φ) = (1 - 0.027) / (0.027 × cos(0.2618))

φ = tan-¹(0.9730 / 0.0265)

φ = 88.46 degrees

Therefore, the cutting ratio is 0.027, and the shear angle is 88.46 degrees.

To determine the work in a thermodynamic process of a two-phase liquid-vapor mixture of H2O, use the provided formula considering initial and final conditions, enabling calculation of the energy transferred.

In this thermodynamic process, the work done can be calculated using the area under the constant pressure line on a P-v diagram. Given the initial and final conditions, the work for the process can be determined.

To calculate the work, use the formula: W = m*(P_final*V_final - P_initial*V_initial)/(1-q), where 'm' is the mass of the substance, 'P' is the pressure, 'V' is the specific volume, and 'q' is the quality of the mixture.

Substitute the values into the formula, convert units as necessary, and calculate the work to find the energy transferred during the process in Btu per lb of H2O present.

Answer:

False

Explanation:

In electric heater electric energy is converted into heat energy. In heater wires are present which have resistance and current is flow in heater when we connect the heater to supply.

And we know that whenever current is flow in any resistance then heat is produced so in electric heaters electric energy is converted into heat energy

So this is a false statement

The roof temperature is calculated to be 93°C without radiation or 86.67°C when accounting for radiation heat transfer with the surroundings.

Given: q = 800 W/m2 h = 12 W/m2∙K T∞ = 293 K

Convection heat transfer equation: q = hA(Ts - T∞)

Plug in values: 800 W/m2 = 12 W/m2∙K (Ts - 293 K)

Distribute the 12 W/m2∙K: 800 W/m2 = 12 W/m2∙K * Ts - 12 W/m2∙K * 293 K

Group the Ts terms: 800 W/m2 = 12 W/m2∙K * Ts - 3,516 W/m2

Add 3,516 W/m2 to both sides: 4,316 W/m2 = 12 W/m2∙K * Ts

Divide both sides by 12 W/m2∙K: Ts = 4,316 W/m2 / 12 W/m2∙K

Calculate:

Ts = 366 K = 93°C

(a) Without radiation: Heat transfer equation: q = hA(Ts - T∞)

Plug in values: 800 W/m2 = 12 W/m2∙K (Ts - 293 K) 800 = 12(Ts - 293) Ts = 800/12 + 293 Ts = 366 K = 93°C

(b) With radiation: Heat transfer equation:

q = hA(Ts - T∞) + εσA(Ts4 - Tsur4)

Given: T∞ = 293 K ε = 0.8

σ = 5.67x10-8 W/m2∙K4

Radiation heat transfer equation: q = εσA(Ts4 - Tsur4)

Assume: Tsur = T∞ = 293 K

Plug in values: q = 0.8 * 5.67x10-8 * A * (Ts4 - (293)4)

(293)4 evaluation: (293)4 = 293 x 293 x 293 x 293 = 2.97 x 108

Plug this into equation:

q = 0.8 * 5.67x10-8 * A * (Ts4 - 2.97x108)

800 = 12(Ts - 293) + 0.8*5.67x10-8(Ts4 - 2.97x108)

Solve for Ts: Ts = 359.8 K = 86.67°C

Therefore, with radiation the roof temperature is 86.67°C.

Part A: The temperature of the roof under steady-state conditions without considering radiation exchange is 86.67°C.

Part B: The temperature of the roof under steady-state conditions considering radiation exchange with an emissivity of 0.8 is 81.17°C.

Part A: Neglecting Radiation Exchange

Step 1

Under steady-state conditions, the heat absorbed by the roof [tex](\( Q_{\text{absorbed}} \))[/tex] is equal to the heat lost through convection [tex](\( Q_{\text{convection}} \))[/tex].

Given:

- Solar radiant flux, [tex]\( q_{\text{solar}} = 800 \, \text{W/m}^2 \)[/tex]

- Convection coefficient, [tex]\( h = 12 \, \text{W/m}^2 \cdot \text{K} \)[/tex]

- Ambient air temperature, [tex]\( T_{\infty} = 20^\circ \text{C} \)[/tex]

The absorbed heat:

[tex]\[ Q_{\text{absorbed}} = q_{\text{solar}} \][/tex]

The heat loss by convection:

[tex]\[ Q_{\text{convection}} = h (T_s - T_{\infty}) \][/tex]

At steady state:

[tex]\[ q_{\text{solar}} = h (T_s - T_{\infty}) \][/tex]

Step 2

Solving for the surface temperature [tex]\( T_s \)[/tex]:

[tex]\[ 800 = 12 (T_s - 20) \][/tex]

[tex]\[ T_s - 20 = \frac{800}{12} \][/tex]

[tex]\[ T_s - 20 = 66.67 \][/tex]

[tex]\[ T_s = 86.67^\circ \text{C} \][/tex]

So, the temperature of the roof under steady-state conditions without radiation exchange is 86.67°C.

Part B: Considering Radiation Exchange

Step 1

When considering radiation exchange, the roof loses heat through both convection and radiation. The net radiative heat loss is given by the Stefan-Boltzmann law.

Given:

- Emissivity, [tex]\( \varepsilon = 0.8 \)[/tex]

- Stefan-Boltzmann constant, [tex]\( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \)[/tex]

The total heat loss (convection + radiation):

[tex]\[ Q_{\text{total}} = Q_{\text{convection}} + Q_{\text{radiation}} \][/tex]

For convection:

[tex]\[ Q_{\text{convection}} = h (T_s - T_{\infty}) \][/tex]

For radiation:

[tex]\[ Q_{\text{radiation}} = \varepsilon \sigma (T_s^4 - T_{\infty}^4) \][/tex]

At steady state:

[tex]\[ q_{\text{solar}} = h (T_s - 20) + \varepsilon \sigma (T_s^4 - 293.15^4) \][/tex]

[tex]\[ 800 = 12 (T_s - 20) + 0.8 \times 5.67 \times 10^{-8} (T_s^4 - 293.15^4) \][/tex]

Step 2

To simplify:

[tex]\[ 800 = 12 (T_s - 20) + 4.536 \times 10^{-8} (T_s^4 - 293.15^4) \][/tex]

This equation is nonlinear and needs to be solved iteratively. Let's outline the steps to solve it without detailed iteration steps:

1. Initial Guess: Start with an initial guess for [tex]\( T_s \)[/tex].

2. Iteration: Adjust [tex]\( T_s \)[/tex] until both sides of the equation are equal.

Through iteration or numerical methods, we find:

[tex]\[ T_s \approx 81.17^\circ \text{C} \][/tex]

In summary, the temperature of the roof is 86.67°C without considering radiation, and it drops to approximately 81.17°C when radiation exchange is taken into account.

Answer:

Velocity in inch per second will be 52.11 inch/sec

Explanation:

We have given velocity = 7943 cm/min

We have to convert this velocity into inches/sec

We know that 1 cm = 0.3937 inch

So 7943 cm = 7943×0.3937=3127.1193inch

And 1 minute = 60 sec

So [tex]7943cm/min =\frac{7943\times 0.3937inch}{60sec}=52.11inch/sec[/tex]

So velocity in inch per second will be 52.11 inch/sec

Answer:

2. Predictive Maintenance

Explanation:

Although definitions differ among authors, it is generally accepted that predictive maintenance uses different kinds of techniques to monitor critical machines to prevent them from failing unexpectedly and causing losses in production (or a service), and many more unpleasant events.

Among, thermography, tribology, ultrasonics, and others, vibration analysis is one of the techniques into predictive maintenance, and since most plant types of equipment are mechanical, this is the primary maintenance technique in predictive maintenance.

In general, vibration analysis first needs to acquire data (making use of vibration monitoring using transductors, like accelerometers). Then, the time-domain data is converted into frequency-domain data using a mathematical technique called Fast Fourier Transform (FFT).

Consequently, for each machine's anomaly, there will be a unique 'signature' in the frequency-domain data that corresponds to it.

For example, if the machine presents some imbalance, then there will be a typical frequency (primary frequency) and multiples of it (harmonics), in that frequency-domain data, unique for this imbalance, and so for other machine elements' anomalies, like misalignment, rolling-element bearings high vibrations, bent shafts, and many more.

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of  vaporisation .

Sensible heat for water Q

[tex]Q=mC_p\Delta T[/tex]

For water

[tex]C_p=4.178\ KJ/Kg.K[/tex]

Q=4.178 x 40 x 100 KJ

Q=16,712 KJ

So total heat

Total heat =100 x 333.23+16,712 + 60 x 2257 KJ

Total heat =185,455 KJ

Approx Total heat = 185,500 KJ

So the answer C is correct.

Answer:

Principal Plane: It is that plane in a stressed body over which no shearing stresses act. As we know that in a stressed body on different planes 2 different kind of stresses act normal stresses acting normal to the plane ans shearing stresses acting in the plane. The special planes over which no shearing stresses act and only normal stresses are present are termed as principal planes.

Principal Stress: The stresses in the principal planes are termed as normal stresses.

Anelasticity: It is the behavior of a material in which no definite relation can found to exist between stress and strain at any point in the stressed body.

Yield Point: It is the point in the stress-strain curve of a body at which the stress in the body reaches it's yield value or the object is just about to undergo plastic deformation if we just increase value of stress above this value. It is often not well defined in high strength materials or in some materials such as mild steel 2 yield points are observed.

Ultimate tensile strength: It is the maximum value of stress that a body can develop prior to fracture.

Hardness: it is defined as the ability of the body to resist scratches or indentation or abrasion.

Toughness: It is the ability of the body to absorb energy and deform without fracture when it is loaded. The area under the stress strain curve is taken as a measure of toughness of the body.

Elastic limit: The stress limit upto  which the body regains it's original shape upon removal of the stresses is termed as elastic limit of the body.

Answer:

a)Resistance to flow

Explanation:

Viscosity of fluid:

  Resistance to flow is called as viscosity of fluid.It is also know as fluid friction.It try to stop the flow of fluid.Viscosity of fluid is a property of fluid and it does mot depends on type of flow .

If fluid viscosity is high it means that it have very low flow ability.And opposite to viscosity is called fluidity.If fluidity is high then it means that it have low viscosity.

Viscosity are of two type

1.Dynamic viscosity

2.Kinematic viscosity

Answer:

22800 tonne

Explanation:

Given:

Length of the container, L = 240 m

Width of the container, B = 22 m

Depth inside the water, H = 10 m

Mass of the ship, m = 30000 tonnes

Now,

Total immersed volume of the ship = LBH = 240 × 22 × 10 = 52800 m³

From the Archimedes principle, we have

Total mass of the ship (i.e mass of the ship along with the load carried)

= Mass of the volume of water displaced by ship

= 52800 × Density of water

also,

Density of water = 1000 kg/m³

thus,

Total mass of the ship (i.e mass of the ship along with the load carried)

= 52800 × 1000 kg

also,

1 tonne = 1000 kg

thus,

Total mass of the ship (i.e mass of the ship along with the load carried)

= 52800 tonne

Therefore,

the load carried by the ship = Total mass of the ship  - mass of ship

or

the load carried by the ship = 52800 - 30000 = 22800 tonne

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

[tex]C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})[/tex]

[tex]C_{1}=1.2606\times10^{-5}[/tex] $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

[tex]C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})[/tex]

[tex]C_{2}=3.334\times10^{-5}[/tex] $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

[tex]C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})[/tex]

[tex]C_{3}=1.66\times10^{-5}[/tex] $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

[tex]a=18.66\ m/s^2[/tex]

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

Answer:

The kinetic energy will be 687.186 BTU

Explanation:

We have given mass of car = 2500 lbm

We know that 1 lbm = 0.4535 kg

So 2500 lbm = [tex]2500\times 0.4535=1133.75kg[/tex]

Speed = 80 mph

We know that 1 mile = 1609.34 meter

1 hour = 3600 sec

So [tex]80mph=\frac{80\times 1609.34}{3600}=35.763m/sec[/tex]

We know that kinetic energy [tex]E=\frac{1}{2}mv^2=\frac{1}{2}\times 1133.75\times 35.76^2=725.033KJ[/tex]

We know that 1 KJ = 0.9478 BTU

So 725.033 KJ = 725.033×0.9478 = 687.186 BTU

Explanation:

Renewable energy -

The energy source that does not get exhaust after using it , and can naturally replenish themselves .

These source of energy is naturally available and can be used with out limitation of being getting over .

The major types of renewable energy sources are  as follows -

Ocean , tide and wave are not the main renewable source , because , these are available only for certain time period , as tidal energy can be used only during high tides , similarly with the ocean and wave .

Answer:

Tidal energy and wave energy are considered renewable resources because tides are controlled by the moon, and the moon will constantly raise and lower the water. This is why tidal energy and wave energy are considered renewable resources.

Explanation:

Answer:

shaft power 0.2034 MW

Explanation:

given details

radius of turbine = 25 m

average wind velocity = 12 m/s

density of air = 1.20 kg/m^2

Total power is calculated as

[tex]P = \frac{1}{2} \rho AV^3[/tex]

  [tex]= \frac{1}{1} \rho \pir^2 V^3[/tex]

  [tex]= \frac{1}{2} 1.20\times \pi \times 625\times 12^3 = 2034,720 watt[/tex]

P = 2.034 MW

shaft power [tex] = \eta \times P[/tex]

                    [tex]= 0.10 \times 2.034[/tex]

                    = 0.2034 MW

Answer:[tex]\mu =9\times 10^{-3}Pa-s[/tex]

Explanation:

Distance between Plates(dy)=4.5 mm

Relative Velocity(du)=5 m/s

We know shear stress is given by [tex]\tau =10 Pa[/tex]

[tex]\tau =\frac{\mu du}{dy}[/tex]

where du=relative Velocity

dy=Distance between Plates

[tex]10=\frac{\mu \times 5}{4.5\times 10^{-3}}[/tex]

[tex]\mu =9\times 10^{-3}Pa-s[/tex]

Answer:

79 kW.

Explanation:

The equation for enthalpy is:

H2 = H1 + Q - L

Enthalpy is defined as:

H = G*(Cv*T + p*v)

This is specific volume.

The gas state equation is:

p*v = R*T (with specific volume)

The specific gas constant for air is:

287 K/(kg*K)

Then:

T1 = 60 + 273 = 333 K

T2 = 200 + 273 = 473 K

p1*v1 = 287 * 333 = 95.6 kJ/kg

p2*v2 = 287 * 473 = 135.7 kJ/kg

The Cv for air is:

Cv = 720 J/(kg*K)

So the enthalpies are:

H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW

H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW

Ang the heat is:

Q = 34 kW

Then:

H2 = H1 + Q - L

381 = 268 + 34 - L

L = 268 + 34 - 381 = -79 kW

This is the work from the point of view of the air, that's why it is negative.

From the point of view of the machine it is positive.

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

[tex]\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U[/tex][tex]15.5=-3.5+0+\bigtriangleup U[/tex]

[tex]\bigtriangleup U=19[/tex] kj.

Step3

Specific internal energy is calculated as follows:

[tex]\bigtriangleup u=\frac{\bigtriangleup U}{m}[/tex]

[tex]\bigtriangleup u=\frac{19}{5.4}[/tex]

[tex]\bigtriangleup u=3.5[/tex] kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

The density of water in SI units, converted from 1.94 slug/ft^3, is approximately 998.847 kg/m^3 when expressed to three significant figures.

The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:

(1.94 slug/ft3)
* (14.5939 kg/slug)
* ((1 ft/0.3048 m)3)

This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.

Water's density conversion to SI units is 1000 kg/m³.

The density of water in SI units can be expressed as 1000 kg/m³. This conversion is based on the fact that the density of water is exactly 1 g/cm³, equivalent to 1000 kg/m³.

Water's density conversion to SI units is 1000 kg/m³.

The density of water in SI units can be expressed as 1000 kg/m³. This conversion is based on the fact that the density of water is exactly 1 g/cm³, equivalent to 1000 kg/m³.

The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:

(1.94 slug/ft3) * (14.5939 kg/slug) * ((1 ft/0.3048 m)3)

This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.

The student has asked to convert the density of water from slug/ft3 to SI units. To convert from slug/ft3 to kg/m3, we need to use the appropriate conversion factors. One slug is equivalent to 14.5939 kilograms, and there are 0.3048 meters in a foot. Therefore, the conversion is as follows:

(1.94 slug/ft3)
* (14.5939 kg/slug)
* ((1 ft/0.3048 m)3)

This equals 1.94 * 14.5939 * (1/0.3048)3 kg/m3, which simplifies to 998.847 kg/m3 when rounded to three significant figures. This is the density of water in SI units.

It is convenient to model a structural element like a beam when a significant amount of forces produce the stress called flexion.

Flexion occurs when an element is supported on one or more supports and a force is presented between them, driving a bending moment in the element.

Answer:50 , 20

Explanation:

Given

Diametrical Pitch[tex]\left ( P_D\right )=\frac{T}{D}[/tex]

where T= no of teeths

D=diameter

module(m) of gears must be same

[tex]m=\frac{D}{T}=\frac{1}{P_D}=0.1[/tex]

Let [tex]T_1 & T_2[/tex]be the gears on two gears

Therefore Center distance is given by

[tex]m\frac{\left ( T_1+T_2\right )}{2}=3.5[/tex]

thus

[tex]0.1\frac{\left ( T_1+T_2\right )}{2}=3.5[/tex]

[tex]T_1+T_2=70----1[/tex]

and Velocity ratio is given by

[tex]VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear} [/tex]

[tex]2.5=\frac{T_1}{T_2}----2[/tex]

From 1 & 2 we get

[tex]T_1=50, T_2=20[/tex]

Answer:

[tex]4\ R=\sqrt 3\ a[/tex]

Explanation:

Given that

Lattice constant = a

Radius of unit cell cell =R

Atom is in BCC structure.

In BCC unit cell (Body centered cube)

1.Eight atoms at eight corner of cube which have 1/8 part in each cube.

2.One complete atom at the body center of the cube

So the total number of atoms in the BCC

 Z= 1/8 x 8 + 1 x 1

Z=2

In triangle ABD

[tex]AB^2=AD^2+BD^2[/tex]

[tex]AB^2=a^2+a^2[/tex]

[tex]AB=\sqrt 2\ a[/tex]

In triangle ABC

[tex]AC^2=AB^2+BC^2[/tex]

AC=4R

BC=a

[tex]AB=\sqrt 2\ a[/tex]

So

[tex]16R^2=2a^2+a^2[/tex]

[tex]4\ R=\sqrt 3\ a[/tex]

So the relationship between lattice constant and radius of unit cell

[tex]4\ R=\sqrt 3\ a[/tex]

Answer:

stress  = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement[tex] \delta = 1.25 \mu m =\frac{1.25}{1000} =  0.00125 mm[/tex]

stress due to displacement in structural steel can be determined by using following relation

[tex]E =\frac{stress}{strain}[/tex]

[tex]stress = E \times strain[/tex]

where E is young's modulus of elasticity

E for steel is 200 GPa

[tex]stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}[/tex]

stress  = 50MPa

Answer:

a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

Explanation:

The stack of plywood has a certain mass. The weight will depend on that mass.

w = m * g

There will be a normal reaction between the stack and the bed of the truck, this will be:

nr = -m * g * cos(θ)

Being θ the tilt angle of the bed.

The static friction force will be:

ffs = - m * g * cos(θ) * fJK

The dynamic friction force will be:

ffd = - m * g * cos(θ) * fik

These forces would produce accelerations

affs = -g * cos(θ) * fJK

affd = -g * cos(θ) * fik

affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)

affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)

These accelerations oppose to movement and must be overcome by another acceleration to move the stack.

The acceleration of the truck is horizontal, the horizontal component of these friction forces is:

affs = -3.92 * cos(θ)^2

affd = -2.94 * cos(θ)^2

The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

Assuming the bed has a lenght L.

The horizontal movement will be over a distance cos(θ) * L because L is tilted.

Movement under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.

If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.

L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2

2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)

a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)