Micrometers with a vernier graduation are capable of taking readings to the nearest 0.0001 in. a)- True b)- false

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.

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:

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

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:

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.

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

Answer:

No we cannot carry 1 cubic meter of liquid water.

Explanation:

As we know that density of water is 1000 kilograms per cubic meter of water hence we infer that 1 cubic meter of water will have a weight of 1000 kilograms of 1 metric tonnes which is beyond the lifting capability of strongest man on earth let alone a normal human being who can just lift a weight of 100 kilograms thus we conclude that we cannot lift 1 cubic meter of liquid water.

No, I cannot carry 1 cubic meter (1 m³) of liquid water. To understand why, let's calculate the weight of 1 cubic meter of water.

1 cubic meter (m³) of water is equivalent to 1000 liters (L). The density of water is approximately 1 kilogram per liter (kg/L). Therefore, the weight of 1 cubic meter of water can be calculated as:

[tex]\[ 1 \, \text{m}^3 \times 1000 \, \text{L/m}^3 \times 1 \, \text{kg/L} = 1000 \, \text{kg} \][/tex]

So, 1 cubic meter of water weighs 1000 kilograms, or about 2204.62 pounds.

This weight is far beyond the carrying capacity of an average human. For comparison, most people can carry only a few tens of kilograms comfortably for a short period, so carrying 1000 kilograms is not feasible for any human.

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:

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:

The radius of curvature is 979 meter

Explanation:

We have given velocity of the canon ball v = 98 m/sec

Acceleration due to gravity [tex]g=9.81m/sec^2[/tex]

We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity

Acceleration due to gravity is given by [tex]a_c=\frac{v^2}{r}[/tex], here v is velocity and r is radius of curvature

So [tex]\frac{98^2}{r}=9.81[/tex]

r = 979 meter

So the radius of curvature is 979 meter

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: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:

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:

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]

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.

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

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:

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:

Explanation:

The pressures given are relative

p1 = 2000 psi

P1 = 2014 psi = 13.9 MPa

p2 = 4 psi

P2 = 18.6 psi = 128 kPa

Values are taken from the steam pressure-enthalpy diagram

h2 = 2500 kJ/kg

If the output of the turbine has a quality of 85%:

t2 = 106 C

I consider the expansion in the turbine to adiabatic and reversible,  therefore, isentropic

s1 = s2 = 6.4 kJ/(kg K)

h1 = 3500 kJ/kg

t2 = 550 C

The work in the turbine is of

w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg

The thermal efficiency of the cycle depends on the input heat.

η = w/q1

q1 is  not a given, so it cannot be calculated.

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

Explanation:

From Stefan's formula

P=A&T^4

T=(P/A&)^1/4

T=(632000W/1.6m^2 x 5.67E-8W/m^2K^4)^1/4

T=

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:

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

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.

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:

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:

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.

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:

The rate of heat supplied to the engine is 71.7 kJ/min

Explanation:

Data

Engine hot temperature, [tex] T_H [/tex] = 900 K

Engine cold temperature, [tex] T_C [/tex] = 300 K

Refrigerator cold temperature, [tex] T'_C [/tex] = -15 C + 273 =  258 K

Refrigerator hot temperature, [tex] T'_H [/tex] = 300 K

Heat removed by refrigerator, [tex] Q'_{in} [/tex] = 295 kJ/min

Rate of heat supplied to the heat engine, [tex] Q_{in} [/tex] = ? kJ/min

See figure

From Carnot refrigerator coefficient of performance definition

[tex] COP_{ref} = \frac{T'_C}{T'_H - T'_C} [/tex]

[tex] COP_{ref} = \frac{258}{300 - 258} [/tex]

[tex] COP_{ref} = 6.14 [/tex]

Refrigerator coefficient of performance is defined as

[tex] COP_{ref} = \frac{Q'_{in}}{W} [/tex]

[tex] W = \frac{Q'_{in}}{COP_{ref}} [/tex]

[tex] W = \frac{295 kJ/min}{6.14} [/tex]

[tex] W = 48.04 kJ/min [/tex]

Carnot engine efficiency is expressed as

[tex] \eta = 1 - \frac{T_C}{T_H}[/tex]

[tex] \eta = 1 - \frac{300 K}{900 K}[/tex]

[tex] \eta = 0.67[/tex]

Engine efficiency is defined as

[tex] \eta = \frac{W}{Q_{in}} [/tex]

[tex] Q_{in} = \frac{W}{\eta} [/tex]

[tex] Q_{in} = \frac{48.04 kJ/min}{0.67} [/tex]

[tex] Q_{in} = 71.7 kJ/min [/tex]

Rounding to one decimal place, the rate of heat supplied to the engine is 147.5   kJ/min.

First, we need to calculate the coefficient of performance (COP) of the Carnot refrigerator using the formula:

[tex]\[ \text{COP} = \frac{T_C}{T_H - T_C} \][/tex]

where:

[tex]- \( T_C \)[/tex]  is the absolute temperature of the cold sink (300 K)

[tex]- \( T_H \)[/tex]  is the absolute temperature of the heat source (900 K)

Substituting the given values, we get:

[tex]\[ \text{COP} = \frac{300}{900 - 300} = \frac{300}{600} = 0.5 \][/tex]

Next, we use the COP of the refrigerator to find the rate of heat supplied to the engine:

[tex]\[ \text{Rate of heat supplied to the engine} = \text{COP} \times \text{Rate of heat removed by the refrigerator} \][/tex]

Given that the rate of heat removed by the refrigerator is 295 kJ/min, we can calculate the rate of heat supplied to the engine:

[tex]\[ \text{Rate of heat supplied to the engine} = 0.5 \times 295 = 147.5 \, \text{kW} \][/tex]

Rounding to one decimal place, the rate of heat supplied to the engine is 147.5 kJ/min.

The complete question is here.

A carnot heat engine receives heat at 900K and rejects the waste heat to the enviroment at 300K. The entire work output of the heat engine is used to drive a carnot refrigerator that removes heat from the cooled space at -150C at a rate of 250 kJ/min and rejects it to the same enviroment at 300 K. Determine (a) the rate of heat supplied to the heat engine and (b) the total rate of heat rejection to the enviroment.