A 600 MW power plant has an efficiency of 36 percent with 15
percent of the waste heat being released to the atmosphere as stack
heat and the other 85 percent taken away in the cooling water.
Instead of drawing water from a river, heating it, and returning it
to the river, this plant uses an evaporative cooling tower wherein
heat is released to the atmoshphere as cooling water is
vaporized.
At what rate must 15C makeup water be provided from the river to
offset the water lost in the cooling tower?

Answer:

power = 20943.95 watts

power = 28.086 horsepower

Explanation:

given data

torque = 200 N

speed = 1000 rpm

to find out

What is the mechanical power in watts and horse power

solution

we know that mechanical power formula that is

power = torque × speed   ...................1

here we have given both torque and speed

we know speed = 1000 rpm = [tex]\frac{2* \pi *1000}{60}[/tex] = 104.66 rad/s

so put here value in equation 1

power = 200 × 104.719

power = 20943.95 watts

and

power = [tex]\frac{20943.95}{745.7}[/tex]

power = 28.086 horsepower

Answer:

Part 1) Power required for motor = 20944 watts.

Part 2) Power required for motor in Horsepower equals = 28.075H.P

Explanation:

Power is defined as the rate of consumption of energy. For rotational motion power is calculated as

[tex]Power=Torque\times \omega[/tex]

where,

[tex]\omega [/tex] is the angular speed of the motor.

Since the rotational speed of the motor is given as 1000 rpm, the angular speed is calculated as

[tex]\omega =\frac{N}{60}\times 2\pi[/tex]

where,

'N' is the speed in rpm

Applying the given values we get

[tex]\omega =\frac{1000}{60}\times 2\pi=104.72rad/sec[/tex]

hence the power equals

[tex]Power=200\times 104.72=20944Watts[/tex]

Now since we know that 1 Horse power equals 746 Watts hence 20944 Watts equals

[tex]Power_{H.P}=\frac{20944}{746}=28.075H.P[/tex]

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = [tex]\frac{\pi }{4} d^2[/tex]    ..................1

here d is diameter

put the value in equation 1

area = [tex]\frac{\pi }{4} 0.15^2[/tex]

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = [tex]\frac{Fnet}{g}[/tex]

mass  = [tex]\frac{350}{9.8}[/tex]

mass = 35.7 kg

so 35.7 kg lid we put

Answer:

1) The initial velocity must be 22 feet/sec.

2) The time required to change the velocity is 8 seconds.

Explanation:

This problem can be solved using third equation of kinematics

According to the third equation of kinematics we have

[tex]v^{2}=u^{2}+2as[/tex]

where

'v' is the final velocity of object

'u' is the initial velocity of object

'a' is acceleration of the object

's' is the distance in which this velocity change is brought

Since the final velocity is 45 mph converting this to feet per seconds we get

45 mph =66 feet/sec

Applying Values in the above equation and solving for 'u' we get

[tex]66^{2}=u^{2}+2\times 5.5\times 352\\\\u^{2}=484\\\\\therefore u=22feet/sec[/tex]

2)

The time required to attain this velocity can be found using first equation of kinematics as

[tex]v=u+at[/tex]

where

't' is the time in which the velocity change occurs

'v', 'u', 'a' have the usual meaning

Thus applying given values we get

[tex]66=22+5.5\times t\\\\\therefore t=\frac{66-22}{5.5}=8seconds[/tex]

Applying the given values we get

Answer:

Mass flow in kg /sec will be 1.511 kg/sec

Explanation:

We have given mass flow = 200 lbm/min

We have to convert lbm/min into kg /sec

We know that 1 lbm = 0.4535 kg

So for converting lbm to kg we have multiply with 0.4535

So 200 lbm = 200×0.4535 =90.7 kg

Now we know that 1 minute = 60 sec

So 200 lbm/min [tex]=\frac{200\times 0.4535kg}{60sec}=1.511kg/sec[/tex]

So mass flow in kg /sec will be 1.511 kg/sec

Answer:

The correct answer is option 'a': 0.046 meters.

Explanation:

We know that the exit velocity of a jet of water is given by Torricelli's law as

[tex]v=\sqrt{2gh}[/tex]

where

'v' is velocity of head

'g' is acceleration due to gravity

'h' is the head under which the jet falls

Now since the jet rises to a head of 90 meters above ground thus from conservation of energy principle it must have fallen through a head of 90 meters.

Applying the values in above equation we get the exit velocity as

[tex]v=\sqrt{2\times 9.81\times 90}=42.02m/s[/tex]

now we know the relation between discharge and velocity as dictated by contuinity equation is

[tex]Q=V\times Area[/tex]

Applying values in the above equation and solving for area we get

[tex]Area=\frac{Q}{v}=\frac{2}{42.02}=0.0476m^{2}[/tex]

The circular area is related to diameter as

[tex]Area=\frac{\pi D^{2}}{4}\\\\\therefore D=\sqrt{\frac{4\cdot A}{\pi }}=\sqrt{\frac{4\times 0.0476}{\pi }}=0.246m[/tex]

Thus the diameter of the nozzle is 0.246 meters

Answer:

Horsepower unit was first time used by James Watt in 1782. The  story refers that James Watt uses worked with pony to charge coal from the mines. According to that story,  he have the need of a unit to measure the force from one of this animals. He founds that they can move 22.000 lbs per minute, so he (arbitrarily) increase this measure in 50% been the unit Horsepower in 33.000 lb/feet per minute.

Explanation:

This measure unit can measure "work" or "force". In the SI correspond to move up 75 Kg, to 1 meter high,  in one second.

1 HP = (330 lb) x (100 feet)/1min = 33000 lb x feet/min

Why would anyone wish to express power in HP?

Is a practical unit, because reduce the amount of digits in a specific value. Also, his use is very spread specially in mechanical applications.

1 HP= 746 W (0,746 kW).

The mass of the air in the tire is 1040.192 grams.

We must know the concept of the Ideal Gas Equation to solve this question.

The Ideal Gas Equation shows the empirical relationship between the Volume, Pressure, Temperature, and the number of moles in a given substance.

Using Ideal Gas Equation:

PV = nRT

From the given information:

[tex]\mathbf{200 kPa \times 800 \ L = n \times 8.31446 L.kPa.K^{-1}.mol^{-1} \times 296 \ K }[/tex]

[tex]\mathbf{n = \dfrac{200 kPa \times 800 \ L}{8.31446 \ L.kPa.K^{-1}.mol^{-1} \times 296 \ K}}[/tex]

[tex]\mathbf{n =65.012 \ mol}[/tex]

From the relation;
Number of moles = mass/molar mass.

Mass of air (O₂) = number of moles × molar mass

Mass of air (O₂) = 65.012 mol × 16 g/mol

Mass of air (O₂) = 1040.192 grams

Learn more about ideal gas equation here:

https://brainly.com/question/25290815

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure   to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = [tex]-\frac{dp}{dV/V}[/tex]  ................1

And [tex]-\frac{dV}{V} = \frac{d\rho}{\rho}[/tex]   ............2

here ρ is density

and we know ρ  for 20°C = 998 kg/m³

and ρ  for 80°C = 972 kg/m³

so from equation 2 put all value

[tex]-\frac{dV}{V} = \frac{d\rho}{\rho}[/tex]

[tex]-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}[/tex]

dV = 0.0130 m³

so now  % change in volume will be

dV % = [tex]-\frac{dV}{V}[/tex]  × 100

dV % = [tex]-\frac{0.0130}{500*10^{-3} }[/tex]  × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = [tex]\frac{\pi }{4} *d^2*l(i)[/tex]    ................3

final volume v2 = [tex]\frac{\pi }{4} *d^2*l(f)[/tex]    ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 =  [tex]\frac{\pi }{4} *d^2*(l(f)-l(i))[/tex]

dV = [tex]\frac{\pi }{4} *d^2*dl[/tex]

put here all value

0.0130 = [tex]\frac{\pi }{4} *2^2*dl[/tex]

dl = 0.004138 m

so water level rise is 4.138 mm

Answer:

Specific weight of the pop, [tex]w_{s} = 8619.45 N/m^{3}[/tex]

Density of the pop, [tex]\rho_{p} = 8790.76 kg/m^{3}[/tex]

[tex]g_{s} = 8.79076[/tex]

[tex]w_{w} = 9782.36 N/m^{3}[/tex]

Given:

Volume of pop, V = 360 mL = 0.36 L = [tex]0.36\times 10^{-3} m^{3}[/tex]

Mass of a can of pop , m = 0.369 kg

Weight of an empty can, W = 0.153 N

Solution:

Now, weight of a full can pop, W

W' = mg = [tex]0.369\times 9.8 = 3.616 N[/tex]

Now weight of the pop in can is given by:

w = W' - W = 3.616 - 0.513 = 3.103 N

Now,

The specific weight of the pop, [tex]w_{s} = \frac{weight of pop}{volume of pop}[/tex]

[tex]w_{s} = \frac{3.103}{0.36\times 10^{- 3}} = 8619.45 N/m^{3}[/tex]

Now, density of the pop:

[tex]\rho_{p} = \frac{w_{s}}{g}[/tex]

[tex]\rho_{p} = \frac{86149.45}{9.8} = 8790.76 kg/m^{3}[/tex]

Now,

Specific gravity, [tex]g_{s} = \frac{\rho_{p}}{density of water, \rho_{w}}[/tex]

where

[tex]g_{s} = \frac{8790.76}{1000} = 8.79076[/tex]

Now, for water at [tex]20^{\circ}c[/tex]:

Specific density of water = [tex]998.2 kg/m^{3}[/tex]

Specific gravity of water = [tex]0.998 kg/m^{3}[/tex]

Specific weight of water at [tex]20^{\circ}c[/tex]:

[tex]w_{w} = \rho_{20^{\circ}}\times g = 998.2\times 9.8 = 9782.36 N/m^{3}[/tex]

Answer:

Total change in energy = 31 KJ.

Explanation:

Mass m=25 kg

Height h = 1 m

Initial velocity = 0

Final velocity = 50 m/s

Energy at initial condition

[tex]E_1=mgh+\dfrac{1}{2}mv^2[/tex]

[tex]E_1=25\times 10\times 1+0[/tex]

[tex]E_1=250\ J[/tex]

Energy at final condition

[tex]E_2=0+\dfrac{1}{2}\times 25\times 50^2[/tex]

[tex]E_2=31250\ J[/tex]

So the change in energy = 31250 -250 J

The total change in energy = 31000 J

Answer:

Change in  energy will be 31.25 KJ

Explanation:

We have given mass of the piston = 25 kg

Initial velocity u = 0 m/sec

Final velocity v = 50 m/sec

Kinetic energy is given by [tex]KE=\frac{1}{2}mv^2[/tex]

We have to find the change in energy

So change in energy = final KE - initial KE

[tex]\Delta E=\frac{1}{2}m(v^2-u^2)[/tex]

[tex]\Delta E=\frac{1}{2}25\times (50^2-0^2)=31250j=31.25KJ[/tex]

So change in  energy will be 31.25 KJ

Answer:

The answer is "b" 3.5W and 37.5w.

Explanation:

Domestic use normally refers to refrigeration equipment that is not subject to high thermal loads, for example, refrigerators should only extract heat from food, and domestic air conditioners should only extract heat from small spaces and a number reduced people.

Answer:

Final velocity will be 36.11 m/sec

Time required by ball to heat the ground = 1.297 sec

Explanation:

We have given height = 50 feet

Initial velocity u = 18 ft/sec

Acceleration due to gravity [tex]g=32.8ft/sec^2[/tex]

We have to find final velocity v

From third equation of motion

[tex]v^2=u^2+2gh[/tex]

So [tex]v^2=18^2+2\times 32.8\times 50=3543.82[/tex]

v = 59.53m/sec

From first equation of motion we know that

v= u+gt

So [tex]59.53=18+32.8\times t[/tex]

t = 1.297 sec

Answer:

b)False

Explanation:

Higher molecular weight will increase the mechanical properties because if molecular weight is more then it will require more energy break the molecule .It means that mechanical properties is also improve.

Higher molecular weight will also improve the resistance to corrosion and reduce the chances of material from oxidation.

Higher molecular weight will also increase the viscosity of material and reduce the fluidity.

Answer:

250 kpsi

Explanation:

Given:

Width of the specimen, w = 0.45 in

Thickness of the specimen, t = 0.20 in

length of the specimen between supports, L = 2.5 in

Failure load, F = 1200 lb

Now,

The transverse rupture strength [tex]\sigma_t=\frac{1.5FL}{wt^2}[/tex]

on substituting the respective values, we get

[tex]\sigma_t=\frac{1.5\times1200\times2.5}{0.45\times0.2^2}[/tex]

or

[tex]\sigma_t=250,000\ psi\ =\ 250 kpsi[/tex]

Explanation:

Drag is the force which is generated parallel and also in opposition to direction of the travel for the object which is moving through the fluid.

Lift is the force which is generated perpendicular to direction of the travel for the object moving through the fluid.

Both of the two forces which are the lift and the drag force act through center of the pressure of object.

Use Pascal's law and the hydrostatic pressure formula to solve for the height of the oil column, considering the specific gravity of the oil and the mass applied on the piston.

The question requires the application of principles from fluid mechanics to calculate the height h2 to which an oil rises in a tube, using the specific gravity of the oil and the mass of a load on a piston. Assuming the system is in equilibrium and by using Pascal's law, the pressure applied by the mass onto the piston is transmitted equally throughout the fluid. Therefore, the pressure exerted by the mass on the larger piston is balanced by the hydrostatic pressure of the oil column of height h2 in the smaller tube.

To solve for h2, we can use the formula for hydrostatic pressure P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column. Given the specific gravity S of the oil, we can find its density (ρ) by multiplying S by the density of water (1000 kg/m³). Then, by setting the hydrostatic pressure of the oil column equal to the pressure exerted by the mass and solving for h2, we find the height of the oil in the tube.

Answer:

0 to 4294967295

Explanation:

Unsigned integers have only positive numbers and zero. Their range goes from zero to (2^n)-1.

In the case of a 32-bit unsigned integer this would be.

(2^32) - 1 = 4294967296 - 1 = 4294967295

So the range goes from 0 to 4294967295 or form zero to about 4.3 billion.

The minus one term is because the zero takes one of the values.

Answer:

The weight is 4.492 lb

The density is [tex]0.5614 lbm/ft^{3}[/tex]

Solution:

As per the question:

Volume of spacecraft component, [tex]V_{s} = 8ft^{3}[/tex]

Mass of the component of spacecraft, [tex]m_{s} = 25 lb[/tex]

Acceleration of gravity at a point on Earth, [tex]g_{E} = 31.0 ft/s^{2}[/tex]

Acceleration of gravity on Moon, [tex]g_{M} = 5.57 ft/s^{2}[/tex]

Now,

The weight of the component, [tex]w_{c} = m_{c}\times \frac{g_{M}}{g_{E}}[/tex]

[tex]w_{c} = 25\times \frac{5.57}{31.0}[/tex]

[tex]w_{c} = 4.492 lb[/tex]

Now,

Average density, [tex]\rho_{avg}[/tex]

[tex]\rho_{avg} = \farc{w_{s}}{V_{s}}[/tex]

[tex]\rho_{avg} = \farc{4.492}{8} = 0.5614 lbm/ft^{3}[/tex]

Answer:

density=3.125pounds/ft^3

weight=4.35lbf

Explanation:

Density is a property of matter that indicates how much mass a body has in a given volume.

It is given by the following equation.

ρ=m/v   (1)

where

ρ=density

m=mass

v=volume

The weight on the other hand is the force which the earth (or the moon) attracts to a body with mass, this force is given by the following equation

W=mg (2)

W=weight

m=mass

g=gravity

to solve this problem we have to calculate the mass of the component using the ecuation number 2

W=mg

m=w/g

w=25lbf=804.35pound .ft/s^2

g=32.2ft/s^2

m=804.35/32.2=25 pounds

density

ρ=m/v   (1)

ρ=25pounds/8ft^3

ρ=3.125pounds/ft^3 =density

weight in the moon

W=mg

W=(25pounds)(5.57ft/s^2.)=139.25pound .ft/s^2=4.35lbf

Answer:

18.7 mm

Explanation:

Fourier's law for heat conduction on a plate is:

q = -k/t * ΔT

Where

q: heat conducted per unit of time and surface

k: thermal conductivity

t: thickness of the plate

ΔT: temperature difference

Rearranging:

t = -k/q * ΔT

t = -1.4/(-1200) * 16 = 0.0187 m = 18.7 mm (q is negative because it is heat leaving the plate)

The maximum thickness of the concrete slab is 18.7 mm.

Answer with Explanation:

There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:

1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

Answer:

Relative humidity 48%.

Dew point 74°F

humidity ratio 118 g of moisture/pound of dry air

enthalpy 41,8 BTU per pound of dry air

Explanation:

You can get this information from a Psychrometric chart for water, like the one attached.

You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.

To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F

for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air

For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air

Answer:40.603 MPa

Explanation:

Given

Car weighs (m)4000 lbs [tex]\approx 1814.37 kg[/tex]

diameter of cable(d)[tex]=0.75 in.\approx 19.05 mm[/tex]

Hill angle [tex]\theta =15^{\circ}[/tex]

Now

tension in cable will bear the weight of car acting parallel to rope which is [tex]mgsin\theta [/tex]

Thus

[tex]T=mgsin\theta [/tex]

[tex]T=1814.37\times 9.81\times sin(15)=11,574.45 N[/tex]

T=11.57 kN

thus stress([tex]\sigma [/tex])=[tex]\frac{T}{A}[/tex]

where A=cross section of wire[tex]=285.059 mm^2[/tex]

[tex]\sigma =\frac{11574.45}{285.059}=40.603 MPa[/tex]

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

[tex]E(t)=E_o[1-a\frac{T}{T_m}][/tex]

where,

E(t) is the modulus of elasticity at any temperature 'T'

[tex]E_o[/tex] is the modulus of elasticity at absolute zero.

[tex]T_{m}[/tex] is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

Answer:

Star

Explanation:

When all hosts are connected to a central hub it is known as a star topology.

The advantages of the star network is that it is very easy to add new devices, a failure in a host will not cause problems on the rest of the network, it is appropriate for large networks and works well under heavy loads,

The disadvantages are that a failure of the central hub brings the whole networks down and it is expensive due to the amount of cables needed.

Answer:

The average force F exherted by the nail over the hammer is 178.4 lbf.

Explanation:

The force F exherted by the nail over the hammer is defined as:

F = |I|/Δt

Where I and Δt are the magnitude of the impact and the period of time respectively. We know that the impact can be calculated as the difference in momentum:

I = ΔP = Pf - Pi

Where Pf and Pi are the momentum after and before the impact. Recalling for the definition for momentum:

P = m.v  

Where m and v are the mass and the velocity of the body respectively. Notice that final hummer's momentum is zero due to the hammers de-acelerate to zero velocity. Then the momentum variation will be expressed as:

ΔP =  - Pi = -m.vi

The initial velocity is given as 50 mph and we will expressed in ft/s:

vi = 50 mph * 1.47 ft/s/mph = 73.3 ft/s

By multiplyng by the mass of 1.8 lbs, we obtain the impulse I:

|I|= |ΔP|= |-m.vi| = 1.8 lb  * 73.3 ft/s = 132 lb.ft/s

Dividing the impulse by a duration of 0.023 seconds, we finally find the force F:

F =  132 lb.ft/s / 0.023 s = 5740 lb.ft/s^2  

Expressing in lbf:

F = 5740 lb.ft/s^2 * 0.031 lbf/lb.ft/s^2  =  178.4 lbf

Answer:

r=0.228m

Explanation:

The equation that defines the states of a gas according to its thermodynamic properties is given by the general equation of ideal gases

PV=nRT

where

P=pressure =5bar=500.000Pa

V=volume

n=moles=10

R = universal constant for ideal gases = 8.31J / (K.mol)

T=temperature=80F=299.8K

solvig For V

V=(nRT)/P

[tex]V=(\frac{(10)(8.31)(299.8)}{500000} )\\V=0.0498m^3[/tex]

we know that the volume of a sphere is

[tex]V=\frac{4\pi r^3}{3} \\[/tex]

solving for r

[tex]r=\sqrt[3]{ \frac{3 V}{4\pi } }[/tex]

solving

[tex]r=\sqrt[3]{ \frac{3 (0.049)}{4\pi } }\\r=0.228m[/tex]

Answer:

M =2.33 kg

Explanation:

given data:

mass of piston - 2kg

diameter of piston is 10 cm

height of water 30 cm

atmospheric pressure 101 kPa

water temperature = 50°C

Density of water at 50 degree celcius is 988kg/m^3

volume of cylinder is  [tex]V = A \times h[/tex]

                                       [tex]= \pi r^2 \times h[/tex]

                                       [tex]= \pi 0.05^2\times 0.3[/tex]

mass of available in the given container is

[tex]M = V\times d[/tex]

  [tex] = volume \times density[/tex]

[tex]= \pi 0.05^2\times 0.3 \times 988[/tex]

M =2.33 kg

Answer:(a)[tex]45,300.24 lb/in/^2[/tex]

(b)0.255

Explanation:

Given

Strain hardening exponent(n)=0.22

Strength coefficient(k)[tex]=54000 lb/in/^2[/tex]

and we know

[tex]\sigma =k\left ( \epsilon \right )^n[/tex]

where

[tex]sigma =true\ stress[/tex]

[tex]\epsilon =true\ strain[/tex]

(a)True strain=0.45

[tex]\sigma =54000\times 0.45^{0.22}[/tex]

[tex]\sigma =45,300.24 lb/in^2[/tex]

(b)true stress[tex]=40,000 lb/in^2[/tex]

[tex]40000=54000\times \epsilon ^{0.22}[/tex]

[tex]\epsilon ^{0.22}=0.7407[/tex]

[tex]\epsilon =0.7407^{4.5454}=0.255[/tex]

Answer:

184.6 BTU

Explanation:

The thermal efficiency for a Carnot cycle follows this equation:

η = 1 - T2/T1

Where

η: thermal efficiency

T1: temperature of the heat source

T2: temperature of the heat sink

These temperatures must be in absolute scale:

1000 F = 1460 R

50 F = 510 R

Then

η = 1 - 510/1460 = 0.65

We also know that for any heat engine:

η = L / Q1

Where

L: useful work

Q1: heat taken from the source

Rearranging:

Q1 = L / η

Q1 = 120 / 0.65 = 184.6 BTU

In this exercise we will deal with the Carnot cycle and calculate how much heat was initially placed, so we have that:

The heat input to the engine was 184.6 BTU

The theoretical Carnot cycle is formed by two isothermal transformations and two adiabatic transformations. One of the isothermal transformations is used for the temperature of the hot source, where the expansion process takes place, and the other for the cold source, where the compression process takes place.

The thermal efficiency for a Carnot cycle follows this equation:

[tex]\eta= 1 - T_2/T_1[/tex]

Where:

These temperatures must be in absolute scale:

[tex]\eta = 1 - 510/1460 = 0.65[/tex]

Rearranging:

[tex]Q_1 = L / \eta\\Q_1 = 120 / 0.65 = 184.6 BTU[/tex]

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Answer:

Therefore, height of instrument is 324.577 ft

Therefore, elevation of point x is 330 m

Explanation:

Given that

Plus sight on BM = 12.03 ft

Minus sight is = 5.43 ft

Elevation = 312.547 ft

Height of instrument is H.I

H.I = elevation on bench mark + plus sight

    =  312.547 + 12.03 = 324.577 ft

Therefore, height of instrument is 324.577 ft

Elevation at point x is = H.I - minus sight

                                    = 324.577 - (- 5.43)

                                     = 330.00 m

Therefore, elevation of point x is 330 m