At the beginning of the compression process of an air-standard Otto cycle, p1 = 1 bar and T1 = 300 K. The compression ratio is 8.5 and the heat addition per unit mass of air is 1400 kJ/kg. Determine the maximum temperature of the cycle in Kelvin (input a number ONLY). Do not assume specific heats are constant. There is a ±5% tolerance.

Answer:

Rt=70.00 ohm

Explanation:

ohm's theory tells us that the connected resistors in series add up directly

Rt=R1+R2+R3+R4.......

Therefore, for this case, the only thing we should do is add the resistance directly

Rt=R1+R2

Rt=40.00 ohm+30.00 Ohm.

Rt=70.00 ohm

the total resistance of this configuration is 70.00 ohm

The expressions squared or cubed with appropriate metric prefixes are: (a) 184.9 Mg^2, (b) 4 x 10^-18 kg^2, and (c) 12.167 km^3. Metric prefixes are used for clarity in representing large quantities.

The question deals with the operation of squaring and cubing given quantities and expressing them with the appropriate metric prefixes. Let's evaluate each expression step by step:

It is important to use the proper metric prefixes when
expressing large numbers to ensure clarity and avoid confusion.

Answer:

Network is a collection of computers

Explanation:

A network is a gathering of at least two gadgets that can convey. Practically speaking, a system is included various distinctive PC frameworks associated by physical or potentially remote associations.  

The scale can run from a solitary PC sharing out fundamental peripherals to huge server farms situated far and wide, to the Internet itself. Despite extension, all systems permit PCs or potentially people to share data and assets.

Answer:

convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.

Explanation:

convection because it make the surrounding air warm. hence make you feel warm without getting physically connect to it.

Convection is the transition of heat between distinct temperature fields by moving fluid (liquid or gas). Dry air is less thick than moist air and in the presence of a temperature difference, convection currents can form.

Answer:

Horizontal acceleration will be [tex]2.07m/sec^2[/tex]

Explanation:

We have given radius of track = 180 m

velocity = 30 m/sec

Total acceleration [tex]a_t=7.07m/sec^2[/tex]

The centripetal;l acceleration which is always normal to the velocity also known as normal acceleration is given by [tex]a_n=\frac{v^2}{r}=\frac{30^2}{180}=5m/sec^2[/tex]

So the tangent acceleration will be [tex]a_t=7.07-5=2.07m/sec^2[/tex]

Answer:

a₁= 1.98 m/s²   : magnitud of the normal acceleration

a₂=0.75  m/s²  : magnitud of the tangential acceleration

Explanation:

Formulas for uniformly accelerated circular motion

a₁=ω²*r : normal acceleration     Formula (1)

a₂=α*r:    normal acceleration     Formula (2)

ωf²=ω₀²+2*α*θ                             Formula (3)

ω : angular velocity

α : angular acceleration

r  : radius

ωf= final angular velocity

ω₀ : initial angular velocity

θ :   angular position theta

r  : radius

Data

r =0.4 m

ω₀= 1 rad/s

α=0.3 *θ , θ= 2π

α=0.3 *2π= 0,6π rad/s²

Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.

We calculate ωf with formula 3:

ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687

ωf=[tex]\sqrt{24.687}[/tex] =4.97 rad/s

a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²    

a₂=α*r = 0,6π * 0.4 = 0.75  m/s²  

Answer:

Thermal resistance:

  Thermal resistance is the property which oppose the the heat transfer.It is the property for measurement of heat flow.

As we know that heat transfer take place from the high temperature to low temperature.Heat transfer Q given as

[tex]Q=\dfrac{\Delta T}{R_{th}}[/tex]

Where Δt is the temperature difference

[tex]{R_{th}}[/tex] is the thermal resistance.

Connection:

1. Series connection

 [tex]R_{total}={R_1}+{R_2}+{R_1}----{R_n}[/tex]

2. Parallel connection

[tex]\dfrac{1}{R_{total}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}-----\dfrac{1}{R_n}[/tex]

Answer:

a)38.65%

b)1221KJ/kg

Explanation:

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

first we calculate the enthalpies in all states

h1=3264kJ/Kg

h2=2043kJ/Kg

h2=h3=105.4kJ/Kg

a)we use the ecuation for efficiency

Efficiency = (h1-h2) / (h1-h4)

Efficiency = (3264-2043) / (3264-105.4)

=0.3865=38.65%

b)we use the ecuation for Wout

Wout = m (h1-h2)

for work ratio=

w = (h1-h2)

w=(3264-2043)=1221KJ/kg

Answer:

a)Humidity ratio =0.013  kg/kg

b) Enthalpy=60.34 KJ/kg

c) Density = 1.16 [tex]Kg/m^3[/tex]

d) Dew point temperature = 64.9°F

e) Wet bulb temperature = 69.53°F

Explanation:

Given that

Dry bulb temperature = 80°F

relative humidity = 60%

Given air is at sea level it means that total pressure will 1 atm.

So P= 1 atm.

As we know that psychrometric charts are always drawn at constant pressure.

Now from charts

We know that dry bulb temperature line is vertical and relative humidity line is curve and at that point these two line will meet ,will us property all property.

a)Humidity ratio =0.013  kg/kg

b) Enthalpy=60.34 KJ/kg

c) Density = 1.16 [tex]Kg/m^3[/tex]

d) Dew point temperature = 64.9°F

e) Wet bulb temperature = 69.53°F

Answer:

The fraction of isotope 1 is 50.7 % and fraction  fraction of isotope 1 is 49.2 %.

Explanation:

Given that

Weight of isotope 1 = 78.918 amu

Weight of isotope 2 = 80.916 amu

Average atomic weight= 79.903 amu

Lets take fraction of isotope 1 is x then fraction of isotope 2 will be 1-x.

The total weight will be summation of these two isotopes

79.903 = 78.918 x + 80.916(1-x)

By solving above equation

80.916 - 79.903 = (80.916-78.918) x

   x=0.507

So the fraction of isotope 1 is 50.7 % and fraction  fraction of isotope 1 is 49.2 %.

Answer:

Height of oil is 7.06 meters.

Explanation:

The situation is shown in the attached figure

The pressure at the bottom of the tank as calculated by equation of static pressure distribution is given by

[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(\rho _{water}h_{water}+\rho_{oil}h_{oil})=P_{bottom}-P_{top}[/tex]

Applying the given values we get

[tex]P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(1000\times 2.5+800\times h_{oil})=80\times 10^{3}\\\\\therefore 800\times h_{oil}=\frac{80\times 10^{3}}{9.81}-2500\\\\\therefore 800\times h_{oil}=5654.94\\\\\therefore h_{oil}=7.06m[/tex]

Answer:n=0.973

Explanation:

Given

When True strain[tex]\left ( \epsilon _T_1\right )=0.171[/tex]

at [tex]\sigma _1=263.8 MPa[/tex]

When True stress[tex]\left ( \sigma _2\right )[/tex]=346.2 MPa

true strain [tex]\left ( \epsilon _T_2\right )[/tex]=0.226

We know

[tex]\sigma =k\epsilon ^n [/tex]

where [tex]\sigma [/tex]=True stress

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

n=strain hardening exponent

k=constant

Substituting value

[tex]263.8=k\left ( 0.171\right )^n------1[/tex]

[tex]346.2=k\left ( 0.226\right )^n-----2[/tex]

Divide 1 & 2 to get

[tex]\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n[/tex]

[tex]1.312=\left ( 1.3216\right )^n[/tex]

Taking Log both side

[tex]ln\left ( 1.312\right )=nln\left ( 1.3216\right )[/tex]

n=0.973

Answer:

Fouling :

  When rust or undesired material deposit in the surface of heat exchanger,is called fouling of heat exchanger.

Effect of heat exchanger are as follows"

1.It decreases the heat transfer.

2.It increases the thermal resistance.

3.It decreases the overall heat transfer coefficient.

4.It leads to increase in pressure drop.

5.It increases the possibility of corrosion.

Answer:

52.2538 psia

Explanation:

The absolute pressure at depth of 27 inches can be calculated by:

Pressure = Local pressure + Gauge pressure

Also,

[tex]P_{gauge}=\rho\times g\times h[/tex]

Where,

[tex]\rho[/tex] is the density of glycerin ([tex]\rho=74.9\ lbm/ft^3[/tex])

g is the gravitational acceleration = 32.1741 ft/s²

h = 27 in

Also, 1 in = 1/12 ft

So,

h = 27 / 12 ft = 2.25 ft

So,

[tex]P_{gauge}=74.9\times 32.1741\times 2.25\ lbf/ft^2=5422.1402\ lbf/ft^2[/tex]

Also,

1 ft = 12 inch

1 ft² = 144 in²

So,

[tex]P_{gauge}=5422.1402\ lbf/ft^2=\frac {5422.1402\ lbf}{144\ in^2}=37.6538\ lbf/in^2=37.6538\ psia[/tex]

Local pressure = 14.6 psia

So,

Absolute pressure = 14.6 psia + 37.6538 psia=52.2538 psia

Answer:

Cut off ratio=2.38

Explanation:

Given that

[tex]T_1=300K[/tex]

[tex]P_1=100KPa[/tex]

[tex]P_2=P_3=7200KPa[/tex]

[tex]T_3=2250K[/tex]

Lets take [tex]T_1[/tex] is the temperature at the end of compression process

For air γ=1.4

[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\dfrac{\gamma -1}{\gamma}}[/tex]

[tex]\dfrac{T_2}{300}=\left(\dfrac{7200}{100}\right)^{\dfrac{1.4-1}{1.4}}[/tex]

[tex]T_2=1070K[/tex]

At constant pressure

[tex]\dfrac{T_3}{T_2}=\dfrac{V_3}{V_2}[/tex]

[tex]\dfrac{2550}{1070}=\dfrac{V_3}{V_2}[/tex]

[tex]\dfrac{V_3}{V_2}=2.83[/tex]

So cut off ratio

[tex]cut\ off\ ratio =\dfrac{V_3}{V_2}[/tex]

Cut off ratio=2.38

Answer:

T=340. 47 K

Explanation:

Given that

Volume of tank =0.5 [tex]m^3[/tex]

Pressure P=120 KPa

Temperature T=300 K

Added heat ,Q= 20 KJ

Given that air is treated as ideal gas and specific heat is constant.

Here tank is rigid so we can say that it is constant volume system.

We know that specific heat at constant volume for air

[tex]C_v=0.71\ \frac{KJ}{kg.K}[/tex]

We know that for ideal gas

P V = m R T

For air R=0.287 KJ/kg.K

P V = m R T

120 x 0.5 = m x 0.287 x 300

m=0.696 kg

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

Lets take final temperature of air is T

Now by putting the values

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

[tex]20=0.696\times 0.71\times (T-300)[/tex]

T=340. 47 K

So the final temperature of air will be 340.47 K.

Answer:

Explanation:

Given

Initial Thickness=45 mm

Final thickness=20 mm

Roll diameter=600 mm

Radius(R)=300 mm

coefficient of friction between rolls and strip ([tex]\nu [/tex])=0.15

maximum draft[tex](d_{max})=\nu ^2R[/tex]

[tex]=0.15^2\times 300=6.75 mm[/tex]

Minimum no of passes[tex]=\frac{45-20}{6.75}=3.70\approx 4[/tex]

(b)draft per each pass

[tex]d=\frac{Initial\ Thickness-Final\ Thickness}{min.\ no.\ of\ passes}[/tex]

[tex]d=\frac{45-20}{4}=6.25 mm[/tex]

Answer:

The acceleration is [tex]2220.00m/s^{2}[/tex]

Explanation:

We know that the acceleration is given by

[tex]a=v\frac{dv}{ds}........................(i)[/tex]

The velocity as a function of position is given by [tex]v=14\cdot s^{7/6}[/tex]

Thus the acceleration as obtained from equation 'i' becomes

[tex]a=14\cdot s^{7/6}\times \frac{d}{ds}(14\cdot s^{7/6})\\\\a=14\cdot s^{7/6}\times \frac{7}{6}\times 14\cdot s^{1/6}\\\\a=\frac{686}{3}\cdot s^{8/6}[/tex]

Hence acceleration at s = 5.5 equals

[tex]a(5.5)=\frac{686}{3}\times (5.5)^{8/6}\\\\\therefore a=2220.00mm/s^{2}[/tex]

Answer:

factor of safety for A36 structural steel is 0.82

Explanation:

given data:

side of column = 3.5 inches

wall thickness = 0.225 inches

load P = 22 kip

Length od column = 9 ft

we know that critical stress is given as

[tex]\sigma_{cr} = \frac{\pi^2 E}{(l/r)^2}[/tex]

where

r is radius of gyration[tex] = \sqrt{\fra{I}{A}}[/tex]

Here I is moment od inertia [tex]= \frac{b_1^2}{12} - \frac{b_2^2}{12}[/tex]

[tex] I == \frac{3.5^2}{12} - \frac{3.05^2}{12} = 5.294 in^4[/tex]

For hollow steel area is given as [tex]A = b_1^2 -b_2^2[/tex]

[tex]A = 3.5^2 -3.05^2 = 2.948 in^2[/tex]

critical stress [tex]\sigma_{cr} = = \frac{\pi^2\times 29\times 10^6}{((9\times12)/(1.34))^2}[/tex]

[tex]\sigma_{cr} =  44061.56 lbs/inc^2[/tex]

considering Structural steel A36

so A36[tex] \sigma_y = 36ksi[/tex]

factor of safety  [tex]= \frac{yield\ stress}{critical\  stress}[/tex]

factor of safety =[tex]\frac{36\times10^3}{44061.56} = 0.82[/tex]

factor of safety for A36 structural steel is 0.82

The student's query involves calculating the factor of safety for a hollow steel column, but without complete material strength data, the calculation cannot be precisely done.

The student's question pertains to determining the factor of safety for a hollow steel column with specified dimensions, material properties (steel E), and loading conditions. To calculate the factor of safety for the column, we would need to compare the column's actual stress under load to its maximum allowable stress. However, due to insufficient data about material yield strength or ultimate strength and considering the provided information does not match the examples given, this calculation cannot be accurately completed without further specifics on the steel's properties.

Answer:

Open system

Explanation:

In refrigerator there is a interaction between refrigerator and environment.

In refrigerator heat moves from the system to the outer environment means there is transfer of heat from one system to environment.

We know that whenever there is transfer of energy or mass from system then it is known as pen system

So refrigerator is a open system

Answer:

Air cooling.

Explanation:

Low power motors are supposed to be low cost, and they dissipate little heat. Therefore a low cost solution is ideal.

Air cooling can be achieved with very little cost. Fins can be added to a cast motor casing and a fan can be places on the shaft to use a small amount of the motor power to move air to cool it.

Measurement error is the difference between observed and calculated values, including both random and systematic errors. Adjusting parameters such as K can help fit experimental data to models, with potential errors addressed through improved methods and investigation of discrepancies.

The difference between the observed and calculated values is known as measurement error or observational error. This error can be classified into two types: random error, which occurs naturally and varies in an unpredictable manner, and systematic error, which is consistent and typically results from a flaw or limitation in the equipment or the experimental design. To identify the sources of these errors, one can compare the experimental data with calculated models and adjust parameters, such as the constant K, to find the best fit.

If the experimental values do not match the accepted values, sources of error should be investigated and procedures modified to improve accuracy. For example, in a physics experiment, one could compare the experimental acceleration to the standard acceleration due to gravity (9.8 m/s²) and identify factors contributing to any discrepancy. Common sources of error might include environmental factors, instrument calibration issues, or procedural mistakes.

Answer:

Explanation:

CO, carbon monoxide is a toxic gas. It casues asphixiation on people and animals by interfering with hemoglobin, not allowing blood to transport oxygen to the cells in the body.

The normal emissions resulting from the combustion  of fussil fuels are CO2 (carbon dioxide) and H2O (water). Carbon monoxide is formed by an incomplete combustion of fossil fuels or carbon containing fuels in general, this not only produces toxic gas, but also is an inefficient combustion that wastes energy.

Answer:

304.42 cc

Explanation:

When the aluminum expands the markings will be further apart. If the 300 cc mark was at a distance l0 of the origin at 20 C, at 100 C it will be

l = l0 * (1 + a * (t - t0))

l = l0 * (1 + 23*10^-6 * (100 - 20))

l = l0 * 1.0018

The volume of water would have expanded by

V = V0 (1 + a * (t - t0))

V = 300 (1 + 2.07*10^-4 * (100 - 20))

V = 304.968 cc

Since the markings expanded they would measure

304.968/1.0018 = 304.42 cc

Answer:

final pressure is 6847.41 kPa

Explanation:

given data:

[tex]P_1 =  10,000 kPa[/tex]

[tex]T_1 =520\ degree\ celcius = 793 K[/tex]

[tex]T_2  = 270 degree celcius = 543 K[/tex]

as we can see all temperature are more than 100 degree, it mean this condition is refered to superheated stream

for ischoric process we know that

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]

[tex]\frac{10*10^6}{793} = \frac{P_2}{543}[/tex]

[tex]P_2 = 6.84741*10^6 Pa[/tex]

final pressure is 6847.41 kPa

Answer:

9%

Explanation:

An ideal gas is one that has its molecules widely dispersed and does not interact with each other, studies have shown that air behaves like an ideal gas, so the state change equation for ideal gases can be applied.

P1V1T2 = P2V2T1

where 1 corresponds to state 1 = 320kPa

and 2 is state 2 = 349kPa.

Given that the volume remains constant the equation is:

P1T2=P2T1

SOLVING for T2/T1

[tex]\frac{T2}{T1} =\frac{P2}{P1} =\frac{349}{320} =1.09\\\\[/tex]

The equation to calculate the percentage increase is as follows

%ΔT=[tex]\frac{(T2-T1)100}{T1} =(\frac{T2}{T1} -1)100=(1.09-1)100=(0.09)100=9%[/tex]

Answer:

[tex]V = 0.208 m^3[/tex]

saturated pressure  = 0.91 bar 0r 91 kPa

Explanation:

Given data:

Mass of liquid water 200 kg

Steel tank temperature is 97 degree celcius = 97°C

At T = 97 degree celcius,  saturated pressure  = 0.91 bar 0r 91 kPa

And also for saturated liquid

Specific volume [tex]\nu[/tex] is 0.00104 m^3/kg

Volume of tank is given as

[tex]V = \nu * m[/tex]

[tex]V = 0.00104 * 200[/tex]

[tex]V = 0.208 m^3[/tex]

Answer:

The mass flow rate of jet =69.44 kg/s

Explanation:

Given that

velocity of jet v= 60 m/s

Power  P=250 KW

As we know that force offered by water F

[tex]F=\rho\ A \ v^2[/tex]

Power P= F.v

So now power given as

[tex]P=\rho\ A \ v^3[/tex]

We know that mass flow rate = ρAv

[tex]P=mass\ flow\ rate\ \times v^2[/tex]

250 x 1000 = mass flow rate x 3600

mass flow rate = 69.44 kg/s

So the mass flow rate of jet =69.44 kg/s

Answer:

maximum tensile load Pmax is 11.91 ksi

Explanation:

given data

area = 0.65 in²

design factor of safety = 3

to find out

what is the maximum tensile load Pmax

solution

we know here area is 0.65 in² and FOS = 3

so by steel table for rod of annealed AISI 1018 steel table we know σy = 55 ksi

so

we use here design factor formula that is

[tex]\frac{ \sigma y}{FOS}  = \frac{Pmax}{area}[/tex]  .............1

put here all these value we get Pmax in equation 1

[tex]\frac{55}{3}  = \frac{Pmax}{0.65}[/tex]

Pmax = 11.91 ksi

so maximum tensile load Pmax is 11.91 ksi

Answer:

this is a typical stress-strain curve for a ductile material

Explanation:

A: proportional limit

B:Elastic limit

C:upper yield point

D:lower yield point

E:ultimate strength

F:rupture strength