What is the mass in both slugs and kilograms of a 1000-lb beam?

Explanation:

The Challenger accident made them aware of the risks. They had been a little naive, since it was never believed that such a thing could happen. This, together with Space Shuttle Columbia disaster revealed the risks of the space shuttle program, which also had very high maintenance costs.

Answer:

d = 2.69 mm

Explanation:

Assuming the cable is rated with a factor of safety of 1.

The stress on the cable is:

σ = P/A

Where

σ = normal stress

P: load

A: cross section

The section area of a circle is:

A = π/4 * d^2

Then:

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]

Replacing:

[tex]d = \sqrt{4*800 / (\pi*\90000)} = 0.106 inches[/tex]

0.106 inches = 2.69 mm

Answer:

Basically there are two principal differences between the convection and conduction heat transfer

Explanation:

The conduction heat transfer is referred to the transfer between two solids due a temperature difference, while for, the convective heat transfer is referred to the transfer between a fluid (liquid or gas) and a solid. Also, they used different coefficients for its calculation.

We can include on the explanation that conduction thermal transfer is due to temperature difference, while convection thermal transfer is due to density difference.

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]

[tex]d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm[/tex]

Answer:

Heat transferred from  the refrigerated space = 95.93 kJ/kg

Work required = 18.45 kJ/kg

Coefficient  of performance = 3.61

Quality at the beginning of the heat  addition cycle = 0.37

Explanation:

From figure  

[tex] Q_H [/tex] is heat rejection process

[tex] Q_L [/tex] is heat transferred from the refrigerated space

[tex] T_H [/tex] is high temperature = 50 °C + 273 = 323 K

[tex] T_L [/tex] is low temperature = -20 °C + 273 = 253 K  

[tex] W_{net} [/tex] is net work of the cycle (the difference between compressor's work and turbine's work)

Coefficient of performance of a Carnot refrigerator [tex] (COP_{ref}) [/tex] is calculated as

[tex] COP_{ref} = \frac{T_L}{T_H - T_L} [/tex]

[tex] COP_{ref} = \frac{253 K}{323 K - 253 K} [/tex]

[tex] COP_{ref} = 3.61 [/tex]

From figure it can be seen that heat rejection is latent heat of vaporisation of R-12 at 50 °C. From table

[tex] Q_H = 122.5 kJ/kg [/tex]

From coefficient of performance definition

[tex] COP_{ref} = \frac{Q_L}{Q_H - Q_L} [/tex]

[tex] Q_H \times COP_{ref} = (COP_{ref} + 1) \times Q_L[/tex]

[tex] Q_L = \frac{Q_H \times COP_{ref}}{(COP_{ref} + 1)} [/tex]

[tex] Q_L = \frac{122.5 kJ/kg \times 3.61}{(3.61 + 1)} [/tex]

[tex] Q_L = 95.93 kJ/kg [/tex]

Energy balance gives

[tex] W_{net} = Q_H - Q_L [/tex]

[tex] W_{net} = 122.5 kJ/kg - 95.93 kJ/kg [/tex]

[tex] W_{net} = 26.57 kJ/kg [/tex]

Vapor quality at the beginning of the heat addition cycle is calculated as (f and g refer to saturated liquid and saturated gas respectively)

[tex] x = \frac{s_1 - s_f}{s_g - s_f} [/tex]

From figure

[tex] s_1 = s_4 = 1.165 kJ/(K kg) [/tex]

Replacing with table values

[tex] x = \frac{1.165 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)}{1.571 kJ/(K \, kg) - 0.9305 kJ/(K \, kg)} [/tex]

[tex] x = 0.37 [/tex]

Quality can be computed by other properties, for example, specific enthalpy. Rearrenging quality equation we get

[tex] h_1 = h_f + x \times (h_g - h_f) [/tex]

[tex] h_1 = 181.6 kJ/kg + 0.37 \times 162.1 kJ/kg [/tex]

[tex] h_1 = 241.58 kJ/kg [/tex]

By energy balance, [tex] W_{t} [/tex] turbine's work is

[tex] W_{t} = |h_1 - h_4| [/tex]

[tex] W_{t} = |241.58 kJ/kg - 249.7 kJ/kg| [/tex]

[tex] W_{t} = 8.12 kJ/kg [/tex]

Finally, [tex] W_{c} [/tex] compressor's work is

[tex] W_{c} = W_{net} + W_{t}[/tex]

[tex] W_{c} = 26.57 kJ/kg + 8.12 kJ/kg[/tex]

[tex] W_{c} = 34.69 kJ/kg [/tex]

Answer:

(a) dynamic viscosity = [tex]1.812\times 10^{-5}Pa-sec[/tex]

(b) kinematic viscosity = [tex]1.4732\times 10^{-5}m^2/sec[/tex]

Explanation:

We have given temperature T = 288.15 K

Density [tex]d=1.23kg/m^3[/tex]

According to Sutherland's Formula  dynamic viscosity is given by

[tex]{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}[/tex], here

μ = dynamic viscosity in (Pa·s) at input temperature T,

[tex]\mu _0[/tex]= reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

[tex]T_0[/tex] = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

[tex]\mu _0=4\pi \times 10^{-7}[/tex]

[tex]T_0[/tex] = 291.15

[tex]\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s[/tex]when T = 288.15 K

For kinematic viscosity :

[tex]\nu = \frac {\mu} {\rho}[/tex]

[tex]kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec[/tex]

Answer:

[tex]\sigma _1=10.33MPa[/tex]

[tex]\sigma _2=5.16MPa[/tex]

Explanation:

Given that

Diameter(d)=62 mm

Thickness(t)= 300 μm=0.3 mm

Internal pressure(P)=100 KPa

Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .

The hoop stress

[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]

Longitudinal stress

[tex]\sigma _l=\dfrac{Pd}{4t}[/tex]

Now by putting the values

[tex]\sigma _h=\dfrac{Pd}{2t}[/tex]

[tex]\sigma _h=\dfrac{100\times 62}{2\times 0.3}[/tex]

[tex]\sigma _h=10.33MPa[/tex]

[tex]\sigma _l=5.16MPa[/tex]

So the principle stress are

[tex]\sigma _1=10.33MPa[/tex]

[tex]\sigma _2=5.16MPa[/tex]

Answer:

Molar mass of the gas will be 60.65 gram/mole

Explanation:

We have given mass of gas = 3.7 gram

Gas contains [tex]3.7\times 10^{22}[/tex]

We know that any gas contain [tex]6.022\times 10^{23}[/tex] molecules in 1 mole

So number of moles [tex]=\frac{3.7\times 10^{22}}{6.022\times 10^{23}}=0.061[/tex]

We know that number of moles [tex]n=\frac{mass\ in\ gram}{molar\ mass}[/tex]

So [tex]0.061=\frac{3.7}{molar\ mass}[/tex]

Molar mass = 60.65 gram/mole

Answer:

heat produced by 150 watt bulb is 510 Btu

Explanation:

Given data:

Power of bulb is 150 W

Duration for light is 20 hr

We know that 1 Watt = 3.4 Btu

hence using above conversion value we can calculate the power in Btu unit

so for [tex]150 W\ bulb  =  150\times 3.4 = 510 Btu[/tex]

therefore, 150 W bulb consist of 510 Btu power for 20 hr

output heat by 150 watt bulb is 510 Btu

Answer:

0.2 m/s

Explanation:

The velocity of a point on the edge of a disk rotating disk can be calculated as:

[tex]v=\omega*r[/tex]

Where [tex]\omega[/tex] is the angular velocity and r the radius of the disk. This leads to:

[tex]v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s[/tex]

Answer:

0.2 m/s

Explanation:

Step 1: identify the given parameters

angular velocity, ω = 0.5 rad/s

radius of the disk, r = 0.4m

Note: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.

Step 2: calculate the velocity of the disk at the outer edge in m/s

Velocity = Angular velocity (rad/s) X radius (m)

Velocity = 0.5 rad/s X 0.4 m

Velocity = 0.2 m/s

Answer with Explanation:

Hydraulic diameter is a term analogous to the diameter of the circular sectional pipe but used for the cases when the cross sectional shape of the pipe is non circular.

It serves as an equivalent diameter that is used to calculate the Reynolds number for the flow.

The hydraulic diameter is 4 times the hydraulic radius of any section.

For a rectangular duct as shown in the attached figure

[tex]R_{h}=\frac{Wetted_{Area}}{Wetted_{perimeter}}\\\\R_h=\frac{d\times b}{2(d+b)}\\\\\therefore D_{h}=4\times R_{h}=4\times \frac{db}{2(d+b)}=\frac{2db}{(d+b)} [/tex]

Where

[tex]D_{h}[/tex] is the hydraulic diameter of the duct with depth 'd' and width 'b'

- XxItzAdiXx

Answer:

Engine not possible

Explanation:

source temperature T1 = 300 K

sink temperature T2= 283 K

therefore, carnot efficiency of the heat engine

η= 1- T_2/T_1

[tex]\eta= 1-\frac{T_1}{T_2}[/tex]

[tex]\eta= 1-\frac{283}{300}[/tex]

= 0.0566

= 5.66%

claims of work produce W = 100 kW,  mass flow rate = 20 kg/s

[tex]Q=mc_p(T_1-T_2)[/tex]

[tex]Q=20\times4.18(300-283)[/tex]

= 1421.2 kW

now [tex]\eta= \frac{W}Q}[/tex]

now [tex]\eta= \frac{100}{1421.2}[/tex]

=7%

clearly, efficiency is greater than carnot efficiency hence the engine is not possible.

Answer:

428°F

Explanation:

The equation to convert degrees Celsius to degrees fahrenheit is

°F (degrees fahrenheit) = (9/5 * °C (degrees celsius) ) + 32

°F = (9/5 * 220 °C (degrees celsius)) +32 = 428 °F (degrees fahrenheit)

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice [tex]C_p=2.03\ KJ/kg.K[/tex]

Latent heat for ice H=336 KJ/kg

Specific heat for ice [tex]C_p=4.187\ KJ/kg.K[/tex]

We know that sensible heat given as

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

Heat for -15F to 32 F:

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

[tex]Q=9\times 2.03(32+15) KJ[/tex]

Q=858.69 KJ

Heat for 32 Fto 200 F:

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

[tex]Q=9\times 4.187(200-32) KJ[/tex]

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So   7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

Answer:

Specific gravity is defined as the ratio of the Densities of the two substances.

Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]

Explanation:

Specific gravity is defined as the ratio of the Densities of the two substances.

For the standardization, the density ration of densities is calculated with respect to the density of water i.e the denominator is the density of water.

Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water}}[/tex]

Also,

At STP density of water is 1000 Kg/m³

Therefore the relation between the specific gravity and density is,

The Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{Density of water at STP}}[/tex]

or

Specific gravity = [tex]\frac{\textup{Density of substance}}{\textup{1000}}[/tex]

Answer:

final temperature is 424.8 K

so correct option is e 424.8 K

Explanation:

given data

pressure p1 = 1 bar

pressure p2 = 5 bar

index k = 1.3

temperature t1 = 20°C = 293 k

to find out

final temperature  t2

solution

we have given compression is reversible and has an index k

so we can say temperature is

[tex]\frac{t2}{t1}= [\frac{p2}{p1}]^{\frac{k-1}{k} }[/tex]  ...........1

put here all these value and we get t2

[tex]\frac{t2}{293}= [\frac{5}{1}]^{\frac{1.3-1}{1.3} }[/tex]

t2 = 424.8

final temperature is 424.8 K

so correct option is e

Answer:

1st payment = $9350

4th payment = $9650

last 10th payment = $10250

Explanation:

given data

loan = $85000

time = 10 year

each payment larger = $100

time value of money = 10% per year

to find out

first, the fourth, and the last payment

solution

we find here actual value at end of 1st year that is

actual value = loan amount + 10% of loan

actual value N = 85000 + (10% ×85000 )

actual value N = $93500

so

1st payment is = [tex]\frac{actual value}{total time}[/tex]

1st payment is = [tex]\frac{93500}{10}[/tex]

1st payment = $9350

and

4th payment = 1st payment + 3× each payment larger

4th payment = 9350 + 3×100

4th payment = $9650

and

last 10th payment = 4th payment + 6× each payment larger

last 10th payment = 9650 + 6× 100

last 10th payment = $10250

Answer:

The correct answer is option 'c': 13.8 kNm

Explanation:

We know that moment of a force equals

[tex]Moment=Force\times Arm[/tex]

The hydro static force is given by [tex]Force=Pressure\times Area[/tex]

We know that the hydrostatic pressure on a rectangular surface in vertical position is given by [tex]Pressure=\rho \times g\times h_{c.g}[/tex]

For the given rectangular surface we have [tex]h_{c.g}=\frac{h}{2}=\frac{1.5}{2}=0.75m[/tex]

Thus applying the values we get force as

[tex]Force=1000\times 9.81\times 0.75\times 1.5\times 2.5=27.59kN[/tex]

This pressure will act at center of pressure of the rectangular plate whose co-ordinates is given by h/3 from base

Thus applying the calculated values we get

[tex]Moment=27.59\times \frac{1.5}{3}=13.8kN.m[/tex]

Answer:

The heat transferred to water equals 1600 kJ

Explanation:

By the conservation of energy we have

All the kinetic energy of the moving vehicle is converted into thermal energy

We know that kinetic energy of a object of mass 'm' moving with a speed of 'v' is given by

[tex]K.E=\frac{1}{2}mv^{2}[/tex]

Thus

[tex]K.E_{car}=\frac{1}{2}\times 2000\times 40^{2}=1600\times 10^{3}Joules[/tex]

Thus the heat transferred to water equals [tex]1600kJ[/tex]

Answer:

pressure at depth 3 m is 33.255 kPa

Explanation:

given data

temperature = 25°C

depth = 3 m

to find out

pressure

solution

we know pressure formula that is

pressure = ρ g h   ................1

and we know specific gravity of Ethylene glycol is 1.13

so

1.13 = [tex]\frac{\rho (e)}{\rho (water)}[/tex]

ρ (Ethylene glyco) = 1130 Kg/m³

so

pressure will be by equation 1

pressure = 1130 × 9.81 × 3

pressure = 33255.9 Pa

so pressure at depth 3 m is 33.255 kPa

Answer:

119.35 mm

Explanation:

Given:

Inside diameter, d = 100 mm

Tensile load, P = 400 kN

Stress = 120 MPa

let the outside diameter be 'D'

Now,

Stress is given as:

stress = Load × Area

also,

Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-d^2)[/tex]

or

Area of hollow pipe = [tex]\frac{\pi}{4}(D^2-100^2)[/tex]

thus,

400 × 10³ N = 120 × [tex]\frac{\pi}{4}(D^2-100^2)[/tex]

or

D² = tex]\frac{400\times10^3+30\pi\times10^4}{30\pi}[/tex]

or

D = 119.35 mm

Answer:

D =119.35 mm

Explanation:

given data:

inside diameter = 100 mm

load = 400 kN

stress = 120MPa

we know that load is given as

[tex]P = \sigma A[/tex]  

where:

P=400kN = 400000N

[tex]\sigma = 120MPa[/tex]

[tex]A =(\frac{1}{4} \pi D^2 - \frac{1}{4}\pi (100^2)[/tex]

[tex]A=\frac{1}{4} \pi (D^2 - 10000)[/tex]

putting all value in the above equation to get the required diameter value

[tex]400 =  120*\frac{1}{4} \pi (D^2 - 10000)[/tex]

solving for

D =119.35 mm

Answer:

  Diameter pitch is the parallel thread of the imaginary cylinder of the diameter that basically intersect in the surface of thread. It is also known as effective diameter. The diameter pitch is basically used to determine the threated parts.

The pitch diameter is the essential component for determining the basic compatibility in the externally and internally thread parts. The diameter pitch is the sensitive measuring tool for determine the specific measurements.

Answer:

The correct option is 'e': f = 0.05.

Explanation:

The head loss as given by Darcy Weisbach Equation is as

[tex]h_{l}=\frac{flv^{2}}{2gD}[/tex]

where

[tex]h_{l}[/tex] is head loss in the pipe

'f' is the friction factor

'l' is the length of pile

'v' is the velocity of flow in pipe

'D' is diameter of pipe

From equation of contuinity we have [tex]v=\frac{Q}{A}[/tex]

Thus using this in darcy's equation we get

[tex]h_{l}=\frac{flQ^{2}}{2gDA}[/tex]

where

'Q' is discharge in the pipe

'A' is area of the pipe [tex]A=\frac{\piD^2}{4}[/tex]

Applying the given values we get

[tex]h_{l}=\frac{8flQ^{2}}{\pi ^{2}gD^{5}}[/tex]

Solving for 'f' we get

[tex]f=\frac{0.53\times \pi ^{2}\times 9.81\times 0.6^{5}}{1000\times 0.1^{2}\times 8}\\\\f=0.05[/tex]

Answer:

Torque at input shaft will be 176.8695 N-m

Explanation:

We have given input power [tex]P_{IN}=42.6KW=42.6\times 10^3W[/tex]

Angular speed = 2300 rpm

For converting rpm to rad/sec we have multiply with [tex]\frac{2\pi }{60}[/tex]

So [tex]2300rpm=\frac{2300\times 2\pi }{60}=240.855rad/sec[/tex]

We have to find torque

We know that  power is given by [tex]P=\tau \omega[/tex], here [tex]\tau[/tex] is torque and [tex]\omega[/tex] is angular speed

So [tex]42.6\times 10^3=\tau \times 240.855[/tex]

[tex]\tau =176.8695N-m[/tex]

So torque at input shaft will be 176.8695 N-m

Answer:

1) g=31.87ft/s^2

2)m=120

W=119.64lbf

Explanation:

first part

the weight of a body with mass is calculated by the following equation

W=mg

we convert 120lb to slug

m=120lbx1slug/32.147lb=3.733slug

solving for g

g=W/m

g=119/3.733=31.87ft/s^2

second part

the mass is the same

m=120lb=3.733slug

Weight

W=3.733slug*32.05ft/s^2=119.64lbf

Answer:

T=5.3° C

Explanation:

Given that

Power input to the pump = 2 KW

Building loses heat rate = 0.5 KW/C

So rate of heat transfer = 0.5(273+30-T)

rate of heat transfer = 0.5(303-T)

T=Ambient temperature

Building temperature = 30° C

We know that ,heat pump is used to heat the building.

COP of pump = 0.5 COP of Carnot heat pump

[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{273+30}{303-T}[/tex]

[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{303}{303-T}[/tex]

[tex]COP\ of\ pump=\dfrac{303-T}{Power}[/tex]    

[tex]COP\ of\ pump=0.5\times \dfrac{303-T}{2}[/tex]     -----1

And also

[tex]COP\ of\ pump=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex]   ----2

So from now equation 1 and 2

[tex]\dfrac{303-T}{4}=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex]

So T= 278.38 K=5.3° C

T=5.3° C

Ambient temperature =5.3° C.

Answer:

The cassette player will operate at normal power for 3333.33 seconds.

Explanation:

The first step is to identify the operating voltage and the operating current with the purpose to determine the power that the cassette player consumes. Remember that power equals the voltage multiplied by the current [tex]P=V\times I[/tex], where P is in Watts (W), V is in Volts (V) and I is in Amperes (A).

The problem says that four batteries are connected in series to provide a voltage of 6V to the player circuit. So, the operating voltage is 6V, [tex]V=6V[/tex]

Then, the problem says that cassette player draws a constant current of 10mA from the battery pack. So, the operating current is 10mA or 0.01A, [tex]I=0.01A[/tex]

From previous, it could be said that the cassette player consumes 0.06W.

[tex]P=V\times I=(6V)\times (0.01A)=0.06W[/tex]

Now, the idea is to calculate how long the cassette will operate at 0.06W.

The problem says that the battery pack stores [tex]200\, W\cdot s[/tex], it means that the battery pack could provide 200W in a second; after a second, the battery pack will not work properly. However, the battery pack just have to provide 0.06W so, it will last more time. For calculating that, you must divide the total power per time the cell can provide by the power that the cassette player needs to work.

[tex]t=\frac{200W\cdot s}{0.06W}=3333.33s[/tex]

As you can see, the W units are canceled and second remains.

Thus, the cassette player will operate at normal power for 3333.33 seconds.

Answer:

5.9*10^-3 Pa*s

Explanation:

The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.

τ = μ * du/dr

u(r) = 0.4/r - 1000*r

The derivative is:

du/dr = -1/r^2 - 1000

On the radius of the inner cylinder this would be

u'(0.02) = -1/0.02^2 - 1000 = -3500

So:

τ = -3500 * μ

We don't care about the sign

τ = 3500 * μ

That is a tangential force per unit of area.

The area of the inner cylinder is:

A = h * π * D

And the torque is

T = F * r

T = τ * A * D/2

T = τ * h * π/2 * D^2

T = 3500 * μ * h * π/2 * D^2

Then:

μ = T / (3500 * h * π/2 * D^2)

μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s

Answer:

Total work: -5.25 kJ

Total Heat: 52 kJ

Explanation:

V0 = 0.15

P0 = 350 kPa

t0 = 150 C = 423 K

P1 = 105 kPa (isentropical transformation)

Δh1-2 = 52 kJ (at constant pressure)

Ideal gas equation:

P * V = m * R * T

m = (R * T) / (P * V)

R is 0.287 kJ/kg for air

m = (0.287 * 423) / (350 * 0.15) = 2.25 kg

The specifiv volume is

v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg

Now we calculate the parameters at point 1

T1/T0 = (P1/P0)^((k-1)/k)

k for air is 1.4

T1 = T0 * (P1/P0)^((k-1)/k)

T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K

The ideal gas equation:

P0 * v0 / T0 = P1 * v1 / T1

v1 = P0 * v0 * T1 / (T0 * P1)

v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg

V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3

The work of this transformation is:

L1 = P1*V1 - P0*V0

L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg

Q1 = 0 because it is an isentropic process.

Then the second transformation. It is at constant pressure.

P2 = P1 = 105 kPa

The enthalpy is raised in 52 kJ

Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh

And the idal gas equation is:

P1 * v1 / T1 = P2 * v2 / T2

T2 = T1 * P2 * v2 / (P1 * v1)

Replacing:

Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2

Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)

v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)

The Cv of air is 0.7 kJ/kg

v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg

V2 = 2.25 * 0.2 = 0.45 m^3

T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K

The heat exchanged is Q = Δh = 52 kJ

The work is:

L2 = P2*V2 - P1*V1

L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ

The total work is

L = L1 + L2

L = -14.7 + 9.45 = -5.25 kJ

Answer:

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

Explanation:

Drag force:

 Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.

Drag force given as

[tex]F_D=\dfrac{1}{2}\rho\ A\ V^2[/tex]

So we can say that drag force depends on following properties

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).