A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving through the middle of a wind farm when an idea snapped into her head. She thought to herself, what if she mounted a small windmill generator on the roof of her Prius and wired it into the battery? She thought, with a little head wind to get her started she could just drive off under wind power! The faster she drove, the more power she would make, and the faster she could go… What’s wrong with her idea? a. As a side question, Priuses make use of regenerative braking. Why is it called regenerative braking? Why don’t cars advertise regenerative accelerators?

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

To calculate the total lb mol air/h needed to heat the hydrocarbon oil, we can use the principle of heat transfer. First, calculate the heat transfer between the hot air and the oil using the formula Q = mcΔT. Finally, divide the heat transfer by the heat capacity of air to find the total lb mol air/h needed.

To calculate the total lb mol air/h needed to heat the hydrocarbon oil, we can use the principle of heat transfer. First, we need to calculate the heat transfer between the hot air and the oil using the formula Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the temperature difference. We know that the heat exchanger transfers 2200 lbm/h of oil from 100°F to 150°F, so the total heat transfer is Q = 2200 lbm/h * (150°F - 100°F) * 0.45 btu/lbm • °F. Next, we can calculate the lb mol of air needed by dividing the heat transfer by the heat capacity of air, which is 0.24 btu/lbm • °F. Therefore, the total lb mol air/h needed is Q / (0.24 btu/lbm • °F).

Answer:

12024000 lb*ft

Explanation:

The total work will be the sum of the energy consumed by friction plus the kinetic energy the car attained.

L = Ek + Lf

Lf = Ff * d

Ek = 1/2 * m * v^2

Therefore:

L = Ff * d + 1/2 * m * v^2

L = 80 * 300 + 1/2 * 15000 * 40^2 = 12024000 lb*ft

THe total work done on the car is of 12024000 lb*ft

Explanation:

Step1

Float glass is the process of glass manufacturing on the flat surface of metal like tin. In this method molten glass is allowed to float on the surface of metal.

Step2

Float glass gives the uniform and flat surface of glass product. The thickness of the glass produced is uniform throughout. This process of glass making is very cheap and has negligible distortion. Flat glass process is called the float process because of producing high quality flat surface.

Answer:

[tex]p = 15260.643 \ lbf/ft^2[/tex]

Explanation:

person weight is 175 lbm

weight of stake board 4lbm

size of stakeboard = 2ft by 0.8 ft

area of stakeboard is [tex]2*0.8 ft^2 = 1.6 ft^2[/tex]

gauge pressure is given as

[tex]p =\frac{ w_p+w_s}{A} g[/tex]

where is g is acceleration due to gravity =  32.17 ft/sec^2

puttng all value to get pressure value

   [tex]= \frac{175 +4}{1.6}* 32.17[/tex]

[tex]p = 15260.643 \ lbf/ft^2[/tex]

Answer:

The main component of fixed wing aircraft

1.Wings

2.Fuselages

3.Landing gear

4.Stabilizers

5.Flight control surface

Cantilever wing:

  A cantilever wing is directly attached to the fuselages and do not have any external support.

Semi cantilever wing:

A cantilever wing does not directly attached to the fuselages and  have any external support.The external support may be one two.

Answer:

a)Resistance to flow

Explanation:

Viscosity of fluid:

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

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

Viscosity are of two type

1.Dynamic viscosity

2.Kinematic viscosity

Answer:

3.0772 m

Explanation:

Given:

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

Length of elongation, δL = 3.5 mm = 0.0035 m

Applied load, P = 25 KN = 25000 N

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

Now,

we have the relation

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

Now,

Where, A is the area of cross-section

A =  [tex]\frac{\pi}{4}d^2[/tex]

or

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

or

A = 0.000314 m²

L is the length of the member

on substituting the respective values, we get

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

or

L = 3.0772 m

Answer:

FALSE.

Explanation:

the correct answer is FALSE.

Projection is the process of representing the 3 D object on the flat surface.

there are four ways of representing the projection

1) First angle projection  

2) second angle projection

3) third angle projection

4) fourth angle projection.

Generally, people prefer First and third angle projection because there is no overlapping of the projection take place.

In USA people uses the third angle of projection.

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.

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

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

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

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

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

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

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

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

Calculate:

Ts = 366 K = 93°C

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

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

(b) With radiation: Heat transfer equation:

q = hA(Ts - T∞) + εσA(Ts4 - Tsur4)

Given: T∞ = 293 K ε = 0.8

σ = 5.67x10-8 W/m2∙K4

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

Assume: Tsur = T∞ = 293 K

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

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

Plug this into equation:

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

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

Solve for Ts: Ts = 359.8 K = 86.67°C

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

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

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

Part A: Neglecting Radiation Exchange

Step 1

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

Given:

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

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

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

The absorbed heat:

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

The heat loss by convection:

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

At steady state:

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

Step 2

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

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

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

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

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

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

Part B: Considering Radiation Exchange

Step 1

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

Given:

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

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

The total heat loss (convection + radiation):

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

For convection:

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

For radiation:

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

At steady state:

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

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

Step 2

To simplify:

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

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

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

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

Through iteration or numerical methods, we find:

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

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

Answer:

parallel ; Viscosity

Explanation:

Laminar is flow is the flow of fluid layer in which motion of the liquid particle is very slow and there is no intermixing of the layer of fluid takes place.

There no perpendicular movement of particles takes place, no eddies are formed or swirl of fluid.

so, the first option will be Parallel. Fluid particles flow parallel to each other in laminar flow.

This path of flow is by the action of Viscosity of the fluid.

Explanation:

Mid-wing configuration places wings exactly at midline of airplane which means at half of height of fuselage. These airplanes are well balanced and also they have a large control surface area.It is the best option aerodynamically as these planes are streamlined much more and also has low interference drag as compared to the high and the low wing configurations.

The mid-wing has almost neutral roll stability that is further good from prespective of the combat as well as the aerobatic aircraft as mid-wing allows for performance of the rapid roll maneuvers with the minimum yaw coupling.

Explanation:

Step1

Factor of safety is the constant factor which is taken for the safe design of any product. It is the ratio of maximum stress induced in the material or the failure stress from tensile test to the allowable stress.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the type of material, environment condition, strength needed, type of component etc.

The expression for factor of safety is given as follows:

[tex]FOS=\frac{\sigma_{f}}{\sigma_{a}}[/tex]

Here, [tex]\sigma_{f}[/tex] is fracture stress and [tex]\sigma_{a}[/tex] is allowable stress.

Answer:

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

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

Explanation:

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

w = m * g

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

nr = -m * g * cos(θ)

Being θ the tilt angle of the bed.

The static friction force will be:

ffs = - m * g * cos(θ) * fJK

The dynamic friction force will be:

ffd = - m * g * cos(θ) * fik

These forces would produce accelerations

affs = -g * cos(θ) * fJK

affd = -g * cos(θ) * fik

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

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

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

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

affs = -3.92 * cos(θ)^2

affd = -2.94 * cos(θ)^2

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

Assuming the bed has a lenght L.

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

Movement under constant acceleration:

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

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

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

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

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

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

Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

D = L/10

D = 50000/10 = 5000 lb

To maintain steady speed the thrust must equal the drag, so

T = D = 5000 lb

Answer:

First angle projection                      

1.Object is between observer and plane of projection.

2.Projection of object take on first quadrant.

3.Plane of projection is assume non transparent.

Third angle projection:

1.Plane of projection is between observer and object.

2.Projection of object take on third quadrant.

3.Plane of projection is assume transparent.

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:

The large percentage of steel includes less than 0.35% carbon

Explanation:

Carbon is perhaps the most important business alloy of steel. Raising carbon material boosts strength and hardness and enhances toughness. However, carbon often increases brittleness.

The large percentage of steel includes less than 0.35% carbon. Any steel with a carbon content range of 0.35% to 1.86% can be considered as hardened.

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

[tex] {v}^{2} - {u}^{2} = 2as \\ {20}^{2} - {10}^{2} = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a = \frac{300}{200 } = \frac{3}{2} \: m {s}^{ - 2} [/tex]

Now, we substitute v, u, and a in the formula

[tex]v = u + at \\ 20 = 10 + \frac{3}{2} t \\ \frac{3}{2} t = 10 \\ 3t = 20 \\ t = \frac{20}{3} = 6.67 \: seconds[/tex]

Please mark Brainliest if this helps!

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:

a) 100%

b) 300%

c) 301 %

Explanation:

The first wafer has a diameter of 150 mm.

The second wafer has a diameter of 300 mm.

The second wafer has an increase in diameter respect of the first of:

((300 / 150)  - 1) * 100 = 100%

The first wafers has a processable area of:

A1 = π/4 * D1^2

The scond wafer has a processable area of:

A2 = π/4 * D2^2

The seconf wafer has a increase in area respect of the first of:

(A2/A1 * - 1) * 100

((π/4 * D2^2) / (π/4 * D1^2) - 1) * 100

((D2^2) / (D1^2) - 1) * 100

((300^2) / (150^2) - 1) * 100 = 300%

The area of a chip is

Ac = Lc^2

So the chips that can be made from the first wafer are:

C1 = A1 / Ac

C1 = (π/4 * D1^2) / Lc^2

C1 = (π/4 * 150^2) / 10^2 = 176.7

Rounded down to 176

The chips that can be made from the second wafer are:

C2 = A2 / Ac

C2 = (π/4 * D2^2) / Lc^2

C2 = (π/4 * 300^2) / 10^2 = 706.8

Rounded down to 706

The second wafer has an increase of chips that can be made from it respect of the first wafer of:

(C2 / C1 - 1) * 100

(706 / 176 - 1) *100 = 301%

Answer:

80 kW; 11 kW; 8 kW; 0.6

Explanation:

Part 1

Isentropic turbine efficiency:  

[tex]\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = \frac{W_{real}}{W_s} [/tex]

[tex]W_{real} = \eta_t*W_s [/tex]

[tex]W_{real} = 0.8*100 kW [/tex]

[tex]W_{real} = 80 kW [/tex]

Part 2

Coefficient of performance COP is defined by:

[tex]COP = \frac{Q_{out}}{W} [/tex]

[tex]Q_{out} = W*COP[/tex]

[tex]Q_{out} = 5 kW*2.2[/tex]

[tex]Q_{out} = 11 kW[/tex]

Part 3

(a)

Energy balance for a refrigeration cycle gives:

[tex]Q_{in} + W = Q_{out} [/tex]

[tex]3 kW + 5 kW = Q_{out} [/tex]

[tex]8 kW = Q_{out} [/tex]

(b)

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

[tex]COP = \frac{3 kW}{5 kW} [/tex]

[tex]COP = 0.6 [/tex]

Answer:

79 kW.

Explanation:

The equation for enthalpy is:

H2 = H1 + Q - L

Enthalpy is defined as:

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

This is specific volume.

The gas state equation is:

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

The specific gas constant for air is:

287 K/(kg*K)

Then:

T1 = 60 + 273 = 333 K

T2 = 200 + 273 = 473 K

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

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

The Cv for air is:

Cv = 720 J/(kg*K)

So the enthalpies are:

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

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

Ang the heat is:

Q = 34 kW

Then:

H2 = H1 + Q - L

381 = 268 + 34 - L

L = 268 + 34 - 381 = -79 kW

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

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

Answer:

Determine A) The Volume Of The Tank (ft^3) Later A Pump Is Used To Extract ... A rigid, sealed cylinder initially contains 100 lbm of water at 70 degrees F and atmospheric pressure. ... Later a pump is used to extract 10 lbm of water from the cylinder. The water remaining in the cylinder eventually reaches thermal equilibrium ...

Answer:

The pressure will be measured to a precision of 0.073 psi.

Explanation:

Since the relation between the measurement of pressure and height is given by

[tex]dP=\rho gh[/tex]

For water we have

[tex]\rho _{water}=62.4lb/ft^3[/tex]

[tex]g=32.17ft/s^2[/tex]

Applying the given values we get

[tex]dP=62.4\times 32.17\times \frac{1}{16\times 12}=5.433lb/ft^2\\\\=\frac{10.455}{144}lb/in^2=0.073psi[/tex]

Answer:

Time taken by the [tex]1\mu m[/tex] diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = [tex]1\times 10^{- 6} m[/tex]

Diameter of the smaller droplet, R' = [tex]0.5\times 10^{- 6} m[/tex]

Now,

Volume of the droplet, V = [tex]\frac{4}{3}\pi R^{3}[/tex]

Volume of the smaller droplet, V' = [tex]\frac{4}{3}\pi R'^{3}[/tex]

Volume of the droplet ∝ Time taken for complete evaporation

Thus

[tex]\frac{V}{V'} = \frac{t}{t'}[/tex]

where

t' = taken taken by smaller droplet

[tex]\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}[/tex]

[tex]\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}[/tex]

t' = [tex]60\times 10^{- 9} s = 60 ns[/tex]

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

Answer:

1) A=282.6 mm

2)[tex]a_{max}=60.35\ m/s^2[/tex]

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

[tex]V=\omega \sqrt{A^2-x^2}[/tex]

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex]    ------3

[tex]2.5=\omega \sqrt{A^2-0.225^2}[/tex]   --------4

Now by dividing equation 3 by 4

[tex]\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}[/tex]

[tex]1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}[/tex]

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex]

[tex]3.5=\omega \sqrt{0.2826^2-0.15^2}[/tex]

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

[tex]a_{max}=\omega ^2A[/tex]

[tex]a_{max}=14.61 ^2\times 0.2826\ m/s^2[/tex]

[tex]a_{max}=60.35\ m/s^2[/tex]

Time period T

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]T=\dfrac{2\pi}{14.609}[/tex]

T=0.42 sec

Answer:

[tex]\Delta T = 28.57°C[/tex]

Explanation:

given data:

temperature at outlet of turbine = 260°C

Flow rate = 2 kg/min = 0.033 kg/s

output power  = 1 kW

specific heat =  1.05 kJ/kg.K

We know that power generated by turbine is equal to change in enthalpy

[tex]W = h_2 - h_1[/tex]

   [tex]= mCp(\Delta T)[/tex]

[tex]1*10^3 = 0.0333*1.05*10^3 * \Delta T[/tex]

[tex]\Delta T = \frac{10^3}{0.033*1.05*10^3}[/tex]

[tex]\Delta T = 28.57°C[/tex]

Answer:

If the shearing stress is linearly related to shearing strain then the fluid is called as Newtonian fluid.

Explanation:

The other types of fluids are:

1) Non-Newtonian fluids which are further classified as

a) Thixotropic Fluid: Viscosity decreases with shearing stress over time.

b) Rehopactic Fluid: Viscosity increases with shearing stress over time.

c) Dilatant Fluid: Apparent viscosity increases with increase in stress.

d) Pseudoplastic: Apparent viscosity decreases with increase in stress.

Answer:

If the shearing stress is linearly related to shearing strain then the fluid is called as Newtonian fluid.

Explanation: