A design that either partially or wholly integrates the bodywork

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

Air is drawn from the atmosphere into a turbomachine. At the exit, conditions are 500 kPa (gage) and 130oC. The exit speed is 10

finlep [7]

Answer:

P=- 88.41 KW

Negative sign indicates that power is given to the system.

Explanation:

Given that

P₂=500 KPa

T₂=130°C

V₂=100 m/s

mass flow rate ,m= 0.8 kg/s

Lets take inlet condition for air

T₁=25°C

P₁=100 KPa

V₁=0 m/s

We know that

Heat capacity for air Cp=1.005 KJ/kg.k

We know that for air change in enthalpy only depends only on temperature

Now from first law for open system

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}+W$

$1.005\times 298+\dfrac{0^2}{2000}=1.005\times 403+\dfrac{100^2}{2000}+W$

$W=1.005\times 298-1.005\times 403-\dfrac{100^2}{2000}$

W=-110.52 KJ/kg

Shaft power P = m .W

P = -110.52 x 0.8

P=- 88.41 KW

Negative sign indicates that power is given to the system.

3 0

4 years ago

What is the the force available at the roadway surface to perform work?

scoray [572]

Answer:

Tractive force or traction

Explanation:

The main purpose of the tractive forces is to improve the ability to <u>transform the engine's energy into the vehicle's movement.</u> There are several systems that have different qualities and uses. Here we explain how they work and what they are for.

In a traction vehicle with one of its axles, its ability to transmit engine power to the ground is limited for two reasons:

- At least one of the wheels must have adhesion, and this as long as it has a self-locking differential. Otherwise, simply with one lacking grip, we can no longer move forward.

- If there are two wheels that must distribute the power, it will always be easier to saturate the traction capacity of the tire than if we divide the force by four. The example is very simple: if we try to drag an object on the ground by pulling a rubber, it will stretch more than if we pull 4 identical tires, although the force we make is the same.

Contrary to what one might think, all-wheel drive vehicles are nothing recent. What happens is that it did take time to reduce the weight and size, as well as to increase the resistance of the homokinetic joints (they are articulations on the axles to allow the wheel to go up and down with the suspension or turn with the steering) to to adapt these systems to cars.

7 0

3 years ago

Para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al

padilas [110]

Answer:

Se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

Explanation:

Dado que para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío, y en el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos, si se introducen al proceso 1000 kg/h de jugo diluido, para calcular la cantidad de agua evaporada y de jugo concentrado saliente se debe realizar el siguiente cálculo;

1000 x 0.07 = 70

60 = 70

100 = X

100 x 70 / 60 = X

7000 / 60 = X

116.66 = X

Por lo tanto, se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

4 0

3 years ago

Royce has a Bevel protractor with a damaged vernier scale. What limitation will Royce face if he uses this protractor to measure

liraira [26]

Answer:

Limitation in the level of possible accuracy of the that can be obtained to mainly whole degrees

Explanation:

The angle measurement values are located around the circularly shaped bevel protractor to which a Vernier scale can be attached to increase the accuracy of the angle measurement reading

The accuracy of the bevel protractor is up to 5 arc minutes or 1/12° with the space on the Vernier scale having graduations of 1/12°, such that two spaces on the main scale is 5 arc minutes more than a space on the Vernier scale and the dimensions of the coincidence of the two scale will have an accuracy of up to 1/12°

Therefore, whereby the Vernier scale is damaged, the accuracy will be limited to the accuracy main scale reading in whole degrees.

6 0

3 years ago