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3 years ago

A body whose velocity is constant has a. positive acceleration b. negative acceleration g. zero acceleration d. all of the above

Engineering

1 answer:

adoni [48]3 years ago

4 0

Answer:

option (c) is the correct answer which is zero acceleration.

Explanation:

It is given in the question that the velocity is constant.

Now,

the options are provided in relation to the acceleration.

We know,

acceleration is rate of change of velocity per unit time i.e

acceleration = $\frac{dV}{dt}$

since, the change in velocity is given to be zero,

thus, dV/dt = 0

hence,

acceleration = 0

therefore, option (c) is the correct answer which is zero acceleration.

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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

Typical noise associated with failed cv joint

igor_vitrenko [27]

Answer:

A worn inner CV joint often makes a clunking noise during starts and stops.

3 0

2 years ago

What does snow fall from?

Klio2033 [76]

Answer:

Clouds

Explanation:

It is created by trapped dust and water.

4 0

3 years ago

What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?

irga5000 [103]

Answer:

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency increase.

Now from T-S diagram

$W_{net}=W_{out}-W_{in}$

$W_{net}=(h_3-h_4)-(h_2-h_1)$

$Q_{in}=h_3-h_5$

Due to generation $(h_5-h_2)$ amount of energy has been saved.

$Q_{generation}=Q_{saved}$

So efficiency of cycle $\eta =\frac{W_{net}}{Q_{in}}$

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Effectiveness of re-generator

$\varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}$

So the efficiency of regenerative cycle

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

7 0

3 years ago

________ are written to “maximize” or “minimize” a specific value associated with the product needs in order to define the goal

Anna35 [415]

Answer:

Objective statements.

Explanation:

An objective statement can be defined as a short statement that explicitly states or describes what a person wants exactly or is looking out for in a particular item.

Objective statements are written to “maximize” or “minimize” a specific value associated with the product needs in order to define the goal or aim of the design process.

This ultimately implies that, objective statements are used by various manufacturing industries or companies to explicitly define the minimum or maximum requirements for the production of its goods.

4 0

3 years ago

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