Do better then me......................................

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

How would you expect an increase in the austenite grain size to affect the hardenability of a steel alloy? Why?

seraphim [82]

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

3 0

3 years ago

A specimen of commercially pure copper has a strength of 240 MPa. Estimate its average grain diameter using the Hall-Petch equat

romanna [79]

Answer:

3.115× $10^{-3}$ meter

Explanation:

hall-petch constant for copper is given by

$S_0$ =25 MPa

k=0.12 for copper

now according to hall-petch equation

$S_Y$ = $S_0$ + $\frac{K}{\sqrt{D}}$

240=25+ $\frac{0.12}{\sqrt{D}}$

D=3.115× $10^{-3}$ meter

so the grain diameter using the hall-petch equation=3.115× $10^{-3}$ meter

5 0

3 years ago

Products that are in the process of being manufactured but are not yet complete are called:

vazorg [7]

Answer:

Those products are generally called Work in Process WIP

Explanation:

Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.

These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.

6 0

3 years ago

A buret is a device designed to precisely dispense liquids. The scale is calibrated in mL with the zero point at the top with nu

kobusy [5.1K]

Answer:

As there was no attached picture, I will explain how to take the measurement of liquids in any buret which you can then apply to the specific question

Explanation:

A buret is a laboratory apparatus used to precisely measure the volume of liquids (usually alkalise or bases) used in a titration experiment. The standard buret has a capacity of 50 ml and graduated in 0.1ml though burets with smaller capacities exist.

From the question, your buret is filled to the top (0.00ml) with liquid. It is very important when taking buret readings to place the buret below your eye level so that the bottom meniscus (lower part of the liquid) can be read.

To take the buret reading, note your initial buret reading (in this case 0.00ml) then titrate the liquid base in the buret against the acid by opening the tap located at the bottom of the buret.

When the titration or reaction is complete, note the final reading against the calibration of buret. You can do this by observing the lower meniscus of the liquid remaining in the buret. (Remember to keep the buret at eye level to avoid parallax error),

The difference between your final buret reading and the initial buret reading gives you the precise volume of liquid used in the reaction.

6 0

3 years ago