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stealth61 [152]
2 years ago
8

3. Air at 1 atm and 20 0 C flows tangentially on both sides of a smooth flat plate of width W=10 ft and length L=10 ft in the di

rection of flow (x-direction). The velocity outside the boundary layer is constant at 20 ft/s. a) What is the laminar boundary layer thickness at x=1 ft? b) The boundary layer varies as x to what power? c) If the plate is 100 ft long, does the boundary layer thickness change at 1 ft?
Engineering
1 answer:
8_murik_8 [283]2 years ago
6 0

Answer:

ExplanationAir at 1 atm and 20°C flows tangentially on both sides of a thin, smooth flat plate of ... ... Smooth Flat Plate Of Width W = 10 Ft, And Of Length L 3 Ft In The Direction Of The Flow. The Velocity Outside The Boundary Layer Is Constant At 20 Ft/s.

:

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Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t
Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

end

%This brings an end to the program

4 0
3 years ago
Steam at 1 MPa, 300 C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam i
stealth61 [152]

Answer:

\dot m = 2.74 kg/s

Explanation:

given data:

pressure 1 MPa

diameter of pipe  =  30 cm

average velocity = 10 m/s

area of pipe= \frac[\pi}{4}d^2

                 = \frac{\pi}{4} 0.3^2

A = 0.070 m2

WE KNOW THAT mass flow rate is given as

\dot m = \rho A v

for pressure 1 MPa, the density of steam is = 4.068 kg/m3

therefore we have

\dot m = 4.068 * 0.070* 10

\dot m = 2.74 kg/s

7 0
3 years ago
What issues does society try to forget in order to not have to deal with them ?
velikii [3]

Answer: Homelessness, Drug addiction, Mental Illness, Climate change.

Explanation:

Society would rather be ignorant to/ ignore all of these realities so they do not have to stop specific behaviors or actually acknowledge other people and help. Sometimes people are so stressed out about their own lives, they can not bare another persons issues. People are ignorant to climate change because correcting it requires massive changes and society is selfish and unwilling to change.

6 0
2 years ago
The error in the trapezoidal rule is proportional to h^2 True b. False
Lynna [10]

Answer:

b) False

Explanation:

in trapezoidal rule the error is proportional to h^3 and the order of accuracy is proportional to h^3 .

Trapezoidal rule is numerical integration method .Trapezoidal rule is used to find the area of curves.In trapezoidal rule we finds the approximate value of integration.But the real value of integration will not differ to much from the value which finds by using trapezoidal rule.

4 0
3 years ago
6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right
dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0
3 years ago
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