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faust18 [17]
4 years ago
13

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

s: (d) 3.78 kg/s (a) 0.731 kg/s(b) 2.74 kg/s (c) 3.18 kg/s
Engineering
1 answer:
stealth61 [152]4 years ago
7 0

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

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Plane wall of material A with internal heat generation is insulated on one side and bounded by a second wall of material B, whic
viktelen [127]

Sorry❤

Have a nice day ✨

8 0
3 years ago
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change i
ehidna [41]

Answer:

Change in length = 0.1257 mm

Change in diameter= -0.03771mm

Explanation:

Given

Diameter, d = 15 mm

Length of rod, L = 200mm

F = Force= 300N

d = 0.015m

Ep=2.70 GPa, np=0.4.

First, we have to calculate the normal stress using

σ = F/A where F = Force acting on the Cross-sectional area

A = Area

Area is calculated as πd²/4 where d = 0.015m

A = 22/7 * 0.015²/4

A = 0.000176785714285m²

A = 1.768E-4m²

So, stress. σ = 300N/1.768E-4m²

σ = 1696832.579185520Pa

σ = 1.697MPa

Calculating E(long)

E(long) = σ /Ep

E(long) = 1.697E-3/2.70

E(long) = 0.0006285

At this point, we fan now calculate the change in length of the element;

∆L = E(long) * L

∆L = 0.0006285 * 200mm

∆L = 0.1257mm

Calculating E(lat)

E(lat) = -np * E(long)

E(lat) = -4 * 0.0006285

E(lat) = -0.002514

At this point, we can now calculate the change in diameter of the element;

∆D = E(lat) * D

∆L = -0.002514 * 15mm

∆L = -0.03771mm

8 0
3 years ago
Column arrays: Transpose a row array Construct a row array countValues with elements 1 to endValue, using the double colon opera
White raven [17]

Answer:

Matlab code with step by step explanation and output results are given below

Explanation:

We have to construct a Matlab function that creates a row vector "countValues" with elements 1 to endValue. That means it starts from 1 and ends at the value provided by the user (endValue).  

function countValues = CreateArray(endValue)

% Here we construct a row vector countValues from 1:endValue

     countValues = 1:endValue;

% then we transpose this row vector into column vector

     countValues = countValues';

 end

Output:

Calling this function with the endValue=11 returns following output

CreateArray(11)

ans =

    1

    2

    3

    4

    5

    6

    7

    8

    9

   10

   11

Hence the function works correctly. It creates a row vector then transposes it and makes it a column vector.

7 0
3 years ago
The acceleration due to gravity at sea level is g=9.81 m/s^2. The radius of the earth is 6370 km. The universal gravitational co
solmaris [256]

Answer:

Mass of earth will be M=5.96\times 10^{24}kg

Explanation:

We have given acceleration due to gravity g=9.81m/sec^2

Radius of earth = 6370 km =6370\times 10^3m

Gravitational constant G=6.67\times 10^{-11}Nm^2/kg^2

We know that acceleration due to gravity is given by

g=\frac{GM}{R^2}, here G is gravitational constant, M is mass of earth and R is radius of earth

So 9.81=\frac{6.67\times 10^{-11}\times M}{(6370\times 10^3)^2}

M=5.96\times 10^{24}kg

So mass of earth will be M=5.96\times 10^{24}kg

3 0
3 years ago
A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla
creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed =  0.5 m

cohesion = 25 kPa

internal friction angle =  5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for  a pure cohesive soil

N_{\gamma } = 0

and we know formula for N_{\gamma } is

N_{\gamma } = (Nq - 1 ) × tan(Ф)   ..................1

so here Ф is very less  N_{\gamma } should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

7 0
4 years ago
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