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3 years ago

Are there engineering students here?

Engineering

2 answers:

leva [86]3 years ago

7 0

Uh, I’d assume so because Brainly has a whole section of questions for them.

Firdavs [7]3 years ago

3 0

Probably...i mean....your asking in the engineering section

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What is the significance of Saint Venant's principle?

nexus9112 [7]

Answer:

While calculating the stresses in a body since we we assume a constant distribution of stress across a cross section if the body is loaded along the centroid of the cross section , this assumption of uniformity is assumed only on the basis of Saint Venant's Principle.

Saint venant principle states that the non uniformity in the stress at the point of application of load is only significant at small distances below the load and depths greater than the width of the loaded material this non uniformity is negligible and hence a uniform stress distribution is a reasonable and correct assumption while solving the body for stresses thus greatly simplifying the analysis.

7 0

3 years ago

I need help with this question please

solniwko [45]

Answer:

The resultant moment is 477.84 N·m

Explanation:

We note that the resultant moment is given by the moment about a given point

The length of the sides of the formed triangles are;

l = sin(40°) × 4/sin(110°) ≈ 2.736

Taking the moment about the lower left hand corner of the figure, with the convention that clockwise moments are positive, we have;

The resultant moment, ∑m, is given as follow;

∑M = 250 N × 4 m + 400 N × cos(40°) × 4 m - 400 N × cos(40°) × 2 m + 400 N × sin(40°) × 2 m × tan(40°) - 600 N × cos(40°) × 2 m - 600 N× sin(40°) × 2 m × tan(40°) = 477.837084 N·m

Therefore, the resultant moment, ∑m ≈ 477.84 N·m clockwise.

6 0

3 years ago

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness

Bingel [31]

Answer:

a) heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

Explanation:

Assumptions:

Constant properties
Steady state conditions
Negligible effect of radiation
Negligible constant resistance between tube and insulation
one dimensional radial conduction

a) What is the heat gain per unit tube length

$R_{conv,i}'=\frac{1}{2\pi r_1h_i}$

$d_1=36mm$ Therefore $r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}$

$r_2=2mm=2*10^{-3}m$

$k_{st}=14.2W/m.k$

$h_o=6W/m^2$

$h_i=400W/m^2$

$R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W$

$R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W$

$R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W$

$R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W$

heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

$r_3=r_1+r_2+10mm=30mm=0.03m$

$R_{conv,i}'$ and $R_{cond,st}'$ are the same, but $R_{conv,o}'$ changes.

Therefore:

$R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W$

$R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W$

The total resistance $R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W$

heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

8 0

3 years ago

Read 2 more answers

Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)

ArbitrLikvidat [17]

Answer:

$\mathbf{P_x =25 \ watts}$

$\mathbf{x_{rmx} = 5 \ unit}$

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:

$P= \dfrac{A^2}{2}$

For the number of sinosoidial signals;

Power can be expressed as:

$P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...$

As such,

For x(t), Power $P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}$

$P_x = \dfrac{25}{2}+ \dfrac{25}{2}$

$P_x = \dfrac{50}{2}$

$\mathbf{P_x =25 \ watts}$

For the number of sinosoidial signals;

$RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...$

For x(t), the RMS value is as follows:

$x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }$

$x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }$

$x_{rmx }=\sqrt{(\dfrac{50}{2} )}$

$x_{rmx} =\sqrt{25}$

$\mathbf{x_{rmx} = 5 \ unit}$

8 0

3 years ago

Which is one of the aspects in PR game marketing?

mr_godi [17]

C, I took the test already.

7 0

3 years ago

Read 2 more answers

Are there engineering students here?​

Are there engineering students here?