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4 years ago

Calculate the approximate mass of the steel pipe. Length of pipe = 4m. Inside radius =130mm. Outside radius = 150mm

Engineering

1 answer:

Thepotemich [5.8K]4 years ago

6 0

Answer:

555.936 kg

Explanation:

h = Length of pipe = 4 m

r = Inside radius =130 mm

R = Outside radius = 150mm

Volume of the cylinder

v = πR²h - πr²h

⇒v = πh(R²-r²)

⇒v = π4(0.15²-0.13²)

⇒v = 0.0704 m³

Density of steel varies between 7,750 and 8,050 kg/m³

So, let us take the average of the two values so we get density of steel as 7900 kg/m³ as the exact density is not given

Density

$\rho=\frac{m}{v}\\\Rightarrow m=\rho \times v\\\Rightarrow m=7900\times 0.704\\\Rightarrow m= 555.936\ kg$

So, approximate mass of the steel pipe is 555.936 kg

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A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory

Molodets [167]

Answer:

Explanation:

Mean temperature is given by

$T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}$

Tmean = (Ti + T∞)/2

$T_mean = 107.5^{0}$

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

$\mu$ = 221.6 × 10⁻⁷N.s/m²

$\kappa$ = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

= 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/ $\mu$

Substituting for Pv

Re = 4m/(πD $\mu$ )

= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

= 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago

What happens to battery when it produces current to the system

olasank [31]

Answer:

Electricity, as you probably already know, is the flow of electrons through a conductive path like a wire. This path is called a circuit. ... The chemical reactions in the battery causes a build up of electrons at the anode. This results in an electrical difference between the anode and the cathode

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3 years ago

Question 9.1 from the textbook. Consider the following workload: Process Burst Time Priority Arrival Time P1 50 4 0 P2 20 1 20 P

Marizza181 [45]

Answer:

Explanation:

The schedule using shortest remaining time, non-preemptive priority and round Robin with quantum number 30 is shown in the attached file, please kindly go through it to access the answer.

5 0

4 years ago

a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8

djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = $\frac{MC}{I}$ = $\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}$ = 159.07 MPa

B) Determine max transverse shear stress

attached below

ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam = 159.07 / 2 = 79.535 MPa

attached below is the remaining solution

6 0

3 years ago