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3 years ago

The blue sticker on the rear of a vehicle indicates that it's designed

Engineering

2 answers:

Maksim231197 [3]3 years ago

6 0

Its FALSE 123456789012345678987653213e4r5678

GenaCL600 [577]3 years ago

4 0

Answer:

The blue sticker on the rear of a vehicle indicates that it's designed to use compressed natural gas.

Explanation:

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When moving cylinders always remove and make

Karolina [17]

Unless cylinders are firmly secured on a special carrier intended for this purpose, regulators shall be removed and valve protection caps put in place before cylinders are moved. A suitable cylinder truck, chain, or other steadying device shall be used to keep cylinders from being knocked over while in use.

5 0

3 years ago

Design a program that calculates the area and circumstance of rectangle?

Phantasy [73]

Let “w” and “L” be the width and length of the rectangle. “p” and “a” are perimeter and area
For python,
w=int(input(“width”))
l=int(input(“length”))
a= w*l
p=2*w+2*l
print(str(a), str(p)

4 0

2 years ago

It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d

denis23 [38]

The exit temperature is 586.18K and compressor input power is 14973.53kW

Data;

Mass = 50kg/s
T = 288.2K
P1 = 1atm
P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\$

Let's substitute the values into the formula

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K$

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

$P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW$

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

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6 0

2 years ago

2.) A fluid moves in a steady manner between two sections in a flow

Talja [164]

Answer:

$250\ \text{lbm/min}$

$625\ \text{ft/min}$

Explanation:

$A_1$ = Area of section 1 = $10\ \text{ft}^2$

$V_1$ = Velocity of water at section 1 = 100 ft/min

$v_1$ = Specific volume at section 1 = $4\ \text{ft}^3/\text{lbm}$

$\rho$ = Density of fluid = $0.2\ \text{lb/ft}^3$

$A_2$ = Area of section 2 = $2\ \text{ft}^2$

Mass flow rate is given by

$m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}$

The mass flow rate through the pipe is $250\ \text{lbm/min}$

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

$m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}$

The speed at section 2 is $625\ \text{ft/min}$ .

3 0

3 years ago

Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation

lions [1.4K]

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

3 0

3 years ago