A. 50

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 years ago

The A-36 steel pipe has a 6061-T6 aluminum core. It issubjected to a tensile force of 200 kN. Determine the averagenormal stress

sasho [114]

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law:

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:

ε = f / k

When the force is distributed between both materials will stretch the same length:

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

5 0

3 years ago

ANSWER QUICK

luda_lava [24]

Answer:

u will need good car parts and a very well made engine

Explanation:

u need a good engine because if u only work on the outer layer of the car the inner parts will be slow and old and the car will have problems

5 0

3 years ago

The criminal and traffic code requires that a driver must have a valid driver's license in his/her

ankoles [38]

Answer:

true

Explanation:

the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once

5 0

3 years ago

A manometer consists of an inclined glass tube which communicates with a metal cylinder standing upright; liquid fills the appar

Vanyuwa [196]

Answer:

48.61

Explanation:

See attached diagram.

The level rise in the tube is l sin α.

The level drop in the cylinder (let's call it y) is:

π/4 D² y = π/4 d² l

D² y = d² l

y = l (d/D)²

The elevation difference is the sum:

l sin α + l (d/D)²

l (sin α + (d/D)²)

From Bernoulli's principle:

P = ρgl (sin α + (d/D)²)

Divide both sides by density of water (ρw) and gravity:

P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)

h = S l (sin α + (d/D)²)

If we disregard the level change in the cylinder:

h = S l (sin α)

We want the percent error between these two expressions for h to be 0.1% when α = 25°.

[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l (d/D)² ] / [ S l (sin α + (d/D)²)] = 0.001

(d/D)² / (sin α + (d/D)²) = 0.001

(d/D)² = 0.001 (sin α + (d/D)²)

(d/D)² = 0.001 sin α + 0.001 (d/D)²

0.999 (d/D)² = 0.001 sin α

d/D = √(0.001 sin α / 0.999)

When α = 25°:

d/D ≈ 0.02057

D/d ≈ 48.61

7 0

3 years ago