Which of the following tasks are performed by the marketing department:

Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,

solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

P₂= 80 psia

Temperature =°F Temperature is convert to °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67 x -1.817

=- 64.10Btu/lbm

The required work therefore for this isothermal compression is 64.10 Btu/lbm

8 0

3 years ago

A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stret

andrew11 [14]

The modulus of elasticity is 28.6 X 10³ ksi

<u>Explanation:</u>

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ = $\frac{8000 X 10^-^3}{0.7}$

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε = $\frac{0.002}{5}$

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

$E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi$

Thus, modulus of elasticity is 28.6 X 10³ ksi

6 0

3 years ago

8. What are two ways SpaceX plans to change personal travel?

GalinKa [24]

Answer:

as all the people should go near stratosphere

8 0

2 years ago

Assume that the water temperature is 10°C and the depth of the settling tank is 3.0 m (9.80 ft). Calculate the theoretical settl

Lelu [443]

Explanation:

See attached file

6 0

3 years ago

A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of

yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

$V_d=2.5\times 10^{-3} m^3$

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

$P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}$

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0

3 years ago