All of these are true about steel EXCEPT that:

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago

For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals

kvv77 [185]

Answer:

(b)False

Explanation:

Given:

Prandtl number(Pr) =1000.

We know that $Pr=\dfrac{\nu }{\alpha }$

Where $\nu$ is the molecular diffusivity of momentum

$\alpha$ is the molecular diffusivity of heat.

Prandtl number(Pr) can also be defined as

$Pr=\left (\dfrac{\delta }{\delta _t}\right )^3$

Where $\delta$ is the hydrodynamic boundary layer thickness and $\delta_t$ is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

8 0

3 years ago

In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had

kipiarov [429]

Answer:

Moisture/ water content w = 26%

Void ratio , e = 0.73

Explanation:

Initial mass of saturated soil w1 = mass of soil - weight of container

= 113.27 g - 49.31 g = 63.96 g

Final mass of soil after oven w2 = mass of soil - weight of container

= 100.06 g - 49.31 g = 50.75

Moisture /water content, w = $\frac{w1-w2}{w2}$ = $\frac{63.96-50.75}{50.75}$ = 0.26 = 26%

Void ratio = water content X specific gravity of solid

= 0.26 X 2.80 =0.728

5 0

3 years ago

g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0

Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵ * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0

2 years ago

Assume you have four fins, each with a mass of 8.0 grams. What is the total weight of these four fins? (Hint: watch your units!)

soldier1979 [14.2K]

The answer is 32.0 grams

7 0

2 years ago