What are the characteristics of carbon fibre?

In poor weather, you should _______ your following distance.

jasenka [17]

In poor weather, you should <u>double</u> your following distance.

6 0

3 years ago

Ninety-five percent of the acetone vapor in an 85 vol.% air stream is to be absorbed by countercurrent contact with pure water i

Sati [7]

Answer:

Explanation:

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5 0

3 years ago

Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held

77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }$

${T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K$

The heat supplied = $\dot {m}_f$ × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = $\dot m$ · $c_p(T_3 - T_2)$

$\dot m$ = 20 kg/s

The heat supplied = 20* $c_p(T_3 - T_2)$ = 21,350 kJ/s

$c_p$ = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

$\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}$

T₅ = 1208.42*(2/2.333) = 1035.94 K

$C_j = \sqrt{\gamma \times R \times T_5}$ = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = $\dot m$ × $C_j$ = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0

3 years ago

Calculate the pressure drop in a duct (measured by a differential oil manometer) if the differential height between the two flui

Burka [1]

Answer:

The pressure drop is 269.7N/m^2

Explanation:

∆P = ∆h × rho × g

∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2

∆P = 0.032×860×9.8 = 269.7N/m^2

6 0

3 years ago

Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm

Simora [160]

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow. You can also see that $R_{th}$ represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

Wall Thickness. More thickness, more thermal resistance.
Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
Cross-section Area. More cross-section area, less thermal resistance.

A expression to define the thermal resistance for the wall is as follows: $R_{th} =\frac{l}{Ak}$ , where l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.

5 0

3 years ago