Identify the following formulas:

a piping system has an internal air pressure of 1,500 kpa. In addition to being subject to the air pressure, the piping supports

Alik [6]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

5 0

4 years ago

A 50000 N plane has wings with a span of 30 m and a chord of 6 m. How much cargo can this plane carry while cruising at 550 km/h

soldi70 [24.7K]

Answer:

The amount of cargo the plane can carry is 8707.89 N

Explanation:

The surface area of the wings facing the air = 30×6×2 × sin(2.5) = 15.7 m²

The speed of the plane 550 km/h = 152.78 m/s

The volume of air cut through per second = 15.7 × 152.78 = 2399.07 m³

The mass of air = Volume × Density = 2399.07 × 0.37 = 887.65 kg

Weight of air = Mass × Acceleration due to gravity = 887.65 × 9.81 = 8707.89 N

Given that the plane is already airborne, the additional cargo the plane can carry is given by the available lift force of the plane.

The amount of cargo the plane can carry = 8707.89 N

8 0

4 years ago

1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se

Lorico [155]

Based on the Chvorinov's rule, the diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a reference time. This rule states that cooling time (t) is directly proportional to the square of the volume (V), in cubic meters, divided to the surface area (A), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]² (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]² (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D² (3)

Where:

t - Cooling time of the cylindrical casting, in time unit.
t' - Cooling time of the prism casting, in time unit.
C - Cooling factor, in time unit per square meter.
D - Diameter of the cylinder, in meters.
L - Length of the casting, in meters.
T - Width of the cross section of the prism casting, in meters.

If we know that D = 0.3 m, then the thickness of the prism casting is:

$T = \sqrt{\frac{\pi}{12} }\cdot D$

T ≈ 0.153 m

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C (2b)

Lastly we find the percentual difference in the solidification times of the two castings by using the following expression:

r = (1 - t'/t) × 100 %

r = (1 - 0.00218/0.0329) × 100 %

r = 93.374 %

The cooling time of the prism casting is 6.626 % of the solidification time of the cylindrical casting. The diference in the solidification times of the two castings is 14.092 times the solidification time of the prism casting. $\blacksquare$