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3 years ago

System Integration summary

Engineering

1 answer:

kenny6666 [7]3 years ago

6 0

Answer:

System integration can be defined as the progressive linking and testing of system components to merge their functional and technical characteristics into a comprehensive interoperable system.

Explanation:

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PLS HELP ME

Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

40 = X

40 x 100 / 120 = X

4000 / 120 = X

33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

133.333 = X

4 0

3 years ago

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3

goldenfox [79]

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

3 0

3 years ago

Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass

snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0

3 years ago

A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring

ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0

3 years ago

A step-down transformer (turns ratio = 1:7) is used with an electric train to reduce the voltage from the wall receptacle to a v

SOVA2 [1]

Answer:

wait lemme check it out

Explanation:

6 0

3 years ago