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Illusion [34]
3 years ago
12

1. Làm thế nào để đảm bảo tính khả thi của văn bản hành chính ?

Engineering
1 answer:
Dennis_Churaev [7]3 years ago
7 0

Answer:

Explanation:

......................

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Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl
Shkiper50 [21]

Answer:

a) 180 m³/s

b) 213.4 kg/s

Explanation:

A_1 = 1 m²

P_1 = 100 kPa

V_1 = 180 m/s

Flow rate

Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s

Volumetric flow rate = 180 m³/s

Mass flow rate

\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s

Mass flow rate = 213.4 kg/s

3 0
3 years ago
Who was part of dempwolf his firm when he first started
Zinaida [17]

Explanation:

Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with  brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.

5 0
2 years ago
A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of
Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0
3 years ago
21.Why are throttling devices commonly used in refrigeration and air-conditioning<br> applications?
Sloan [31]

Answer is given below

Explanation:

we know that some common types of throttling devices are

  • Hard -throttling devices
  • Capillary valve
  • Constant pressure throttling devices
  • Thermostatic expansion valve
  • Float expansion valve

so here throttling devices commonly used in refrigeration and air-conditioning because

  • To reduce the coolant pressure, the high pressure of the refrigerant from the condenser is necessary to reduce the evaporation to obtain evaporation at the right temperature  
  • To meet the refrigerated load, the throttling valve flows through the coolant to cool the load at high temperatures.
5 0
3 years ago
Why is a metal work enclosure dangerous?
dedylja [7]

Answer:

Why is a metal work enclosure dangerous? Metalworkers are not only exposed to pollutants from metal cut ting and polishing procedures, but they are also exposed to metalworking fluids (MWF).

8 0
2 years ago
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