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Y_Kistochka [10]
3 years ago
12

6.3. A __________ is used to indicate the base material that needs to be beveled.

Engineering
1 answer:
Ivan3 years ago
4 0

Answer:

A <u>break in the arrow</u> is used to indicate the base material that needs to be beveled

A. Break in the arrow

Explanation:

In a weld that involves the welding of a joint that has a, a bevel, flare bevel, or J-groove, the side that requires beveling (chamfering) preparation in the welded joint before welding is indicated by an arrow having a definitive before pointing to the beveled member

However, the break in the arrow could excluded when the side to be chamfered is unmistakable

Therefore, the break in the arrow is the option that is used to indicate which of the material that is to be beveled.

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What scale model proves the initial concept?
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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

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