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3 years ago

12

name the process by which mild steel can be converted into high carbon steel and explain it briefly ?

1 answer:

Fudgin [204]3 years ago

6 0

Answer:

please give me brainlist and follow

Explanation:

Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

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System Integration summary

kenny6666 [7]

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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3 years ago

On aircraft equipped with fuel pumps, when is the auxiliary electric driven pump used?.

pochemuha

In an airplane equipped with fuel pumps, the auxiliary electric fuel pump is used in the event the engine-driven fuel pump fails.. hope this helped !

6 0

2 years ago

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

$= 71.82 \ kN/m^2$

$\sim 72 \ kN/m^2$

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 =$ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

= 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

= 13.14

= 13

8 0

3 years ago

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

A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The infl

german

Answer:

BOD concentration at the outflow = 17.83 mg/L

Explanation:

given data

flow rate of Q = 4,000 m³/day

BOD1 concentration of Cin = 25 mg/L

volume of the pond = 20,000 m³

first-order rate constant equal = 0.25/day

to find out

What is the BOD concentration at the outflow of the pond

solution

first we find the detention time that is

detention time t = $\frac{volume}{flow rate}$

detention time t = $\frac{20000}{4000}$

detention time = 5 days

so

BOD concentration at the outflow of pond is express as

BOD concentration at the outflow = $Cin ( 1 - e^{-kt} )$

here k is first-order rate constant and t is detention time and Cin is BOD1 concentration

so

BOD concentration at the outflow = $25 ( 1 - e^{-0.25(5)} )$

BOD concentration at the outflow = 17.83 mg/L

8 0

3 years ago