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IRINA_888 [86]
3 years ago
8

Explain the underlying physical reason why when we conduct various heat treatments on 1018 steel we expect the modulus of elasti

city to stay about the same for every heat treatment.
Engineering
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

Explanation:

Generally the underlying physical reason why when we conduct various heat treatments on 1018 steel we expect the modulus of elasticity to stay about the same for every heat treatment is  

 

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

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A lake with constant volume 1.1 x 10^6 m^3 is fed by a stream with a non-conservative pollutant of 2.3 mg/L and flow rate 35 m^3
Jet001 [13]

Answer:

12.84 mg/L

Explanation:

We are given;

Volume of lake; V = 1.1 x 10^(6) m³

decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s

Factory rate: Q_f = 4.3 m³/s

Factory concentration: C_f = 100 mg/L

Stream rate: Q_s = 34 m³/s

Stream Concentration: C_s = 2.3 mg/L

Now, to find the steady state concentration of pollutant in the lake, we will use the formula;

(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)

Where C_L is the steady state concentration of pollutant in the lake.

Thus, making C_L the subject, we have;

C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)

Plugging in the relevant values gives;

C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))

C_L = 12.84 mg/L

4 0
3 years ago
A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27
lidiya [134]

Answer:

Q = 62    ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol

T_1 = 27^0 \ C = ( 27+273)K =  300 K

P_1 = 200 \ kPa

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if V_1 = x\\V_2 = 2V_1

Using Charles Law; since pressure is constant

V \alpha T

\frac{V_2}{V_1}  =\frac{T_2}{T_1}

\frac{2V_1}{V_1}  =\frac{T_2}{300}

T_2 = 300*2\\T_2 = 600

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg  × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg  × 40

mass of He = 280 kg

Now; the amount of  Heat  Q transferred = m_{He}Cp_{He} \delta T  + m_{Ar}Cp_{Ar} \delta T

From gas table

Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar}  = 0.5203 \  kJ/Kg/K

∴ Q = 12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)

Q = 62.389 *10^6

Q = 62 MJ

Q = 62    ( since we are instructed not to include the units in the answer)

5 0
3 years ago
Read 2 more answers
A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the
Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

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\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})

Using the Boyle equation we have

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})

Where,

C_p = Specific heat at constant pressure

T_1= Initial temperature of gas

T_2= Final temprature of gas

R = Universal gas constant

v_1= Initial specific Volume of gas

v_2= Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

T_1 = T_2, v_2=v_1

The equation would turn out as

\Delta S = C_p ln1-ln1

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

\Delta S = \frac{Q}{T}

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

\Delta S = \frac{230kJ}{(30+273)K}

\Delta S = 0.76 kJ/K

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

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This manometer is used to measure the difference in water level between the two tanks.
SpyIntel [72]

Answer:

a) True

Explanation:

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3 years ago
A large building will need several different types of workmen to install and repair pipes for water, heating,
siniylev [52]

Answer:

Plumber and pipefitters

Explanation:

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