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My name is Ann [436]
3 years ago
12

Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat

er at the rate of 12.5litres per second. Take the frictional coefficient ,f, as being 0.008​
Engineering
1 answer:
zlopas [31]3 years ago
8 0

Answer: 10.631\times 10^3\ N/m^2

Explanation:

Given

Discharge is Q=12.5\ L

Diameter of pipe d=150\ mm

Distance between two ends of pipe L=800\ m

friction factor f=0.008

Average velocity is given by

\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s

Pressure difference is given by

\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow  \Delta P=10.631\ kPa

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3 years ago
Which state did NOT have people that got sick from the
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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)

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4 years ago
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Answer:

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Explanation:

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3 years ago
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

                   =0.02812 \ Kg/mol

                   =28.12 \ g/mol

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3 years ago
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