Water to a residential area is transported at a rate of 1.5 m 3 /s via 70-cm-internal-diameter concrete pipes with a surface roughness of 3 mm and a total length of 1500 m. In order to reduce pumping power requirements, it is proposed to line the interior surfaces of the concrete pipe with 2-cm-thick petroleum-based lining that has a surface roughness thickness of 004 mm, this is a concern that the reduction of pipe diameter to 66 cm and the increase in average velocity may offset any gains. Taking p = 1000 kg/m 3 and v = 1 × 10-6 m 2 /s for water, determine the percent increase or decrease in the pumping power requirements due to pipe frictional losses as a result of lining time concrete pipes.

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Textbook Question

Chapter 8, Problem 96P

Water to a residential area is transported at a rate of 1.5 m³/s via 70-cm-internal-diameter concrete pipes with a surface roughness of 3 mm and a total length of 1500 m. In order to reduce pumping power requirements, it is proposed to line the interior surfaces of the concrete pipe with 2-cm-thick petroleum-based lining that has a surface roughness thickness of 004 mm, this is a concern that the reduction of pipe diameter to 66 cm and the increase in average velocity may offset any gains. Taking p = 1000 kg/m³ and v = 1 × 10-6 m²/s for water, determine the percent increase or decrease in the pumping power requirements due to pipe frictional losses as a result of lining time concrete pipes.

Expert Solution & Answer

To determine

The percentage change in pumping power requirement due to frictional losses.

Answer to Problem 96P

The percentage decrease in pumping power requirement due to frictional losses is $54.157%$ .

Explanation of Solution

Given information:

The density of the water is $1000 kg/m3$ , kinematic viscosity of the water is $1×10−6 m2/s$ , diameter of the pipe without line interior surface is $70 cm$ , diameter of the pipe with line interior surface is $66 cm$ , roughness of the line interior concrete pipe surfaces is $0.04 mm$ , roughness of the concrete pipe surfaces is $3 mm$ , volume flow rate of water is $1.5 m3/s,$ and length of the pipe is $1500 m$ .

Write the expression for the Reynolds number.

$Re=vDυ$ ........... (I)

Here, Reynolds number is $Re$ , velocity of the water is $v$ , diameter of the pipe is $D,$ and kinematic viscosity of the water is $υ$ .

Write the expression for the volume flow rate of the water.

$Q˙=π4D2v$ ........... (II)

Here, the volume flow rate of the water is $Q˙$ .

Write the expression for the friction factor for turbulent flow.

$1f=−2log(εD3.7+2.51Ref)$ ........... (III)

Here, friction factor for turbulent flow is $f$ and roughness of the plastic pipe surfaces is $ε$ .

Write the expression for the head loss through the pipe.

$hf=fLv22gD$ ........... (IV)

Here, length of the pipe is $L$ , gravitational acceleration is $g,$ and the head loss through the pipe is $hf$ .

Write the expression for the pressure drop through the pipe.

$ΔP=ρghf$ ........... (V)

Here, the pressure drop through the pipe is $ΔP$ and density of the water is $ρ$ .

Write the expression for the pumping power requirement to maintain the flow rate of water.

$W˙pump=Q˙ΔP$ ........... (VI)

Here, pumping power requirement to maintain the flow rate of water is $W˙pump$ .

Substitute $ρghf$ for $ΔP$ and $π4D2v$ for $Q˙$ in Equation (VI).

$W˙pump=(π4D2v)(ρghf)=π4D2vρghf$ ........... (VII)

Write the expression for the pumping power for without line interior surface pipe.

$( W ˙ pump)without line=π4ρg(vhfD2)without line$ ........... (VIII)

Here, subscript for without line interior pipe is without line.

Write the expression for the pumping power for with line interior surface pipe.

$( W ˙ pump)with line=π4ρg(vhfD2)with line$ ........... (IX)

Here, subscript for with line interior pipe is with line.

Write the expression for the percentage change in power due to frictional losses.

$m=( ( W ˙ pump ) with line ( W ˙ pump ) without line)×100%$ ........... (X)

Here percentage change in power due to frictional losses is $m$ .

Substitute $π4ρg(vhfD2)without line$ for $( W ˙ pump)without line$ and $π4ρg(vhfD2)with line$ for $( W ˙ pump)with line$ in Equation (X).

$m=( π 4 ρg ( v h f D 2 ) with line π 4 ρg ( v h f D 2 ) without line )×100%=( ( v h f D 2 ) with line ( v h f D 2 ) without line )×100%$ ........... (XI)

Calculation:

Substitute $1.5 m3/s$ for $Q˙$ , $70 cm$ for $D$ and $vwithout line$ for $v$ in Equation (II).

$(1.5 m 3/s)=π4(70 cm)2(v)without line(1.5 m 3/s)=π4(( 70 cm)[ 1 m 100 cm ])2(v)without line(v)without line=3.897 m/s$

Substitute $3.897 m/s$ for $v$ , $1×10−6 m2/s$ for $υ$ , $70 cm$ for $D$ and $Rewithout line$ for $Re$ in Equation (I).

$Rewithout line=( 3.897 m/s )( 70 cm)( 1× 10 −6 m 2 /s )=( 3.897 m/s )( 70 cm)[ 1 m 100 cm]( 1× 10 −6 m 2 /s )=2.728×106$

Substitute $2.728×106$ for $Re$ , $3 mm$ for $ε$ , $70 cm$ for $D$ and $fwithout line$ for $f$ in Equation (III).

$1 f without line =−2log( ( 3 mm ) ( 70 cm ) 3.7+ 2.51 ( 2.728× 10 6 ) f without line )1 f without line =−2log( ( 3 mm )[ 1 cm 10 mm ] ( 70 cm ) 3.7+ 2.51 ( 2.728× 10 6 ) f without line )fwithout line=0.029$

Substitute $0.029$ for $f$ , $1500 m$ for $L$ , $3.897 m/s$ for $v$ , $9.81 m/s2$ for $g$ , $70 cm$ for $D$ and $(hf)without line$ for $hf$ in Equation (IV).

$( h f)without line=( 0.029)( 1500 m) ( 3.897 m/s )22( 9.81 m/ s 2 )( 70 cm)=( 0.029)( 1500 m) ( 3.897 m/s )22( 9.81 m/ s 2 )( 70 cm)[ 1 m 100 cm]=48.1 m$

Substitute $1.5 m3/s$ for $Q˙$ , $66 cm$ for $D$ and $vwith line$ for $v$ in Equation (II).

$(1.5 m 3/s)=π4(66 cm)2(v)with line(1.5 m 3/s)=π4(( 66 cm)[ 1 m 100 cm ])2(v)with line(v)with line=4.384 m/s$

Substitute $4.384 m/s$ for $v$ , $1×10−6 m2/s$ for $υ$ , $66 cm$ for $D$ and $Rewith line$ for $Re$ in Equation (I).

$Rewith line=( 4.384 m/s )( 66 cm)( 1× 10 −6 m 2 /s )=( 4.384 m/s )( 66 cm)[ 1 m 100 cm]( 1× 10 −6 m 2 /s )=2.893×106$

Substitute $2.893×106$ for $Re$ , $0.04 mm$ for $ε$ , $66 cm$ for $D$ and $fwith line$ for $f$ in Equation (III).

$1 f with line =−2log( ( 0.04 mm ) ( 66 cm ) 3.7+ 2.51 ( 2.893× 10 6 ) f with line )1 f with line =−2log( ( 0.04 mm )[ 1 cm 10 mm ] ( 66 cm ) 3.7+ 2.51 ( 2.893× 10 6 ) f with line )fwith line=0.0117$

Substitute $0.0117$ for $f$ , $1500 m$ for $L$ , $4.384 m/s$ for $v$ , $9.81 m/s2$ for $g$ , $66 cm$ for $D$ and $(hf)with line$ for $hf$ in Equation (IV).

$( h f)with line=( 0.0117)( 1500 m) ( 4.384 m/s )22( 9.81 m/ s 2 )( 66 cm)=( 0.0117)( 1500 m) ( 4.384 m/s )22( 9.81 m/ s 2 )( 66 cm)[ 1 m 100 cm]=26.048 m$

Substitute $3.897 m/s$ for $vwithout line$ , $70 cm$ for $Dwithout line$ , $48.1 m$ for $(hf)without line$ , $4.384 m/s$ for $vwith line$ , $66 cm$ for $Dwith line$ and $26.048 m$ for $(hf)with line$ in Equation (XI).

$m=( ( 4.384 m/s )( 26.048 m ) ( 66 cm ) 2 ( 3.897 m/s )( 48.1 m ) ( 70 cm ) 2 )×100%=( ( 4.384 m/s )( 26.048 m ) ( 66 cm ) 2 ( 3.897 m/s )( 48.1 m ) ( 70 cm ) 2 )×100%=54.157%$

Conclusion:

The percentage decrease in pumping power requirement due to frictional losses is $54.157%$ .

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