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Olenka [21]
3 years ago
14

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f

or the water supply over a distance of 200 m? Hint: Use an iterative approach for the empirical equation.
Engineering
1 answer:
dangina [55]3 years ago
7 0

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

    = \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s

obtained Darcy friction factor  

calculate Reynold number (Re) ,

Re = \frac{\rho V D}{\mu}

where,\rho = density of water

\mu = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

Re = \frac{1000\times 2.015\times 0.318}{0.001}

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}

HL = 1.64 m

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(a) The amount of heat transferred to the air, q_{out} is 215.5077 kJ/kg

(b) The net work output, W_{net}, is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

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T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}}  \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K

From;

\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }

We have;

p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa

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\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86  \times 729.21}{2190.43} = 1458.42 \, K

Process 3 to 4 is isentropic expansion, therefore;

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1458.42= T_{4} \times \left (9.2 \right )^{0.4}

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The amount of heat transferred to the air, q_{out} = 215.5077 kJ/kg

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W_{net} = q_{in} - q_{out}

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\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg

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