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nasty-shy [4]
3 years ago
8

Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur

e gradient, ?P/L, along the pipe.
Engineering
1 answer:
ruslelena [56]3 years ago
3 0

Answer:

\frac{\delta p }{l} = 30.4 lb/ft^3

Explanation:

Given data:

flow rate = 10 gallon per  minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q =  VA

solve for velocity V = \frac{Q}{A}[/tex]

V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}

V = 7.27 ft/sec

we know that Reynold number

Re = \frac{VD}{\nu}

Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}

Re = 3.76 \times 10^4

calculate the \frac{\epsilon }{D}ratio to determine the fanning friction f

\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008

from moody diagram f value corresonding to Re and \frac{\epsilon }{D}is 0.037

for horizontal pipe

\delta p = \frac{f l \rho v^2}{2D}

\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}

where 1.94 slug/ft^3is density of  water

\frac{\delta p }{l} = 30.4 lb/ft^3

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// The method is defined with a void return type

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Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

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T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

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c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

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