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Zanzabum
3 years ago
5

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

ontal pipe from a storage tank open to the atmosphere. The height of the liquid level above the center of the pipe is 4 m. Disregarding the minor losses, determine the flow rate of oil through the pipe.
Engineering
1 answer:
Naddik [55]3 years ago
4 0

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s       and ρ= 850 kg/m³  

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ =  ρv

   =850 x 0.00062

   = 0.527 kg/m·s  

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

Q = \dfrac{\Delta P \pi D^4}{128 \mu L}

Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}

Q = 5.06 x 10⁻⁸ m³/s

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Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

PV=mRT

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

T_1 = 38\°C

T_2 = 14\°C

\eta = 97\%

\dot{v} = 510m^3/kg

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

PV = mRT

\dot{m} = \frac{PV}{RT}

\dot{m} = \frac{0.6626*10^{5}*510}{287*311}

\dot{m} = 37.85kg/min

Therfore the mass flow rate at which water condenses, then

\eta = \frac{\dot{m_v}}{\dot{m}}

Re-arrange to find \dot{m_v}

\dot{m_v} = \eta*\dot{m}

\dot{m_v} = 0.97*37.85

\dot{m_v} = 36.72 kg/min

PART B) Enthalpy is given by definition as,

H= H_a +H_v

Where,

H_a= Enthalpy of dry air

H_v= Enthalpy of water vapor

Replacing with our values we have that

H=m*0.0291(38-25)+2500m_v

H = 37.85*0.0291(38-25)-2500*36.72

H = 91814.318kJ/min

In the conversion system 1 ton is equal to 210kJ / min

H = 91814.318kJ/min(\frac{1ton}{210kJ/min})

H = 437.2tons

The cooling requeriment in tons of cooling is 437.2.

3 0
3 years ago
Hellpppppppppp will give brainliest just help!!!!!!
aliya0001 [1]
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3 years ago
Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

R_b=e^{0.144}=1.154

So here B is more Reliable.

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ValentinkaMS [17]

Answer:

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Explanation:

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Answer:

I don't know. It does the same thing to me to. I had to create a new account with my friends email because it won't let me log in

Explanation:

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