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Nadusha1986 [10]
3 years ago
5

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

ensity is 855 kg/m3 and the viscosity is 2.1 × 10−2 Pa · s. What is the velocity in the pipe?It is desired to maintain the same Reynolds number of 2100 and the same velocity as in part (a) using a second fluid with a density of 925 kg/m3 and a viscosity of 1.5 × 10−2 Pa · s. What pipe diameter should be used?
Engineering
1 answer:
alexdok [17]3 years ago
8 0

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number (Re) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

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

where \rhois the density of the fluid, \mu is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm

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A coil having resistance of 7 ohms and inductance of 31.8 mh is connected to 230v,50hz supply.calculate 1. The circuit current 2
lora16 [44]

(1) The current in the circuit is 18.87 A,

(2) The phase angle is 54.97°

(3) The power factor is 0.574

(4) The power consumed is 2491.2 W

(1) To calculate the current in the circuit, first, we need to find the overall impedance of the circuit.

We can calculate the overall impendence of the circuit using the formula below.

  • Z = √[R²+(2πfL)²]........................ Equation 1

Where:

  • R = resistance of the coil
  • f = Frequency
  • L = Inductance of the coil
  • Z = Overall impedance of the circuit

From the question,

Given:

  • R = 7 ohms
  • L = 31.8 mH = 0.0318 H
  • f = 50 Hz
  • π = 3.14

Substitute these values into equation 1

  • Z = √[7²+(2×3.14×50×0.0318)²]
  • Z = √(49+99.7)
  • Z = √(148.7)
  • Z = 12.19 ohms.

Therefore we use the formula below to calculate the current in the circuit.

  • I = V/Z.................. Equation 2

Where:

  • V = Voltage
  • I = current in the circuit.

Given:

  • V = 230 V.

Substitute into equation 2

  • I = 230/12.19
  • I = 18.87 A

(2) To calculate the phase angle, we use the formula below.

  • ∅ = tan⁻¹(2πfL/R)............... Equation 3

Where:

  • ∅ = Phase angle.


Substitute into equation 3

  • ∅ = tan⁻¹(2×3.14×50×0.0318/7)
  • ∅ = tan⁻¹(9.9852/7)
  • ∅ = tan⁻¹(1.426)
  • ∅ = 54.97°

(3) To calculate the power factor, we use the formula below.

  • pf = cos∅............ Equation 4

Where:

  • pf = power factor.

Substitute the value of ∅ into equation 4

  • pf = cos(54.97°)
  • pf = 0.574.

(4) And Finally to calculate the power consumed we use the formula below.

  • P = V×I×pf................ Equation 5

Where:

  • P = The power consumed

Substitute the values into equation 5

  • P = 230(18.87)(0.574)
  • P = 2491.22 W


Hence, (1) The current in the circuit is 18.87 A, (2) The phase angle is 54.97° (3) The power factor is 0.574 (4) The power consumed is 2491.2 W

Learn more about Impedance here: brainly.com/question/13134405

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2 years ago
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anastassius [24]
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If the slotted arm rotates counterclockwise with a constant angular velocity of thetadot = 2rad/s, determine the magnitudes of t
astraxan [27]

Answer:

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Explanation:

The solution of the problem is given in the attachments

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Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.
nikitadnepr [17]

Answer:

a)  m_e= 3.05 Kg

b)  \rho=1072.3kg/m^3

c)  m_e= 3.05 Kg

Explanation:

From the question we are told that:

Beaker Mass m_b=1.20

Liquid Mass m_l=1.85

Balance D:

Mass m_d=3.10

Balance E:

Mass m_e=7.50

Volume v=4.15*10^{-3}m^3

a)

Generally the equation for Liquid's density is mathematically given by

m_e=m_b+m_l+(\rho*v)

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b)

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

m_d=m- (\rho *v )

m=3.10+ (1072.30 *4.15*10^{-3}m^3 )

m=18.10kg

c)

Generally the equation for E's Reading at A pulled is mathematically given by

m_e=m_b+m_l

m_e = 1.20 + 1.85

m_e= 3.05 Kg

6 0
3 years ago
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