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2 years ago

9

What is the maximum volume flow rate, in m^3/hr, of water at 15.6°C a 10-cm diameter pipe can carry such that the flow will be l

aminar?

1 answer:

WITCHER [35]2 years ago

7 0

Answer:

Q = 0.00017597 m3/sec

Explanation:

given data:

temperature of water 15.6 degree celcius

diameter = 10cm =0.1 m

dynamix viscosity of water = .0011193 kg/m sec at 15.6 degree celcius

density of water at 15.6 degree = 999.07 kg/m3

for laminar flow REYNOLD NUMBER < 2000

maximum flow rate =area*velocity

Reynold number $Re = \frac{\rho vd}{\mu}$

$v = \frac{Re \mu}{\rho d}$ $v = \frac{2000*0.00011193}{999.07 *0.1}$

v = 0.022406 m/sec

$Q =\frac{\pi}{4} d^2 v$

where Q is flow expressed in m^3/s

$Q =\frac{\pi}{4} *0.1^2 *0.022406$

Q =0.00017597 m3/sec

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Answer: 15%

Explanation:

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1 year ago

The modifications of superheat and reheat for a vapor power plant are specifically better for the operation which of the followi

MatroZZZ [7]

The modifications of superheating and reheat for a vapor power plant are specifically better for the operation which of the following components b.Boiler.

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2 years ago

The function below takes a single argument: data_list, a list containing a mix of strings and numbers. The function tries to use

pychu [463]

Answer:

def filter_only_certain_strings(data_list):

new_list = []

for data in data_list:

# fix here. change >= to >

# because we want to return strings longer than 5 characters in length.

if type(data) == str and len(data) > 5:

new_list.append(data)

return new_list

Explanation:

def filter_only_certain_strings(data_list):

new_list = []

for data in data_list:

# fix here. change >= to >

# because we want to return strings longer than 5 characters in length.

if type(data) == str and len(data) > 5:

new_list.append(data)

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3 0

3 years ago

An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h

EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ( $\eta$ ), dimensionless, as:

$\eta = \frac{\dot W_{out}}{\dot W_{in}}$ (Eq. 1)

Where:

$\dot W_{in}$ - Maximum power possible from hydrogen flow, measured in kilowatts.

$\dot W_{out}$ - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

$\dot W_{in} = \dot V\cdot \rho \cdot L_{c}$ (Eq. 2)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$\rho$ - Density of hydrogen, measured in kilograms per cubic meter.

$L_{c}$ - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that $\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}$ , $\rho = 0.0899\,\frac{kg}{m^{3}}$ , $L_{c} = 141790\,\frac{kJ}{kg}$ and $\dot W_{out} = 80\,kW$ , then the efficiency of this fuel cell is:

(Eq. 1)

$\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 99.143\,kW$

(Eq. 2)

$\eta = \frac{80\,kW}{99.143\,kW}$

$\eta = 0.807$

The efficiency of this fuel cell is 80.69 percent.

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2 years ago

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ivolga24 [154]

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