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kompoz [17]
3 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?
Engineering
1 answer:
WITCHER [35]3 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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Water is the working fluid in an ideal Rankine cycle. Superheatedvapor enters the turbine at 10MPa, 480°C, and the condenser pre
choli [55]

Answer:

Explanation:

Given that:

Superheated vapor enters the turbine at 10 MPa, 480°C,

From the tables of superheated steam tables; the following values are obtained

h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state

s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}

s_2 =0.51624 + x(7.82)

s_2 =0.51624 + 7.82x

From the values obtained;

s_1 =s_2= 6.52846 \ kJ/kg.K

Therefore;

6.52846 = 0.51624+7.82x

6.52846 - 0.51624 = 7.82 x

6.01222  = 7.82 x

x = 6.01222/7.82

x = 0.7688

The enthalpy for this process at state (s_2) can be determined as follows:

h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 )  \\ \\ h_2 =2010.4084   \ kJ/kg

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:

n_T = \dfrac{h_1-h_{2a}}{h_1-h_2}

0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}

0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}

0.8 * {1311.6116}= {3322.02-h_{2a}

1049.28928=  {3322.02-h_{2a}

h_{2a}=   {3322.02- 1049.28928

h_{2a}=   2272.73072 kJ/kg

The work pump is calculated by applying the formula:

w_p = v_{f  \  6 kpa} (p_4-p_3)

w_p = 0.0010062 * (10000-6)

w_p = 0.0010062 *9994

w_p = 10.0559628 \  kJ/kg

However;

w_p = h_4 -h_3

From the process;

h_3 = h_{f(6 kpa)} = 150.15 \  kJ/kg

10.0559628 = h_4 - 150.15

10.0559628+  150.15 = h_4

160.2059628= h_4

h_4= 160.2059628 \  kJ/kg

The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:

n_P = \dfrac{h_4-h_{3}}{h_{4a}-h_3}

6 0
3 years ago
Consider the equation y = 10^(4x). Which of the following statements is true?
o-na [289]

Answer: Plot of  \log y vs x would be linear with a slope  of 4.

Explanation:

Given

Equation is y=10^{4x}

Taking log both sides

\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x

It resembles with linear equation y=mx+c

Here, slope of \log y vs x is 4.

5 0
3 years ago
The wet density of a sand was found to be 1.9 Mg/m3 and the field water content was 10%. In the laboratory, the density of solid
Nookie1986 [14]

Answer:

a) 44.4%

b) 72 mm

Explanation:

See attached pictures.

3 0
3 years ago
A strain gage is mounted at an angle of 30° with respect to the longitudinal axis of the cylindrical pressure. The pressure vess
GuDViN [60]

Answer:

1790 μrad.

Explanation:

Young's modulus, E is given as 10000 ksi,

μ is given as 0.33,

Inside diameter, d = 54 in,

Thickness, t = 1 in,

Pressure, p = 794 psi = 0.794 ksi

To determine shear strain, longitudinal strain and circumferential strain will be evaluated,

Longitudinal strain, eL = (pd/4tE)(1 - 2μ)

eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)

eL = 3.64 x 10-⁴ radians

Circumferential strain , eH = (pd/4tE)(2-μ)

eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)

eH = 1.79 x 10-³ radians

The maximum shear strain is 1790 μrad.

4 0
3 years ago
Hey friends.... ajao bat Kare ✌️✌️❤️​
Dvinal [7]

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4 0
3 years ago
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