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kompoz [17]
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?
Engineering
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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Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w
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Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

4 0
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