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3 years ago

5

Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai

m correct? a. Yes b. No

1 answer:

Alika [10]3 years ago

5 0

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

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Liquid benzene and liquid n-hexane are blended to form a stream flowing at a rate of 1700 lbm/h. An on-line densitometer (an ins

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3 years ago

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

Dafna11 [192]

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

3 years ago

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee

Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given </u>

-The mass of C_o2 = 1 kg

-The volume of the tank V_tank = 1 m^3

-The added energy E = 14 W

-The time of adding energy t = 10 s

-The increase in specific internal energy Δu = +10 kJ/kg

-The change in kinetic energy ΔKE = 0 and The change in potential energy

ΔPE =0

<u>Required </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.

(c)The energy transferred by the heat transfer W in kJ and the direction of

the heat transfer.

Assumption

-Quasi-equilibrium process.

<u>Solution</u>

(a) The volume and the mass doesn't change then, the specific volume is constant.

v= V_tank/m ---> 1/1= 1 m^3/Kg

(b) The added work is defined by.

W =E * t ---> 14 x 10 x 3600 x 10^-3 = 504 kJ

(c) From the first law of thermodynamics.

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

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Answer:

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