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kolezko [41]
3 years ago
11

Water from an upper tank is drained into a lower tank through a 5 cm diameter iron pipe with roughness 2 mm. The entrance to the

pipe has minor loss coefficient 0.4 and the exit has minor loss coefficient of 1, both referenced to the velocity in the pipe. The water level of the upper tank is 4 m above the level of the lower tank, and the pipe is 5 m long. You will find the drainage volumetric flow rate. a) What is the relative roughness
Engineering
1 answer:
nignag [31]3 years ago
5 0

Answer:

Relative roughness = 0.04

Explanation:

Given that:

Diameter = 5 cm

roughness = 2 mm

At inlet:

Minor coefficient loss k_{L1} = 0.4

At exit:

Minor coefficient loss k_{L2} = 1

Height h = 4m

Length = 5 m

To find the relative roughness:

Relative roughness is a term that is used to describe the set of irregularities that exist inside commercial pipes that transport fluids. The relative roughness can be evaluated by knowing the diameter of the pipe made with the absolute roughness in question. If we denote the absolute roughness as e and the diameter as D, the relative roughness is expressed as:

e_r = \dfrac{e}{D}

e_r = \dfrac{0.2 }{5}

\mathbf{e_r = 0.04}

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Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.
saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 \mathbf{m^{3}}

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; v_1 = vg_1 = 0.0996 \mathbf{m^{3}} / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 \mathbf{m^{3}} / kg

From temperature T₂ = 200⁰ C

v_f_2 = 0.0016 \ m^3/kg  

vg_2 = 0.127 \ m^3/kg  

Since  vf_2 < v_2 , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}

x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}

x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}

\mathsf{x_2 =0.78}

At temperature T₂, the specific internal energy u_f_2 = 850.6 \ kJ/kg , also ug_2 = 2594.3 \ kJ/kg

Thus,

u_2 = uf_2 + x_2 (ug_2 -uf_2)

u_2 =850.6  +0.78 (2594.3 -850.6)

u_2 =850.6  +1360.086

u_2 =2210.686 \ kJ/kg

<u>At state 3:</u>

Temperature T_3=T_2 = 200 ^0 C ,

V_3 = 2V_1 = 0.06 \ m^3

Specific volume v_3 = 0.2  \ m^3/kg

Thus; vg_3 =vg_2 = 0.127 \ m^3/kg ,

SInce v_3 > vg_3, therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at v_3 = 0.2  \ m^3/kg and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 \ m^3/kg

The specific internal energy u_3 at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

u_2-u_1

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

u_3-u_2

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0
3 years ago
In a black box experiment, when the amount of material exiting a closed system is less than the amount of material entering the
Oksi-84 [34.3K]

When the material that exits is lesser in amount than that of the entering material in a black box experiment, the parts of the system need to be changed.

<h3>What happens in a black box experiment?</h3>

In a black box experiment, the experimenters need to make assumptions regarding the drawing of conclusions. One such conclusion is the amount of material that exits.

If such amount is lesser than the one that enters the system, such experiment concludes that it is the time to change the parts of the system.

Hence, option D holds true regarding the black box experiment.

Learn more about black box experiment here:

brainly.com/question/13403296

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4 0
2 years ago
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

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3 years ago
Design process 8 steps with definition​
Troyanec [42]

Answer:

Step 1: Define the Problem.  

Step 2: Do Background Research. .

Step 3: Specify Requirements. .

Step 4: Brainstorm, Evaluate and Choose Solution.  

Step 5: Develop and Prototype Solution.  

Step 6: Test Solution.

Step 7: Does Your Solution Meet the Requirements?  

Step 8: Communicate Results.

can u tell me the definition tho?

palled correctly as “though” which is an alternate form of “although”) at the end is informal usage. It's better placed before “she seems better today

8 0
3 years ago
• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma
elena-s [515]

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

T(n) = 10\times 2n

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

\frac{T(k)}{T(n)} = 64

\frac{20k}{20n} = 64

k = 64n

Thus the new machine will process 64 inputs in the time duration T

8 0
4 years ago
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