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Anvisha [2.4K]
3 years ago
5

The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the

specimen is soaked in a bucket of water for sev-eral days until it is fully saturated, what should the saturated density be?
Engineering
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

the saturated density should be

Explanation:

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Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
2 years ago
A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging
dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

\alpha = \frac{k}{\rho c}

           = \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s

                   

Temperature at 28 mm distance after t time  = =  50 degree C

we know that

\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})

\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]

0.909 = erf{\frac{3.8245}{\sqrt{t}}}

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008  and erf w = 0.11246 so by interpolation we have

w = 0.08073

erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]

0.08073 = \frac{3.8245}{\sqrt{t}}

solving fot t we get

t = 2244.3 sec

3 0
2 years ago
A gas cylinder is connected to a manometer that contains water. The other end of the manometer is open to the atmosphere, which
horrorfan [7]

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa

4 0
3 years ago
1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g
Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x 10^{22}  m^{-3}

atomic weight = 58.69 g/mol

density = 8.8 g/cm³  

solution

we get here N that is

N  = \frac{N_A \times \rho}{A}   ...........1

N = \frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}

N = 9.030 \times 10^{28}  

and here no of vacancy will be

Nv = N \times e^{\frac{-Qv}{kT}}  .................2

put here value

4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}  

10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}

take ln both side

ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})

-14.468 = \frac{-Qv}{0.0968}  

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0
3 years ago
You need to own your own printer in order to join the 3D printing movement <br><br> True or False
Nata [24]

Answer:

False

Explanation:

7 0
3 years ago
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