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3 years ago

Aspectos de las técnicas escultóricas

1 answer:

3 0

Técnicas y materiales de escultura
Modelado de arcilla. Clay tiene muchas ventajas para un artista. ...
Modelado de cera. El modelado con cera permite a un artista crear un diseño único. ...
Esculpir. ...
Talla de madera. ...
Talla de marfil. ...
Piedras semipreciosas y tallas de conchas. ...
Fundición de bronce. ...
Decoración.

Corrígeme si estoy equivocada.. :)

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Two bodies have heat capacities (at constant volume) c, = a and c2 = bT and are thermally isolated from the rest of the universe

prisoha [69]

Answer:

Explanation:

The answer to the above question is given in attached files.

8 0

3 years ago

Why is it important to review your plan when you write an algorithm?

user100 [1]

An algorithm is itself a general step-by-step solution of your problem. ... The most important point here is that you must use algorithms to solve problem, one way or the other. Most of the time it's better to think about your problem before you jump to coding - this phase is often called design.

7 0

3 years ago

A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p

Stella [2.4K]

Answer:

Cut-off ratio $\dfrac{V_3}{V_2}=6.15$

Cxpansion ratio $\dfrac{V_4}{V_3}=3.25$

The exhaust temperature $T_4=1997.5R$

Explanation:

Compression ratio CR(r)=20

$\dfrac{V_1}{V_2}=20$

$P_2=P_3=920 psia$

$T_1=520 R ,T_{max}=T_3,T_3=3200 R$

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

$\dfrac{T_2}{T_1}=r^{\gamma -1}$

$\dfrac{T_2}{520}=20^{1.4 -1}$

$T_2=1723.28R$

$\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}$

$\dfrac{V_3}{V_2}=\dfrac{3200}{520}$

So cut-off ratio $\dfrac{V_3}{V_2}=6.15$

$\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}$

Now putting the values in above equation

$\dfrac20=\dfrac{V_4}{V_3}\times 6.15$

$\dfrac{V_4}{V_3}=3.25$

So expansion ratio $\dfrac{V_4}{V_3}=3.25$ .

$\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}$

$\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}$

$T_4=1997.5R$

So the exhaust temperature $T_4=1997.5R$

3 0

4 years ago

Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 m/s and leaves at 300 kPa and 350 K. Using variable spec

adoni [48]

Answer: a) Efficiency = 0.92

b) V = 319.19 m/s

c) S = 0.012 kj/kg.k

Explanation: please find the attached files for the solution

3 0

3 years ago

Consider an ideal gas undergoing a constant pressure process from state 1 to state

Radda [10]

Answer:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

$ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}$

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

$\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,$

We obtain the expression to compute the specific entropy change:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Best regards.

6 0

4 years ago