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disa [49]
2 years ago
6

A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False

Engineering
1 answer:
ozzi2 years ago
8 0

Answer:

True

Explanation:

For point in xz plane the stress tensor is given by\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]

where Dx is the direct stress along x ; Dz is direct stress along z ;  tzx and txz are the  shear stress components

We know that the stress tensor matrix is symmetrical which means that tzx = txz  ( obtained by moment equlibrium )

thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane

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What is an air mass?​
kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

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3 years ago
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Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per
kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

                                                             $= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$. Then,

Power $ \propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

            $P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

           $P_2 = \text{const.} \times 1000 \ W$

5 0
2 years ago
Which of the following has special properties that allow forces and pressure to be distributed evenly?
Thepotemich [5.8K]

Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

  • Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
  • Therefore, when pressure is applied to them, it permeates evenly on all parts.
  • Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.
7 0
3 years ago
Are front-end engineers starting to decline in China?
laiz [17]

Yes. They are declining in China. Very fast

7 0
3 years ago
.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves
vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 m^3/kg

Specific volume of saturated liquid=v_f=0.00102 m^3/kg

Now the relation  of total specific volume for a specific dryness value is given as

                                  v=v_f+x(v_g-v_f)

Substituting the values give

v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg

Now the volume of vessel is given as

v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3

So the volume of vessel is 4.7680m^3.

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ

So the quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0
3 years ago
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