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Natasha2012 [34]

3 years ago

15

The density of aluminum is 2.7 × 103 kg/m3 . the speed of longitudinal waves in an aluminum rod is measured to be 5.1 × 103 m/s.

what is the value of young's modulus for this aluminum?

1 answer:

andrey2020 [161]3 years ago

5 0

<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2. So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density. So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave. Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>

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Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

<u>At State 1:</u>

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

<u>At State 1:</u>

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

<u>At State 1;</u>

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$

At $T_1$ = 125° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )$

$=278.93 + ( 7.23) (\dfrac{8}{10} )$

$= 284.714 \ kJ/kg\\$

At $T_1$ = 125° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg$

$U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )$

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

<u>At State 2:</u>

The internal energy is calculated as:

$U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )$

At temperature 110° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )$

$271.69+ (7.24) (0.3)$

$= 273.862 \ kJ/kg\\$

At temperature 110° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg$

$U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )$

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

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3 years ago

Calcular la energía cinética de un cometa cuya masa es de 5×10 elevado a 31 kg y se mueve con velocidad de 216000km/h

PolarNik [594]

The kinetic energy is $9\cdot 10^{40}J$ .

Explanation:

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

K is the kinetic energy of the object

m is the mass of the object

v is the speed of the object

For the comet in this problem, we have:

$m=5\cdot 10^{31} kg$ is its mass

$v=216,000 km/h$ is the speed

First, we convert the speed from km/h to m/s:

$v=216,000 \frac{km}{h} \cdot \frac{1000 m/km}{3600 s/h}=60,000 m/s$

Therefore, the kinetic energy of the comet is

$K=\frac{1}{2}(5\cdot 10^{31})(60,000)^2=9\cdot 10^{40}J$

Learn more about kinetic energy here:

brainly.com/question/6536722

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3 years ago

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