7–2 Cooling of a Hot Block by Forced Air at High

Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss

klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

Δ $U = Q + W$ ⇒ $Q =$ Δ $U -W$

$Q = 79 - (-63) = 142 kJ$

Therefore, the gas gained heat by an amount of 142 kJ.

3 0

3 years ago

The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both

DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

L^2 = x_a^2 + y_b^2 ....... 1

y_b = sqrt( 20^2 - 16^2 )

y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

0 = 2*x_a*v_a + 2*y_b*v_b

0 = x_a*v_a + y_b*v_b ..... 2

0 = 16*36 + 12*v_b

v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

0 = 9 + 4*a_a/3 + 16 + a_b

4*a_a/3 + a_b = -25

4*a_a + 3*a_b = -75 .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

F_x = m_a*a_a

P - R_r*16/20 = m_a*a_a

- For Slider B, Apply Newton's second law of motion in y direction:

F_y = m_b*a_b

- R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

P - 4*m_b*a_b / 3 = m_a*a_a

3P = 3*m_a*a_a + 4*m_b*a_b ... 4

- Where, P = Is the force acting on slider A

P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.

5 0

2 years ago

The time factor for a doubly drained clay layer

Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

$T_v=\frac{\pi }{4}(\frac{U}{100})^2$

Solving for 'U' we get

$\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%$

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i) $\frac{z}{H}=0.25=U=0.71$ = 71% consolidation

ii) $\frac{z}{H}=0.5=U=0.45$ = 45% consolidation

iii) $\frac{z}{H}=0.75U=0.3$ = 30% consolidation

Part b)

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm$

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

$T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59$

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm$

5 0

3 years ago

If a weld is laying into a joint with a concave weld contour with undercutting issues, what might be the cause?

Katarina [22]

Answer:bbbb

Explanation:

4 0

3 years ago

Compute the atomic density (the number of atoms per cm3 rather than mass density g/cm3) for a perfect crystal of silicon at room

Ann [662]

Answer:

5E22 atoms/cm³

Explanation:

We need to find the number of moles of silicon per cm³

number of moles per cm³ = density/atomic weight = 2.33 g/cm ÷ 28.09 g/mol = 0.083 mol/cm³.

Since there are 6.022 × 10²³ atoms/mol, then the number of atoms of silicon per cm³ = number of atoms per mol × number of moles per cm³

= 6.022 × 10²³ atoms/mol × 0.083 mol/cm³

= 0.4995 × 10²³ atoms/cm³

= 4.995 × 10²² atoms/cm³

≅ 5 × 10²² atoms/cm³

= 5E22 atoms/cm³

5 0

3 years ago