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3 years ago

11. Technicians A and B are discussing

Engineering

1 answer:

algol [13]3 years ago

5 0

Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

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3 years ago

An empty metal can is heated to 908C and sealed. It is then placed in a room to cool to 208C. What is the pressure inside the ca

Natali5045456 [20]

The pressure inside the can upon cooling is 0.4 atm.

<u>Explanation:</u>

Given -

Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K

Final Temperature, T2 = 208°C = 208 + 273 K = 481 K

Pressure upon cooling, P2 = ?

Using Gay Lussac's law:

P1/T1 = P2/T2

P2 = P1 X T2 / T1

P2 = 1 atm X 481 / 1181

P2 = 0.4 atm

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3 years ago

The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphe

Sonbull [250]

Answer:

a. The time required for the tank to empty halfway is presented as follows;

$t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$

b. The time it takes for the tank to empty the remaining half is presented as follows;

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The total time 't', is presented as follows;

$t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }$

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

$\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}$

$\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}$

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

$dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh$

The time for the tank to drop halfway is given as follows;

$\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh$

$t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)$

$t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)$

$t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$ The time required for the tank to empty halfway, t₁, is given as follows;

$t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

$\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh$

$t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)$

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The total time, t, to empty the tank is given as follows;

$t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}$

$t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }$

3 0

3 years ago

In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini

storchak [24]

Answer:

attached below

Explanation:

8 0

3 years ago