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3 years ago

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A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of

its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa

1 answer:

lyudmila [28]3 years ago

7 0

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 150 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v L

$V_2$ = final volume of gas = $\frac{v}{3}L$

$150\times v=P_2\times \frac{v}{3}$

$P_2=450kPa$

Therefore, the new pressure of the gas will be 450 kPa.

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Answer:

Among the different types of excavation protection system, as a way of preventing accidents against cave-ins, the sloping involves cutting back the trench wall at an angle inclined away from the excavation. Shoring requires installing aluminum hydraulic or other types of supports to prevent soil movement and cave-ins. Shielding protects workers by using trench boxes or other types of supports to prevent soil cave-ins (OSHA). In addition, the regulations do not allow employees to work on excavations where there is an accumulation of water. If this occurs, water on the site must be constantly removed by suitable equipment preventing water from accumulating. The entry of surface water into the excavations must also be prevented by means of diversion ditches, dam, or other suitable means.

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3 years ago

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

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Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

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$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

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What is the instantaneous center of zero velocity? List two approaches for determining the is the instantaneous center of zero v

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From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

$I_{12},I_{23},I_{31} all\ are\ co-linear.$

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3 years ago

A heat pump cycle whose coefficient of performance is 2.5 delivers energy by heat transfer to a dwelling at a rate of 20kW.

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Answer:

a) 8kW

b) $128

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Given the coefficient of performance of the heat pump cycle to be 2.5

Energy delivered by the heat pump = 20kW

a) net power required to operate the heat pump = Energy delivered / coefficient of performance

Net power required = 20/2.5

= 8kW

b) Given the cost of electricity is $0.08 for 1kWhour

Since net power required to operate heat pump = 8kW

If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour

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2 years ago

An axial compressive load of 708 kN is applied to a cylindrical component, 81 mm in diameter and 418 mm long, made of aluminium.

dalvyx [7]

Answer:

The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)

Explanation:

The compressive stress in a cylindrical component can be calculated aby dividing the compressive force F to the cross sectional area A:

fc= F/A

If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.

For acylindrical component the cross sectional area A is:

A=πR^

If the diameter of the component is 81 mm, the radius is the half:

R=81mm /2 = 40.5 mm

Then A result:

A= 3.14 * (40.5 mm)^2 = 5150.4 mm^2

In square meters:

A= 3.14 * (0.0405 m)^2 = 0.005150 m^2

Replacing 708 kN to the force:

fc= 708 kN / 5150.4 mm^2 = 0.137 kN/mm^2

Using the force in Newtons:

F= 70800 N

Finally the compressive stress in Pa is:

fc= 708000 / 0.005150 m^2 = 137465051 Pa = 137 MPa

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2 years ago