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3 years ago

Which of the following would NOT fall into one of the WHMIS 1988 classes?

Engineering

2 answers:

sweet [91]3 years ago

8 0

A. Compressed Gas Yup

My name is Ann [436]3 years ago

4 0

Answer:

Explanation:

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An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compress

pickupchik [31]

Answer:

<em>The temperature will be greater than 25°C</em>

Explanation:

In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.

mathematically

Change in the internal energy of a system ΔU = ΔQ + ΔW

in an adiabatic process, ΔQ = 0

therefore

ΔU = ΔW

where ΔQ is the change in heat into the system

ΔW is the work done by or done on the system

when work is done on the system, it is conventionally negative, and vice versa.

also W = pΔv

where p is the pressure, and

Δv = change in volume of the system.

In this case,<em> work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C </em>

8 0

3 years ago

Plis 3 conclusiones de este video

vazorg [7]

No hay videos? de cual video estás hablando?

6 0

2 years ago

The substance is steam (H2O). NOTE: The purpose of this problem is to illustrate that there are conditions where water vapor is

Gennadij [26K]

Answer:

See the attached pictures for detailed answer.

Explanation:

See the attached pictures for step by step explanation.

7 0

3 years ago

What is the primary function of NCEES?

joja [24]

National Council of Examiners for engineering and surveying a nonprofit organization

7 0

3 years ago

One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isotherm

maria [59]

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a) Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

6 0

3 years ago