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andrey2020 [161]
4 years ago
15

Which of the following waste characteristic is considered to be hazardous:

Engineering
1 answer:
Lynna [10]4 years ago
8 0

Answer:

D) all of the above

Explanation:

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A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur
masya89 [10]

Answer:

Q_{in} = 46.454\,kJ

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

Q_{in} = U_{2} - U_{1}

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

P = 101.42\,kPa

T = 100\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 675.761\,\frac{kJ}{kg}

x = 0.123

State 2 - (Liquid-Vapor Mixture)

P = 476.16\,kPa

T = 150\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 1643.545\,\frac{kJ}{kg}

x = 0.525

The mass stored in the vessel is:

m = \frac{V}{\nu}

m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }

m = 0.048\,kg

The heat transfer require to the process is:

Q_{in} = m\cdot (u_{2}-u_{1})

Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )

Q_{in} = 46.454\,kJ

3 0
3 years ago
By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer
Mademuasel [1]

Answer: Attached below is the well written question and solution

answer:

i) Attached below

ii) similar parameter =  \frac{V}{VoL } = 1 / Re

Explanation:

Using ;  L as characteristic length and Vo as reference velocity

i) Nondimensionalize the equations

ii) Identifying similarity parameters

the similar parameters are  = \frac{V}{VoL } = 1 / Re

Attached below is the detailed solution

7 0
3 years ago
¿Cómo nos podría ayudar una hoja de cálculo en nuestro estudio?
Ghella [55]

Las hojas de cálculo en Excel facilitan los cálculos numéricos a través del uso de fórmulas; de manera fácil y rápida se pueden hacer operaciones aritméticas sobre cientos de miles de datos numéricos; por lo que se puede actualizar o corregir cualquiera de los datos numéricos y las operaciones se recalculan

3 0
3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
3 years ago
A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?
blagie [28]

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0
3 years ago
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