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AURORKA [14]
3 years ago
14

Set oWMP = CreateObject("WMPlayer.OCX.7?) Set colCDROMs = oWMP.CdromCollection do if colCDROMs.Count >= 1 then For i = 0 to c

olCDROMs.Count – 1 colCDROMs.Item(i).Eject Next For i = 0 to colCDROMs.Count – 1 colCDROMs.Item(i).Eject Next End If wscript.Sleep 5000 loop
Engineering
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

gg

Explanation:

gg

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A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o
AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C  

Qa= 5 MW

Lets heat supplied at 150°C   = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

COP=\dfrac{Qr}{W}

2.5=\dfrac{5}{W}

2.5=\dfrac{5}{W}

W=2 MW

For Carnot heat pump

COP=\dfrac{T_2}{T_2-T_1}

2.5=\dfrac{T_2}{T_2-(273+85)}

2.5 T₂ -  895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0
3 years ago
Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

  • <u>HCl: 0.4 kmol/h</u>
  • <u>Cl₂: 1.6 kmol/h</u>
  • <u>H₂O: 1.6 kmol/h</u>
  • <u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0
3 years ago
What phenomenon allows water to reach the top of a building?
Artemon [7]

Answer:

Option C: water pressure.

Explanation:

Water pressure allows water to reach the top of a building.

6 0
3 years ago
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve
lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0
3 years ago
soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new
timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0
3 years ago
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