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gizmo_the_mogwai [7]
3 years ago
13

(25 points).The heat is turned on in a building in a cold morning, but the heater does not burnthe natural gas properly and emit

s carbon monoxide (CO) at 55mg/hr. CO is detrimental tohumans at concentrations of 5 ppm or higher. If the ventilation of the building is refreshing the air(i.e., it brings fresh air from outside that contains no CO at a flow rate of 100 m3/hr, the buildinghas a volume of 500m3, and CO is slowly oxidized in the air to carbon dioxide (CO2) with a rateconstant of kd=0.0001mg.L-1.hr-1, will the COconcentration ever pass the 5ppm limit or not?Atomic weights: C-12, O-16, air-29; 1 mole of gas occupies 22.4L at the conditions of theproblem (1atm, 250C).
Engineering
1 answer:
nydimaria [60]3 years ago
5 0

Answer: Yes, it will be around 40ppm(at steady state)

Explanation:

The above question can be solved by performing material balance of CO in the building. The material balance equation is given by, Rate of change of CO in the building = Rate of inflow of CO into the  building  - Rate of outflow of CO from the blood

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Answer:

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Explanation:

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3 years ago
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Can you give me examples of reciprocating motion?
tensa zangetsu [6.8K]

Answer:

Other examples of reciprocating motion include piston pumps and compressors, roller pressure and tensioning systems, material testing devices, and insertion machines.

Explanation:

6 0
3 years ago
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

8 0
3 years ago
In casting experiments performed using a certain alloy and type of sand mold, it took 155 sec for a cube-shaped casting to solid
Tems11 [23]

Answer:

A) Cm = 2.232 s/mm²

B) Time taken to solidify = 74.3 seconds

Explanation:

(A) Since a side is 50mm and all sides of a cube are equal, thus, Volume of the cube is;V = 50 x 50 x 50 = 125,000 mm³

There are 6 faces of the cube, thus Surface Area A = 6 x (50 x 50) = 15,000 mm²

So, Volume/Area = (V/A) = 125,000/15,000 = 8.333 mm

Cm is given by the formula; Cm =[Tts] /(V/A)² where Tts is time taken to solidify and it's 155 seconds in the question. Thus;

Cm = 155/(8.333)²= 2.232 s/mm²

(B) For;Cylindrical casting with D = 30 mm and L = 50 mm.;

Volume of cylinder is;

V = (πD²L) /4

So,V = (π x 30² x 50)/4 = 35,343mm³

Surface area of cylinder is;

A = (2πD²)/4 + (πDL)

Thus, A = ((π x 30²)/2) + (π x 30 x 50) = 6126 mm²

Volume/Area is;

V/A = 35,343/6126 = 5.77 mm

Same alloy and mold type was hsed as in a above, thus, Cm is still 2.232 s/mm²

Since Cm =[Tts] /(V/A)²

Making Tts the subject, we have;

Tts =Cm x (V/A)²

Tts = 2.232 x (5.77)² = 74.3 seconds

3 0
3 years ago
Describe why the motion of a follower acted on by a cam is periodic motion.
PolarNik [594]
When you go and use a periodic motion there a reason
6 0
3 years ago
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