Air enters a 28-cm diameter pipe steadily at 200 kPaand 208C with a velocity of 5 m/s. Air is heated as it flows, and leaves the pipe at 180 kPa and 408C. Determine (a) the volume flow rate of air at the inlet, (b) the mass flow rate of air, and(c) the velocity and volume flow rate at the exit.

Answers

Answer 1

Answer:

(a) [tex]\dot V_1[/tex] = 0.308 m³/s

(b) [tex]\dot m[/tex] = 0.732 kg/m³

(c) v₂ = 5.94 m/s.

Explanation:

(a) The volume flow rate is given by the cross sectional area of the pipe × Velocity of flow of air

Diameter of pipe = 28 cm = 0.28 m

The cross sectional area, A, of the pipe = 0.28²/4×π = 0.0616 m²

Volume flow rate = 5 × 0.0616  = 0.308 m³/s

[tex]\dot V_1[/tex] = 0.308 m³/s

(b) From the general gas equation, we have;

p₁v₁ = RT₁ which gives;

p₁/ρ₁ = RT₁

ρ₁ = p₁/(RT₁)

Where:

ρ₁ = Density of the air

p₁ = 200 kPa

T₁ = 20 C =

R = 0.287 kPa·m³/(kg·K)

ρ₁ = 200/(0.287 ×293.15) = 2.377 kg/m³

The mass flow rate = Volume flow rate × Density

The mass flow rate, [tex]\dot m[/tex] = 2.377×0.308 = 0.732 kg/m³

[tex]\dot m[/tex] = 0.732 kg/m³

(c) The density at exit, ρ₂, is found from the the universal gas equation as follows;

ρ₂ = p₂/(RT₂)

Where:

p₂ = Pressure at exit = 180 kPa

T₂ = Exit temperature = 40°C = 273.15 + 40 = 313.15 K

∴ ρ₂ = 180/(0.287×313.15) = 2.003 kg/m³

[tex]\dot m[/tex] = ρ₂×[tex]\dot V_2[/tex]

[tex]\dot V_2[/tex] = [tex]\dot m[/tex]/ρ₂ = 0.732/2.003 = 0.366 m³/s

[tex]\dot V_2[/tex] = v₂ × A

v₂ = [tex]\dot V_2[/tex]/A = 0.366/0.0616 = 5.94 m/s.

v₂ = 5.94 m/s.


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When adjusting your side mirrors, the horizon should be _______ the mirror. A. above the center of B. across the center of C. below the center of D. directly behind

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Answer:

B. ACROSS THE CENTER OF

Trust me I've seen this before

Hope this is correct

HAVE A GOOD DAY!

Answer:

B, across the center

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Royce has a Bevel protractor with a damaged vernier scale. What limitation will Royce face if he uses this protractor to measure angles?

Answers

Answer:

Limitation in the level of possible accuracy of the that can be obtained to mainly whole degrees

Explanation:

The angle measurement values are located around the circularly shaped bevel protractor to which a Vernier scale can be attached to increase the accuracy of the angle measurement reading

The accuracy of the bevel protractor is up to 5 arc minutes or 1/12° with the space on the Vernier scale having graduations of 1/12°, such that two spaces on the main scale is 5 arc minutes more than a space on the Vernier scale and the dimensions of the coincidence of the two scale will have an accuracy of up to 1/12°

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Answers

Answer:

When the expenditure increased, then the consumer's expenditure is increased to Rs.150 and when the price falls of the good it becomes Rs.5. Then, Good X will be Rs.10.

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BHE:FLI:JPM next sequence

Answers

Answer:

NTQ

Explanation:

The given sequence is

BHE : FLI : JPM

If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.

B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.

Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.

H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.

Alphabets on third places are E, I, M. Difference between their place vales is 4.

E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.

Therefore, the next term is NTQ.

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Answers

Answer:

Technician B is the answer

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Answer:

no, it seems that everyone is having the same issue. If you use the app you can still find the answers and see them.

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