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qaws [65]
3 years ago
11

A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flo

wing in the supply line at 600 kPa and 22°C. The valve is opened, and air is allowed to enter the tank until the pressure in the tank reaches the line pressure, at which point the valve is closed. A thermometer placed in the tank indicates that the air temperature at the final state is 77°C. Determine (a) the mass of air that has entered the tank and (b) the amount of heat transfer
Engineering
1 answer:
Finger [1]3 years ago
3 0

Answer:

9.58 Kg of air has entered the tank.

heat entered=3483.76 Kilo.Joule

Explanation:

(A) R=287 Kilo.J/Kg.K

as per initial conditions P=100 Kilo.Pa ,V=2 cubic meter, T=22 C=295.15 K,

using the relation P*V=m*R*T

m=(100*1000*2)/(287*295.15)=2.36 Kg this is the mass that is already present in tank.

after filling tank at 600 Kilo.Pa.

P=600 Kilo Pa T=77 C=350.15 K

P*V=m*R*T

m=(600*1000*2)/(287*350.15)=11.94 Kg

mass that has entered=11.94-2.36=9.58 Kg

(b) using air psychometric  property table

specific heat content initial  100 KILO Pa and 22 C=295.576 Kilo.Joule/Kg

specific heat content final  600 Kilo Pa and 77 C=350.194 Kilo.Joule/Kg

heat at initial stage=295.576*2.36=697.56 Kilo.Joule

heat at final stage=350.194*11.94=4181.32 Kilo.Joule

heat entered=4181.32-697.56=3483.76 Kilo.Joule

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kkurt [141]

Answer:

For block A, a = 9.66 ft/s²

For block B, a = 15 ft/s²

Explanation:

A free body diagram for this force system is attached to this solution

Mass of block A = m₁ = 8 lb

Mass of block B = m₂ = 6 lb

Coefficient of kinetic friction = μ

Normal reaction on the blocks = N

Spring stiffness of the spring btw block A and B = k = 20 lb/ft

Compression of the spring = 0.2 ft

Analysing Block A first

The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 8 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 8/32.2 = 0.248 lbm

(20×0.2) - (0.2 × 8) = (0.248) aₓ

aₓ = 9.66 ft/s²

The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 6 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 6/32.2 = 0.186 lbm

(20×0.2) - (0.2 × 6) = (0.186) aₓ

aₓ = 15 ft/s²

4 0
3 years ago
Air enters a tank through an area of 0.2 ft2 with a velocity of 15 ft/s and a density of 0.03 slug/ft3. Air leaves with a veloci
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Answer:

please find attached.

Explanation:

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3 years ago
A 1000-turn coil of wire 1.0 cm in diameter is in a magnetic field that increases from 0.10 T to 0.30 T in 10 ms. The axis of th
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emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

Magnetic field (Initial)= 0.10 T

Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

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E=0.0785(.2/10*10-3)

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4 0
3 years ago
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Answer:

b) false

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Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

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Most pedestrian fatalities occur
Makovka662 [10]

Answer: Outside an intersections

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6 0
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