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Lady_Fox [76]
3 years ago
14

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. 1) Starting with the most general form of the appropriate mass balance equation, determine the mass of air that has entered the tank (4 points). 2) Starting with the most general form of the appropriate energy balance equation, determine the amount of heat transferred and whether it was heat transferred in or out (8 points). 3) List at least 3 assumptions needed to complete this problem (3 points).

Engineering
1 answer:
Alja [10]3 years ago
4 0

Answer:

Check the explanation

Explanation:

First of all the initial or primary and final masses can be calculated with the use of the ideal gas relations.

The net It transfer is determined from the energy balance. The initial and final internal energies and the enthalpy of the air in the supply line are obtained from A-I] for the given temperatures.  

kindly check the attached image below to see working.

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Find the compressibility factor Z for oxygen at 3 MPa and 160 K.
saveliy_v [14]

Answer:

Z= 0.868

Explanation:

Given that

P= 3 MPa

T = 160 K

We know that

P v= Z R T

P= Pressure

v = specific volume

R= gas constant

T = Absolute temperature

Z=  Compressibility factor

Here specific volume of gas is not given so we assume that specific volume gas

v=0.012\ m^3/kg

We know that for oxygen gas constant

R = 0.259 KJ/kg.K

Now by putting the values

P v = Z R T

3000 x 0.012 = Z x 0.259 x 160

Z= 0.868

So  Compressibility factor is 0.868.

5 0
3 years ago
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in t
vovikov84 [41]

Answer:

Explanation:

r_2=∞

q=4\pi kT_1(T_2-T_1)\\

q=2\pi kD.ΔT--------(1)

q=hA ΔT=4\pi r_1^2(T_2_s-T_1_s)\\

N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9}  }  ------(3)

By equation (1) and (2)

2\pi kD.ΔT=h.4\pi r_1^2ΔT

2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\-------(4)

From equation (3) and (4)

So for sphere R_a→0

6 0
3 years ago
Consider a unidirectional continuous fiber-reinforced composite with epoxy as the matrix with 55% by volume fiber.i. Calculate t
ohaa [14]

Answer:

I)E= 40.95 GPa

II)E=5.29 GPa

Explanation:

I)

Given that

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ = 72.5 GPa   ,V₂=0.55

Longitudinal moduli  given as ;

E= E₁V₁+E₂V₂

E= 2.41 x 0.45 + 72.5 x 0.55 GPa

E= 40.95 GPa

II)

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ =230 GPa   ,V₂=0.55

Transverse moduli given as:

\dfrac{1}{E}=\dfrac{V_1}{E_1}+\dfrac{V_2}{E_2}

\dfrac{1}{E}=\dfrac{0.45}{2.41}+\dfrac{0.55}{230}

E=5.29 GPa

7 0
3 years ago
The mean of 10 numbers is 9, then the sum (total) of these numbers will be​
qwelly [4]

Answer:

90

Explanation:

mean is basically taking the sum of all numbers and then dividing the sum with the number of all given numbers..

here, the mean is 9, total numbers are 10.. so the sum will be 9 multiplied by 10, that is 90.

5 0
3 years ago
Read 2 more answers
Base course aggregate has a target dry density of 119.7 lb/cu ft in place. It will be laid down and compacted in a rectangular s
natita [175]

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

given data

dry density = 119.7 lb/cu ft

area = 2000 ft × 48 ft × 6 in

aggregate = 3.1%

required compaction = 95%

solution

we get  here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5 = 48000 ft³

when here space fill with aggregate of density is

density = 0.95 × 119.7    = 113.72 lb/ft³

and

dry weight of this aggregate will be  is

dry weight = 48000 × 113.72 = 5458320 lbs

and

we consider take percent moisture by weigh so that there weight of moisture in aggregate is express as

weight of moisture = 0.031 × 5458320 = 169208 lbs

and

total weight of aggregate will be

total weight of aggregate = 5458320 + 169208

total weight of aggregate =  5627528 lbs = 2814 tons  

5 0
3 years ago
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