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solmaris [256]
3 years ago
6

A spherical hot air balloon is initially filled with air having 120 kPa pressure and 24 °C temperature. Initial diameter of the

hot air balloon was measured to be equal to 9.905 m. Before operating balloon the operator filled the balloon with air having 120 kPa pressure and 24 °C temperature via an inlet having 1.458 m diameter. It took operator 17.108 minutes to fill balloon to its operation diameter of 16.502 m. Calculate the velocity of air supply from the pump to the balloon through the inlet. Give your answer in m/s having 3 decimal points.
Engineering
1 answer:
tino4ka555 [31]3 years ago
8 0

Answer:

v = 1.076 m /s

Explanation:

Initial volume of balloon = 4/3 x  3.14 x (9.905/2)³

=508.56 m³

Final volume of balloon = 4/3 x 3.14 x (16.502/2)³

= 2351.73 m³

Increase in volume = 1843.17 m³

Cross sectional area of inlet  A  = 3.14 x( 1.458/2)²

A = 1.6687 m²

Volume rate of flow of air = cross sectional area x velocity of inflow

= 1 .6687 V [ V is velocity of inflow ]

Total time taken = Increase in volume / rate of flow of air

17.108 X 60 = 1843.17 / 1.6687 V

V = \frac{1843.17}{1.6687\times17.108\times60}

v = 1.076 m /s

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A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti
Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = \frac{0.03194}{0.315}

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m  

here h is heat required and m is mass flow rate

H = 2680.36  × 0.10139

H =  271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = \frac{H}{0.90}

net boiler heat = \frac{271.75}{0.90}

net boiler heat = 301.94 kW

5 0
3 years ago
A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an
Zigmanuir [339]

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y  ----- ( 1 )

M ( bending moment ) = R * length of span / 2

                                     = (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

                                    = ( 2 )*( d )^3  / 12 =  2d^3 / 12

b = 2 in ,  d = ?

length of span = 4 * 12 = 48 inches

R = P  / 2 =  240 * 10^3 / 2 =   120 * 10^3 Ib

y ( centroid distance ) = d / 2  inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

                = ( 288 * 10^4 * 12 ) / 2d^3 )  * d / 2

940300  = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d  ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188   ∴ Value of d ≈ 3.03 in

8 0
3 years ago
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at
andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:  

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that;  distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using  this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0
2 years ago
Explain when it is appropriate to use Tier II?
garik1379 [7]

????? what do u mean tier 2

3 0
3 years ago
Wright Company deposits all cash receipts on the day when they are received and it makes all cash payments by check. At the clos
alina1380 [7]

Answer:

                                              Wright Company

                                             Bank Reconciliation

                                                 May 31, 2013

Credit side                                                                                   Debit side

Bank statement $26200                 |                          Book balance $27900

<em>Add;                                                    </em>

Deposit on May 31 $6400

Bank error $420

Sub-total=$33020

Deductions;                                        |                       Deduct

ions

Outstanding checks $5800              |                 Bank service charge $120

Adjusted bank balance $27220       |                  NSF check $560

                                                                             Total deduction $680

                                                      Adjusted book balance $27220

8 0
3 years ago
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