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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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Answer: B, repetitive practice! hope this helps. :)

Explanation:

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3 years ago
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Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a
Ann [662]

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

8 0
3 years ago
Which of these fuel injection systems operates with fuel injectors located only in the intake manifold near each intake
Stella [2.4K]

Answer:

C. Multipoint fuel injection

Explanation:

A fuel injection system can be defined as a system found in the engine of most automobile cars, used for the supply of a precise amount of fuel or fuel-air mixture to the cylinders in an internal combustion engine through the use of an injector.

There are different types of fuel injection system and these includes;

I. Central-point injection.

II. Throttle (single point) body injection.

III. Gasoline direct injection.

IV. Multipoint (port) fuel injection.

Multipoint fuel injection is a type of fuel injection system that operates with fuel injectors located only in the intake manifold near each intake valve and sprays fuel toward the valve. As a result, it allows for the supply of a precise amount of fuel and as such creating a better air-fuel ratio for automobile cars.

8 0
2 years ago
I don’t get this it’s hella hard
qwelly [4]

Answer:

V₂ = 20 V

Vt = 20 V

V₁ = 20 V

V₃ = 20 V

I₁ = 10 mA

I₃ = 3.33 mA

It = 18.33 mA

Rt = 1090.91 Ω

Pt = 0.367 W

P₁ = 0.2 W

P₂ = 0.1 W

P₃ = 0.067 W

Explanation:

Part of the picture is cut off.  I assume there is a voltage source Vt there?

First, use Ohm's law to find V₂.

V = IR

V₂ = (0.005 A) (4000 Ω)

V₂ = 20 V

R₁ and R₃ are in parallel with R₂ and the voltage source Vt.  That means V₁ = V₂ = V₃ = Vt.

V₁ = 20 V

V₃ = 20 V

Vt = 20 V

Now we can use Ohm's law again to find I₁ and I₃.

V = IR

I = V/R

I₁ = (20 V) / (2000 Ω)

I₁ = 0.01 A = 10 mA

I₃ = (20 V) / (6000 Ω)

I₃ = 0.00333 A = 3.33 mA

The current It passing through Vt is the sum of the currents in each branch.

It = I₁ + I₂ + I₃

It = 10 mA + 5 mA + 3.33 mA

It = 18.33 mA

The total resistance is the resistance of the parallel resistors:

1/Rt = 1/R₁ + 1/R₂ + 1/R₃

1/Rt = 1/2000 + 1/4000 + 1/6000

Rt = 1090.91 Ω

Finally, the power is simply each voltage times the corresponding current.

P = IV

Pt = (0.01833 A) (20 V)

Pt = 0.367 W

P₁ = (0.010 A) (20 V)

P₁ = 0.2 W

P₂ = (0.005 A) (20 V)

P₂ = 0.1 W

P₃ = (0.00333 A) (20 V)

P₃ = 0.067 W

7 0
3 years ago
"Continuous duty" is defined as
larisa86 [58]

Answer:

operates at a substantially constant load for an indefinitely long time. ... Short-time duty: operates at a substantially constant load for a time that is definite, short, and specified. Varying duty: the loads and intervals of operation change.

Explanation:

4 0
3 years ago
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