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ANTONII [103]
3 years ago
14

Help me is it a b c or d?

Engineering
1 answer:
AleksandrR [38]3 years ago
4 0

b is the correct answer of this question

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Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a
guapka [62]

Answer:

K_v=12.34

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo =   Discharge / area of the orifice

or

Vo = \frac{0.01}{\frac{\pi}{4}0.045^2}

or

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, V_v =   Discharge / area of the orifice

or

V_v = \frac{0.01}{\frac{\pi}{4}0.0675^2}

or

V_v = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

or

25 m = K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}

where,

K_v is the loss coefficient for the gate valve

on substituting the values, we get

25 m = 10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}

or

K_v\frac{2.79^2}{2\times9.81} = 4.898

or

K_v=12.34

3 0
4 years ago
What is the different between isometric view and isometric projection
yanalaym [24]

Answer:

All the dimensions in the isometric drawing are actual while when in the Isometric projection due to this it has to be the isometric scale is to be used.

7 0
2 years ago
A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of
SIZIF [17.4K]

Answer:

<em>number of mole is 31342.36 moles</em>

<em>mass is 125.369 kg</em>

<em></em>

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = \frac{4}{3} \pi r^3

V = \frac{4}{3}* 3.142* 4.5^3 = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n  = 76350000/2436 = <em>31342.36 moles</em>

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m =<em> 125.369 kg</em>

3 0
3 years ago
Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

8 0
3 years ago
Driving alone without the distraction from passengers reduces the risks you face in driving.
vredina [299]

Driving alone without the distraction from passengers reduces the risks you face in driving.

True

3 0
2 years ago
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