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mamaluj [8]
2 years ago
5

A 0.3 m pipe carries water at a velocity of 24.4 m/s. At points A and B measurements of pressure and elevation were respectively

361 KN/m2 and 288 KN/m2 , and 30.5 and 33.5 m. For steady flow, find the loss of head between A and B.?

Engineering
1 answer:
Ksju [112]2 years ago
5 0

Answer:

The answer is attached

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Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i
leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0
2 years ago
The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt
Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

3 0
2 years ago
A horizontal, cylindrical, tank, with hemispherical ends, is used to store liquid chlorine at 10 bar. The vessel is 4 m internal
sp2606 [1]

Answer: 0.021818m =2.18x10^-²m

Explanation: Using Laplace principle

Design pressure is =12 bar= 1200000N/m²

T = Wall thickness

Design stress=110Mn/m²= 110,000,000N/m²

Radius= diameter/2 =4/2=2m

Design stress=Hoop stress =Pr/t where p=internal pressure or internal pressure, r=radius and t= wall thickness.

As Laplace equation stated.

110,000,000= 1,2000,00 x 2/t

t= 2,400,000/110,000,000.

t= 0.021818m

t=2.18x10^-2m.

3 0
2 years ago
Oliver is designing a new children’s slide to increase the speed at which a child can descend. His first design involved steel b
AVprozaik [17]

Answer:

The correct option is;

A) Steel becomes too hot in the Sun and can burn the children

Explanation:

The properties of steel includes;

Low specific heat capacity, high thermal and electrical toughness, high hardness, high tensile strength, high yield strength, appreciable elongation, high fatigue strength, can easily corrode, high malleability and ability to creep

Therefore, due to the low specific heat capacity, which is 0.511 J/(g·°C) and high conductivity of steel which is about 32 W/(m·k), the temperature of the steel can rapidly rise and the hot steel surface can readily conduct the heat, (due to the temperature difference) to other bodies that come in contact

8 0
2 years ago
. A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa a
Delvig [45]

Answer:

Qin = 1857 kJ

Explanation:

Given

m = 0.5 Kg

T₁ = 25°C = (25 + 273) K = 298 K

P₁ = 100 kPa

P₂ = 500 kPa

First, the temperature when the piston starts rising is determined from the ideal gas equations at the initial state and at that state:

T₂ = T₁*P₂/P₁

⇒  T₂ = 298 K*(500 kPa/100 kPa) = 1490 K

Until the piston starts rising no work is done so the heat transfer is the change in internal energy

Qin = ΔU = m*cv*(T₂-T₁)

⇒   Qin = 0.5*3.1156*(1490 - 298) kJ = 1857 kJ

8 0
2 years ago
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