In the figure show, what's the distance from point H to point C?

A water supply agency is planning to add two reservoirs to its system. Water will flow from Reservoir A to Reservoir B via a 10,

NikAS [45]

Attached is the solution to the above question.

3 0

3 years ago

When two or more simple machines are combined they form

Volgvan

Compound machine is the answer

8 0

2 years ago

Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a

Arada [10]

Answer:

a) $h_c = 0.1599 W/m^2-K$

b) $H_{loss} = 5.02 W$

c) $T_s = 302 K$

d) $\dot{Q} = 25.125 W$

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, $T_a = 25^0 C = 298 K$

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

$\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec$

Rate of heat transfer,

$\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W$

a) To calculate the convection coefficient relationship for heat transfer by convection:

$\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K$

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, $k_c = 0.8$

$\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K$

b) Heat loss from the pipe to the environment:

$H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W$

d) The required fan control power is 25.125 W as calculated earlier above

5 0

3 years ago

Two soils are fully saturated with liquid (no gas present) and the soils have the same void ratio. One soil is saturated with wa

mariarad [96]

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol = 0.8 g/cm³

solution

we get here water content that is express as

water content = $S_s \times \gamma _w$ ....................1

here soil is full saturated so $S_s$ is 100% in both case

so put here value for water

water content = 100 % × 1

water content = 1 g

and

now we get for alcohol that is

water content = 100 % × 0.8

water content = 0.8 g

so here water sample have more water content

5 0

3 years ago

The interior wall of a building is made from 2×4 wood studs, plastered on one side. If the wall is 13 ft high, determine the loa

Elanso [62]

Answer:

load = 156 lb/ft

Explanation:

given data

interior wall of a building = 2×4 wood studs

plastered = 1 side

wall height = 13 ft

solution

we get here load so first we get wood stud load and that is

we know here from ASCE-7 norm

dead load of 2 x 4 wood studs with 1 side plaster = 12 psf

and we have given height 13 ft

so load will be = 12 psf × 13 ft

load = 156 lb/ft

7 0

2 years ago