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3 years ago

Can someone pls give me the answer to this?

Engineering

1 answer:

Zolol [24]3 years ago

7 0

During World War ll, the United States and the Soviet Union fought together as the allies against the Axis power. However, the relationship between the two nations was a tense one. Americans had long been wary of Soviet communism and concerned about Russian leader Joseph Stalin’s tyrannical rule of his own country

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Question 2 (Multiple Choice Worth 3 points)

ololo11 [35]

Answer:

the car to the right

Explanation:

its in the name the RIGHT of way hope it helps good luck

6 0

3 years ago

Help please really fast!!

ra1l [238]

Answer:368 hdhtygtÿ

901 vuiøöńč

Explanation:

6 0

3 years ago

2. Determine the surface area of a primary settling tank sized to handle a maximum hourly flow of 0.570 m3/s at an overflow rate

Hitman42 [59]

Answer:

The surface area of the primary settling tank is 0.0095 m^2.

The effective theoretical detention time is 0.05 s.

Explanation:

The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.

Volumetric flow rate = 0.570 m^3/s

Overflow rate = 60 m/s

Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2

Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate

Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3

Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s

7 0

3 years ago

Read 2 more answers

I am trying to test out the software Classroom relay and I am just ask if there is any way kids can stop Classroom relay form se

Inessa05 [86]

Answer:

What is classroom relay?

Plz answer in ch-at

Explanation:

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3 years ago

Read 2 more answers

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