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3 years ago

Define cooling tower "range" as it applies to cooling towers.

Engineering

1 answer:

Arlecino [84]3 years ago

8 0

Answer:

The answer to the range of a cooling tower is the difference between the temperature of the water going in and the temperature of the water going out of the tower.

Explanation:

Cooling towers are used to cool a stream of water using the air of the environment where it's placed or installed. In theory, the lowest temperature the water can be cooled to would be the wet bulb temperature (temperature of the air going in the cooling tower). The range would be the difference (R=Ti-To) between the temperature of the hot water going in and the cooler water going out.

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What type of car engine is best for cold weather.

Komok [63]

Answer:Antifreeze/coolant

Explanation: keeps your engine cool in warm weather and keeps it from freezing up in the winter. A 50-50 mix of full strength coolant and water generally protects to around -30 degrees Fahrenheit. Make sure you check with the supplier or your owner's manual for the correct formulation

5 0

2 years ago

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number ( $Re$ ) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

$Re=\frac{\rho v D}{\mu}$

where $\rho$ is the density of the fluid, $\mu$ is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

$v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s$

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

$D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm$

8 0

3 years ago

Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br

vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation $v = v_o + a_ct$ to determine the velocity of car B.

where;

$v_o =$ initial velocity

$a_c$ = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

$S = d + v_ot + \dfrac{1}{2}at^2$

Then:

$v_B = 60-12t$

The distance traveled by car B in the given time (t) is expressed as:

$S_B = d + 60 t - \dfrac{1}{2}(12t^2)$

For car A, the needed time (t) to come to rest is:

$v_A = 60 - 18(t-0.75)$

Also, the distance traveled by car A in the given time (t) is expressed as:

$S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

Relating both velocities:

$v_B = v_A$

$60-12t = 60 - 18(t-0.75)$

$60-12t =73.5 - 18t$

$60- 73.5 = - 18t+ 12t$

$-13.5 =-6t$

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

$S_B = S_A$

$d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

$d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2$

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

3 0

3 years ago

I'll mark brainliest plz help

Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

7 0

3 years ago

What are the general rules for press fit allowances

Keith_Richards [23]

Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

please rate brainliest if helps and follow

4 0

1 year ago