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Likurg_2 [28]
3 years ago
8

A natural-draft cooling tower receives 250,000 ft3/min of air at standard atmospheric pressure, 70oF, and 45 percent relative hu

midity. The air leaves saturated at 100oF. 3500 gallons per minute of water at 104oF is cooled. Calculate
(a) the temperature of the water at the tower exit, and
(b) the necessary height of the tower if the total pressure losses are 2.00 lbf/ft2

Engineering
1 answer:
notsponge [240]3 years ago
5 0

Find the attachment for complete solution

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Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
3 years ago
6.3.9 A coin was tossed n = 1000 times, and the proportion of heads observed was
4vir4ik [10]

Answer:

No

Explanation:

51 / 100 = 510 / 1000

Chance of getting a head is 1 / 2 of total throws

= 1 / 2 × 1000

= 500 is the probability

and the number of heads was just 10 more the the probability...if the was a greater gap, there would be evidence to say the coin is unfair

4 0
4 years ago
Q2: The average water height of an ocean area is 2.5 m high and each wave lasts for an average period of 7 s. Determine (a) the
navik [9.2K]

Answer:

(a) 561.12 W/ m² (b) 196.39 MW

Explanation:

Solution

(a) Determine the energy and power of the wave per unit area

The energy per unit are of the wave is defined as:

E = 1 /16ρgH²

= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²

=3927. 83 J/m²

Thus,

The power of the wave per unit area is,

P = E/ t

= 3927. 83 J/m² / 7 s = 561.12 W/ m²

(b) The average and work power output of a wave power plant

W = E * л * A

= 3927. 83 J/m² * 0.35 * 1 *10^6 m²

= 1374.74 MJ

Then,

The power produced by the wave for one km²

P = P * л * A

= 5612.12 W/m² * 0.35 * 1* 10^6 m²

=196.39 MW

8 0
3 years ago
Please help me do this with my exam tomorrow.
blagie [28]

Answer:

ddddddddddddddddddddddddddddd

Explanation:

cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

6 0
2 years ago
The alternator produces direct current, which is changed to alternating current by the diodes.
exis [7]

Answer:

False

Explanation:

<u>Alternat</u>or produces <u>alternat</u>ing current  ....diodes convert this to direct current

8 0
2 years ago
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