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evablogger [386]
3 years ago
12

Flue gases from an incinerator are released to atmosphere using a stack that is 0.6 m in diameter and 10.0 m high. The outer sur

face of the stack is at 40°C and the surrounding air is at 10°C. Determine the rate of heat transfer from the stack, assuming
(a) there is no wind and
(b) the stack is exposed to 20 km/h winds.
Engineering
1 answer:
Goshia [24]3 years ago
7 0

Answer:

(a) Rate of heat transfer assuming there is no wind is 152.70J/s

(b) Rate of heat transfer assuming the stack is exposed to 20km/h winds is 151.49J/s

Explanation:

Q/t = KA(T2 - T1)/d

Range of heat transfer coefficient (K) for flue gases is 60-180W/mk

Assuming K=180W/mk, diameter (d) =0.6m, A=3.142×0.6^2/4 = 028278m^2, distance (d) = 10m, T2=40°C, T1=10°C

(a) Q/t = 180×0.28278×(40-10)/10 = 180×0.28278×30/10 = 152.70J/s

(b) V= 20km/h=20×1000/3600m/s= 5.6m/s, t=d/v=10/5.6=1.8s

Q = KA(T2 - T1)/V = 180×0.28278×(40 - 10)/5.6 = 180×0.28278×30/5.6 = 272.68J

Q/t = 272.68J/1.8s = 151.48J/s

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The "view factor" Fij depends on surface emissivity and surface geometry. a) True b) False
Alex

Answer:

(B) FALSE

Explanation:

view factor F_{ij} depends on the surface emissivity and the surface of geometry  view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by  F_{ij}

4 0
3 years ago
This is an essential safety procedure that protects workers from injury while working on or near electrical circuits and equipme
Mice21 [21]

Answer:

Option B

Lockout/ Tagout

Explanation:

lockout and tagout is a safety procedure used in most industries and work places. This ensures that during any maintenance work, that dangerous machines are properly shut off and not able to be started up again prior to the completion of maintenance or repair work.

The procedure includes:

1. Turns off and disconnects the machinery or equipment from its energy source(s) before performing service or maintenance.

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8 0
4 years ago
In primary processing their are 3 different steps, what are the steps?
Nana76 [90]

Answer:

Primary processing involves cutting, cleaning, packaging, storage and refrigeration of raw foods to ensure that they are not spoilt before they reach the consumer.

8 0
3 years ago
An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu
Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

T_{heater} = 120 Degree F

T_{room} = 60 Degree F

A = 30 ft^2

\eta = 100%

Heat loss before previous final value = \frac{A \Delta T}{R}

                                                              =\frac{30\times *(120-60)}{5}

                                                              = 360 Btu/hr

Heat loss after new value= \frac{30\times \times (120-60)}{15} = 120 Btu/hr

saving would be = 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year

                           = 616.1782 kw hr/yr

cost = 616.1782 \times 0.06$

         = 36.917 $\yr

6 0
4 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Inessa05 [86]

Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}

Now we can use the equation that defines the percentage of increase, in this case for speed

i=(\frac{V2}{V1} -1) 100

Now we use the equation obtained in the previous step, and replace values

i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65

the percent increase in the velocity of air is 25.65%

6 0
3 years ago
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