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Andru [333]
3 years ago
14

An object of irregular shape has a characteristic length of � = 1.00 [�] and is maintained at a uniform surface temperature of �

. = 400 [�]. When placed in atmospheric air at a temperature of �- = 300 [�] and moving with a velocity of � = 100 [ 3 . ], the average heat flux from the surface to the air is 13,000[ 2 3&]. If a second object of the same shape, but with a characteristic length of � = 5.00 [�], is maintained at a surface temperature of �. = 400 [�] and is placed in atmospheric air at �- = 300 [�], what will the value of the average convection coefficient be if the air velocity is � = 20 [ 3 . ]?
Engineering
1 answer:
Pani-rosa [81]3 years ago
5 0

The correct question;

An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?

Answer:

h'_2 = 40 W/K.m²

Explanation:

We are given;

L1 = 1m

L2 = 5m

T_s = 400 K

T_(∞) = 300 K

V = 100 m/s

q = 20,000 W/m²

Both objects have the same shape and density and thus their reynolds number will be the same.

So,

Re_L1 = Re_L2

Thus, V1•L1/v1 = V2•L2/v2

Hence,

(h'_1•L1)/k1 = (h'_2•L2)/k2

Where h'_1 and h'_2 are convection coefficients

Since k1 = k2, thus, we now have;

h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)

Thus,

h'_2 = [20,000/(400 - 300)]•(1/5)

h'_2 = 40 W/K.m²

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3 years ago
The water requirement for Class H cement is 38% (i.e.,water (%) by weight of cement),whereas the water requirement for barite is
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Answer:

weight of barite = 398.4355 kg

Explanation:

Solution:- The values given in the question are as follows:

water requirement for H class cement = 38% by weight of cement

water requirement for barite = 2.4 gal / 100 lbm

H class cement slurry = 15.7 lbm/gal

one sack of cement = 50 kg or 110.231 lbm

one sack of cement require water = (38/100)*110.231

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water required 100 lbm barite = 2.4 gal

or water required barite = 2.4% by weight of barite

H class cement slurry = (weight of cement + weight of barite)/total weight of water

15.7 =(110.231 + weight of barite)/(water required one sack of cement + 2.4%*weight of barite)

15.7 = 110.231 + (weight of barite)/(41.887 + 0.024*weight of barite)

15.7*41.8877 + 15.7*0.024*weight of barite = 110.231 + weight of barite

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7 0
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mass flow rate = 21.55 Ibm/s

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using given data to obtain values from table F7.1

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Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm

from table F.3

specific constant of glycerin C_{p} = 0.58 Btu/Ibm-R

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h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )

where ; T4 = 50 F

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3 0
3 years ago
1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturati
alexandr1967 [171]

Answer:

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16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}

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solving both  equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

6 0
3 years ago
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