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Fudgin [204]
3 years ago
6

A water jet jump involves a jet cross-sectional area of 0.01 m2 , and a jet velocity of 30 m/s. The jet is surrounded by entrain

ed water. The total cross-sectional area associated with the jet and the entrained steams is 0.075 m2 . Determine the pumping rate (i.e. the entrained fluid flowrate) involved in liters/s

Engineering
1 answer:
oee [108]3 years ago
5 0

Answer: 300ltrs/sec

Explanation:

Calculation and explanation is shown below

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A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s
Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})

where

'H' is the initial depth of the layer

C_c is the Compression index

e_o is the inital void ratio

\bar{\sigma_o} is the initial effective stress at the depth

\Delta \bar{\sigma_o} is the change in the effective stress at the given depth

Applying the given values we get

\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04

3 0
3 years ago
Who developed the process of blueprinting?
VikaD [51]
Answer: C.) John Herschel
3 0
3 years ago
Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest
docker41 [41]

Answer:

a)We know that acceleration a=dv/dt

So dv/dt=kt^2

dv=kt^2dt

Integrating we get

v(t)=kt^3/3+C

Puttin t=0

-8=C

Putting t=2

8=8k/3-8

k=48/8

k=6

5 0
3 years ago
Electric current originates from which part of an atom? *
yanalaym [24]

Answer: Electric current originates from positively charged protons negatively charged electrons of an atom.

Explanation:

The movement of ions (positive or negative) from one point to another is called electric current.

An atom has three sub-atomic particles. These are protons, neutrons and electrons.

Protons are positively charged, neutrons have no charge and electrons are negatively charged. Protons and neutrons reside inside the nucleus of an atom whereas electrons revolve around the nucleus.

So, protons and electrons are responsible for originating electric current form an atom as these are the charged particles.

Thus, we can conclude that electric current originates from positively charged protons negatively charged electrons of an atom.

3 0
2 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
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