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Tasya [4]
2 years ago
7

A liquid jet vj of diameter dj strikes a fixed cone and deflects back as a conical sheet at the same velocity. find the cone ang

le θ for which the restraining force f = (3/2)rhoajvj^2 .
Engineering
1 answer:
dmitriy555 [2]2 years ago
7 0

Answer:

lol i cant real it sorry

Explanation:

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For Laminar flow conditions, what size pipe will deliver 90 gpm of medium oil at 40°F (υ = 6.55 * 10^‐5)?
MrRissso [65]

Answer:

1.693 feet

Explanation:

We have given Q = 90 gpm

We know that 1 cubic feet per second = 443.833 gpm

So 90 gpm will be equal to \frac{90}{443.833}=0.202\frac{ft^3}{sec}

Let d is the diameter of the pipe then V_{avg}=\frac{Q}{A}=\frac{0.2}{\frac{\pi }{4}d^2}=\frac{0.255}{d^2}

We know that for pipe flow critical Reynolds number =2300

So \frac{V_{avg}d}{\nu }=2300

Value of \nu is 6.55\times 10^{-5} given in question

So \frac{0.255\times d}{d^2\times 6.5\times 10^{-5}}=2300

d=1.693 feet  

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10.) A solid circular titanium control rod at 7,000 lb axial tension force where stress must not
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2 years ago
A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The
wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

q=\frac{Ak}{L}dT

\frac{q}{A}=\frac{k}{L}dT

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

{k_{fi}= 0.03 W/m.K

{L_{fi}= 5 cm = 0.05 m

{T_{fi}= 25 \°C

{k_{cb} = 0.5 W/m.K

{L_{cb}= 20 cm = 0.20 m

{T_{cb}= 0 \°C

{T_{m}= ? \°C = temperature at the interface

Solving for {T_{m} will give the temperature at the interface:

\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})

\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})

15 -0.6T_{m}=2.5T_{m}

3.1T_{m}=15

T_{m}=4.8

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