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galina1969 [7]
3 years ago
13

Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read

ing of the pressure gage, in lbf/in^2. (gage). The densities of the oil and mercury are 55 and 845, respectively, each in lb/ft^3. Let g=32.2ft/s^2.
Engineering
1 answer:
damaskus [11]3 years ago
3 0

Answer:

P_2-P_1=27209h

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

P_1+p_1gh_1=p_2_gh_2+P_2

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h

P_2-P_1=27209h

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Explanation:

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3 years ago
A 24-tooth gear has AGMA standard full-depth involute teeth with diametral pitch of 12. Calculate the pitch diameter, circular p
torisob [31]

Answer:

Explanation:

Given:

Tooth Number, N = 24  

Diametral pitch pd = 12

pitch diameter, d = N/pd = 24/12 = 2in

circular pitch, pc = π/pd  = 3.142/12 = 0.2618in

Addendum, a  = 1/pd = 1/12 =0.08333in

Dedendum, b = 1.25/pd = 0.10417in

Tooth thickness, t = 0.5pc = 0,5 * 0.2618  = 0.1309in

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5 0
3 years ago
Read 2 more answers
Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by
irakobra [83]

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

5 0
3 years ago
An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio
saveliy_v [14]

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}

Put the value into the formula

K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}

K_{H}=41.9\ m/d

We need to calculate the vertical conductivity

Using formula of vertical conductivity

K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}

Put the value into the formula

K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}

K_{V}=37.2\ m/d

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

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VLD [36.1K]

Answer:

Cylinder sleeve

Explanation:

7 0
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