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Lina20 [59]
3 years ago
12

A manometer is used to measure the air pressure in a tanlc The fluid used has a specific gravity of 1.25, and the differentialhe

ight between the two arms of the manometer is 28 in. If the local atmospheric pressure is 127 psia, determine theabsolute pressure in the tank for the cases of the manometer arm with the (a) higher and (b) lower fluid level beingattached to the tank.
Engineering
1 answer:
Alenkasestr [34]3 years ago
8 0

Answer:

(a) 11.437 psia

(b) 13.963 psia

Explanation:

The pressure exerted by a fluid can be estimated by multiplying the density of the fluid, acceleration due to gravity and the depth of the fluid. To determine the fluid density, we have:

fluid density = specific gravity * density of water = 1.25 * 62.4 lbm/ft^3 = 78 lbm/ft^3

height = 28 in * (1 ft/12 in) = 2.33 ft

acceleration due to gravity = 32.174 ft/s^2

The change in pressure = fluid density*acceleration due to gravity*height = 78*32.174*(28/12) = 5855.668 lbm*ft/(s^2 * ft^2) = 5855.668 lbf/ft^2

The we convert from lbf/ft^2 to psi:

(5855.668/32.174)*0.00694 psi = 1.263 psi

(a) pressure = atmospheric pressure - change in pressure = 12.7 - 1.263 = 11.437 psia

(b) pressure = atmospheric pressure + change in pressure = 12.7 + 1.263 = 13.963 psia

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3 years ago
A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft
tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ;  f_{Lw}=6.6 mi/h

5' on right   ;    f_{Lc} = 0.4 mi/hr

3' on left   ;      no adjustment

3 lanes in each direction    f n = 3 mi/h

v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}

f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}

f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}

          = 0.877

v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}

       = 1,555 veh/hr/lane

FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}

      = (55 + 5) - 6.6 - 0.4 -3 -0

      = 50 mi/h

D = \dfrac{V_P}{s}

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4 years ago
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Answer:

masters

associate

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Explanation:

8 0
3 years ago
Read 2 more answers
There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter
weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

                                        u_e = 9.30777 * 10^-8  J/m^3

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3 years ago
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Answer:

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(e) Specific volume = 0.86 m^3/Kg of dry air

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