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3 months ago

6

On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading

be when you carry it up to the observation deck 500 ft above the base of the monument?

1 answer:

Mrrafil [318]3 months ago

8 0

Answer:

The barometer will read 29.43 in

Explanation:

Using the pressure variation equation,

p2 - p1 = -yair * H

= 7.65 * $10^{-2} \frac{lb}{ft^{3} } * 500 ft\\$

= 38.5 $\frac{lb}{ft^{2} }$

According to how pressure relates to the height of the mercury column,

p = yHg * h --> with yHg and h being the barometer reading

yHg $(\frac{29.97}{12} ft)$ - yHg * h1 = 38.5 $\frac{lb}{ft^{2} }$

h1 = ( $\frac{29.97}{12} ft$ ) - $\frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }$

$[(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in$

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It is not feasible to install the wires within the conduit. Explanation: The given dimensions show that the total area is 2.04 square inches while the wires occupy 0.93 square inches. The maximum allowable fill for the conduit is 40%. To determine if placement is possible, compute the conduit’s area at 40% which equates to 0.816 square inches, less than the required area of the wires.

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2 months ago

The rigid beam is supported by a pin at C and an A992 steel guy wire AB of length 6 ft. If the wire has a diameter of 0.2 in., d

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Answer:

Change in length = 0.0913 in

Explanation:

Given data:

Length = 6 ft

Diameter = 0.2 in

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∑M(c) = 0.............1

This can be used to find the force in AB.

10× 200 × ( 5) - (T cos(30)) × 10 = 0

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And the area will be

Area = $\frac{\pi }{4}\times 0.2^2$

The change in length is expressed as

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Substituting values results in

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2 months ago

Steam flows at steady state through a converging, insulated nozzle, 25 cm long and with an inlet diameter of 5 cm. At the nozzle

iogann1982 [368]

Answer:

The velocity at exit U_2 is 578.359 m/s

The exit diameter d_e is 1.4924 cm

Explanation:

Provided data includes:

Length of the nozzle L = 25 cm

Inlet diameter d_i = 5 cm

At the nozzle entrance (state 1): Temperature T_1 = 325 °C, Pressure P_1 = 700 kPa, Velocity U_1 = 30 m/s, Enthalpy H_1 = 3112.5 kJ/kg, Volume V_1 = 388.61 cm³/gAt the nozzle exit (state 2): Temperature T_2 = 250 °C, Pressure P_2 = 350 kPa, Velocity U_2, Enthalpy H_2 = 2945.7 kJ/kg, Volume V_2 = 667.75 cm³/g To determine:a. Exit Velocity U_2b. Exit Diameter d_e

a.The Energy Equation can be represented by:ΔH + ΔU² / 2 + gΔz = Q + WAssuming Q = W = Δz = 0Substituting the values yields:

(H_2 - H_1) + (U²_2 - U²_1) / 2 = 0From which we can derive U_2 = sqrt((2* (H_1 - H_2 )) + U²_1) with the calculations leading to U_2 = sqrt ( 2 * 10^3 * (3112.5 -2945.7) + 900) yielding U_2 = 578.359 m/s

b.

Using mass balance approach, we have U_1 * A_1 / V_1 = U_2 * A_2 / V_2

Here, A = π*d² / 4

This leads to U_1 * d_i² / V_1 = U_2 * d_e² / V_2, thus d_e = d_i * sqrt((U_1 / U_2) * (V_2 / V_1)). Hence, d_e = 5 * sqrt((30 / 578.359) * (667.75 / 388.61)) computes to d_e = 1.4924 cm

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2 months ago

The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re

mote1985 [299]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, $g = 9.81 m/s^2$

mass flow rate, $\dot{m} = 60.6 kg/s$

flow density, $\rho = 1000 kg/m^3$

efficiency of the pump, $\eta = 0.74$

output gauge pressure, $p_o = 344.75 kPa$

input gauge pressure, $p_i = 68.95 kPa$

cross-sectional area of output pipe, $A_o = 0.069 m^2$

cross-sectional area of input pipe, $A_i = 0.093 m^2$

height of discharge, $z_o = 1.22 m - 0.61 m = 0.61 m$ (evaluated at pump’s maximum height of 1.22 m)

input height, $z_i = 0 m$

hydraulic power of the pump, $P =? kW$

Initially, the volumetric flow (Q) needs to be determined

$Q = \frac{\dot{m}}{\rho}$

$Q = \frac{60.6 kg/s}{1000 kg/m^3}$

$Q = 0.0606 m^3/s$

Next, compute the velocity (v) for both input and output

$v_o = \frac{Q}{A_o}$

$v_o = \frac{0.0606 m^3/s}{0.069 m^2}$

$v_o = 0.88 m/s$

$v_i = \frac{Q}{A_i}$

$v_i = \frac{0.0606 m^3/s}{0.093 m^2}$

$v_i = 0.65 m/s$

Subsequently, the total head (H) can be calculated

$H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}$

$H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}$

$H = 28.74m$

Finally, the computation of pump power is done as follows

$P = \frac{Q \, \rho \, g \, H}{\eta}$

$P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}$

$P = 23.09 kW$

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