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coldgirl [10]
3 years ago
13

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

a building are 730 and 755 mmHg, respectively, determine the height of the building. Assume an average air density of 1.18 kg/m^3.
Engineering
1 answer:
bixtya [17]3 years ago
7 0

Answer:

h = 287.1 m

Explanation:

the density of mercury \rho =13570 kg/m3

the atmospheric pressure at the top of the building is

p_t = \rho gh  = 13570*908*0.73 = 97.08 kPa

the atmospheric pressure at bottom

p_b = \rho gh  = 13570*908*0.75 = 100.4 kPa

\frac{w_{air}}{A} =p_b -p_t

we have also

(\rho gh)_{air} = p_b - p_t

1.18*9.81*h = (100.4 -97.08)*10^3

h = 287.1 m

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The technique of smoothing out joint compound on either side of a joint is known as which of the following
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The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.
zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity  G = 11 × 10³ Ksi =  11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft J\tau

shaft J\tau = \dfrac{\pi}{2}r^4

where ;

r = 1 in /2

r = 0.5 in

shaft J \tau = \dfrac{\pi}{2} \times 0.5^4

shaft J\tau = 0.098218

Now; the angle of twist at  B with respect to D  is calculated by using the expression

\phi_{B/D} = \sum \dfrac{TL}{JG}

\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

where;

T_{CD} \ \  and \ \  L_{CD} are the torques at segments CD and length at segments CD

{T_{BC} \  \ and  \ \ L_{BC}} are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

L_{BC}} = 2.5  in

J\tau = 0.098218

G =  11 × 10⁶ lb/in²

T_{BC = -60 lb.ft

T_{CD = 0 lb.ft

L_{CD = 5.5 in

\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}

\phi_{B/D} = \dfrac{-21600}{1079980}

\phi_{B/D} = − 0.02 rad

To degree; we have

\phi_{B/D}  = -0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{B/D}  = -1.15^0}

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For  the angle of twist of C with respect to D; we have:

\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{C/D} = \dfrac{21600}{1079980}

\phi_{C/D} = 0.02 rad

To degree; we have

\phi_{C/D}  = 0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{C/D}  = 1.15^0}

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dalvyx [7]

Answer:

123.9 Btu

Explanation:

The energy balance on the air is:

∆E = E2 − E1 = ∆KE + ∆PE + ∆U = Q + W

ignore  ∆KE and ∆PE,

W = ∆U − Q = m(u2 − u1) − Q;                               (u2 − u1 = 51.94 Btu/lb)

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W = (2 lb)(143.98 − 92.04) Btu/lb − (− 20 Btu) = 123.9 Btu

u2 − u1 ≈ cv(T2 − T1) = (0.173 Btu/lb°R)(840 − 540) °R = 51.9 Btu/lb

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