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Mnenie
2 months ago
15

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

uilding are 675 and 695 mm Hg, respectively, determine the height of the building in m. Take the densities of air and mercury to be 1.18 kg/m3 and 13,600 kg/m3.
Engineering
1 answer:
pantera1 [306]2 months ago
8 0

Answer:

230.51 m

Explanation:

Pb = 695 mmHg

Pt = 675 mmHg

Pb - Pt = 20 mmHg

To find dP:

dP = p * g * H = (13600)*(9.81)*(20/1000) = 2668.320 Pa

To determine the building height, since dP is the same for any liquid medium:

dP = p*g*H = 2668.320

H = 2668.32 / (1.18 * 9.81) = 230.51 m

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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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