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KatRina [158]
3 years ago
11

A smoking lounge is to accommodate 19 heavy smokers. The minimum fresh air requirement for smoking lounges is specified to be 30

L/s per person (ASHRAE, Standard 62, 1989). Determine the minimum required flow rate of fresh air that needs to be supplied to the lounge and the diameter of the duct if the air velocity is not to exceed 8 m/s. (Round the final answers to two and three decimal places, respectively.) The minimum required flow rate of fresh air that needs to be supplied to the lounge is m3/s. The diameter of the duct if the air velocity is not to exceed 8 m/s is m.
Engineering
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer 1: minimum required flow rate of fresh air is 0.57 m^3/ses

Explanation: since the minimum requirement per person is 30 L/sec

Converting to m^3 it becomes

30/1000 = 0.03 m^3/sec

For 19 heavy smoker we will require

19 * 0.03 = 0.57m^3/sec

Answer 2: diameter of the duct will be 0.3m

Explanation: since flow rate is

Q =0.57m^3/sec

Also

Q = AV (continuity equation)

Where A is the duct area and V is the velocity of air flow in m/sec

0.57m^3/sec = A * 8m/sec

A = 0.57/8 = 0.071m^2

Area of the duct is that of a circle

A = 3.142 *(d^2 ÷4)

d^2 = (0.017 * 4)/3.142 = 0.09

d is square root of 0.09

d = 0.3m

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A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
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The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.

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