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nydimaria [60]
3 years ago
11

If you measured a pressure difference of 50 mm of mercury across a pitot tube placed in a wind tunnel with 200 mm diameter, what

is the velocity of air in the wind tunnel? What is the Reynolds number of the air flowing in the wind tunnel? Is the flow laminar or turbulent? Assume air temperature is 25°C.
Engineering
1 answer:
mart [117]3 years ago
4 0

Answer:

V=33.66 m/s

Re=448.8\times 10^6

Re>4000, The flow is turbulent flow.

Explanation:

Given that

Pressure difference  = 50 mm of Hg

We know that density of Hg=136000Kg/m^3

ΔP= 13.6 x 1000 x 0.05 Pa

ΔP=680 Pa

Diameter of tunnel = 200 mm

Property of air at 25°C

ρ=1.2Kg/m^3

Dynamic viscosity

\mu =1.8\times 10^{-8}\ Pa.s

Velocity of fluid given as

V=\sqrt{\dfrac{2\Delta P}{\rho_{air}}}

V=\sqrt{\dfrac{2\times 680}{1.2}}

V=33.66 m/s

Reynolds number

Re=\dfrac{\rho _{air}Vd}{\mu }

Re=\dfrac{1.2\times 33.66\times 0.2}{1.8\times 10^{-8}}

Re=448.8\times 10^6

Re>4000,So the flow is turbulent flow.

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Superheated water vapor at a pressure of 20 MPa, a temperature of 500oC, and a flow rate of 10 kg/s is to be brought to a satura
katrin2010 [14]

Answer:

1.96 kg/s.

Explanation:

So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;

=> Superheated water vapor at a pressure = 20 MPa,

=> temperature = 500°C,

=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."

=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."

K1 = 3241.18, k2 = 93.28 and 2725.47.

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10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.

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7 0
3 years ago
Find the rate of heat transfer through a 6 mm thick glass window with a cross-sectional area of 0.8 m2 if the inside temperature
kiruha [24]

Answer:

6.9

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3 years ago
A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he
sergejj [24]

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

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{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

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T₂ =93.77  °C

8 0
3 years ago
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