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3 years ago

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An air conditioner is designed to bring 10,000 ft3/min of outside air (90°F, 29.8 inches Hg, 88% relative humidity) to 40°F, the

reby condensing a portion of the water vapor, and then to reheat the air before releasing it into a room at 65°F. Calculate the rate of condensation (gallons H2O/min) and the volumetric flow rate of the air delivered to the room. (Suggestion: On the flowchart, treat the cooling–condensation and the reheating as separate process steps.)

1 answer:

jarptica [38.1K]3 years ago

3 0

Answer:

rate of condensation= 1.81 gal/min

volumetric flow rate= 9223 ft³/min

Explanation:

See attached pictures.

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A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe ha

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Answer :

The force needed to move the cylinder is 25.6 N

<h2>Further explanation </h2>

Given that,

Length of the cylinder, l = 0.5 m

Outer diameter of the cylinder, d = 8 cm = 0.08 m

Outer radius of the cylinder, $r=0.04\ m$

Inside diameter of the pipe, d = 8.5 cm = 0.085 m

Inside radius of the pipe, $r=0.0425\ m$

Specific gravity of the oil, $\rho=0.92$

Density of oil, $d=\rho\times \rho_w$

Kinematic viscosity of the oil, $v=5.57\times 10^{-4}\ m^2/s$

Velocity of the cylinder, u = 1 m/s

We need to find the force needed to move the cylinder. Let the force is F.

Specific gravity is defined as the ratio of the density of the substance to the density of water.

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

$v=\dfrac{\mu}{d}$

Where, $d$ = density of oil

And $d=\rho\times \rho_w$ (density of oil = specific gravity × density of water )

$d=0.92\times 10^3\ kg/m^3$

So,

$\mu=v\times d$ ..............(1)

$\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3$

$\mu=0.512\ Pa-s$

The separation between the cylinder and pipe is given by :

$dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m$

$d_p\ and\ d_c$ are diameter of pipe and cylinder respectively.

The mathematical expression for the Newton's law of viscosity can be written as:

$\tau\propto\dfrac{du}{dy}$

$\tau=\mu\times \dfrac{du}{dy}$ ..........(2)

Where

$\tau$ = Shear stress, $\tau=\dfrac{F}{A}$ ............(3)

$\mu$ = viscosity

$\dfrac{du}{dy}$ = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :

$\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}$ ...............(4)

A is the area of the cylinder, $A=2\pi rl$

Equation (4) becomes :

$F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl$ ..............(5)

$A=\pi d\times l$

$A=\pi \times 0.08\ m\times 0.5\ m$

$A=0.125\ m^2$

Now, equation (5) becomes :

$F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl$

$F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2$

F = 25.6 N

<h2>Learn more </h2>

Kinematic viscosity : brainly.com/question/12947932

<h2>Keyword : </h2>

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.

7 0

3 years ago