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3 years ago

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A liquid refrigerant (sg=1,2) is flowing at a weight flow rate of 20,9 N/h. Refrigerant flashes into a vapor and its specific we

ight become 11,5 N/m^3. If the weigt flow rate remains at 20,9 N/h, compute the volume flow rate[10^-4 m^3/sec]. Express results in 10^-4 m^3/sec.

1 answer:

Iteru [2.4K]3 years ago

5 0

Answer:

Explanation:

volume of 20.9 N

= 20.9 / 11.5 m³

= 1.8174 m³

In one hour 1.8174 m³ flows

in one second volume flowing = 1.8174 / 60 x 60

= 5 x 10⁻⁴ m³

Rate of volume flow = 5 x 10⁻⁴ m³ / s .

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A rectangular concrete (n=0.013) channel 20 feet wide, on a 2.5% slope, is discharging 400 ft3 /sec into a stilling basin. The s

Sonbull [250]

Answer:

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Explanation:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

we will assume the depth of flow as "d"

Q = 1/n (R)^2/3 (slope)^1/2 A ( here R is the hydraulic Radius)

by substituting the given data in above formula we get:

400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d

R = A/P

here, A is the flow area and P is the wetted perimeter

400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d

<u>d = 1.42 feet</u>

<u></u>

Depth of stilling channel before the jump will be = d1 = 8 feet

Depth of stilling channel after the jump will be = d2 = 1.42 feet

Length of stilling basin = 5(d2 - d1)

= 5( 8 - 1.42)

<u>Length of stilling basin = 32.9 feet </u>

Now calculating the height of end sill:

Jump height = (8 - 1.42)

<u>Height of end sill = 6.58 feet</u>

8 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

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4 years ago

2. How many amperes of current will flow through a circuit that has 3 ohms of resistance if

Dmitriy789 [7]

Answer:

4A

Explanation:

From Ohms law ;

R= V/I where R is resistance, V is voltage and I is current

Given that;

R= 3Ω

V= 12

Using the values in the formula as;

R= V/I

3=12/I

I= 12/3

I= 4 A

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