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3 years ago

13

The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in

Appendix A to obtain a conversion

1 answer:

amm18123 years ago

3 0

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be $Q\frac{gallons}{minute}$

Now replacing the gallon and the minute by the above values we get

$Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}$

Thus $Q'=0.631Q\frac{liters}{second}$

Now since we know that 1 liter = $0.0353ft^{3}$

Using this in above relation we get

$Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q$

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

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Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?

zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

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<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

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4 0

1 year ago

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve

Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature $T_1 = -16^0\ C$

Quality $x_1 = 0.2$

Outlet:

Temperature $T_2 = -16^0 C$

Quality $x_2 = 1$

The following data were obtained at saturation properties of R134a at the temperature of -16° C

$v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg$

$v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg$

$m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}$

$\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}$

3 0

2 years ago

Air flows through a convergent-divergent duct with an inlet area of 5 cm² and an exit area of 3.8 cm². At the inlet section, the

Luda [366]

Answer:

The mass flow rate is 0.27 kg/s

The exit velocity is 76.1 m/s

The exit pressure is 695 KPa

Explanation:

Assuming the flow to be steady state and the behavior of air as an ideal gas.

The mass flow rate of the air is given as:

Mass Flow Rate = ρ x A1 x V1

where,

ρ = density of air

A1 = inlet area = 3.8 cm² = 3.8 x 10^-4 m²

V1 = inlet velocity = 100 m/s

For density using general gas equation:

PV = nRT

PV = (m/M)RT

PM/RT = ρ

ρ = (680000 N/m²)(0.02897 kg/mol)/(8.314 J/mol.k)(60 + 273)k

ρ = 7.11 kg/m³

Therefore,

Mass Flow Rate = (7.11 kg/m³)(3.8 x 10^-4 m²)(100 m/s)

<u>Mass Flow Rate = 0.27 kg/s = 270 g/s</u>

Now, for steady flow, the mass flow rate remains constant throughout the flow. Hence, flow rate at inlet will be equal to the flow rate at outlet:

Mass Flow Rate = ρ x A2 x V2

where,

ρ = density of air = 7.11 kg/m³ (Assuming in-compressible flow)

A2 = exit area = 5 cm² = 5 x 10^-4 m²

V2 = exit velocity = ?

Therefore:

0.27 kg/s = (7.11 kg/m³)(5 x 10^-4 m²) V2

<u>V2 = 76.1 m/s</u>

Now, for exit pressure, we use Bernoulli's equation between inlet and exit, using subscript 1 for inlet and 2 for exit:

P1 + (1/2) ρ V1² + ρ g h1 = P2 + (1/2) ρ V2² + ρ g h2

Since, both inlet and exit are at same temperature.

Therefore, h1 = h2, and those terms will cancel out.

P1 + (1/2) ρ V1² = P2 + (1/2) ρ V2²

P2 = P1 + (1/2) ρ V1² - (1/2) ρ V2²

P2 = P1 + (1/2) ρ (V1² - V2²)

P2 = 680000 Pa + (0.5)(7.11 kg/m³)[(100m/s)² - (76.1 m/s)²]

P2 = 680000 Pa + 14962.25 Pa

<u>P2 = 694962.25 Pa = 695 KPa</u>

4 0

3 years ago

Determine the speed of sound in air at 400 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3

Reika [66]

Answer:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

Explanation:

For this case we have given the following data:

$T= 400 K$ represent the temperature for the air

$v = 310 m/s$ represent the velocity of the air

$k = 1.4$ represent the specific heat ratio at the room

$R = 0.287 KJ/ Kg K$ represent the gas constant for the air

And we want to find the velocity of the air under these conditions.

We can calculate the spped of the sound with the Newton-Laplace Equation given by this equation:

$\alpha = \sqrt{\frac{K}{\rho}}=\sqrt{k RT}$

Where K = is the Bulk Modulus of air, k is the adiabatic index of air= 1.4, R = the gas constant for the air, $\rho$ the density of the air and T the temperature in K

So on this case we can replace and we got:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

The Mach number by definition is "a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound" and is defined as:

$Ma=\frac{v}{\alpha}$

Where v is the flow velocity and $\alpha$ the volocity of the sound in the medium and if we replace we got:

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

And since the Ma<0.8 we can classify the regime as subsonic.

7 0

3 years ago

To measure the current through a resistor, a _______________ must be placed in _______________ with the resistor.

fgiga [73]

Answer:

Explanation:

To measure the current through a resistor, the Ammeter must be placed in series with the resistor so that the same current that passes through the resistor also passes through the Ammeter. Ammeter is a measuring device used to measure current.

5 0

3 years ago