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2 years ago

Which of the following uses pressure and flow to transmit power from one location to another?

Engineering

1 answer:

lord [1]2 years ago

5 0

Answer:

fluid power

Explanation:

fluids commonly used in fluid power are Oil, Water, Air, CO², and Nitrogen gas, fluid power is commonly confused with hydraulic power, which only uses liquids, fluid power uses either liquids or gases

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Air is pumped from a vacuum chamber until the pressure drops to 3 torr. If the air temperature at the end of the pumping process

malfutka [58]

Answer:

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr = $3\times 0.133\ kPa$

At the end of pumping, temperature of air, $T = 5^{\circ}C = 278 K$

After the rise in the air temperature, $T' = 20^{\circ}C = 293 K$

Now, we know the ideal gas eqn:

PV = mRT

$P = \frac{m}{V}RT$

$P = \rho_{a}RT$ (1)

where

P = Pressure

V = Volume

$\rho_{a} = air\ density$

R = Rydberg's constant

T = Temperature

Using eqn (1):

$P = \rho_{a}RT$

$\rho_{a} = \frac{P}{RT}$

$\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3}$

Now, at constant volume the final pressure, P' is given by:

$\frac{P}{T} = \frac{P'}{T'}$

$P' = \frac{P}{T}\times T'$

$P' = \frac{3}{278}\times 293 = 3.16 torr$

7 0

4 years ago

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe

fgiga [73]

Answer:

The rate of fuel required to drive the air conditioner $Q_h = 6.061 kW$

The flow rate of the cold air is $\r m = 0.30765 kg/s$

Explanation:

From this question we are told that

The efficiency is $\eta = 33$ % = 0.33

Temperature for the hot day is $T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)$

Temperature after cooling is $T_c = 5^oC = 278K$

The input power is $P_{in} = 2kW$

The rate of fuel required to drive the air conditioner can be mathematically represented as

$Q_h = \frac{P_{in}}{\eta}$

$= \frac{2}{0.33} = 6.061 kW$

From the question the air condition is assumed to be half as a Carnot refrigeration unit

This can be Mathematically interpreted in terms of COP(coefficient of performance) as

$\beta_{air} = 0.5 \beta$

where $\beta$ denotes COP and is mathematically represented as

$\beta = \frac{Q_c}{P_{in}}$

= > $Q_c = \beta P_{in}$

Where $Q_c$ is the rate of flue being burned for cold air to flow

Now if the COP of a Carnot refrigerator is having this value

$\beta_{Carnot } = \frac{T_c}{T_h - T_c}$

$= \frac{278}{308-278}$

$\beta_{Carnot} = 9.267\\$

Then

$\beta_{air} = 0.5 * 9.2667$

$= 4.6333$

Now substituting the value of $\beta$ to solve for $Q_c$

$Q_c = \beta P_{in}$

$= 4.6333 *2$

$9.2667kW$

The equation for the rate of fuel being burned for the cold air to flow

$Q_c = \r mc_p \Delta T$

Making the flow rate of the cold air

$\r m = \frac{Q_c}{c_p \Delta T}$

$= \frac{9.2667}{1.004}* (308 - 278)$

$= 0.30765 kg/s$

4 0

3 years ago

A baseband signal with a bandwidth of 100 kHz and an amplitude range of±1 V is to be transmitted through a channel which is cons

soldi70 [24.7K]

Answer:

Baseband is a signal that has a near zero frequency range. In telecommunication and signal processing, baseband are transmitted without modulation.

Explanati

7 0

3 years ago

Read 2 more answers

All machines have three fundamental hazards: moving parts, point of operation, and?

OlgaM077 [116]

Answer:

All machines have three fundamental hazards: moving parts, point of operation, and the power transmission.

Explanation:

The unit that supplies power to the machine is a critical hazard due to high energy sources being potential fatal if proper protocols are not followed. This is why lockout tagout (LOTO) measures are put in place in order to protect people while they work on equipment.

3 0

2 years ago

Serratia marcescens bacteria are used for the production of threonine. The maximum specific oxygen uptake rate of S. marcescens

Sati [7]

Answer:

The rate of cell metabolism is limited by mass transfer since the value of maximum cell concentration obtained (38 g/l) is lower than 50 g l-1, the value planed.

Explanation:

Data

kLa = 0.17/s

Solubility of oxygen = 8 × 10^-3 kg / m^3

The maximum specific oxygen uptake rate = 4 mmol O2 / g h.

Concentration of oxygen = 0.5 × 10^-3 kg/ m^3

**The maximum cell density = 50 g/l

___________________

The calculated maximum cell concentration:

xmax= kLa · CAL*/ qo

CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate

Replacing the data given

xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 4 mmol O2 / g h

4 mmol O2 / g h to kg O2/ g s

$4 \frac{mmol}{gh} \frac{1 gmol}{1000mmol}\frac{1h}{3600s}\frac{32 g}{gmol} \frac{1Kg}{1000g}$

= 3.56 x 10^-3 kg O2/ g s

So then,

xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 3.56 x 10^-3 o kg O2/ g s

xmax= 3. 8 x 10^4 g/ m^3 = 38 g/l

_____________________

5 0

3 years ago