What is the difference between absolute and gage pressure?

Answers

Answer 1

Answer:

Explanation:

Step1

Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.

The expression for absolute pressure is given as follows:

$P_(ab)=P_(g)+P_(atm)$

Here, $P_(ab)$ is absolute pressure, $P_(g)$ is gauge pressure and $P_(atm)$ is atmospheric pressure.

Step2

Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.

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An electric current of 237.0 mA flows for 8.0 minutes. Calculate the amount of electric charge transported. Be sure your answer has the correct unit symbol and the correct number of significant digits x10

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

The percentage loss of the window is $E = 97.3$ %

Explanation:

From the question we are told that

The area of pane of glass is $A = 0.15 m^2$

The thickness is $d = 5mm = (5)/(1000) = 0.005m$

The thickness of the wall is $D = 0.15m$

The area of the wall is $a = 10m^2$

Generally the heat lost as a result of conduction of the window is

$Q_(window) = (j_(glass) * A * (\Delta T) )/(d)$

Where $j_(glass)$ is the thermal conductivity of glass which has a constant value of

$j_(glass) = 0.80 J/(s \cdot m \cdot C^o)$

Substituting values

$Q_(window) = ( 0.80 * 0.15 * (\Delta T) )/(0.005)$

$Q_(window) = 24 \Delta T$

Generally the heat lost as a result of conduction of the wall is

$Q_(wall) = (j_(styrofoam) * A * (\Delta T) )/(d)$

$j_(styrofoam)$ s the thermal conductivity of Styrofoam which has a constant value of $j_(styrofoam) = 0.010J / (s \cdot m \cdot C^o)$

Substituting values

$Q_(wall) = ( 0.010 * 10 * (\Delta T) )/(0.15)$

$Q_(wall) = 0.667 \ \Delta T$

Now the net loss of heat is

$Q_(net) = Q_(window) + Q_(wall)$

Substituting values

$Q_(net) = 24 + 0.667$

$Q_(net) = 24.667$

Now the percentage loss by the window is

$E = (Q_(window) )/(Q_(net)) * 100$

Substituting value

$E = (24)/(24 .667) * 100$

$E = 97.3$ %

How much computer memory (in bytes) in minimum would be required to store 10 seconds of a sensor signal sampled by a 12-bit A/D converter operating at a sampling rate of 5 kHz?

Answers

Answer:

73.24 K byte

Explanation:

Assuming that

N = total number of samples

N = 10 * 5kHz

N = 50*10^3

Also, the total number of bits, T

T = 12 * N

T = 12 * 50*10^3

T = 600 * 10^3

And then, finally, the total number of byte,

B = 600*10^(3/8)

B = 75*10^3 byte

75*10^3 byte = 75*10^3/1024 kilo byte

And on converting to decimal, we will have

= 73.24 K byte

Therefore, the memory required = 73.24 K byte

An activated sludge plant is being designed to handle a feed rate of 0.438 m3 /sec. The influent BOD concentration is 150 mg/L and the cell concentration (MLVSS) is 2,200 mg/L. If you wish to operate the plant with a food-to-microorganism ratio of 0.20 day-1 , what volume of aeration tank should you use? Please give your answer in m3 .

Answers

Answer:

Volume of aeration tank = 1.29 x 10^4 m³

Explanation:

Food/Micro- organism Ratio = 0.2/day

Feed Rate (Q) = 0.438 m³/s

Influent BOD = 150 mg/L

MLVSS = 2200 mg/L

The above mentioned parameters are related by the equation

F/M = QS₀/VX

where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get

V = 0.4380 x 150/0.2 x 2200

V = 0.1493 (m³/s) x day

V = 0.1493 x 24 x 60 x 60

V = 1.29 x 10^4 m³

Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 seconds. In Eager Mode, the app utilizes a faster timer resolution for its computations, so the execution time in Eager Mode is 2 seconds (i.e., Eager Mode execution time is 60% of Lazy Mode execution time).After finishing computation, the app sends some data to the cloud, regardless of the mode itâ€™s in. The data size sent to the cloud is 600 MB. The bandwidth of communication is 15 MBps for WiFi and 5 MBps for 4G. Assume that the communication radio is idle during the computation time. Assume that the communication radio for WiFi has a power consumption of 75 mW when active and 15 mW when idle. Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization. Assume that the power consumption of the CPU is a linear function of its utilization. In other words: P = (Idle Power) + (Utilization)*(Power per unit Utilization). A configuration of the mobile app involves choosing a timer resolution (Lazy or Eager) and choosing a type of radio (WiFi or 4G). For example, faster timer resolution (Eager) and 4G network is a configuration, while slower resolution (Lazy) and WiFi is another. There are four possible configurations in all.

Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?

Answers

The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Why reducing leads to increasing wages?

Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.

Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.

Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.

Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Learn more about average power on:

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A rectangular channel with a width of 2 m is carrying 15 m3/s. What are the critical depth and the flow velocity

Answers

Answer:

The critical depth of the rectangular channel is approximately 1.790 meters.

The flow velocity in the rectangular channel is 4.190 meters per second.

Explanation:

From Open Channel Theory we know that critical depth of the rectangular channel ( $y_(c)$ ), measured in meters, is calculated by using this equation:

$y_(c) = \sqrt[3]{(\dot V^(2))/(g\cdot b^(2)) }$ (Eq. 1)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$b$ - Channel width, measured in meters.

If we know that $\dot V = 15\,(m^(3))/(s)$ , $g = 9.807\,(m)/(s^(2))$ and $b = 2\,m$ , then the critical depth is:

$y_(c) = \sqrt[3]{(\left(15\,(m^(3))/(s) \right)^(2))/(\left(9.807\,(m)/(s^(2)) \right)\cdot (2\,m)^(2)) }$

$y_(c) \approx 1.790\,m$

The critical depth of the rectangular channel is approximately 1.790 meters.

Lastly, the flow velocity ( $v$ ), measured in meters, is obtained from this formula:

$v = (\dot V)/(b\cdot y_(c))$ (Eq. 2)

If we know that $\dot V = 15\,(m^(3))/(s)$ , $b = 2\,m$ and $y_(c) = 1.790\,m$ , then the flow velocity in the rectangular channel is:

$v = (15\,(m^(2))/(s) )/((2\,m)\cdot (1.790\,m))$

$v = 4.190\,(m)/(s)$

The flow velocity in the rectangular channel is 4.190 meters per second.

Please define the specific heat of material?

Answers

Answer and Explanation:

SPECIFIC HEAT :

Specific heat is denoted by $c_v$
It is the heat required for increasing the temperature of a substance which has mass of 1 kg.
Its SI unit is joule/kelvin
It is physical property
It can be calculated by $c_v=(Q)/(m\Delta T)$ , here Q is heat energy m is mass of gas and $\Delta T$ is change in temperature.

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