A heating torch is usually referred to as what? Stick Flower Rose-bud Lighter

Technician A says amperage cannot exist without both voltage and resistance. Technician B says if amperage is high, then you kno

Ivan

Answer:

Technician A

Explanation:

Ohms law: I= E/R so rest resistance must be present along with E/potential difference. Even if just wire shorted together there is resistance but very little.

Tech B: Again ohms law. Current flow is directly proportional to the voltage and inversely proportional to R (resistance or impedance).

8 0

3 years ago

All aspects of the Kirby-Bauer test are standardized to assure reliability. What might be the consequence of pouring the plates

EleoNora [17]

Answer:

it would affect the distance the antiantibodies diffuse from the disk

Explanation:

7 0

2 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

3 years ago

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at (100%) (125

Anna [14]

Where loads are likely to be on continuously, the calculated load for branch circuits and feeders must be figured at 125%.

Section 210.19(A)(1) permits the bigger of the two values listed below to be utilized as the connectors 's ultimate size for sizing an ungrounded branch circuit conductor:

Without any extra adjustments or corrections, either 125% of the continuous load, OR

When adjustment and corrective factors are applied, the load is 100% (not 125% as stated previously).

This will be the same in the 2020 NEC. The introduction of new exception 2 is what has changed. To comprehend this new exception, one must study it very carefully. A part of a branch circuit connected to pressure connectors (such as power distribution blocks) that complies with 110.14(C)(2) may now be sized using the continuous load plus the noncontiguous load instead of 125% of the continuous load thanks to the new exception.

To know more about connectors click here:

brainly.com/question/16987039

#SPJ4

4 0

1 year ago

Question text

lisabon 2012 [21]

Answer:

That's a really nice question sadly I don't know the answer I'm replying to you cuz I'm tryna get points so... Sorry

3 0

3 years ago