All categories

3 years ago

This style is closest to the cuisine enjoyed by French royalty and nobility.

Engineering

1 answer:

liraira [26]3 years ago

8 0

Answer: С

. Haute cuisine

Explanation:

Haute Cuisine is typically associated with higher level establishments due to the time and cost it takes to prepare as it involves meticulous preparation and usually uses exotic ingredients which make it quite expensive.

It is therefore no wonder that it came to be in order to feed the 'privileged' people in France being the French Royals and Nobles in the 17th century. It differed from normal cooking at the time by using exotic ingredients such as tongue and caviar and as such, gave those who ate it a feeling of superiority.

You might be interested in

A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass

Illusion [34]

Https://www.slader.com/discussion/question/an-insulated-rigid-tank-is-divided-into-two-equal-parts-by-a-partition-initially-one-part-contains-4/

there will be the answer

6 0

3 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

4 years ago

What type of bond guarantees that if a contractor goes bankrupt on a project the surety will pay the necessary amount to complet

Morgarella [4.7K]

Answer: Payment bond.

Explanation:

A payment bond is a type of guaranteed payment provided for a contractor which ensure that all subcontractors, suppliers, workers etc involved in a project will be duly paid. Payment bond usually comes along with performance bond especially in the construction industry, and it is obtained before the commencement of a construction project. This serves as a form of protection for subcontractors.

7 0

4 years ago

Technician A says that a 12 Volt light bulb that draws 12 amps has a power output of 1 watt. Technician B says that a motor that

Artemon [7]

Answer:

Technician B

Explanation:

Resistance, $R=\frac {V}{I}$ where V is the voltage and I is the current in amps

Therefore, $R=\frac {12}{12}=1 ohm$

Power=VI=12*12=144 W

Therefore, the power is 144 W and resistance is 1 Ohm. This implies that technician A is wrong while technician B is correct

6 0

4 years ago

An air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 99°F. The rat

IrinaVladis [17]

Answer:

a) the coefficient of performance of the air conditioner is 3.5729

- the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner is 18.2759

Explanation:

Given the data in the question;

Lower Temperature $T_L$ = 70°F = ( 70 + 460 )R = 530 R

Higher Temperature $T_H$ = 99° F = ( 99 + 460 )R = 559 R

Cooling Load $Q_L$ = 30000 Btu/h

we know that 1 hp = 2544.43 Btu/h

Net power input P = 3.3 hp = ( 3.3 × 2544.43 )Btu/h = 8396.619 Btu/h

Coefficient of performance of the air conditioner;

$COP_{air-condition$ = Cooling Load $Q_L$ / power P

we substitute

$COP_{air-condition$ = 30000 Btu/h / 8396.619 Btu/h

$COP_{air-condition$ = 3.5729

Therefore, the coefficient of performance of the air conditioner is 3.5729

- Power input required ( in hp )

$Q_L$ / $P_{required$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

30000 Btu/h / $P_{required$ = 530 R / ( 559 R - 530 R )

30000 Btu/h / $P_{required$ = 530 R / 29 R

we solve for $P_{required$

$P_{required$ = ( 30000 Btu/h × 29 R ) / 530 R

$P_{required$ = ( 870000 Btu/h / 530 )

$P_{required$ = 1641.5094 Btu/h

we know that; 1 hp = 2544.43 Btu/h

so;

$P_{required$ = ( 1641.5094 / 2544.43 ) hp

$P_{required$ = 0.645 hp

Hence, the power input required for a reversible air conditioner is 0.645 hp

- the coefficient of performance for the reversible air conditioner;

$COP_{rev-air-condition$ = $T_L$ / ( $T_H$ - $T_L$ )

we substitute

$COP_{rev-air-condition$ = 530 R / ( 559 R - 530 R )

$COP_{rev-air-condition$ = 530 R / 29 R

$COP_{rev-air-condition$ = 18.2759

Hence, the coefficient of performance for the reversible air conditioner is 18.2759

3 0

3 years ago