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4 years ago

13

Which option is an example of a physical model?

1 answer:

Burka [1]4 years ago

6 0

Answer:

A. A clay ball with a slice cut out, showing the layers of the earth

Explanation:

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Read 2 more answers

Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

3 years ago

What is the volume of the rectangular prism shown.

Likurg_2 [28]

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7 0

3 years ago

Steam enters an adiabatic condenser (heat exchanger) at a mass flow rate of 5.55 kg/s where it condensed to saturated liquid wat

Evgen [1.6K]

Answer:

The minimum mass flow rate will be "330 kg/s".

Explanation:

Given:

For steam,

$m_{s}=5.55 \ kg/s$

$\Delta h=2491 \ kg/kj$

For water,

$\Delta T=10^{\circ}C$

$(Cp)_{w}=4.184 \ kJ/kg^{\circ}C$

They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,

⇒ $m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T$

On putting the estimated values, we get

⇒ $5.55\times 2491=M_{w}\times 4.184\times 10\\$

⇒ $13825.05=M_{w}\times 41.84$

⇒ $M_{w}=330 \ kg/s$

7 0

3 years ago