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igomit [66]
3 years ago
12

A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

n average temperature of 25o C. (a) If the house loses heat to the atmosphere at the rate of 60,000 kJ/h, determine the minimum power supplied to the heat pump (in kW) that can maintain the interior temperature of the house at 25o C. (b) Suppose the heat pump absorbs energy from
Engineering
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

a) \dot W = 1.062\,kW

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}

COP_{HP} = 15.692

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

\dot Q_{H} = 60000\,\frac{kJ}{h}

The minimum power supplied to the heat pump is:

\dot W = \frac{\dot Q_{H}}{COP}

\dot W = \frac{\left(60000\,\frac{kJ}{h}  \right)\cdot \left(\frac{1\,h}{3600\,s}  \right)}{15.692}

\dot W = 1.062\,kW

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Answer:

Explanation:

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3 years ago
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0
3 years ago
Propane burns at an equivalence ratio (ER) of 0.6, determine actual air-fuel ratio. If excess air is 5%, what will be the actual
dimaraw [331]

Answer:

when 5% excess air is supplied, moles of air supplied/moles of fuel = 23.81\times 1.05 =25

Explanation:

Equivalence ratio = 0.6

Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR

combustion reaction of propane is

C_3H_8+ 5O_2 ----->3CO_2+4H_2O

From above reaction,  1 mole of propane, from the reaction, 5  moles of oxygen required,  

we know that air contains 21% O_2 and 79% N_2,

Therefore, moles of air based on stoichiometry = \frac{5}{0.21} = 23.81

Theoretical air to fuel ratio = \frac{23.81}{1} = 23.81

Given\frac{AFR}{SFR} = 0.6

Actual Air Fuel Ratio = 23.81\times 0.6 = 14.3

when 5% excess air is supplied, moles of air supplied/moles of fuel = 23.81\times 1.05 =25

6 0
3 years ago
N2 flows through an insulated horizontal tube at steady state. With an inlet temperature of 150°C and a velocity change from 2 m
slava [35]

Answer:

The exit temperature is

Explanation:

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3 years ago
A rotating shaft is subjected to a steady torsional stress of 13 ksi and an alternating bending stress of 22 ksi.
mixas84 [53]

Answer:

A) б1 = 28 ksi and  б2 = -6.02 ksi

B) 1.25

Explanation:

Given data :

Torsional stress = 13 ksi

Alternating bending stress = 22ksi

A) determine yielding factor of safety  according to the distortion energy theory

б1,2 = \frac{22}{2} ± √(22/2)² + 13²

       = 11  ± 17

therefore б1 = 28 ksi  hence б2 = -6.02 ksi

B) determine the fatigue factor of safety  

with properties ;  Se = 35ksi, Sy = 60 ksi, Sut = 85 ksi

( б1 - б2 )²  + ( б2 - б3 )² + ( б3 - б1 )²  ≤  2 ( Sy / FOS ) ²

( 28 + 6.02 ) ² + ( 6.02 - 0 )² + ( 0 - 28 )² ≤  2 ( 60 / FOS ) ²

solving for FOS = 1.9

Next we can determine FOS with the use of Goodman criterion

бm / Sut  + бa / Se  =  1 / FOS

= 0 / 85 + 28/35 = 1 / FOS

making FOS the subject of the equation ; hence  FOS = 1.25

3 0
3 years ago
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