If 180 g of potassium Iodide is dissolved in 100 CM three of water at 30°C and blank solution is formed

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

Difference between velocity profile and velocity distribution

Irina18 [472]

Answer:

Profile is a graphical representation of velocity distribution

7 0

3 years ago

Mobility refers to the ability to?

aev [14]

Answer:

Drive

Explanation:

Drive is a great example

4 0

3 years ago

The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of 0.75 in2, Determine the mag

algol13

Answer:

0.029N

Explanation:

Pressure = displacement × density of steel × acceleration due to gravity

displacement = 0.1in = 0.1×0.0254m = 0.00254m, density of steel = 7900kg/m^3, acceleration due to gravity = 9.8m/s^2

Pressure = 0.00254×7900×9.8 = 196.65N/m^2

Force(P) = pressure × area

Area of 1 pin-connected A-36 steel member = 0.75in^2 = 4.84×10^-6m^2

Area of 3 pin-connected A-36 steel member = 3×4.84×10^-6 = 1.452×10^-5m^2

Force(P) = 196.65 × 1.4652×10^-5 = 0.0029N

8 0

4 years ago

A 280-km-long pipeline connects two pumping stations. If 0.56m3/s are to be pumped through a 0.62-m-diameter line, the discharge

ratelena [41]

Answer:

Explanation:

(a) we take the free surface of the lake to point 1 and the free surfaces of the storage tank to point 2. We also take the lake surface as the reference level (z1= 0), and thus the potential energy at points 1and 2 are pe1= 0 and pe2= gz2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere (P1= P2= Paytm). Further, the kinetic energy both points is zero (ke1= ke2= 0) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 kg/s70/s)m070.0)(kg/m1000(33===V&&rmkJ/kg196.0/sm1000kJ/kg1m)20)(m/s(9.8122222===gzpe

Then the rate of increase of the mechanical energy of water becomes kW13.7kJ/kg)6kg/s)(0.1970()0()(22inmech,outreach,fluid mech,===-=-=DpempemeemE&&&&The overall efficiency of the combined pump-motor unit is determined from its definition,67.2%or 0.672

NB:

Please check the attached file for clearer work.

4 0

3 years ago