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notka56 [123]
3 years ago
11

5.16 A power cycle operating between hot and cold reservoirs at 500 K and 300 K, respectively, receives 1000 kJ by heat transfer

from the hot reservoir. The magnitude of the energy discharged by heat transfer to the cold reservoir must satisfy (a) QC > 600 kJ, (b) QC ≥ 600 kJ, (c) QC = 600 kJ, (d) QC ≤ 600 kJ.

Engineering
1 answer:
Umnica [9.8K]3 years ago
8 0

Answer:

(b) QC ≥ 600 kJ

Explanation:

Go through the attachments for the step by step solution.

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Carbon dioxide gas enters a pipe at 3 MPa and 500 K at a rate of 2 kg/s. CO2 is cooled at constant pressure as it flows in the p
Viktor [21]

Answer:

a) 0.0629 m^{3}/s,ρ1=31.76 kg/m^{3},0.05688 m^{3}/s

b) 0.06171 m^{3}/s,32.40966 kg/m^{3}0.05498 m^{3}/s

c) 0.004%

given data:

n=2 kg/s

R=188.92 J/kgK

T_{1}=500 K

T_{2}=450 K

solution:

a) as we know ideal gas relation

PV=nRT..........(1)

The volume flow rate of inlet is

V_{1}= \frac{nRT_{1} }{P}..................(2)

putting the value in eq(2)

V_{1}  =0.0629 m^{3}/s

to find destiny

ρ1=\frac{n}{V_{1} }...........................................(3)

putting value of n and V1 is eq 3

ρ1=31.76 kg/m^{3}

V_{2} =\frac{nRT_{2} }{P}............(A)

putting value in eq A

    =0.05688 m^{3}/s

to find the rates from the compressibility factors we have to find the reduced pressure and temperature in both cases

P(reduced 1)=\frac{P_{1} }{P_{crit} }=\frac{3.10^{6} }{7.39 . 10^{6}}=0.41

T(reduced 1)= \frac{T_{1} }{T_{crit} }=\frac{500K}{304.25K}=1.64

b) After looking at the chart we obtain Z_{1}=0.98 now the volume flow rate at the inlet is:  

V_{1} =\frac{nRT_{1} Z_{1} }{P} ....................................................(4)

Putting the values in eq(4)      

    =0.06171 m^{3}/s        

Now the density is

ρ=\frac{n}{V_{1} }

  =32.40966 kg/m^{3}

now the reduced pressure and compressibility factor in second case will be

T(reduced 2)=\frac{T_{2} }{T_{crit} }=\frac{450}{304.25}=1.48

P(reduced 2)=P(reduced 1)=0.41

Z_{2}=0.97

V_{2} =\frac{nRT_{2} Z_{2} }{P}............................................(5)

putting value in eq (5)

    =0.05498 m^{3}/s

c) now find the error

Δn/n=modulus(ρ1V_{1}-n/n)*100%.....................(6)

putting the value in eq 6

      =0.004%

6 0
3 years ago
How to recolor objects that are all the same color in affinity designer
LUCKY_DIMON [66]

Answer:

You can adjust the stroke width as small as you like. Then expand the stroke. Apply Layer > Expand Stroke. It makes the curve a filled area which looks the same as the curve.

Explanation:

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3 years ago
4. What are the basic scientific principles the engineers who designed the digital scales would have needed to understand to des
tamaranim1 [39]

Answer:

Electric currents

Physical properties of metals

Influence of gravitational force on objects

Explanation:

The engineer who designed the digital scale must have a good working knowledge of the electric current, and its principle of flow through various materials, and also, factors that can affect its flow through a material (especially metals).

The physical properties of metal would have to be known by the engineer, in order to design the digital scale. The engineer must be able to predict some of the effect of the changes in the physical properties of a metal, and then use this to get some function that he wants from the digital scale, especially the effect of these physical changes on the electrical conducting ability of a metal.

The knowledge of the effect of gravitational force on objects is a must-know for the engineer. The process of scaling here on earth is very much dependent on the gravitational pull on objects downwards to the earth. This knowledge can be used to predict how much deflection this gravity forces will induce on an object towards the earth.

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3 years ago
What are the 5 basic types of propulsion systems​
Basile [38]
Propeller , turbine engine
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3 years ago
A 120-volt fluorescent ballast has an input current of 0.34 ampere and an input power rating of 22 watts. The power factor of th
melisa1 [442]

Answer:

PF= .54

Explanation:

Power Factor equals working/real power (W) over apparent power (VA). 1.0 PF is an efficient equipment. PF= 22/(120*.34)

4 0
3 years ago
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