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kolbaska11 [484]
3 years ago
13

An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

he condenser at 100 kPa. The temperature is 4508C at the entrance of the high-pressure and lowpressure turbines. The mass flow rate through the cycle is 1.74 kg/s. Determine the power used by pumps, the power produced by the cycle, the rate of heat transfer in the reheater, and the thermal efficiency of this system
Engineering
1 answer:
solniwko [45]3 years ago
7 0

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the w_{p} to h₁, where  w_{p}  is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute  w_{p} To computer  

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water v_{f} = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ =  0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

    = 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

    =  417.5 + 0.001043 (14900)

    = 417.5 + 15.5407

    = 433.04 kJ/kg

Find h₃  

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = s_{f} + x_{4} s_{fg}

x_{4} = s_{4} -s_{f} /s_{fg}

x_{4} = 6.14 -  2.45 / 3.89

x_{4}   = 0.9497

The enthalpy h₄:

h₄ = h_{f} +x_{4} h_{fg}

    = 908.4 + 0.9497(1889.8)

    =  908.4 + 1794.7430

    = 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅  = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to  So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = s_{f} + x_{6} s_{fg}

x_{6} = s_{6} -s_{f} /s_{fg}

x_{6} = 7.29 - 1.3028 / 6.0562

x_{6}   = 0.988

The enthalpy h₆:

h₆ = h_{f} +x_{6} h_{fg}

    = 417.51 + 0.988 (2257.5)

    = 417.51 + 2230.41

  h₆ =  2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

P_{p} is found by using:

mass flow rate = m =  1.74 kg/s

Volume of water = v₁ =  0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

P_{p}  = ( m ) ( v₁ ) ( p₂ - p₁ )

     = (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

     = (1.74 kg/s) (0.001043 m³/kg) (14900)

     = 27.04

P_{p} = 27 kW

Compute heat added q_{a} and heat rejected q_{r}  from boiler using computed enthalpies:

q_{a} = ( h₃ - h₂ ) + ( h₅ - h₄ )

      = ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

      = 2726 + 655

      = 3381  kJ/kg

q_{r} =  h₆ - h₁

  = 2648 kJ/kg - 417.5 kJ/kg

  = 2232 kJ/kg

Compute net work

W_{net} = q_{a} - q_{r}

       = 3381  kJ/kg - 2232 kJ/kg

       = 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m =  1.74 kg/s

W_{net} = 1150 kJ/kg

P = m * W_{net}

  = 1.74 kg/s * 1150 kJ/kg

  = 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

   =  1.74 kg/s * 655

   =  1140 kW

Compute Thermal efficiency of this system

μ_{t} = 1 - q_{r} /  q_{a}

   = 1 - 2232 kJ/kg / 3381  kJ/kg

   = 1 - 0.6601

   = 0.34

   = 34%

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The optimum cutting speed for the minimum cost

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On comparing with the Taylor equation VT^n=C,

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T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)

Given that,  

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Batch setting time = 2 hr

Part handling time: T_h=2.5 min

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Part length: l=250 mm=250\times 10^{-3} m

Feed: f=0.30 mm/rev= 0.3\times 10^{-3} m/rev

Depth of cut: d=3.5 mm

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Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, C_e= \$20/15+2=\$3.33/edge

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C= 80 m/min

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(a) From equation (i), cutting speed for the minimum cost:

V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}

\Rightarrow 47.7 m/min

(b) From equation (ii), the tool life,

T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}

\Rightarrow T=53.4 min

(c) Cycle time: T_c=T_h+T_m+\frac{T_t}{n_p}

where,

T_m= Machining time for one part

n_p= Number of pieces cut in one tool life

T_m= \frac{l}{fN} min, where N=\frac{V_{opt}}{\pi D} is the rpm of the spindle.

\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc

So, the number of parts produced in one tool life

n_p=\frac {T}{T_m}

\Rightarrow n_p=\frac {53.4}{4.01}=13.3

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So, the cycle time

T_c=2.5+4.01+\frac{3}{13}=6.74 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc

(e) Total time to complete the batch= Sum of setup time and production time for one batch

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(f) The proportion of time spent actually cutting metal

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Cost per edge,

C_e= \$8/6=\$1.33/edge

Tool changing time, T_t=1min

C= 650 m/min

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(a) Cutting speed for the minimum cost:

V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min [from(i)]

(b) Tool life,

T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min [from(ii)]

(c) Cycle time:

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T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc

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(e) Total time to complete the batch=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.53}{275.5}=0.0962=9.62\%

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C_e= \$10/6=\$1.67/edge

T_t-1min

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n=0.6

(a) Cutting speed:

V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}

\Rightarrow V_{opt}=2105 m/min

(b) Tool life,

T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc

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T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

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A clear details for the question is also attached.

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