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rjkz [21]
3 years ago
8

A cylinder fitted with a movable piston contains water at 3 MPa with 50% quality, at which point the volume is 20 L. The water n

ow expands to 1.2 MPa as a result of receiving 600 kJ of heat from a large source at 300◦C. It is claimed that the water does 124 kJ of work during this process. Is this possible?

Engineering
2 answers:
WINSTONCH [101]3 years ago
7 0

Answer:

Yes it is possible

Explanation:

Attached is the solution

Marysya12 [62]3 years ago
5 0

Answer:

The process is possible:

Explanation:

We are going to find out if the entropy generated is greater than 0, if it is greater than 0, then the process is feasible. If it is not, the process is not feasible.

P_{1} = 3 MPa

x_{1} = 50 % = 0.5

V_{1} = 20 L = 0.02 m^{3}

P_{2} = 1.2 MPa

T_{H} = 300^{0} C = 573 K

Received heat energy, Q_{12} = 600 kJ

Work done, W_{12} = 124 kJ

At state 1, using the steam table:

T_{1} = T_{s} = 233.9^{0} C\\v_{f1} = 0.001216 m^{3} /kg\\v_{fg1} = 0.06546m^{3} /kg\\u_{f1} = 1004.76 kJ/kg\\u_{fg1} = 1599.34 kJ/kg\\s_{f1} = 2.6456 kJ/kg-K\\s_{fg1} = 3.5412kJ/kg-K

v_{1} = v_{f1} + x_{1} * v_{fg1}

v_{1} = 0.001216 + 0.5*(0.06546)\\v_{1} = 0.03395 m^{3} /kg

M = \frac{V_{1} }{v_{1} } \\M = 0.02/0.03395\\M = 0.5892 kg

u_{1} = u_{f1} + x_{1} * u_{fg1}\\u_{1} = 1004.76 + 0.5*1599.34\\u_{1} = 1804.43 kJ/kg

s_{1} = s_{f1} + x_{1} * s_{fg1}\\s_{1} = 2.6456 + 0.5*3.5412\\s_{1} = 4.4162 kJ/kg

Q_{12} = m(u_{2} - u_{1} ) + W_{12} \\600 = 0.5892(u_{2} -1804.43) + 124\\

Solving for u₂

u_{2} = 2612.3 kJ/kg

Since P₂ = 1.2 MPa, u₂ = 2612.2 kJ/kg,

then from steam table, T₂ = 200°C, S₂ = 6.5898 kJ/kg-K

The entropy generated will be:

\triangle S = m(S_{2} -S_{1} ) - \frac{Q_{12} }{T_{H} }\\  \triangle S= 0.5892(6.5898 - 4.4162) - \frac{600 }{573 }\\ \triangle S =0.233 kJ/K

Since ΔS > 0, this process is possible

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Water is the working fluid in an ideal Rankine cycle. Superheatedvapor enters the turbine at 10MPa, 480°C, and the condenser pre
choli [55]

Answer:

Explanation:

Given that:

Superheated vapor enters the turbine at 10 MPa, 480°C,

From the tables of superheated steam tables; the following values are obtained

h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state

s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}

s_2 =0.51624 + x(7.82)

s_2 =0.51624 + 7.82x

From the values obtained;

s_1 =s_2= 6.52846 \ kJ/kg.K

Therefore;

6.52846 = 0.51624+7.82x

6.52846 - 0.51624 = 7.82 x

6.01222  = 7.82 x

x = 6.01222/7.82

x = 0.7688

The enthalpy for this process at state (s_2) can be determined as follows:

h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 )  \\ \\ h_2 =2010.4084   \ kJ/kg

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:

n_T = \dfrac{h_1-h_{2a}}{h_1-h_2}

0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}

0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}

0.8 * {1311.6116}= {3322.02-h_{2a}

1049.28928=  {3322.02-h_{2a}

h_{2a}=   {3322.02- 1049.28928

h_{2a}=   2272.73072 kJ/kg

The work pump is calculated by applying the formula:

w_p = v_{f  \  6 kpa} (p_4-p_3)

w_p = 0.0010062 * (10000-6)

w_p = 0.0010062 *9994

w_p = 10.0559628 \  kJ/kg

However;

w_p = h_4 -h_3

From the process;

h_3 = h_{f(6 kpa)} = 150.15 \  kJ/kg

10.0559628 = h_4 - 150.15

10.0559628+  150.15 = h_4

160.2059628= h_4

h_4= 160.2059628 \  kJ/kg

The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:

n_P = \dfrac{h_4-h_{3}}{h_{4a}-h_3}

6 0
2 years ago
What have you learned about designing solutions? How does this apply to engineering? Think of some engineering solutions that st
Andrew [12]

Answer:

In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.

Explanation:

4 0
2 years ago
A fluid of specific gravity 0.96 flows steadily in a long, vertical 0.71-in.-diameter pipe with an average velocity of 0.90 ft/s
KengaRu [80]

Answer:

0.00650 Ib s /ft^2

Explanation:

diameter ( D ) = 0.71 inches = 0.0591 ft

velocity = 0.90 ft/s ( V )

fluid specific gravity = 0.96 (62.4 )  ( x )

change in pressure ( P ) = 0 because pressure was constant

viscosity =  (change in p - X sin∅ ) D^{2} / 32 V

              = ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2  / 32 * 0.90

              = - 59.904 sin (-90) * 0.0035 / 28.8

              = 0.1874 / 28.8

  viscosity = 0.00650 Ib s /ft^2

8 0
3 years ago
Read 2 more answers
The underground storage of a gas station has leaked gasoline into the ground. Among the constituents of gasoline are benzene, wi
vovangra [49]

Answer:

a) benzene = 910 days

b) toluene = 1612.67 days

Explanation:

Given:

Kd = 1.8 L/kg (benzene)

Kd = 3.3 L/kg (toluene)

psolid = solids density = 2.6 kg/L

K = 2.9x10⁻⁵m/s

pores = n = 0.37

water table = 0.4 m

ground water = 15 m

u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s

a) For benzene:

R=1+\frac{\rho * K_{d}  }{n}, \rho = 2.6\\ R=1+\frac{2.6*1.8}{0.37} =13.65

The time will take will be:

t=\frac{xR}{a} , x=12,a=0.18\\t=\frac{12*13.65}{0.18} =910days

b) For toluene:

R=1+\frac{2.6*3.3  }{0.37} = 24.19

t=\frac{12*24.19}{0.18} =1612.67days

6 0
2 years ago
Read 2 more answers
A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2
IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length l for the deformation, we use the following relation;

l = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

l = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

l = 3161.025289 / 7240

l = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0
3 years ago
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