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3 years ago

13

Mercury flows inside a copper tube 9m long with a 5.1сm inside diameter at an average velocity of 7.0 m/s. The inside surface te

mperature of the tube is kept at 38 C and the mean temperature of the mercury is 66 C. Assuming that the velocity and temperature profiles are fully developed throughout, calculate the rate of heat transfer by convection. Justify any equations you may use.

1 answer:

svp [43]3 years ago

6 0

Answer:

rate of heat transfer = 9085708.80 W

Explanation:

Given:

Inside diameter, D = 5.1 cm

= 5.1 x $10^{-2}$ m

Average velocity, V = 7 m/s

Mean temperature, T = (66+38) /2

= 52°C

Therefore kinematic viscosity at 52°C is ν = 0.104 X $10^{-6}$ $m^{2}$ / s

Prandtl no., Pr = 0.021

We know Renold No. is

Re = $\frac{V\times D}{\nu }$

Re = $\frac{7\times 5.1\times 10^{-2}}{0.104\times 10^{-6}}$

= 3.432 X $10^{6}$

Therefore the flow is turbulent.

Since the flow is turbulent and the ratio of L/D is greater than 60 we can use Dittua-Boelter equation.

Nu = 0.023 $Re^{0.8}$ . $Pr^{0.3}$

= 0.023 x $(3.432 \times10^{6})^{0.8}$ x $(0.021)^{0.3}$

= 1221.52

Since Nu = $\frac{h.D}{k}$

h = $\frac{k\times Nu}{D}$

= $\frac{9.4\times 1221.52}{5.1\times 10^{-2}}$

= 225143.3

Therefore rate of heat transfer, q = h.A(T- $T_{\infty }$

q= 225143.3 x 2πrh ( 66-38)

= 225143.3 X 2π X $\frac{5.1\times10^{-2}}{2}\times 9\times 28$

= 9085708.80 W

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3 years ago

A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The

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Answer:

0.31

126.23 kg/s

Explanation:

Given:-

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- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

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- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

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s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )

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$w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}$

- From the following relation we can determine ( h2 ) as follows:

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Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

P3 = 8 MPa

T3 = ? ( assume fluid exist in the saturated vapor phase )

h3 = hg-P3 = 2758.7 KJ/kg

s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

$q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}$

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

P4 = 10 KPa

s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

$x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947$

- Determine the isentropic ( h4s ) at this state as follows:

$h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}$

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

$h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\$

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

$w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}$

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

$W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}$

Answer: The mass flow rate of the steam would be 126.23 kg/s

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$n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31$

Answer: The thermal efficiency of the cycle is 0.31

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Answer:

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