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ioda
3 years ago
13

an inclined manometer is connected to a pitor tube to measure the velocity at the center of a circular duct. If the inclined man

ometer measured 2.3 inches of water and the minimum division on the manometer scale is 0.04 inches of water. What is the uncertainty of the velocty at the center of the pipe in m/s
Engineering
1 answer:
jekas [21]3 years ago
8 0
57.5 m/s
I did 2.3/0.04
I’m not sure if it’s correct though
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A steam power plant operates on an ideal Rankine cycle with two stages of reheat and has a net power output of 120 MW. Steam ent
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Answer:

a) 40.6%

b)72.19kg/s

Explanation:

The Rankine cycle with two reheat stages has 9 stages in total.

The maximum pressure will be at the first inlet stage of the HP turbine which is stage 3. The minimum pressure will be the exit stage of the condenser because the condenser operates under vacuum pressure which is stage 1.

The following assumptions can be made:

1 - Each component in the cycle is analyzed as an open system operating at steady-state.

  2 - All of the processes are internally reversible.

  3 - The turbine and pump operate adiabatically and are internally reversible, so they are also isentropic.

   4 - Condensate exits the condenser as saturated liquid.

  5 - The effluent from the HP turbine is a saturated vapor.

  6 - No shaft work crosses the system boundary of the boiler or condenser.

   7 - Changes in kinetic and potential energies are negligible

a) The thermal efficiency of the cycle is defined as the work of the cycle divided by the total heat input to the system. The stages that have heat input is stages 2-3, 4-5, 6-7.

For stage 2:

s₁=s₂ assuming isentropic

s_1=0.4762 @ P_1=15MPa

enthalpy will be a compressed liquid so after interpolation

h_2=97.93+(0.4762-0.2932)((180.77-97.93)/(0.5666-0.2932))=153.38kJ/kg

For stage 3:

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h_3=3310.8kJ/kg

Stage 4:

superheated vapor

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s₃=s₄=6.3480 kJ/kg, we must use interpolation to find h₄

h_4=2925.7-(6.348-6.2111)((3069.3-2925.7)/(6.4516-6.2111))=3007.44kJ/kg

Stage 5:

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h_5=3434.7kJ/kg

Stage 6:

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s₅=s₆

s_6=6.9781

We find h₆ using interpolation from the steam tables:

h_6=2943.1-(6.9781-6.9265)((3051.6-2943.1)/(7.1246-6.9265))=2970.67kJ/kg

Stage 7:

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The heat into the cycle is:

=(h_3-h_2)+(h_5-h_4)+(h_7-h_6)

=(3310.8-153.38)+(3434.7-3007.44)+(3479.1-2970.67)=4108.74kJ/kg

We can determine the work out by the condenser from stage 9 to stage 1:

Stage 1:

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Stage 9:

We assume that its a saturated liquid with quality of 1 at 5kPa and

s₇=s₉ and after interpolation

h_9=2568.53kj/kgK

Qout = [/tex]2568.53-137.75=2430.79kJ/kg[/tex]

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b)

The mass flow rate can be calculated from the Wnet:

W_n=W_t-W_p

Work of the turbines minus the work of the pumps:

W_n=m((h_3-h_4)+(h_5-h_6)+(h_7-h_9)-(h_1-h_2)

120000=m(1662.33)

m=72.19

mass flow rate of steam is 72.19 kg/s

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