solve 6.71 plz Figure P6.71 provides steady-state operating data for a well-insu

Question

solve 6.71 plz

Figure P6.71 provides steady-state operating data for a well-insulated device having steam entering at one location and exiting at another. Neglecting kinetic and potential energy effects, determine (a) the direction of flow and (b) the power output or input, as appropriate, in kJ per kg of steam flowing. Steam enters a well-insulated nozzle operating at steady state at 540 degree C, 3 MPa and a velocity of 4 m/s. At the nozzle exit, the pressure is 150 kPa and the velocity is 300 m/s. Determine the rate of entropy production, in kJ/K per kg of steam flowing. Air at 500 kPa, 980 K enters a turbine operating at steady state and exits at 200 kPa, 680 K, Heat transfer from the turbine occurs at

Explanation / Answer

Note- you will need to have access to steam tables to solve this problem.
a) the flow will be from right to left as in the diagram, an explanation can be seen from the values obtained in part B.

b)Using steam tables, let's get the data first.
for Pi = 2 MPa = 2000 kPa
T = 280 C.

h = 3024
s = 6.7684

Now for P = 200 kPa, we need to find the properties for s = 6.7684

Sf = 1.5302
Sfg = 5.5968

we know that s = 6.7684 = sf + x * sfg

solving for a vapor quality of 93.59 % or x = 0.936

So,
we can now calculate h

h = hf + x * hfg
hfg =  2201.6
hf = 504.71
x= 0.936

so, h = 504.71 + 0.97*2201 = 2639.68 hJ/kg

isoentropic power = (m/s)(h2-h1) = (1)(3024 -2639.68 ) = 384.32 kJ/s = 384.32 kW

Kindly upvote as it encourages us a lot. Please feel free to ask any doubts.

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