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jekas [21]
3 years ago
8

Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure

on: (a) Pump work input; (b) Turbine work output; (c) Heat supplied; (d) Heat rejected; (e) cycle efficiency; (f) Moisture content at turbine exit
Physics
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer:

The effect of lowering the condenser pressure on different parameters is explained below.

Explanation:

The simple ideal Rankine cycle is shown in figure.

Effect of lowering the condenser pressure on

(a). Pump work input :- By lowering the condenser pressure the pump work increased.

(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.

(c). Heat supplied :- Heat supplied increases.

(d). Heat rejected :- The heat rejected may increased  or decreased.

(e). Efficiency :- Cycle  efficiency is increased.

(f). Moisture content at turbine exit :- Moisture content increases.

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3 years ago
I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)
Helga [31]
Hello! I can help you with this! 

4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.

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4 0
3 years ago
A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she
klasskru [66]
Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer: 
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0
3 years ago
PLEASE ANSWER QUICK!
svet-max [94.6K]
I think it’s C sorry if it’s wrong :(
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3 years ago
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