Question

Q Tb = 500 K Water 1 T1 = 20°C p1 = 1 bar 2 T2 = 95°C p2 = 1 bar (a) Waste heat Water 1 T1 = 20°C p1 = 1 bar 2 T2 = 95°C p2 = 1 bar + – (b) Electrical resistor

          Q
Tb = 500 K
Water 1
T1 = 20°C
p1 = 1 bar
2
T2 = 95°C
p2 = 1 bar
(a) Waste heat
Water 1
T1 = 20°C
p1 = 1 bar
2
T2 = 95°C
p2 = 1 bar
+ –
(b) Electrical resistor

Added by Kimberly H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/15/2023

Step 1

In this case, there is no useful work output, so Wout = 0. The work input is also zero since we are not using any mechanical or electrical energy. Show more…

Show all steps

Thanks for your feedback!

Figure P7.104 provides two options for generating hot water at a steady state. In (a), water heating is achieved by utilizing industrial waste heat supplied at a temperature of 500 K. In (b), water heating is achieved by an electrical resistor. For each case, devise and evaluate an exergetic efficiency. Compare the calculated efficiency values and comment. Stray heat transfer and the effects of motion and gravity are negligible. Let To = 20"C and Po = 1 bar.