Soln_Fall2008_HW6

Soln_Fall2008_HW6 - SOLUTIONS- HW6 1. In a steam power...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
SOLUTIONS- HW6 1. In a steam power plant 2 MW is added in the boiler, 1.2 MW is taken out in the condenser and the pump work is 0.1 MW. Find the plant thermal efficiency. If everything could be reversed find the coefficient of performance as a refrigerator. Solution: CV. Total plant: Energy Eq.: Q . H + W . P,in = W . T + Q . L W . T = 2 + 0.1 – 1.2 = 0.9 MW ηTH = (W . T – W . P,in )/Q . H = (900– 100)/2000 = 0.4 β = Q . L ( W . T – W . P,in )= 1200/(900 – 100) = 1.5 2. R-410a at 50°C, x = 0.2 flowing at 3 kg/s is brought to saturated vapor in a constant- pressure heat exchanger. The energy is supplied by a heat pump with a coefficient of performance of β′ = 3. Find the required power to drive the heat pump. Solution: C.V. Heat exchanger m . 1 = m . 2 ; m . 1h1 + Q . H = m . 1h2 Given coefficient of performance β′ = Q . H / W . = 3 Table B.4.1: h1 = hf + x1hfg = 143.65 + 0.2 × 135.73 = 170.836 kJ/kg, h2 = hg = 279.58 kJ/kg Energy equation for line 1-2: Q . H = m . R-12(h2 - h1) = 326.232 kW W . = Q . H / β′ = 326.232/3 = 108.743 kW
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4 A certain solar-energy collector produces a maximum temperature of 50°C. The energy
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/29/2009 for the course ME 530.230 taught by Professor Katz during the Spring '09 term at Johns Hopkins.

Page1 / 4

Soln_Fall2008_HW6 - SOLUTIONS- HW6 1. In a steam power...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online