finala_spr05

# finala_spr05 - PHY 206 FINAL EXAM May 4 2005 NAME SIGNATURE...

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Unformatted text preview: PHY 206 FINAL EXAM May 4, 2005 NAME: SIGNATURE: ID: 1. 2. 3. 4. 5. Some relations you may need: p + ρgy + 1 2 ρv 2 = const. dQ = mCdT Δ V = βV Δ T pV = nRT dQ = dU + dW dW = pdV e = W Q H n a sin θ a = n b sin θ b n a s + n b s = n b- n a R γ = 1 p 1- u 2 /c 2 For relative motion along x, x : x = γ ( x- ut ) t = γ ( t- ux/c 2 ) v x = v x- u 1- uv x /c 2 Δ t = γ Δ t L = L γ f = r c + u c- u f (source approaching) 203 (130) 1. Small hydroelectric plants in the mountains sometimes bring the water from a reservoir down to the power plant through enclosed tubes. In the plant depicted in the figure, the intake tube (cross-sectional area A 1 ) in the base of the dam is a distance d below the reservoir surface. The water drops through the tube before flowing into the turbine through a nozzle with cross-sectional area A 2 . Take atmospheric pressure to be p a , independent of altitude. a) Compute the water speed into the turbine. b) Find the intake pressure (eliminate any velocities in your answer). c) For what value of the ratio A 2 /A 1 will the intake pressure be p a ? (Your answer should depend only on d and H ). H A 1 d A 2 turbine intake Solution: a) We have the usual assumption, for a large body of water and small output, that the velocity of the reservoir surface is zero. Using the Bernouilli equation to compare points at the top surface of the reservoir and at the turbine (with y = 0 at the turbine): p a + ρgH = p a + 1 2 ρv 2 2 ⇒ v 2 = p 2 gH b) Now Bernouilli again comparing points at the reservoir surface and at depth d below the surface at the intake: p a + ρgH = p intake + 1 2 ρv 2 1 + ρg ( H- d ) p intake = p a + ρgd- 1 2 v 2 1 The continuity equation, v 1 A 1 = v 2 A 2 combined with the result from a) lets us eliminate v 1 in terms of the A ’s and H : p intake = p a + ρgd- ρ A 2 A 1 ¶ 2 gH c) By inspection, p intake = p a when, ρgd- ρ A 2 A 1 ¶ 2 gH = 0 ⇒ A 2 A 1 = r d H 2. Two metal blocks, isolated from the environment, are brought into contact, exchange heat, and reach equilibrium. The first block has mass m , specific heat (per mass) C , and initial temperature T 1 . The second block has mass 2 m , specific heat C/ 4, and initial temperature...
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finala_spr05 - PHY 206 FINAL EXAM May 4 2005 NAME SIGNATURE...

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