Homework 1 Solutions

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Mechanical Engineering

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Apr 3, 2024

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Homework 1 Solutions 1. Give the missing property of P, T, ν, and x for these states. a. R-134a T = -20°C P = 150 kPa b. R-134a P = 300 kPa ν = 0.072 m 3 /kg c. CH 4 T = 155 K ν = 0.04 m 3 /kg d. CH 4 T = 350 K ν = 0.25 m 3 /kg Solution : a. From Table B.5.1, P > P sat at -20°C. R-134a is a compressed liquid . We assume the liquid is incompressible: ν ≈ ν(T) = ν f (-20°C) = 0.000738 m 3 /kg Since we are not in a saturated state, the quality x is undefined . b. From Table B.5.1, ν > ν g at 300 kPa. R-134a is a superheated vapor . From Table B.5.2, we use linear interpolation to solve for T: (T – 10) / (0.072 – 0.07111) = (20 – 10) / (0.07441 – 0.07111) T = 12.70°C Since we are not in a saturated state, the quality x is undefined . c. From Table B.7.1, ν f < ν < ν g at 155 K. CH 4 is a saturated liquid-vapor mixture . P sat (155 K) = 1295.6 kPa x = (v – v f ) / v fg = 0.806 d. From Table B.7.1, T > T sat when ν g = 0.25 m 3 /kg. CH 4 is a superheated vapor . From Table B.7.2, we use linear interpolation to solve for P: (P – 600) / (0.25 – 0.30067) = (800 – 600) / (0.22510 – 0.30067) P = 734.1 kPa Since we are not in a saturated state, the quality x is undefined . T C.P. v a b P = cons P v a T b c d c d

P C.P. v T C.P. v T P = 857 kPa 20 C 2. Ammonia at 20°C with a quality of 50% and total mass 2 kg is in a rigid tank with an outlet valve at the bottom. How much liquid (mass) can be removed through the valve until the liquid is gone, assuming the temperature stays constant? Solution : Initially, there is 1 kg of saturated liquid and 1 kg of saturated vapor in the tank. The total volume of the tank is calculated using Table B.2.1: V tot = m liq ν liq + m vap ν vap = 0.001638 + 0.14922 = 0.15086 m 3 The liquid is removed until only saturated vapor remains. The mass of vapor remaining in the tank is: m vap = V tot / ν vap = 0.15086 / 0.14922 = 1.011 kg Therefore, the mass of liquid removed from the tank is: m liq = m tot – m vap = 2 – 1.011 = 0.989 kg Comment: The mass out and the final state do not depend on which phase is removed. However, the heat transfer will be significantly different if you take vapor from the top. The liquid must be vaporized which requires a significant amount of energy versus taking liquid from the bottom, in which case only a small amount of vaporization is needed. 3. Saturated vapor R-410A at 60°C has its pressure increased to decrease the volume by 20% while the state is still saturated vapor. Find the new temperature and pressure. Solution : The pressure and specific volume for state 1 are found in Table B.4.1: P 1 = P sat = 3836.9 kPa ν 1 = ν g = 0.00497 m 3 /kg Use the fact that the system is a control mass to find the specific volume in state 2: ν 2 = 0.8 ν 1 = 0.003976 m 3 /kg

x 2 = 1 From Table B.4.1, we use linear interpolation to find the saturation temperature and pressure corresponding to ν g = ν 2 : (T 2 – 65) / (0.003976 – 0.00399) = (70 – 65) / (0.00286 – 0.00399) (P 2 – 4278.3) / (0.003976 – 0.00399) = (4763.1 – 4278.3) / (0.00286 – 0.00399) T 2 = 65.06°C P 2 = 4284.3 kPa P C.P. v T C.P. v T 3.84 MPa 60 C o 65 60 4. An engine has a piston/cylinder with 0.05 kg air at 4 MPa, 1800 K after combustion and this is expanded in a polytropic process with n = 1.5 to a volume 10 times larger. Find the expansion work and heat transfer using the specific heat value in Table A.5. Solution : From the definition of a polytropic process: PV n = constant P 1 V 1 n = P 2 V 2 n P 2 = P 1 (V1/V2) n = 4000 (0.1) 1.5 = 126.5 kPa The ideal gas assumption is used to determine the temperature at state 2: P 1 V 1 / T 1 = mR = P 2 V 2 / T 2 T 2 = P 2 V 2 T 1 / P 1 V 1 = T 1 (V 1 /V 2 ) n-1 = 569.2 K The energy equation is used to find work and heat transfer: E 2 – E 1 = U 2 – U 1 = 1 Q 2 – 1 W 2 1 W 2 = ∫ P dV = (P 2 V 2 – P 1 V 1 ) / (1 – n) = mR(T 2 – T 1 ) / (1 – n) = 35.32 kJ 1 Q 2 = U 2 – U 1 + 1 W 2 = mC ν0 (T 2 – T 1 ) + 1 W 2 = – 8.80 kJ

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P 1 2 T P = C T 3 P v 1 2 T v 1 2 T T 2 1 T = C v - 0.5 P = C v - 1.5 5. Two rigid tanks are filled with water. Tank A is 0.2 m 3 at 200 kPa, 150°C and tank B is 0.3 m 3 at saturated vapor 300 kPa. The tanks are connected by a pipe with a closed valve. We open the valve and let all the water come to a single uniform state while we transfer enough heat to have a final pressure of 300 kPa. Give the two property values that determine the final state and find the heat transfer. B A Solution : From Tables B.1.2 and B.1.3, we can find properties of each tank in state 1: ν 1,A = 0.95964 m 3 /kg u 1,A = 2576.87 kJ/kg ν 1,B = 0.60582 m 3 /kg u 1,B = 2543.55 kJ/kg The total volume and mass in the system is the sum of tanks A and B: V = V A + V B = 0.5 m 3 m A = V A / ν A = 0.2084 kg m B = V B / ν B = 0.4952 kg m = m A + m B = 0.7036 kg We use this information to calculate specific volume for state 2: ν 2 = V / m = 0.71062 m 3 /kg State 2 is fully specified by pressure and specific volume (can also find temperature): [P 2 , ν 2 ] = [300 kPa, 0.71062 m 3 /kg] T 2 = 196.6°C From Table B.1.3, we use linear interpolation to solve for u 2 : (u 2 – 2570.79) / (0.71062 – 0.63388) = (2650.65 – 2570.79) / (0.71629 – 0.63388) u 2 = 2645.2 kJ/kg The energy equation is used to find heat transfer (no work is done by the system): U 2 – U 1 = 1 Q 2 – 1 W 2 1 Q 2 = mu 2 – (m A u 1,A + m B u 1,B ) = 64.6 kJ