ME 235 EXAM I , Oct 19, 2011
K. Kurabayashi and D. Siegel, ME Dept.
Exam Rules: Open Book and one page of notes allowed.
I have observed the honor code and have neither given nor received aid on this exam.
1. [30 points] Determine the phase of the following substances and find the values of the unknown
(a) Water, P = 400 kPa, T = 40°C, v = ?, u = ?
(b) Nitrogen, T = 120K, h = 60 kJ/kg, P = ?, u =?
(c) R-410a, T = -20 °C, v = 0.0648 m /kg, P = ?, u = ?
(d) Water, P = 500 kPa, u = 2802.91 kJ/kg, T = ?, v = ?
2. [30 points] Carbon dioxide (ideal gas) in a piston cylinder is at 300 kPa, 300K with a volume of
0.2m (State1). It is now compressed via a polytropic process with PV = constant (n ≠ 1) to a final
temperature of 500K (State 2). The work done during the process is -80 kJ.You may use a
constant specific heat to solve this problem C v0= 0.653 kJ/kg -K
(a) What is the mass of carbon dioxide?
(b) Find the value of n
(c) Find the heat transfer in the process
(e) Show the P-v diagram for the process
3. [40 points] A 500 Watt heater is used to melt 2 kg of solid ice at −10ºC to liquid at +5ºC at a
constant pressure of 150 kPa.
a) Find the change in the total volume of the water. (Note that “water” = ice + liquid)
b) Calculate the work done on the water during this process.
c) Find the energy the heater must provide.
d) Find the time the process will take assuming uniform T in the water. 1.
(a) Table B.1.2 T < T = 143.63°C at 400kPa => compressed liquid.
Approximate properties by saturated liquid same T.
v = v (40°C) = 0.001008 m /kg … 3
u = u (40°C) = 167.53 kJ/kg …
(b) Table B.6, h < v atg120K = 74.30 kJ/kg => saturated liquid + saturated vapor.
At T = 120 K, P= P sat= 2513 kPa,
x = (h − h)/f = fg8444
u = u fx×u = 4fg29 kJ/kg