Written Assignment 8 Solutions.pdf

# M 1 and m 2 can be found by using the formula v v m

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m 1 and m 2 can be found by using the formula: v V m State 1 is saturated vapor at 100kPa; the relevant properties can be found in table A-5. v 1 = v g at 100 kPa = 1.6941 m 3 /kg State 2 is at 300kPa and 200 ° C and is therefore a superheated vapor; the relevant properties can be found in table A-6. v 2 = 0.71643 m 3 /kg The container is a rigid tank and the volume V stays constant at 2 m 3 during this process. kg kg m m v V m 791619558 . 2 / 0.71643 2 3 3 2 2 kg kg kg m m m i 8 1.61105155 180568 . 1 8 2.79161955 1 2 (b) By using the first law relation for a transient process, the enthalpy of the supplied steam h sup can be found. 𝑄 𝑐𝑣 − 𝑊 𝑐𝑣 = ∑ 𝑚 𝑒 𝑒 𝑒𝑥𝑖?? − ∑ 𝑚 𝑖 𝑖 𝑖𝑛𝑙𝑒?? + (𝑚 2 𝑢 2 − 𝑚 1 𝑢 1 ) 𝑐𝑣 (1) Energy leaves the system during cooling, therefore, Q cv = 700 kJ. There is no work done on/by the system, W cv = 0. Steam only enters the tank, so m e = 0 and h i = h sup . kg kg m m v V m 180568 . 1 / 6941 . 1 2 3 3 1 1

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