2. A rigid tank contains 2kg water at 100°C with quality of 0.3189. Then the wat

Question

2. A rigid tank contains 2kg water at 100°C with quality of 0.3189. Then the water is heated up to 200°C. (20 point)
a. What is the final state (subcooled, saturated or superheated) and final pressure in kPa or Pa?
b. What is the heat transfer in kJ or J for the process?

Explanation / Answer

note : the values i found will be written in red , there rest is given for this kind of question you will need to use the thermodynamics tables , i will do it for you and just write the answers in here . first of all you shuold notice that a rigid tanks involves no work input or output since there is no change in the volume , and hance v1=v2 , and work is zero water state 1 : 2kg 100 C x1 = 0.3189 since quailty is given then you should use the saturated water vapor mixture to find any property you need v1 = vf + x ( vfg ) = 0.534 m3/ kg - this is the specific volume at state 1 V1 = mv1 = 1.068 m3 and this is the total volume of the rigid tank u1 = uf + x ( ufg ) = 1084.64 Kj/Kg - this is the specific internal enegy at state 1 water state 2 : 200 C V2=V1 =1.068 m3 m2 = m1 = 2 kg v2 = V2 / m2 = 0.534 and this is the specific volume at state 2 u2 = 2646.83 Kj/Kg - this is the specific internal enegy at state2 so we have the tempreture and the specific volume , looking at the tables again : v2 > vfg ... so it is a superheated vapor abd the pressure is ( again from the tables ) : P2 = 400 Kpa state : superheated vapor b ) from thermodynamics first law : Q-W = m( ?u + ?ke + ? Pe ) notice that w =0 ? KE & ? pe = 0 .. since the boundariy is stationary and not moving so Q = 2 ( 2646.83 - 1084.64 = 3124.38 kj

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.