Below are some information for water. Mark the correct answer in the box and the

Question

Below are some information for water. Mark the correct answer in the box and then find the internal energy(is) that is appropriate corresponding to the given state: P 200kPa, T = 120.21 degree C Superheated Saturated sub cooled u_f = u_g = u_fg = u = P = 50 kPa, T = 75 degree C Superheated Saturated sub cooled u_f = u_g = u_fg = u = P 200kPa, T = 200 degree C Superheated Saturated sub cooled u_f = u_g = u_fg = u = Water at 100 kPa, 20degree C steadily enters a tank where it is heated by a 50 kW electric heater as shown. The mass flow rate water is 0.5 kg/s. Assume that the inlet and exit velocities are negligible and there is no elevation difference in the flow. Do the following: Write the corresponding energy equation by applying the 1^st Law of thermodynamics for this system. Calculate the exit temperature of the water, degree C. [For this problem, do not interpolate for properties. Select the closest available data property table].

Explanation / Answer

a)at 200Kpa and 120.21 deg,c, the steam is saturated ( from steam tables). The internal energy of saturated vapor is 2529 Kj/Kg

b) Saturation temperauture corresponding to 50 kpa is 61.35 deg.c. since the given temperature is 75 deg.c, the steam is super heated. The intenral energy at 75 deg.c= 348 K is 2811.5 Kj/Kg

c)200 Kpa correspond to a saturation temperaure of 120.21deg.c and when temperautre is 200 deg.c, it is super heated and its internal energy = 2887.2Kj/Kg

2. First law of thermodynamics applied to flow processs for steady state is

m*{delta(H+ 1/2u2+gZ}= Q+W

m* =mass flow rate, deltaH= change in enthalpy, detlau2/2 = change in kinetic energy , g deltaZ= change in potential energy

Q= heat supplied and W =work done

when change in kinetic energy and potential energy are neglected, and there is no work done

Change in enthalpy*m= Q

0.5*1000g/m*4.18 j/g*(T-20)= 10*1000 j/s, T is the outlet temperature of water

0.5*4.18*(T-20)= 10 , T =24.78 deg.c

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