A cylindrical tank with a cross-sectional area of 1 m^2 is originally filled wit

Question

A cylindrical tank with a cross-sectional area of 1 m^2 is originally filled with 2500 kg of saturated liquid water at 20 degreeC. It is then drained through a side duct at the bottom of the tank whose cross sectional area is 3Times10^-4 m^2. The final mass of water remaining in the tank is 900 kg. The velocity of the water exits varies according to (2gz)^1/2, where z is the water level in [m], and g is gravity (9.807 m/s^2). Determine: a) The time it takes in [min] to empty the tank if it heats up during the emptying process and the final temperature of the saturated liquid water is a uniform 40degreeC. (Use an averaging approach with the flow rates leaving the tank) b) The total amount of heat added to the tank during the process in [kJ]. (Again, use an averaging approach for the energy leaving the tank) Show all unit conversions and calculations to receive full points. Moreover, you will need to make approximations in order to solve the questions. Ignore KE within the tank. However. PE within the tank and KE exiting are important.

Explanation / Answer

Solution:

initial mass = 2500 kg

Area of tank A= 1 m^2

Area of duct a= 3 x 10^4 m^2

final mass = 900 kg

A)

density of water is 10^3 kg/m^3

density = mass/volume

10^3 = 2500/V

V= 2.5 m^3

2.5 = A h

hi = 2.5 m

similarly

10^3 = 900/V

V= 0.9 m^3

0.9 = A h

hf = 0.9 m

Velocity is given by Vi = sqrt (2 x g x hi)

= sqrt (2 x 9.8 x 2.5)

= 7 m/s

Similarly Vf = sqrt (2 x g x hf)

= sqrt (2 x 9.8 x 0.9)

= 4.2 m/s

Average velocity V = (Vi + Vf)/2

= 5.6 m/s

Mass flow is given by = density x a x average velocity

= 10^3 x 3 x 10^4 x 5.6

= 168 x 10^6 kg/s

(2500-900)/t = 168 x 10^6

t = 105000 s

t = 1750 min

B)

initial temp = 20

final temp = 40

Cp = 4.18 kJ/kg K

Q = average mass x Cp X change in temp

= (2500-900)/2 x 4.18 x (20)

= 66880 kJ

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