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5A) A one inlet one outlet steam piping segment is being analyzed for a variety of stress tests, and is sketched below. The system is operated at steady state, and is well insulated. The inlet and outlet pipes both have circular cross sections. The total work done by the system is 25 kW. Kinetic and potential energy effects can be ignored. The steam enters the system at 0.7 bars, a temperature of 200 oC, and a velocity of 10 m/s. The steam leaves at a pressure of 1 bar, 160 oC, and a velocity of 15 m/s. Determine:
(a) The input mass flow rate, in kg/s
(b) The diameter of the inlet and outlet pipes, in cm
a) By approaching this problem from an energy rate balance since we are given heat and work information, will allow for the solution of the mass flow rate for the system. The basic equation for energy balance for a control volume is:
𝑑𝐸𝑐𝑣 𝑑𝑡 =^ 𝑄𝑐𝑣^ − 𝑊𝑐𝑣^ +^ 𝑚𝑖^ ℎ𝑖^ +^
𝑣𝑖^2
2 +^ 𝑔𝑧𝑖^ −^ 𝑚𝑜^ ℎ𝑜^ +^
𝑣𝑜^2
2 +^ 𝑔𝑧𝑜
Since we are at steady state, with no heat transfer (well insulated), and negligible kinetic and potential energy effects we have:
Ti = 200 oC
Pi = 0.7 bar
To = 160 oC
Po = 1.0 bar
We now need to solve for the specific enthalpies at the inlet and outlet. Since for each part we are given the pressure and temperature, we can look up these values on the given table. Doing so yields:
ℎ𝑖 = 2876.7 𝑘𝐽/𝑘𝑔
ℎ𝑜 = 2796.2 𝑘𝐽/𝑘𝑔
Using these values in our governing equation we have:
b) To determine the diameter of the inlet and outlet pipes we first need to determine the cross sectional areas of these pipes. This may be determined from the mass flow rate. Since we are at steady state we have:
𝑚𝑖 = 𝑚𝑜
And
𝑚 = 𝜌𝐴𝑣
Therefore we need to determine the density of the steam at the inlet and outlet pipes. Since density is the inverse of the specific volume of the steam, we can read off the specific volume for the inlet and outlet pipes from the tables. Doing so we find:
𝑣𝑖 = 3.108 𝑚^3 /𝑘𝑔
𝑣𝑜 = 1.984 𝑚^3 /𝑘𝑔
Therefore:
Using this equation we have:
𝑑𝑖 = 35.1 𝑐𝑚
𝑑 0 = 22.8 𝑐𝑚
5B) A one inlet one outlet steam piping segment is being analyzed for a variety of stress tests, and is sketched below. The system is operated at steady state. The only work done by this system is flow work. The inlet and outlet pipes both have circular cross sections. Heat is lost from this system at a rate of 35kW. Kinetic and potential energy effects can be ignored. The steam enters the system at 0.7 bars, a temperature of 360 oC, and a velocity of 20 m/s. The steam leaves at a pressure of 1 bar, 320 oC, and 15 m/s. Determine:
(a) The output mass flow rate, in kg/s
(b) The diameter of the inlet and outlet pipes, in cm
a) By approaching this problem from an energy rate balance since we are given heat and work information, will allow for the solution of the mass flow rate for the system. The basic equation for energy balance for a control volume is:
𝑑𝐸𝑐𝑣 𝑑𝑡
𝑣𝑖^2
𝑣𝑜^2
To = 320 oC
Po = 1.0 bar
Ti = 360 oC
Pi = 0.7 bar
𝑣𝑜 = 2.732 𝑚^3 /𝑘𝑔
Therefore:
𝜌𝑖 =
𝑣𝑖^ = 0.240^ 𝑘𝑔/𝑚
3
𝑣𝑜^ = 0.366^ 𝑘𝑔/𝑚
3
Using these values we can solve for the inlet and outlet cross sectional areas as follows:
𝑚 = 𝜌𝐴𝑣
𝐴 =
𝐴𝑖 = 0.089𝑚^2 𝑥
(100𝑐𝑚)^2
1 𝑚^2
𝐴𝑖 = 890 𝑐𝑚^2
0.366 𝑘𝑔/𝑚^3 (15 𝑚𝑠 )
𝐴𝑜 = 0.078𝑚^2 𝑥
(100𝑐𝑚)^2
1 𝑚^2
𝐴𝑜 = 780 𝑐𝑚^2
Now that we have the cross sectional areas we can solve for the diameters as follows:
𝐴 = 𝜋𝑟^2 = 𝜋(
2
Using this equation we have:
𝑑𝑖 = 33.7 𝑐𝑚
𝑑 0 = 31.5 𝑐𝑚
