Please be clear. A cylinder fitted with a freely floating piston contains 0.5 kg
ID: 2994674 • Letter: P
Question
Please be clear.
A cylinder fitted with a freely floating piston contains 0.5 kg water at 3 MPa absolute pressure and a quality of 0.25. The water is then heated until its temperature reaches 700 K. On a P-v diagram, show this process, the vapor dome, and the isotherms passing through the initial and final state. Find the work done by the water. Find the heat transfer during this process. Air enters an adiabatic nozzle steadily at 300 kPa, 470K, and 30 m/s and leaves at 100 kPa and 180 m/s. The inlet area of the nozzle is 80 cm2. Use an energy correction factor of 1.05. Determine The mass flow rate through the nozzle The exit temperature of the air. Use Table C.2 for h(T). (Answer: 453.8 K) The exit area of the nozzle A well-insulated valve is used to throttle steam from 8 MPa and 500 degree C to 6 MPa. Determine the final temperature of the steam. A throttling valve is a restriction in a pipe used to reduce the flow pressure without work output. Since the device is small, heat transfer can be neglected. Also neglect kinetic and potential energy. (Answer: 490 degree C) The water table of a housing development is 400 feet below the surface. You have to install a well pump that will deliver 15 gal/min of water (use p=8.33 Ibm/gal and v=0.016044 ft3/lbm) at a pressure of 30 psig (gage pressure) at the surface. What horsepower motor should you use? Assume incompressible flow (p= constant) and neglect changes in water temperature. The pressure at the water surface in the well is atmospheric. (Answer: 1.78 hp)Explanation / Answer
1.
From steam properties,
At P1 = 3 MPa and quality = 0.25, we get h1 = 1460 kJ/kg, v1 = 0.0176 m^3/kg
At P2 = P1 = 3 MPa and T2 = 700 K, we get h2 = 3290 kJ/kg, v2 = 0.104 m^3/kg
b)
Work done W = m*P*(v2 - v1) = 0.5*3*10^6*(0.104 - 0.0176) = 129.6 kJ
c)
Heat transfer Q = m*(h2 - h1) = 0.5*(3290 - 1460) = 915 kJ
2.
a)
m = rho*A*V
But rho = P/(RT)
rho = 300*10^3 / (287*470) = 2.224 kg/m^3
m = 2.224*(80*10^-4)*30 = 0.5338 kg/s
b)
h1 + k*V1^2 /2 = h2 + k*V2^2 /2
For air, At P1 = 300 kPa and T1 = 470 K, we get h1 = 473000 J/kg
473000 + 1.05*30^2 /2 = h2 + 1.05*180^2 /2
h2 = 456462.5 J/kg
From air properties, at P2 = 100 kPa and h2 = 456462.5 J/kg, we get T2 = 453.8 K
c)
rho = P/(RT) = 100*10^3 / (287*453.8) = 0.7678 kg/m^3
m = rho*A*v
0.5338 = 0.7678*A*180
A = 0.00386 m^2 = 38.6 cm^2
3.
From steam properties, at P1 = 8 MPa and T1 = 500 deg C, we get h1 = 3400 kJ/kg
During throttling h2 = h1.
From steam properties, at P2 = 6 MPa and h2 = h1 = kJ/kg, we get T2 = 490 deg C.
Hence, T2 = 490 deg C.
4.
1 gal = 0.1337 ft^3
H = 400 ft
Q = 15 gal/min
rho = 8.33 lb/gal = 8.33 / 0.1337 lb/ft^3 = 62.3 lb/ft^3
P = 30 psig
Mass flowrate m = rho*Q = 8.33*15 = 124.95 lb/min
Equivalent head for 30 psig = 30*12^2 / 62.3 = 69.338 ft
Total head = 400 + 69.338 = 469.338 ft
Power = m*Total head
= 124.95*469.338
= 58643.8 lb-ft/min
= 1.78 hp (1 hp = 3.03*10^-5 ft-lb/min)