Fundamentals Of Engineering Thermodynamics
9th Edition
ISBN: 9781119391388
Author: MORAN, Michael J., SHAPIRO, Howard N., Boettner, Daisie D., Bailey, Margaret B.
Publisher: Wiley,
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Chapter 3, Problem 3.39CU
To determine
For a polytropic process,
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A 300-lb iron casting, initially at 1500°F, is quenched in a tank filled with 2121 lb of oil, initially at 80°F. The iron casting and oil can be
modeled as incompressible with specific heats 0.10 Btu/lb · °R, and 0.45 Btu/lb · °R, respectively.
(a) For the iron casting and oil as the system,determine the final equilibrium temperature, in °F.
Ignore heat transfer between the system and its surroundings.
Tf = i
°F
(b) For the iron casting and oil as the system,determine the amount of entropy produced within the tank, in Btu/°R.
Ignore heat transfer between the system and its surroundings.
O =
i
Btu/°R
3. Air is contained within a piston-cylinder assembly The cross sectional area of the piston is 0.01 m².
Initially the piston is at 1 bar and 25°C, 10 cm above the base of the cylinder. In this state, the spring exerts
no force on the piston. The system is then reversibly heated to 100°C. As the spring is compressed, it
exerts a force on the piston according to: F=-kx where k= 50,000 N/m and x is the displacement length from
its uncompressed position. Determine the work done.
a. -166 J
b. -216 J
c. 166 J
d. 216 J
A 300-lb iron casting, initially at 1500°F, is quenched in a tank filled with 2121 Ib of oil, initially at 80°F. The iron casting and oil can be
modeled as incompressible with specific heats 0.10 Btu/lb - °R, and 0.45 Btu/lb - °R, respectively.
(a) For the iron casting and oil as the system,determine the final equilibrium temperature, in °F.
Ignore heat transfer between the system and its surroundings.
T= i
°F
(b) For the iron casting and oil as the system,determine the amount of entropy produced within the tank, in Btu/°R.
Ignore heat transfer between the system and its surroundings.
Btu/°R
Chapter 3 Solutions
Fundamentals Of Engineering Thermodynamics
Ch. 3 - Prob. 3.1ECh. 3 - Prob. 3.2ECh. 3 - Prob. 3.3ECh. 3 - Prob. 3.4ECh. 3 - Prob. 3.6ECh. 3 - Prob. 3.7ECh. 3 - Prob. 3.8ECh. 3 - Prob. 3.9ECh. 3 - Prob. 3.10ECh. 3 - Prob. 3.11E
Ch. 3 - Prob. 3.12ECh. 3 - Prob. 3.13ECh. 3 - Prob. 3.1CUCh. 3 - Prob. 3.2CUCh. 3 - Prob. 3.3CUCh. 3 - Prob. 3.4CUCh. 3 - Prob. 3.5CUCh. 3 - Prob. 3.6CUCh. 3 - Prob. 3.7CUCh. 3 - Prob. 3.8CUCh. 3 - Prob. 3.9CUCh. 3 - Prob. 3.10CUCh. 3 - Prob. 3.11CUCh. 3 - Prob. 3.12CUCh. 3 - Prob. 3.13CUCh. 3 - Prob. 3.14CUCh. 3 - Prob. 3.15CUCh. 3 - Prob. 3.16CUCh. 3 - Prob. 3.17CUCh. 3 - Prob. 3.18CUCh. 3 - Prob. 3.19CUCh. 3 - Prob. 3.20CUCh. 3 - Prob. 3.21CUCh. 3 - Prob. 3.22CUCh. 3 - Prob. 3.23CUCh. 3 - Prob. 3.24CUCh. 3 - Prob. 3.25CUCh. 3 - Prob. 3.26CUCh. 3 - Prob. 3.27CUCh. 3 - Prob. 3.28CUCh. 3 - Prob. 3.29CUCh. 3 - Prob. 3.30CUCh. 3 - Prob. 3.31CUCh. 3 - Prob. 3.32CUCh. 3 - Prob. 3.33CUCh. 3 - Prob. 3.34CUCh. 3 - Prob. 3.35CUCh. 3 - Prob. 3.36CUCh. 3 - Prob. 3.37CUCh. 3 - Prob. 3.38CUCh. 3 - Prob. 3.39CUCh. 3 - Prob. 3.40CUCh. 3 - Prob. 3.41CUCh. 3 - Prob. 3.42CUCh. 3 - Prob. 3.43CUCh. 3 - Prob. 3.44CUCh. 3 - Prob. 3.45CUCh. 3 - Prob. 3.46CUCh. 3 - Prob. 3.47CUCh. 3 - Prob. 3.48CUCh. 3 - Prob. 3.49CUCh. 3 - Prob. 3.50CUCh. 3 - Prob. 3.51CUCh. 3 - Prob. 3.52CUCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Prob. 3.3PCh. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Prob. 3.9PCh. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Prob. 3.12PCh. 3 - Prob. 3.13PCh. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - Prob. 3.23PCh. 3 - Prob. 3.24PCh. 3 - Prob. 3.25PCh. 3 - Prob. 3.26PCh. 3 - Prob. 3.27PCh. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - Prob. 3.30PCh. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - Prob. 3.33PCh. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - Prob. 3.40PCh. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Prob. 3.43PCh. 3 - Prob. 3.44PCh. 3 - Prob. 3.45PCh. 3 - Prob. 3.46PCh. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Prob. 3.75PCh. 3 - Prob. 3.76PCh. 3 - Prob. 3.77PCh. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - Prob. 3.95PCh. 3 - Prob. 3.96PCh. 3 - Prob. 3.97PCh. 3 - Prob. 3.98PCh. 3 - Prob. 3.99P
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- A 300-lb iron casting, initially at 600°F, is quenched in a tank filled with 2121 lb of oil, initially at 80°F. The iron casting and oil can be modeled as incompressible with specific heats 0.10 Btu/lb. °R, and 0.45 Btu/lb. °R, respectively. (a) For the iron casting and oil as the system,determine the final equilibrium temperature, in °F. Ignore heat transfer between the system and its surroundings. T₁ = i (b) For the iron casting and oil as the system,determine the amount of entropy produced within the tank, in Btu/°R. Ignore heat transfer between the system and its surroundings. J = °F Mi Btu/ºRarrow_forwardExample 3.5 If the processes of Ex. 3.4 are carried out irreversibly but so as to accomplish exactly the same changes of state-the same changes in P, T, Uig, and Hig then different values of Q and W result. Calculate Q and W if each step is carried out with a work efficiency of 80% Example 3.4 A gas in its ideal-gas state undergoes the following sequence of mechanically reversible processes in a closed system: (a) From an initial state of 70°C and 1 bar, it is compressed adiabatically to 150°C. (b) It is then cooled from 150 to 70°C at constant pressure. (c) Finally, it expands isothermally to its original state. Calculate W, Q, AUig, and AHig for each of the three processes and for the entire cycle. Take CV ig = 12.471 and CP ig = 20.785 J-mol-1-K-1.arrow_forwardArgon (molar mass 40 kg/kmol) compresses reversibly in an adiabatic system from 5 bar, 25 0C to a volume of 0.2 m3. If the initial volume occupied was 0.9 m3, calculate the work input in MJ to 3 decimal places. Assume nitrogen to be a perfect gas and take cv = 0.3122 k J / k g K.arrow_forward
- Give the formula for ∆U, ∆H, Q , W, ∆S for the following thermodynamic processes:1. adiabatic2. polytropicarrow_forwardsolve the following thermodynamic questionarrow_forwardCalculate the Q, deltaU and W for the following processes. If pressure is not constant in the process, represent work in its integral form. Isothermal compression of a gas from 150cm^3 to 50 cm^3, where P = 1.5 atm Isochoric cooling of 1 mol CO2 from 400 degrees celsius to 40 degrees celsius where V = 2.5 L C. adiabatic compression of a gas from 850 mLto and 275 mL at P = 200 kPaarrow_forward
- Water vapor undergoes the following processes: a. From a state at 200 kPa and 600 ° C it expands isothermally up to 100 kPa. B. Immediately isobarically compressed to 2 m3 / kg. C. From this last state it follows an isochoric process until reaching 600 ° C. Perform: a. Plotting the processes on the exact P-ν diagram. You must use the diagram P-νarrow_forwardA fluid at 0.5 bar occupying 0.08 m3 compressed reversibly to a pressure of 9.8 bar and specific volume of 0.6 m³/kg according to the law pvn = c. The fluid then expands reversibly according to the law pv2 = c to 1.3 bar. A reversible cooling at constant volume then restores the fluid back to initial state. Calculate the net work for the process in Joules to round figure. No mega or Kilo for units.arrow_forward1. As shown in the figure below, Refrigerant 134a enters a condenser operating at steady state at 70 lbf/in2, 160 °F and is condensed to saturated liquid at 60 lbf/in on the outside of tubes through which cooling water flows. In passing through the tubes, the cooling water increases in temperature by 20 'F and experiences no significant pressure drop. Cooling water can be modeled as incompressible with v-0.0161 ft'/lb and c = 1 Btu/lb R. The mass flow rate of the refrigerant is 3100 lb/h. Neglecting kinetic and potential energy effects and ignoring heat transfer from the outside of the condenser, determine: (a) The volumetric flow rate of the entering cooling water, in gal/min (b) The rate of heat transfer, in Btu/h, to the cooling water from the condensing refrigerant (5 points) Refrigerant 134a P= 70 in. T= 160 F 3100 heh 7,-7,-20F-20R Reirigerant 134a [P-60 lbin V Saturated liquidarrow_forward
- Five kilograms of a two-phase liquid-vapor mixture of water initially at 300\deg C and x1 = 0.5 is heated from the initial state to a saturated vapor state while the volume remains constant. The process is brought about by heat transfer from a thermal reservoir at 380\deg C. The temperature of the water at the location where the heat transfer occurs is 380\deg C.Let To = 300 K, Po = 1 bar, and ignore the effects of motion and gravity.arrow_forwardSimple question on Thermodynamics: A 30-lb iron casting, initially at 600°F, is quenched in a tank filled with 2121 lb of oil, initially at 80°F. The iron casting and oil can be modeled as incompressible with specific heats 0.10 Btu/lb · °R, and 0.45 Btu/lb · °R, respectively. (a) For the iron casting and oil as the system,determine the final equilibrium temperature, in °F.(Ignore heat transfer between the system and its surroundings.) (b) For the iron casting and oil as the system,determine the amount of entropy produced within the tank, in Btu/°R.(Ignore heat transfer between the system and its surroundings.)arrow_forwardA fluid at 0.7 bar occupying 0.1 m3 is compressed reversibly to a pressure of 10 bar and specific volume of 0.6 m3/kg according to the law pvn = c. The fluid then expands reversibly according to the law pv2 = c to 2.2 bar. A reversible cooling at constant volume then restores the fluid back to initial state. Calculate the value of n in the processarrow_forward
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