This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 370 Thermodynamics Fall 2010 Course Number: 14319 Instructor: Larry Caretto Unit Seven Homework Solutions, October 26, 2010 1 A 600 MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40%. Determine the rate of heat transfer to the river water. Will the actual heat transfer rate be higher or lower than this value? Why? From the basic definition of cycle efficiency, H H Q W Q W = = , we can compute MW MW W Q H 1500 4 . 600 = = = . From the cycle relationship between |Q H |, |Q L |, and |W|, |Q H | = |Q L | + |W|, we can write, for heat rates and power, W Q Q L H + = . Thus we find that the heat rejected to the river is W Q Q H L- = =1500 MW 600 MW = 900 MW . In the actual power plant there will be other sources of heat loss. These include heat transfer to the surrounding air from the power plant and leaks of the working fluid. In addition, power plant efficiency is usually determined as the power output divided by the equivalent heat from the fuel. There is a significant contribution to the inefficiency from the exhaust gases. 2 An automobile engine consumes fuel at a rate of 28 L/h and delivers 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm 3 , determine the efficiency of the engine. We start with the basic definition of cycle efficiency, H H Q W Q W = = , and we have to compute the heat input rate as the product of the fuel rate, and the heating content of the fuel. (We need the density to convert from a volume flow rate of fuel to a mass flow rate of fuel. Thus, we compute the heat input as follows. h kJ L cm g kg kg kJ h L cm g HV V HV m Q fuel fuel fuel fuel fuel H 600 , 985 000 , 1 000 , 1 1 000 , 44 28 8 . ) ( ) ( 3 3 = = = = We divide this into the power (work rate) of 60 kW to compute the efficiency. h s s kW kJ h kJ kW Q W H 600 , 3 1 600 , 985 60 = = = 21.9% Jacaranda (Engineering) 3519 Mail Code Phone: 818.677.6448 E-mail: email@example.com 8348 Fax: 818.677.7062 3 A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min. Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air. From the basic definition of coefficient of performance, W Q W Q cop L L = = , we can compute s kJ s kW kJ cop Q W L 60 min 1 1 2 . 1 min 60 = = = 0.83 kW .....
View Full Document
- Spring '08