Heat Chap01-121

# Heat Chap01-121 - Chapter 1 Basics of Heat Transfer Review...

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Chapter 1 Basics of Heat Transfer Review Problems 1-121 Cold water is to be heated in a 1200-W teapot. The time needed to heat the water is to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the teapot and the water are constant. 3 Heat loss from the teapot is negligible. Properties The average specific heats are given to be 0.6 kJ/kg. ° C for the teapot and 4.18 kJ/kg. ° C for water. Analysis We take the teapot and the water in it as our system that is a closed system (fixed mass). The energy balance in this case can be expressed as E E E E U U U in out system in system water tea pot - = = = + Then the amount of energy needed to raise the temperature of water and the teapot from 18 ° C to 96 ° C is E mC T mC T in water teapot kg)(4.18 kJ / kg. C)(96 18) C + kg)(0.6 kJ / kg. C)(96 18) C = 853 kJ = + = ° - ° ° - ° ( ) ( ) ( . ( . 2 5 08 The 1500 W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for this heater to supply 853 kJ of heat is determined from min 11.8 = s 710 kJ/s 1.2 kJ 853 transfer energy of Rate ed transferr energy Total transfer in = = = = E E t Discussion In reality, it will take longer to accomplish this heating process since some heat loss is inevitable during the heating process. 1-61 Water 18 ° C Heater 1200 W

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Chapter 1 Basics of Heat Transfer 1-122 The duct of an air heating system of a house passes through an unheated space in the attic. The rate of heat loss from the air in the duct to the attic and its cost under steady conditions are to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141 ° C and 3.77 MPa. 2 Steady operating conditions exist since there is no change with time at any point and thus m E CV CV and = = 0 0 . 3 The kinetic and potential energy changes are negligible, ke pe 2245 2245 0 . 4 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is C p = 1.007 kJ/kg· ° C (Table A-15). Analysis We take the heating duct as the system. This is a control volume since mass crosses the system boundary during the process. There is only one inlet and one exit and thus m m m 1 2 = = . Then the energy balance for this steady-flow system can be expressed in the rate form as ( ) E E E E E mh Q mh Q mC T T in out in out out p - = = = = + 2245 2245 = - Rate of net energy transfer by heat, work, and mass system (steady) Rate of change in internal, kinetic, potential, etc. energies out (since ke pe 0) n n n n n n n n nn n n n n 0 1 2 1 2 0 The density of air at the inlet conditions is determined from the ideal gas relation to be
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## This note was uploaded on 02/21/2011 for the course CHEMICAL E 0905343 taught by Professor Yehiakhraishe during the Spring '11 term at University of Jordan.

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Heat Chap01-121 - Chapter 1 Basics of Heat Transfer Review...

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