ENGINEERING MECHANIC - EXAMPLE 15.3 : A biochemical trolley of mass 15 kg is towing a trailer of mass 5 kg along a straight horizontal pathway. The trailer and the trolley are connected by a light inextensible tow-bar. The engine of the trolley exerts a driving force of magnitude 100 N. The trailer and the trolley experience resistances of magnitude 10 N and 30 N respectively. (a) Form 2 equations with unknowns T and a, that represents the equilibrium for the 2 systems of the trolley and trailer. (b) Solve the simultaneous equations from the 2 equations that are obtained in part (a) of this question. T is the tension of the tow-bar and a is the acceleration.
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ENGINEERING MECHANIC - ANSWER 15.3 : (a) Trolley : 100 - 30 - T = 70 - T = 15a as Equation (1). Trailer : T - 10 = 5a as Equation (2). (b) Trailer : 3 times Equation (2) : 3T - 30 = 15a as Equation (3). Equation (1) - Equation (3) = 100 - 4T = 0 then T = 100 / 4 = 25 N. From Equation (2), 5a = T - 10 then a = (T - 10) / 5 = (25 - 10) / 5 = 3 N / kg. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).
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