Chapter2_A - Chapter 2 - Section A - Mathcad Solutions 2.1...

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t 1 20 degC := C P 4.18 kJ kg degC := M H2O 30 kg := t 2 t 1 U total M H2O C P + := t 2 20.014degC = Ans. (d) For the restoration process, the change in internal energy is equal but of opposite sign to that of the initial process. Thus Q U total := Q 1.715 kJ = Ans. (e) In all cases the total internal energy change of the universe is zero. 2.2 Similar to Pb. 2.1 with mass of water = 30 kg. Answers are: (a) W = 1.715 kJ (b) Internal energy change of the water = 1.429 kJ (c) Final temp. = 20.014 deg C (d) Q = -1.715 kJ Chapter 2 - Section A - Mathcad Solutions 2.1 (a) M wt 35 kg := g 9.8 m s 2 := z5 m := Work M wt g ⋅∆ z := Work 1.715kJ = Ans. (b) U total Work := U total 1.715kJ = Ans. (c) By Eqs. (2.14) and (2.21): dU d PV () + C P dT = Since P is constant, this can be written: M H2O C P dT M H2O dU M H2O P dV + = Take Cp and V constant and integrate: M H2O C P t 2 t 1 U total = 9
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Q 34 800 J := W 34 300J := Ut 34 Q 34 W 34 + := Ut 34 500 J = Ans. Step 1 to 2 to 3 to 4 to 1: Since U t is a state function, U t for a series of steps that leads back to the initial state must be zero. Therefore, the sum of the U t values for all of the steps must sum to zero. Ut 41 4700J := Ut 23 Ut 12 −∆ Ut 34 Ut 41 := Ut 23 4000 J = Ans. Step 2 to 3: Ut 23 4 10 3 × J = Q 23 3800 J := W 23 Ut 23 Q 23 := W 23 200 J = Ans. For a series of steps, the total work done is the sum of the work done for each step. W 12341 1400 J := 2.4 The electric power supplied to the motor must equal the work done by the motor plus the heat generated by the motor. i 9.7amp := E 110V := Wdot mech 1.25hp := Wdot elect iE := Wdot elect 1.067 10 3 × W = Qdot Wdot elect Wdot mech := Qdot 134.875W = Ans. 2.5 Eq. (2.3): U t QW + = Step 1 to 2: Ut 12 200 J := W 12 6000 J := Q 12 Ut 12 W 12 := Q 12 5.8 10 3 × J = Ans. Step 3 to 4: 10
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U1 2 kJ := Q U := Q1 2 kJ = Ans. 2.13 Subscripts: c, casting; w, water; t, tank. Then m c U c m w U w + m t U t + 0 = Let C represent specific heat, CC P = C V = Then by Eq. (2.18) m c C c ⋅∆ t c m w C w t w + m t C t t t + 0 = m c 2kg := m w 40 kg := m t 5kg := C c 0.50 kJ kg degC := C t 0.5 kJ kg degC := C w 4.18 kJ kg degC := t c 500 degC := t 1 25 degC := t 2 30 degC := (guess) Given m c C c t 2 t c () m w C w m t C t + t 2 t 1 = t 2 Find t 2 := t 2 27.78degC = Ans.
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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter2_A - Chapter 2 - Section A - Mathcad Solutions 2.1...

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