Review of Mathematics

Transcription

1 Review for Exam #1

2 Review of Mathematics 2

3 Weighted Mean A certain property of material 1 is P 1 and that of material 2 is P 2 If x 1 amount (weight or volume) of material 1 is mixed with x 2 amount of material 2, what is the corresponding property of the resulting mixture? Resulting property = Note: Most (not all) properties follow the rule of weighted mean 3

4 Weighted Mean (contd) If the fraction of material 1, 2, and 3 in a product are c 1, c 2, and c 3 respectively and their respective properties (say density) are P 1, P 2, and P 3 respectively, then the resulting property of the mixtures is given by: P = c 1 (P 1 ) + c 2 (P 2 ) + c 3 (P 3 ) Note: c 1 + c 2 + c 3 = 10 4

5 Linear Interpolation Linear interpolation involves assumption of a linear relationship between x and y If y = y 1 when x = x 1 & y = y 2 when x = x 2, then for any value x 3 (that lies between x 1 & x 2 ), the value of y is determined as follows using the concept that the slope of a straight line is the same no matter which two points are used in determining the slope: In the above equation, all values except y 3 are known Hence, y 3 can be calculated 5

6 Linear Interpolation (Graphical Visualization) y 2 (x 2, y 2 ) y y 3 y 1 (x 1, y 1 ) y 3 = slope (x 3 x 1 ) + y 1 x 1 x x 3 x 2 Note: Always check to ensure that the value of y 3 always lies between the values of y 1 and y 2 6

7 Logarithms ln (a b) = ln (a) + ln (b) ln (a/b) = ln (a) - ln (b) Note: ln (a + b) ln (a) + ln (b) Note: ln (x n ) = n ln (x) ln (e x ) = x ln (e) = 1 e ln(x) = x [Log to ln: multiply by 2303; ln to log: divide by 2303] 7

8 Logarithms (Application) Consider the equation: y = x n In order to linearize the above equation and solve for n, we take the natural logarithm on both sides This results in: ln (y) = n ln (x) We can then solve for n as n = [ln(y)] / [ln(x)] 8

9 Units and Dimensions 9

10 Primary and Secondary Dimensions Primary dimensions Mass (M) Length (L) Time(T) Temperature (K) Secondary dimensions Combination of primary dimensions Example: Velocity = Distance/Time Thus, its dimensions are L/T OR LT -1 10

11 Secondary Dimensions Force (Units: N) o Force = (mass) (acceleration) = (mass) (velocity)/(time) Dimensions: M LT -1 / T = MLT -2 Energy or work done (Units: Joules) o Work = (force) (distance) Dimensions: (MLT -2 ) (L) = ML 2 T -2 Power (Units: J/s or Watts) o Power = (work)/(time) Dimensions: (ML 2 T -2 )/(T) = ML 2 T -3 Specific heat (Units: J/kg K) o Energy reqd to inc temp of 1 kg of material by 1 C Dimensions: (ML 2 T -2 )/(M K) = L 2 T -2 K -1 11

12 SI Units Mass: kg (g is NOT SI unit) Length: m (cm is NOT SI unit) Time: s Temperature: K Electric current: A Amount of a substance: mol Luminous intensity: Cd 12

13 Conversion of Temperature 0 C = 32 F = 273 K 100 C = 212 F = 373 K -40 C = -40 F = 233 K C = ( F 32) * 5/9 F = ( C) * 9/ K = C C change in temp = 18 F change in temp = 1 K change in temp 1 C 18 F 1 C 1 K 13

14 Conversion of Specific Heat 1 J/kg C = 1 J/kg K = (1/18) J/kg F 1 J/kg F = 18 J/kg C = 18 J/kg K Note 1: There is NO factor of 32 to account for as we are NOT converting temperature, but ONLY temperature change Note 2: Recall that specific heat was the energy required to change the temperature of unit mass of a substance by one unit 14

15 Mass and Energy Balances 15

16 Terminology (contd) Sensible heat transferred to a given mass of material (Q: J or J/s) Energy transferred from hot to cold object For a batch system: Q = m c p T For a continuous system: Q = m c p T Q is in J Q is in J/s or W Latent heat transferred to a given mass of material (Q: J or J/s) Energy transferred during phase change without any temperature change For a batch system: Q = m For a continuous system: Q = m Latent heat of vaporization ( vap : J/kg) Q is in J Q is in J/s or W Energy reqd to convert 1 kg of a liquid to vapor phase w/o temperature change It equals energy released when vapor condenses to liquid at that temperature For heating foods, the energy of condensing steam is made use of Latent heat of vaporization of water at 100 C = kj/kg Latent heat of fusion ( fus : J/kg) Energy required to convert 1 kg of a solid to liquid phase w/o temperature change Latent heat of fusion of ice at 0 C = 3332 kj/kg 16

17 Terminology (contd) Saturation temperature and pressure Any substance can exist in more than one phase at one time For any pressure, there is a temperature at which a liquid can coexist with the vapor phase This pressure and the corresponding temperature are called the saturation pressure and saturation temperature respectively Example: Water at 1013 kpa (atmospheric pressure) has a saturation temperature of 100 C At higher pressure, saturation temperature is higher Saturated liquid and saturated vapor When only liquid phase exists at the saturation pressure and temperature, the liquid is called saturated liquid or condensate (its enthalpy is denoted by H c ) When only vapor exists at the saturation pressure and temperature, the vapor is called saturated vapor (its enthalpy is denoted by H v ) 17

18 Phase Diagram of Water Pressure Subcooled liq Liq + Vap Liq 2 Vap * * * ** 5 * Saturated steam H v 6 Superheated steam 1: Sub-cooled liquid 2: Saturated liquid 3: Sat steam (low qual) 4: Sat steam (high qual) 5: Saturated vapor 6: Superheated vapor Within the dome, water exists as steam, which is a mixture of liquid and vapor Here, temperature and pressure are constant and are called the saturation temperature and pressure respectively Left of dome: Subcooled liq; Within dome: Saturated steam; Right of dome: Superheated steam As we move from left to right within the dome, more and more of the water is in vapor phase The fraction that is in vapor phase is called steam quality (denoted by x ) It varies from 0 to 1 OR 0% to 100% Latent heat of vap ( vap ) at any temp or pr = H v H c at that temp or pr 18

19 Steam Quality (x) The term Steam generally refers to saturated steam and not superheated steam Steam is a mixture of liquid (condensate) and vapor The enthalpy of steam (H s ) is a weighted mean of enthalpy of condensate (H c ) and enthalpy of vapor (H v ) H s = x H v + (1 x) H c Rearranging, x = (H s H c )/(H v H c ) Note that x = 0 when H s = H c Also, x = 1 when H s = H v Higher the steam quality, higher the value of H s 0 x 1 OR 0% x 100% Note: H c & H v at saturation temperature & pressure are determined from steam tables 19

20 Q versus T 100 T (Temperature in C) 0 S: Solid (ice) L: Liquid (water) V: Vapor (superheated steam) L+V: Saturated steam S S to L Latent L L to V Latent V Q 1 = fusion(ice) = 3332 kj/kg Q 2 = c p(water) T = kj/kg Q 3 = vap(water) = kj/kg Q 1 Q 2 Q 3 Q (Energy Supplied) 20

21 Energy Content Thermal energy content of a solid or liquid: m c p (T Kelvin - 273) OR m c p (T Celsius -0) This is based on the assumption that the energy content of the substance is zero at 0 C (= 273 K) Thermal energy content (or enthalpy) of steam is given by: m s (H s ) Note: H s = x (H v ) + (1 - x) H c 21

22 Direct Heating with Steam (Continuous System) m f, x f, c p(f), T f Cold product Direct contact HX m p, x p, c p(p), T p Warm diluted product m s, quality of x Steam Pressure, P OR Temperature, T m, x, c p, T: Mass flow rate, solids fraction, specific heat, temperature resp Subscripts: f for feed; p for product Overall mass balance: m f + m s = m p Solids balance: m f x f = m p x p Energy balance: m f {c p(f) } T f + m s (H s ) = m p {c p(p) } T p H s = (x) H v + (1 x) H c with H v and H c being determined from steam tables at the steam pressure of P or steam temperature of T 22

23 Indirect Heating with Steam (Continuous System) m f, x f, c p(f), T f Cold product Indirect contact HX m p, x p, c p(p), T p Warm undiluted product m s, quality of x Steam Pressure, P OR Temperature, T 100% Condensate m, x, c p, T: Mass flow rate, solids fraction, specific heat, temperature resp Subscripts: f for feed; p for product Overall mass balance: m f = m p Solids balance: m f x f = m p x p Thus, x p = x f Energy balance: m f {c p(f) } T f + m s (H s ) = m p {c p(p) } T p + m s (H c ) H s = (x) H v + (1 x) H c with H v and H c being determined from steam tables at the absolute (not gauge) steam pressure of P or steam temperature of T 23

24 Indirect Cooling with Water (Continuous System) m f, x f, c p(f), T f Hot feed Indirect contact Cooler T 1 Cold water m cw m p, x p, c p(p), T p m, x, c p, T: Mass flow rate, solids fraction, specific heat, temperature resp Subscripts: f for feed; p for product; 1 for inlet water; 2 for outlet water Overall mass balance: m f = m p Solids balance: m f x f = m p x p Thus, x p = x f Energy balance: m f {c p(f) }T f + m cw {c p(1) }T 1 = m cw {c p(2) }T 2 + m p {c p(p) }T p T 2 Cold product c p(1) and c p(2) are determined from tables containing the properties of water at temperatures T 1 and T 2 respectively When approximations are used, c p(1) = c p(2) = 4180 J/kg K 24