Question

The time, t, it takes to pour aliquid from a container depends on several factors, including the kinematic viscosity, v, of the liquid. In some laboratory tests, various oils having the same density but different viscosities were poured at a fixed tipping rate from small 150-ml beakers. The time required to pour 100 ml of the oil was measured, and it was found that an approximate equation for the pouring time in seconds was t 1 + 9 x 10' + 8 x 10 v2 with v in m/s. (a) Is this a general homogeneous equation? Explain. (b) Find the kinematic viscosity in m/s for SAE 30 oil at o C and 40 °C from the figure below. (c) Calculate the pouring times in seconds it would take to pour 100 ml of SAE 30 oil from a 150-ml beaker at 0°C and 40°C. 110

Answer 1

Solution: (a) In the given differential equation t-1479x102 5 + 8x10382 the terms t, 1 9x10²s & 8x 103 r² are Seponated by tf = and where t is an independent variable. So, it is a non- homogeneous differential equation. (6) from the graph of kinematic viscosity and 1 елиродоLлло "до сде 30 oil : Kinematic viscosity at 0°C = 2.2 x 103 m²/s Kinematic viscosity at 40°C = 1x10-4 m/s. (c) Pouring time at 0°C to pour 100ml of SAE 30 oil t = 1+ 9x1024(2.2x10 3 + 8x103< (2:2x1033) 7 t = 1+ 1.98 +0.0387 7 t = 3.0187 seconds

Pouring time at 40°C to pour 100 ml of SAE 30 oil! t = 1+ 9X10%x(1x104) + 8x10% (1X104)2 t= 1+ 0.09 + 0.00008 a. t = 1.09008 seconds
Summary: The given differential equation is Non- homogeneous.(b) From the graph ,kinematic viscosity at 0℃ is 2.2x10^-3 nd at 40℃ it is 1x10^-4 m2/s.(c) Pouring time at 0℃ for |SAE30 oil is 3.0187 seconds and at 40℃ it is 1.09008 seconds.