A shell and tube heat exchanger is designed as a counter to...

A shell and tube heat exchanger is designed to cool a mass flow rate of 60 kg/h of engine oil (cp = 1060 J/kg°C) from 300°C to 150°C. The cooling water, cp = 4182 J/kg°C, enters the tube side (Length is 1.75 m and Outer diameter is 2.54 cm) at a temperature of 20°C. If the overall heat transfer coefficient, U, is 10.83 W/m²°C and the Capacity Ratio of the Heat Exchanger is 2.4, calculate the following: a) The heat transfer rate, b) The mass flow rate of the cooling water in the units of (kg/h), c) The surface area of the heat exchanger.

Transcript

00:01 So for this question, first let's note all the given data that we have received from the question.

00:09 M0 is equal to 60 kgs per hour, which is the mass flow rate.

00:21 Cp .0 is equal to 1060 joules per kg degree celsius.

00:33 Initial temperature is 300 degrees celsius and it is reduced to 150 degree celsius.

00:55 Temperature of initially of the tube is 20 degrees celsius and cpw for cooling water.

01:18 It is 4182 joules per kg degree celsius.

01:25 Now considering the tube, the dimensions, its length, considering the length as 1 .75 meters, the diameter as 2 .54 centimeters.

02:05 U, which is the heat transfer coefficient, it is given as 10 .83 weber per meter square, celsius and the capacity ratio of heat exchanger is 2 .4 so capacity ratio is equal to max minimum is equal to 2.

03:07 Now for party we need to calculate the heat transfer rate this can be given by q is equal to mac, cbo initial temperature.

03:48 Putting the values into this equation we get 60 .3 6 .6 .000 multiplied by 1 -060, multiplied by 1 -3 -0 -60, multiplied by 1 -3 -0 -0 minus 150.

04:16 Multiplied by 300 minus 150 so that is equal to 2650 watt so the heat transfer rate is equal to 2680 now for part two of this question we need to calculate the mass flow rate of cooling water need to calculate the mass flow rate of cooling water so heat capacity which is equal to mcb for h from equilibrium we can say that q of hot is equal to q for cold heat transfer rate for hot would be equal to the heat transfer rate for cool would be equal to the heat transfer rate for cool opening these equations we can write heat transfer formulas here for cool it would be mw cpw this is the equation one for counter flow heat exchanger the maximum temperature attained by the cold water can be maximum 150 degree ratio of heat capacity given but which one is higher that is not given so we need to decide for information so from equation 1 we can put the values as m c for water therefore sea oil multiplied by 150 is equal to sea water multiplied by tc not minus 20 from the above equation tc maximum can be 150 degrees so this would be equal to 150 minus 20 which is equal to 150 130 degree celsius so sea water is more as compared to the sea oil so we can conclude from above that the specific heat capacity of water is more than the specific heat capacity of oil so so c max on c minimum is equal to 2 .4 which is given the question which means mass of water and specific capacity of water divided by mass of oil specific capacity of oil is equal to 2 .4 mass of water multiplied by the mass rate of water multiplied by 4182 divided by the mass rate of oil which is 16 into 1060 is equal to 2 .4.

09:56 Now to calculate the mass flow rate of water we get as equal to 36 .5 kg per hour...

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