A heat exchanger uses medium pressure steam at 10 bar and 350 C to heat cyclohexane vapor from 100 C up to 275 C as shown in the figure below. The steam is available at a flowrate of 32 kg/h and it condenses completely in the heat exchanger and exits as saturated liquid condensate at 10 bar. a) How much heat (in kW) can be provided by the steam? b) How much cyclohexane can be processed in kg/h? Vapor Cyclohexane T = 100 C P = 1 atm Steam T = 350 C = 10 bar Condensate P Vapor Cyclohexane T = 275 C P = 1 atm P = 10 barUse given tabulated data for answer . **Table B.2: Heat Capacities**This table presents the heat capacities for various compounds. It provides the information in three different units: \(C_P\) [kJ/(mol K)], [Btu/(mol °R)], and [cal/(mol K)]. The formula to calculate heat capacity at a specific temperature is provided.**Formula:** \(C_P = E + FT + GT^2 + HT^3 + J/T^2\)**Example:** \(C_P\) [kJ/(mol K)] = 34.760 + 4.360 \(\times\) 10\(^-2\)T - 2.005 \(\times\) 10\(^5\)T\(^-2\) where T is in °C.Additional note indicates that these equations are sufficiently accurate for pressures low enough for the ideal-gas equation of state to apply.| Range (Temp in °C) | Compound | Formula | Mol. Wt. | Temp. Unit | State Form | \(E \times 10^5\) | \(F\) | \(G \times 10^3\) | \(H \times 10^6\) | \(J \times 10^5\) ||--------------------|----------|-------------|----------|------------|------------|------------------|-----|-----------------|-----------------|------------------|| 600-1500 | Acetone | C\(_3\)H\(_6\)O | 58.08 | °C | g | 12.683 | 0 | 31.525 | -3.495 | 0 || 0-1200 | Air | | 28.97 | °C | g | 28.090 | -0.196 | 0.48060 | | || 0-1200 | Argon | Ar | 39.948 | °C | g | 20.786 | 4.850 | 0 | | || | | | | | 0.016 | 0 | | || 10-2000 | Column 1 | 1 | | |

Given, the mass flow rate of steam, m3 = 32 kg/h The pressure of steam, P3 = 10 bar The temperature of steam, T3 = 350 °C The initial temperature of cyclohexane, T1 = 100 °C = 373 K The final temperature of cyclohexane, T2 = 275 °C = 548 KDraw the diagram for heat exchanger as follows:(a) We need to calculate the amount of heat supplied by the steam. From the steam tables, The initial enthalpy of the steam, H3 = 3157.67 kJ/kg The final enthalpy of the steam, H4 = 780.753 kJ/kg Calculate the heat transfer as follows: Qsteam = m3H3-H4= 32 kghr×3157.67 kJkg-780.753 kJkg = 76061.344×kJhr×1 hr3600 s×kW1 kJ/s= 21.13 kW Therefore, the required solution is 21.13 kW. (b) We need to calculate the mass flow rate of cyclohexane. Calculate the enthalpy change of cyclohexane as follows: ∆Hcyclohexane = ∫T1T2CpcyclohexanedT= ∫100 °C275 °C94.14×10-3+ 49.62×10-5T -31.9×10-8T2 +80.63×10-12T3dT=94.14×10-3T+ 49.62×10-5T22 -31.9×10-8T33 +80.63×10-12T44100 °C275 °C= 94.14×10-3275 °C-100 °C+ 49.62×10-52752-10022 -31.9×10-82753-10033 +80.63×10-12×2754-10044=16.4745+16.2815-2.1051+0.1133 kJ/molK= 30.7642 kJ/molAssume that there is no heat loss to the surroundings. So the heat lost by the steam is gained by the cyclohexane. Write the expression for the heat balance as follows: Qsteam = QcyclohexaneQsteam = mcyclohexane∆Hcyclohexane Substitute the corresponding values in the above equation and calculate the mass flow rate of cyclohexane. Qsteam = mcyclohexane∆Hcyclohexane21.13 kW = mcyclohexane×30.7642kJmol×mol84.16 gmcyclohexane = 21.13 kW 0.36554 kJ×1 kJ/s1 kW g= 57.8 gs×1 kg1000 g×3600 s1 hr= 208.08 kg/hr Therefore, the required solution is 208.08 kg/hr.