Problems-and-solutions-heat-transfer

1.) A surface condenser serving 50,000kW steam turbo-generator unit receives exhaust steam at the rate of 196,000 kg/hr. Vacuum in condenser is 702 mmHg. Sea water cooling enters at 29.5°C and leaves at 37.5°C. For steam turbine condenser, manufacturers consider 950 Btu/lb of steam turbine condensed as heat given up to cooling water. Calculate the logarithmic temperature difference. Solution: Let LMTD = Log mean temperature difference LMTD =

(∆𝑡)𝑚𝑎𝑥 − (∆𝑡)𝑚𝑖𝑛 𝑙𝑛

(∆t)max (∆t)min

Where: P(condenser) = 101.325 kPa – [702 𝑝𝑠𝑖 (

101.325 𝑘𝑃𝑎 760 𝑝𝑠𝑖 )]

= 7.733 kPa (tsat = 40.86°C) ∆tmax = 40.86 – 29.5 = 11.36°C ∆tmin = 40.86 – 37.5 = 3.36°C Then;

LMTD =

11.36°C−3.36°C 𝑙𝑛

11.36°C 3.36°C

Thus;

LMTD = 6.57°C

2.) The stack gas from a chemical operation contains noxious vapors that must be condensed by lowering its temperature from 315°C to 35°C. The gas flow rate is 0.70m3/s. Water is available to 10°C at 1.26kg/s. A two shell and 4 tube pass, counter flow heat exchanger

will be used with a water flowing through the tubes. The gas has a specific heat of 1.10 kJ/kg-K and a gas constant of 0.26kJ/kg-K. Calculate the logarithmic mean temperature difference. Solution: Average gas temperature =

315°C+35°C

2

= 175°C Density of gas (r): r = r =

𝑃 𝑅𝑇 101.325 𝑘𝑃𝑎 0.26𝑘𝐽 ( 𝑘𝑔−𝐾 )(175+273𝐾)

r = 0.867kg/m3 Mass flow rate of gas: mg = (0.7m3/s)(0.867kg/m3) = 0.607 kg/s Heat gained by cooling water = Heat Lost by the gases

MwCpg∆tw = MwCpg∆tg (1.26kg/s) (4.187) (t-10°C) = (0.607kg/s) (1.10) (315°C -35°C) t = 45.5°C Solving for (∆t)max and (∆t)min:

(∆t)max = 315°C – 45.5°C = 269.5°C (∆t)min = 35°C - 10°C = 25°C Then; LMTD =

(∆𝑡)𝑚𝑎𝑥 − (∆𝑡)𝑚𝑖𝑛 𝑙𝑛

(∆t)max (∆t)min

=

269.5°C−25°C 𝑙𝑛

269.5°C 25°C

Thus;

LMTD = 102.8°C

3.) Exhaust steam at 7kPa at the rate of 75kg/s enters a single pass condenser containing 5,780 pcs copper tubes with a total surface area of 2950m2. The steam has a moisture content of 10% and the condensate leaves saturated liquid at steam temperature. The

cooling water flow rate is 4413 liters per second entering at 20°C. Size of tubes, 25mm O.D. by 3mm thick wall. Find the overall heat transfer coefficient. Solution: Enthalpy of steam entering the condenser: h1 = (hf + xhfg)7 kPa = 163.4kJ/kg + 0.90kJ/kg(2572.5kJ/kg) = 2478.65kJ/kg Enthalpy and temperature of condensate: h2 = hf @7kPa = 163.4kJ/kg, Tsat @ 7kPa = 29°C Enthalpy and temperature of the condensate: Qwater = Qsteam MwCpg∆tw = Ms(h1-h2) [4,413 L/s (1kg/L)] (4.187)(t-20) = 75kg/s(2478.65kJ/kg – 163.4kJ/kg) t = 29.40°C

(∆t)max = 39°C – 20°C = 19°C (∆t)min = 39°C – 29.4°C = 9.6°C LMTD =

(∆𝑡)𝑚𝑎𝑥 − (∆𝑡)𝑚𝑖𝑛 𝑙𝑛

(∆t)max (∆t)min

=

19°C−9.6°C 𝑙𝑛

19°C 9.6°C

= 13.77°C

From: Q = A U (LMTD) Where: Q = ms (h1-h2) = 75 (2478.65 – 163.4) = 173,643.75 kW Then: 173,643.75 = 2950 U (13.77) 𝒌𝑾

U = 4.275𝒎𝟐−𝑲 or

𝑾

U = 4275𝒎𝟐−𝑲

4.) A heat exchanger has an overall coefficient of heat transfer of 900W/m 2-K. The mean temperature difference is 20°C and heat loss is 15,000W. Calculate the heat transfer area.

Solution: Q = AUθ 15,000 = A (900W/m2-K) (20°C)

A = 0.833m2

5.) Waters enters the condenser at 20°C and leaves at 35°C. What is the log mean temperature difference if the condenser temperature is 45°C. Solution: LMTD =

(∆𝑡)𝑚𝑎𝑥 − (∆𝑡)𝑚𝑖𝑛 𝑙𝑛

(∆t)max (∆t)min

(∆t)max = 45°C – 20°C = 25°C (∆t)min = 45°C – 35°C = 10°C LMTD =

25°C−10°C 𝑙𝑛

25°C 10°C

LMTD = 16.37°C