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QUESTION: 1

Assumptions made for calculation of logarithmic mean temperature difference are

(i) Constant overall heat transfer coefficient

(ii) The kinetic and potential energy changes are negligible

(iii) There is no conduction of heat along the tubes of heat exchanger

Identify the correct statements

Solution:

These assumptions are made for simplicity. During heat exchange between two fluids, the temperature of the fluids change in the direction of flow and consequently there occurs a change in the thermal head causing the flow of heat.

QUESTION: 2

A cold fluid (specific heat 2.95 k J/kg K) at 10 kg/min is to be heated from 25 degree Celsius to 55 degree Celsius in a heat exchanger. The task is accomplished by extracting heat from hot water (specific heat 4.186 k J /kg K) available at mass flow rate 5 kg/min and inlet temperature 85 degree Celsius. Identify the type of arrangement of the heat exchanger

Solution:

m _{h }c _{h }(t _{h 1} – t _{h 2}) = m _{c }c _{c }(t _{c 2} – t _{c 1}).

QUESTION: 3

In a food processing plant, a brine solution is heated from – 12 degree Celsius to – 65 degree Celsius in a double pipe parallel flow heat exchanger by water entering at 35 degree Celsius and leaving at 20.5 degree Celsius. Let the rate of flow is 9 kg/min. Estimate the area of heat exchanger for an overall heat transfer coefficient of 860 W/m^{2} K. For water c _{P} = 4.186

Solution:

Q = m c _{P }d t = 9104.5 J/s. A = Q/ U α _{m}.

QUESTION: 4

Exhaust gases (c _{P }= 1.12 k J/kg K) flowing through a tubular heat exchanger at the rate of 1200 kg/hr are cooled from 400 degree Celsius to 120 degree Celsius. This cooling is affected by water (c_{P }= 4.18 k J/kg K) that enters the system at 10 degree Celsius at the rate of 1500 kg/hr. If the overall heat transfer coefficient is 500 k J/m^{2} hr degree, what heat exchanger area is required to handle the load for parallel flow arrangement?

Solution:

m _{h }c _{h }(t _{h 1} – t _{h 2}) = m _{c }c _{c }(t _{c 2} – t _{c 1}).

QUESTION: 5

A steam condenser is transferring 250 k W of thermal energy at a condensing temperature of 65 degree Celsius. The cooling water enters the condenser at 20 degree Celsius with a flow rate of 7500 kg/hr. Calculate the log mean temperature difference

Solution:

Q = m _{c }c _{c }(t _{c 2} – t _{c 1}) and log mean temperature difference = α _{1 }– α _{2 }/ log (α _{1}/α _{2}).

QUESTION: 6

Consider the above problem, find what error would be introduced if the arithmetic mean temperature difference is used rather than the log-mean temperature difference? Take overall heat transfer coefficient for the condenser surface as 1250 W/m2 K

Solution:

α = α _{1 }+ α _{2}/2. Error = 7.08 – 6.52/7.08 = 7.91%.

QUESTION: 7

For what value of end temperature difference ratio, is the arithmetic mean temperature difference 5% higher than the log-mean temperature difference?

Solution:

α _{1}/ α _{2} = 2.2.

QUESTION: 8

A company is heating a gas by passing it through a pipe with steam condensing on the outside. What percentage change in length would be needed if it is proposed to triple the heating capacity?

Solution:

Explanation: Present capacity, Q _{1} = U _{1 }A _{1 }α_{ 1} and new capacity, Q _{2} = U _{2 }A_{2 }α_{ 2}. According to the given condition, U _{2 }A _{2 }α_{ 2} = 3 U _{1 }A _{1 }α _{1}.

QUESTION: 9

A steam condenser is transferring 250 k W of thermal energy at a condensing temperature of 65 degree Celsius. The cooling water enters the condenser at 20 degree Celsius with a flow rate of 7500 kg/hr. If overall heat transfer coefficient for the condenser surface is 1250 W/m^{2}K, what surface area is required to handle this load?

Solution:

Q = U A α _{m}. So, A = 7.08 m^{2}.

QUESTION: 10

Exhaust gases (c _{P }= 1.12 k J/kg K) flowing through a tubular heat exchanger at the rate of 1200 kg/hr are cooled from 400 degree Celsius to 120 degree Celsius. This cooling is affected by water (c_{P }= 4.18 k J/kg K) that enters the system at 10 degree Celsius at the rate of 1500 kg/hr. If the overall heat transfer coefficient is 500 k J/m^{2} hr degree, what heat exchanger area is required to handle the load for counter flow arrangement?

Solution:

m _{h }c _{h }(t _{h 1} – t _{h 2}) = m _{c }c _{c }(t _{c 2} – t _{c 1}).

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