Question

Saturated steam coming off the turbine of a steam power plant at 40°C condenses on the outside of a 3-cm-outerdiameter, 35-m-long tube at a rate of 63 kg/h. Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe. The properties of water at the saturation temperature of 40°C are hfg

Answer 1

Q = 30.07 kJ/sec

Step-by-step explanation:

GIVEN DATA:

temperature of steam = 40 degree celcius

mass flow rate = 63 kg/h

from saturated water tables,

from temperature 40 °, enthalpy of evaporation
`h_f` value is 2406 kj/kg

rate of heat transfer (Q) can be determine by using following relation

`Q = \dot{m} h_f`

putting all value to get Q value

Q = 45 *2406

Q = 108270 kJ/h

`Q = 108270 *(1)/(3600) kJ/s`

Q = 30.07 kJ/sec

Answer 2

Step-by-step explanation:

According to the water table, the value of enthalpy of evaporation at a temperature of
`40^(o)C` is 2406.0 kJ/kg.

Hence, we will calculate the rate of heat transfer by using the formula as follows.

Q =
`m * h_(fg)`

where, m = mass

`h_(fg)` = enthalpy of evaporation

Putting the given values into the above formula as follows.

Q =
`m * h_(fg)`

=
`63 kg/h * 2406.0 kJ/kg`

= 151578 kJ/h

or, =
`151578 * (1)/(3600) kJ/s`

= 42.105 kW

Thus, we can conclude that rate of heat transfer from the steam to the cooling water flowing through the pipe is 42.105 kW.