For low-pressure steam at 0.3 MPa with a mass flow rate of 2 t/h and constant V, what is the mass flow rate when it is converted to high-pressure steam at 1.4 MPa?

Understanding logic is more important than knowing calculation formulas or using software. For boilers and steam, there are established techniques, terminology, and logical principles. One must understand these and express them in terms that everyone can comprehend; it’s not acceptable to create one’s own logical rules. What assumption is it that \"the volume of steam remains constant\"? Two assumptions are made: first, the power of the boiler remains constant; it previously produced steam at 0.3 Mpa but now produces it at 1.4 Mpa, and since the enthalpy of the steam increases, the net weight of steam produced per hour decreases. The mass flow rate is expressed in Nm3 and is related to the mass in kg/h. The second assumption is that the mass flow rate of steam leaving the boiler remains unchanged, and this is achieved by increasing the heating power of the boiler

How can one make a mistake on such a simple question that can even be calculated with a calculator? ? ? Aren’t you assuming that if the flow velocity in the pipe remains constant, then the volumetric flow rate will also remain constant? How did it change later on? ? According to your assumption, this calculation simply involves multiplying the ratio of the specific volumes of the two types of steam by the original mass flow rate; you can just use a calculator for this. M2 = M1 * (0.61/0.143) = 2 * 4.29 = 8.58 tons/h. Steam cannot be treated as an ideal gas. What’s frustrating is that you’ve already found the specific volume of steam and calculated its density correctly – so how can there still be a mistake? ? ?

Well, I have a younger brother who has just started working in this field. I would appreciate your guidance. Under condition 1, that is, when the power of the boiler remains unchanged, if steam was previously produced at 0.3 Mpa but now at 1.4 Mpa, the enthalpy of the steam increases; as a result, the net weight of steam produced per hour decreases. How should I calculate this?

Well, I have a younger brother who has just started in this field. I’ll keep a straight face and thank you for your explanation: lol

It is indeed my description that is incorrect; I assumed that the power of the boiler remained constant. Previously, it produced steam at 0.3 Mpa, while now it produces it at 1.4 Mpa. Since the enthalpy of the steam increases, the net weight of steam produced per hour decreases. What formula should I use for this calculation?

For this problem (the power of the boiler remains unchanged; it previously produced steam at 0.3 Mpa and now produces it at 1.4 Mpa. The low-pressure steam has a pressure of 0.3 Mpa and a mass flow rate of 2 t/h. What is the mass flow rate for the high-pressure steam at 1.4 Mpa?), what formula should I use to carry out the calculation?:D

This post was last edited by arpcd on 2019-5-7 14:06. You can understand this problem in this way. This boiler produces steam at 1.4 MPa and 200°C; then, a valve is used to directly reduce the pressure to 0.3 MPa. Under isenthalpic conditions during this throttling process (with those minor heat losses ignored), the parameters of the steam resulting from throttling are 0.3 MPa and 170°C as superheated steam. By adding some cold water (since the temperature of the superheated steam is higher than the desired steam temperature), the steam with the desired parameters of 0.3 MPa and 140°C is obtained. This process also corresponds to the actual process, in which medium-pressure steam is cooled and depressurized to produce low-pressure steam. Based on 1 ton of medium-pressure steam (1.4 Mpa, 200°C), 1.023 tons of low-pressure steam (0.3 Mpa, 140°C) can be obtained; in fact, this is the ratio of the enthalpies of these two types of steam. Pressure P = 1.40000000 MPa, Temperature T = 200.00 ℃, Specific enthalpy H = 2802.98 KJ/Kg. Pressure P = 0.30000000 MPa, Temperature T = 140.00 ℃, Specific enthalpy H = 2739.36 KJ/Kg. 2739.36/2802.98 = 97.7%; applying a 2% discount, the output of the boiler when low-pressure steam is produced is 2 tons per hour, so the output of medium-pressure steam would be approximately 1.954 tons per hour. In actual production, taking into account factors such as heat loss on top of a 10% discount, an additional 5% discount is applied, resulting in a rate of around 1.86 tons per hour.

Okay, thank you very much:handshake

If this is the data you’re looking for, the reply from the 9th floor should be correct; with his level of expertise, he can’t get such a question wrong. It’s just a bit complicated. To understand it more simply, refer to the enthalpy values at http://www.docin.com/p-283209473.html: at 0.4 Mpa, the enthalpy is 2738.5 kJ/kg. The boiler’s power can be calculated as 2738.5 * 2000 / 3600 = 1521 Kw. At 1.5 Mpa, the enthalpy is 2795.1 kJ/kg. Applying a 15% discount as suggested in the comment above, the power becomes 1521 * 3600 * 0.95 / 2795.1 = 1861 kg/h